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Steinberg groups for Jordan pairs

Ottmar Loos

Erhard Neher

1 November 2018

Ottmar LoosFakultat fur Mathematik und InformatikFernUniversitat in HagenD-58084 Hagen, [email protected]

Erhard NeherDepartment of Mathematics and StatisticsUniversity of OttawaOttawa, Ontario K1N 6N5, [email protected]

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Chapter I. Groups with commutator relations . . . . . . . . . . . . . . . . . . . . . . . . . . 1

§1. Nilpotent sets of roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2§2. Reflection systems and root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 9§3. Groups with commutator relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21§4. Categories of groups with commutator relations . . . . . . . . . . . . . . . . 32§5. Weyl elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Chapter II. Groups associated with Jordan pairs . . . . . . . . . . . . . . . . . . . . . . . 70

§6. Introduction to Jordan pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71§7. The projective elementary group I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92§8. The projective elementary group II . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107§9. Groups over Jordan pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Chapter III. Steinberg groups for Peirce graded Jordan pairs . . . . . . . . . . . 133

§10. Peirce gradings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133§11. Groups defined by Peirce gradings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145§12. Weyl elements for idempotent Peirce gradings . . . . . . . . . . . . . . . . 153§13. Groups defined by sets of idempotents . . . . . . . . . . . . . . . . . . . . . . . . 159

Chapter IV. Jordan graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

§14. 3-graded root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175§15. Jordan graphs and 3-graded root systems . . . . . . . . . . . . . . . . . . . . . 194§16. Local structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209§17. Classification of arrows and vertices . . . . . . . . . . . . . . . . . . . . . . . . . . 220§18. Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233§19. Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

Chapter V. Steinberg groups for root graded Jordan pairs . . . . . . . . . . . . . . 251

§20. Root gradings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252§21. Groups defined by root gradings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258§22. The Steinberg group of a root graded Jordan pair . . . . . . . . . . . . 276§23. Cogs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288§24. Weyl elements for idempotent root gradings . . . . . . . . . . . . . . . . . . 309§25. The monomial group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326§26. Centrality results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

Chapter VI. Central closedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

§27. Statement of the main result and outline of the proof . . . . . . . . . 344§28. Invariant alternating maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351§29. Vanishing of the binary symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363§30. Vanishing of the ternary symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376

iii

iv CONTENTS

§31. Definition of the partial sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392§32. Proof of the relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

Notation Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

Introduction

Traditionally, the study of the classical elementary and Steinberg groups ofmatrices was done, roughly speaking, in four cases: the linear, symplectic, unitary,and orthogonal case. It was A. Bak [6] who took a first unifying step by introducingthe concept of form ring which made it possible to treat these cases in a uniformmanner.

It is the aim of the present volume to take a second unifying step by makingthe theory independent of explicit matrix realizations and base it instead on twofoundations: an algebraic one, the theory of Jordan pairs, and a combinatorial one,the theory of 3-graded locally finite root systems. Here is our basic observation.

A look in [6, §3] and [32, 5.3A] shows that the matrices of the elementaryunitary group G = EU2n(A, J, Λ, ε) of a form ring (A, J, Λ, ε) decompose naturallyinto four blocks (

a bc d

)where a, b, c, d are n×n-matrices with entries from A satisfying suitable conditions.Moreover [32, 5.3.18], for n> 3, G is generated by the “triangular subgroups”

U− =

(1 0V − 1

), U+ =

(1 V +

0 1

),

where V − and V + consist essentially of skew-hermitian matrices over A with acondition on the diagonal entries involving Λ. There is a third naturally occurringsubgroup G0 of G consisting of the diagonal block matrices. Finally, G contains afamily (Uα)α∈R of subgroups indexed in a natural way by the classical root systemR = Cn. The commutator relations satisfied by these subgroups serve to define theunitary Steinberg groups in a fairly complicated manner, see [6, §3], [32, 5.3B].

It is this set-up which we intend to put on a more general footing, using theframework of Jordan pairs and 3-graded root systems.

The spaces V + and V − have the following algebraic structure: they form abeliangroups under addition, and the quadratic-linear maps sending (x, y) ∈ V +×V − toxyx and (y, x) ∈ V − × V + to yxy take their values in V + and V −, respectively.The non-linear compositions V + × V − → V + and V − × V + → V − thus definedsatisfy algebraic identities which say that V = (V +, V −) is a Jordan pair in thesense of [52].

In addition, V + and V − carry a combinatorial structure which can be describedin terms of the root system R. We realize R as the set ±εi±εj : i, j = 1, . . . , n inEuclidean n-space with standard basis ε1, . . . , εn. Then the subgroups U−, G0, U

+

of G are reflected in a 3-grading of R, a decomposition R = R−1 ∪ R0 ∪ R1, givenby R1 = εi+εj : i, j = 1, . . . , n and R−1 = −R1. Moreover, V ± is the direct sumof spaces (V ±γ )γ∈R1 where V ±2εi consists of the diagonal matrices with entries in the

(i, i)-position and zeroes elsewhere, while for i 6= j, the matrices in V ±εi+εj have zero

entries outside of the (i, j)- and (j, i)-positions. The V ±γ obey composition ruleswith respect to the Jordan pair structure, called a root grading of V .

This suggests that one should be able to develop, in a unified manner, a theoryof elementary groups and Steinberg groups, based on an arbitrary Jordan pairendowed with a root grading. It is the purpose of this book to show that this isindeed possible.

v

vi INTRODUCTION

∗Let us point out the main novelties of our approach. The first is the use of

locally finite root systems. These are defined like finite root systems, except thatfiniteness is replaced by local finiteness: the intersection of the root system withevery finite-dimensional subspace of the ambient space is finite. Their classificationis known and presents no surprises: besides the five exceptional ones, there arenatural, possibly infinite, analogues of the five classical root systems [63]. By usinglocally finite root systems we are able to bypass the traditional way of dealing withthe infinite elementary and Steinberg groups. Instead of first considering the finitecase and then obtaining the infinite case by a limiting process, we deal with bothcases at the same time. The root systems we consider have to possess a 3-grading.With the exception of the root systems E8, F4 and G2, all locally finite root systems,in particular, the infinite ones, possess a 3-grading. In view of the fact that themain interest is on the infinite case, this restriction seems not serious.

A second novel aspect is the use of Jordan structures in the theory of elementaryand Steinberg groups which not only introduces new techniques but also allowsus to describe the groups in a uniform manner by relatively few relations. Thisis particularly noticeable when one compares the relations defining the unitarySteinberg groups in Bak [6, Lemma 3.16] or Hahn [32, 5.3B] with our description.

A preliminary version of the main results of this book has been published in[64].

∗The book comprises 32 sections, arranged in six chapters. The diagram below

indicates the interdependence of the various sections. The combinatorics of 3-graded root systems are developed in sections 1 and 2 and sections 14 – 19, whilethe algebraic background on Jordan pairs is presented in sections 6 and 10.

1–2

JJJJJJJJJ

3–5 6

xxxxxxxxx

7–9 10

xxxxxxxx

14–19

JJJJJJJJJ 11–13

20–26

27–32

We now give a rough outline of the contents and refer to the summaries precedingeach chapter for more details.

INTRODUCTION vii

Chapter I on groups with commutator relations does not require Jordan theoryand can be read independently from the rest of the text. We study groups gener-ated by a family (Uα)α∈R of subgroups, indexed by a generalized root system Rand satisfying commutator relations of the type

(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) for so-called

nilpotent pairs (α, β) of roots. Here the bold parentheses denote both the groupcommutator and a combinatorial analogue in R.

The first step in the program outlined above is taken in Chapter II. For an ar-bitrary Jordan pair V , we define the group which replaces the elementary unitarygroup in the example discussed before. Since matrices are not available, this groupmust be constructed entirely from the given Jordan pair. With V we associate a Z-graded Lie algebra L(V ), the Tits-Kantor-Koecher algebra [98, 41, 45]. The spacesV ± are contained in L(V ), and there are natural injective group homomorphismsexp from V ± into the automorphism group of L(V ). The projective elementarygroup PE(V ) is defined as the subgroup of the automorphism group of L(V ) gen-erated by exp(V +) and exp(V −). We then consider groups over V , that is, groupsG with a pair U+ and U− of generating subgroups and a surjective homomorphismπ: G→ PE(V ) which induces isomorphisms U± ∼= exp(V ±).

Chapter III paves the way for the study of general root graded Jordan pairs anddeals with the special case where the root system is of type C2.

In Chapter IV we review 3-graded root systems and show that they are equiva-lent to a class of mixed graphs, called Jordan graphs (Theorem 15.11). The graphΓ associated with a 3-graded root system has vertex set R1 and edges, possiblydirected, depending on the Cartan integers of the roots involved. The remainderof this chapter deals with further graph-theoretical properties of these graphs.

Having prepared the combinatorial tools, we study in Chapter V Steinberggroups for root graded Jordan pairs. A root grading R of a Jordan pair V by aJordan graph Γ is a decomposition V =

⊕γ∈Γ Vγ where the subpairs Vγ satisfy

appropriate composition rules in terms of the Jordan pair structure. Given a groupG over V , we define, for all α in the root system R associated with Γ , subgroupsUα of G, and consider the category of those groups which satisfy R-commutatorrelations with respect to the Uα. This category has an initial object St(V,R),defined by generators and relations, called the Steinberg group of (V,R).

Our eventual aim is to show that St(V,R) is the universal central extension of theprojective elementary group PE(V ). This needs further assumptions, and the keyto a positive result is to require the existence of sufficiently many Weyl elements. AWeyl element for the root α is an element w ∈ U−α Uα U−α such that conjugationby w implements the reflection sα in the root α in the sense that wUβ w

−1 =Usα(β) for all β ∈ R. Weyl elements are intimately related to idempotents inthe Jordan pair V . In the example of the elementary unitary groups, they arefurnished by the fact that the coordinate ring A has a unit element. We introducesystems of idempotents, called cogs, compatible with the given root grading. Thenthe Steinberg group St(V,R) is a central extension of the projective elementarygroup, provided V has a sufficiently large cog giving rise to Weyl elements, and theirreducible components of R have infinite rank (Theorem 26.5).

The final chapter VI is the heart of the book. Our main results are as fol-lows. Suppose that every irreducible component of R has rank at least five, andassume the existence of a sufficient supply of Weyl elements. Then the Steinberggroup St(V,R) is centrally closed (Theorem 27.4). If all irreducible components

viii INTRODUCTION

of R have infinite rank then St(V,R) is the universal central extension of PE(V )(Corollary 27.6).

Here is a quick outline of the main steps of the proof. Let p: E → G = St(V,R)be a central extension. Since G is in particular a group over V , it is generated byabelian subgroups U± ∼= V ±. A first step is to establish that the inverse images ofU± under p in E are abelian (§28 and §29). We then show in §30 and §31 that theinduced central extensions p−1(U±)→ U± admit sections s±. Finally, §32 containsthe proof that the s± satisfy the defining relations of G and thus yield a sections: G→ E.

There are many topics traditionally studied for elementary and Steinberg groupsover rings whose analogues in the Jordan setting are not touched upon in the presenttext. We mention two of them. One is the normal subgroup structure of thesegroups. Under suitable assumptions, it is well known that the normal subgroups ofthe elementary group of a ring A is determined by the ideal structure of A. So anatural question is: can the normal subgroups of PE(V ) described in terms of theideal structure of the Jordan pair V ?

A second topic concerns the analogues of the K2-groups, that is, the kernel ofthe canonical map π: St(V,R) → PE(V ). Can one find generators of Ker(π) likethe Steinberg symbols in the case of linear Steinberg groups? This seems to be anarea wide open for further research.

Acknowledgment. During the more than two decades of work on this book,each author visited the other author’s department on numerous occasions. Theyboth thank their respective departments for providing a fruitful working environ-ment during these visits. Partial support for these visits was provided by the secondauthor’s NSERC grant.

The authors gratefully acknowledge the hospitality of the MathematischesForschungs-Institut Oberwolfach during a stay in the “Research in Pairs” pro-gramme in November-December 2010 where part of the work on the first two chap-ters was done.

Particular thanks go to H. P. Petersson for valuable discussions and his constantencouragement.

CHAPTER I

GROUPS WITH COMMUTATOR RELATIONS

Summary. The chapter begins with two sections developing some of the combinatorics

required for the rest. Instead of restricting ourselves to root systems in the usual sense and with a

view towards later applications we base at least part of the theory on “sets in free abelian groups”,that is, pairs (R,X) consisting of a free abelian group X and a subset R of X generating X and

containing 0. With an obvious definition of morphisms, they form a category SF.The key notion of §1 is that of a nilpotent pair. For non-zero elements α, β of R we define(((((((

α, β)))))))

= R ∩ (N+α+ N+β),[[[[[α, β

]]]]]=(((((((α, β

)))))))∪ (R ∩ N+α) ∪ (R ∩ N+β),

called the open and closed root interval from α to β, respectively. Here N+ denotes the positive

natural numbers. We say (α, β) is a nilpotent pair if[[[[[α, β

]]]]]is finite and does not contain zero.

Of course, more than these bare bones are needed to develop an interesting theory. Therefore,

the next section §2 introduces reflection systems, based on [65]. Prime examples are the usual

finite, not necessarily reduced, root systems, but also locally finite root systems [63], as well asthe roots of a Kac-Moody algebra, the root systems occurring in the theory of extended affine Lie

algebras, and the roots of classical Lie superalgebras.

The following sections §§3–5 introduce and study groups with commutator relations. Theseare (abstract) groups G with a family U = (Uα)α∈R of subgroups for an arbitrary (R,X) ∈ SF.

We say G has R-commutator relations with root groups Uα if the Uα generate G and satisfy

U0 = 1, Uα ⊂ Uβ if α ∈ Nβ,(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) if (α, β) is a nilpotent pair.

Here(((((((Uα, Uβ

)))))))denotes the subgroup generated by all group commutators

(((((((a, b)))))))

= aba−1b−1 for

a ∈ Uα, b ∈ Uβ , and for any subset S of R, US is the subgroup of G generated by all Uγ , γ ∈ S.Groups with R-commutator relations form a category gcR, the morphisms being group homo-

morphisms ϕ: G→ G′ preserving root groups: ϕ(Uα) ⊂ U ′α for all α. Such a morphism is said to

be a covering if it is bijective on root groups and satisfies ϕ(U[[[α,β]]]) = U ′[[[α,β]]] for all nilpotent pairs

(α, β). We say G is simply connected if it admits no proper coverings. Every G ∈ gcR admits

an essentially unique simply connected covering, called its Steinberg group. This is a consequenceof the more precise statement that the simply connected groups form a coreflective subcategoryof gcR (Theorem 4.9). The Steinberg group is constructed as an inductive limit, following and

generalizing Tits’ approach to Steinberg groups in the Kac-Moody setting [102]. Under a mild

assumption on G, it can also be described in more down-to-earth fashion by generators and re-lations (Theorem 4.14). This notion of Steinberg group specializes to the well-known finite or

infinite Steinberg groups when G is the elementary linear group of a ring (4.15). In the remainder

of §4, we show that the categories gcR form the fibres of an opfibration gc over SF.The last §5 introduces Weyl elements in groups with commutator relations. Here we assume

that R is a reflection system. Essentially, this means that, for all α in a suitable subset Rre of

R, reflections sα of R are defined with properties similar to the well-known reflections of ordinaryroot systems.

A Weyl element for α ∈ Rre in a group G ∈ gcR is an element w = xyz ∈ U−α Uα U−α

with the property that conjugation by w in G implements the reflection sα in the sense that

wUβ w−1 = Usα(β) for all β ∈ R. We then say that (x, y, z) is a Weyl triple for α. As a variant

of the result in §4, we show that it is possible to define Steinberg groups with a prescribed set

of Weyl elements for a given set of Weyl triples (Theorem 5.12). This yields the usual Steinberg

group St2(A) of a ring A, starting from the elementary group of 2 × 2-matrices and the set of

all Weyl triples (5.14). The remainder of this section contains further results on Weyl elements

and Weyl triples and in particular studies the connection with rank one groups in the sense of

Timmesfeld [95].

1

2 GROUPS WITH COMMUTATOR RELATIONS [Ch. I

§1. Nilpotent sets of roots

1.1. N-free subsets. In this section, X is a free abelian group. For a subsetA of X we use the notations

Z[A] = spanZ(A) and N[A]

for the subgroup and the submonoid of (X,+) generated by A, respectively. More-over, we let N(A) be the free abelian monoid generated by A, i.e., the set of allmaps v: A → N with finite support. Depending on the context, it may be moreconvenient to think of an element of N(A) as a family (nα)α∈A, where nα ∈ N andnα = 0 except for finitely many α. We denote by κ: N(A) → X the canonical mapsending v to

∑α∈A v(α)α and put

N+[A] := κ(N(A) 0

)=

∞⋃n=1

(A+ · · ·+A)︸︷︷︸n

.

Note that N[A] = κ(N(A)).

A subset A of X is called N-free [8] if 0 /∈ N+[A]; in other words, if for all(nα)α∈A ∈ N(A), the relation

∑α∈A nα · α = 0 implies nα = 0 for all α. Clearly

subsets of N-free sets are N-free, and N-free sets do not contain 0. An N-free set Adefines a partial order <A on X by

x<A y ⇐⇒ x− y ∈ N[A]. (1)

This is easily verified. The notation x A y means x<A y and x 6= y.The following fact will be useful.

1.2. Lemma. Let V be a finite-dimensional real vector space and let VQ ⊂ Vbe a rational form of V , i.e., a Q-vector subspace such that VQ ⊗Q R ∼= V underthe natural map sending x⊗ r to xr. We endow V with its natural topology and letU ⊂ V be an open subset. Then

U 6= ∅ =⇒ U ∩ VQ 6= ∅. (1)

Proof. After choosing a basis of V contained in VQ, we may identify V with Rnand VQ with Qn, so the claim follows from density of Qn in Rn.

Since X is free abelian, the natural map X → XQ = X ⊗ Q, x 7→ x ⊗ 1, isinjective, and we identify X with its image under this map. Likewise, XQ may beidentified with its natural image in XR = X ⊗ R.

1.3. Lemma. Let A ⊂ X be a finite non-empty N-free subset. We denote byVQ = spanQ(A) the rational span of A and put V = VQ ⊗Q R. We endow V withits natural topology and embed A into V canonically.

(a) The convex hull K of A in V is compact and does not contain 0.

(b) The pointed convex cone C spanned by A in V is closed and proper andspanned by its extremal rays. An extremal ray has the form R+ · α for suitable

§1] Nilpotent sets of roots 3

α ∈ A; in particular, A contains elements α such that R+ · α is an extremal ray ofC.

Proof. (a) Recall that the convex hull of A consists of all real linear combi-nations x =

∑α∈A rα · α where rα > 0 with the property that

∑rα = 1 [14, II,

§2.3, Corollary 1 of Proposition 8]. Assume to the contrary that x = 0 has such arepresentation. Consider the finite-dimensional rational vector space QA with basis(eα)α∈A and the canonical map f : QA → VQ sending eα to α. Let fR: RA → V bethe R-linear extension of f , let WQ = Ker(f) ⊂ QA and W = Ker(fR) ⊂ RA. Sincethe exact sequence

0 // WQ // QAf // VQ // 0

remains exact upon tensoring with R, we have WQ ⊗ R ∼= Ker(fR) = W , so WQ isa rational form of W .

Let U = W ∩ RA++. Then U is open in W because RA++ is open in RA, andby our assumption, (rα)α∈A ∈ U . By (1.2.1), U ∩WQ is not empty as well. Sothere exists u = (qα) ∈ QA++ such that 0 = f(u) =

∑α∈A qα · α. By multiplying

this relation with the product of the denominators of the qα we obtain a non-trivialrelation 0 =

∑α∈A nα ·α where nα ∈ N+. This contradicts the fact that A is N-free.

Finally, K is compact by [14, II, §2.6, Corollary 1 of Proposition 15].

(b) The pointed convex cone C (with vertex 0) spanned by A is

C = R+[A] =∑α∈A

rα · α : rα ∈ R+

,

and this is clearly the same as the smallest pointed cone which contains K. By[14, II, §7.3, Proposition 6], C is proper and closed in V . Hence by [14, II, §7.2,Proposition 5], C is the closed convex hull of the union of its extremal rays; inparticular, such rays exist. One sees easily (cf. [63, B.1]) that an extremal ray ofC has the form R+ · α for some α ∈ A, so A contains elements α such that R+ · αis an extremal ray of C.

By a positive functional for a subset A of X we mean a homomorphismh: Z[A]→ Z of abelian groups taking strictly positive values on A.

1.4. Proposition. Any subset of X admitting a positive functional is N-free.Conversely, a finite N-free subset admits a positive functional.

Proof. Suppose h is a positive functional for A, and let (nα)α∈A in N(A) with∑α∈A nα · α = 0. Applying h yields

∑α∈A nα · h(α) = 0, and since all h(α) are

positive, it follows that all nα = 0.

Conversely, suppose A is finite and N-free. We use the notations of Lemma 1.3.By that lemma, K is compact and does not contain 0. By [14, II, §5.3, Prop. 4],there exists a hyperplane separating 0 and K strictly. Thus there exists a linearform g ∈ V ∗, the dual of V , and c ∈ R such that g(0)− c < 0 and g(x)− c > 0 forall x ∈ K; in particular, g(α) > 0 for all α ∈ A.

V ∗ has the rational form V ∗Q = f ∈ V ∗ : f(VQ) ⊂ Q, and for each α ∈ A,the set Uα = f ∈ V ∗ : f(α) > 0 is open in V ∗. Since A is finite, U :=

⋂α∈A Uα

is open as well, and g ∈ U by the above. By (1.2.1), U ∩ V ∗Q 6= ∅, so there existsf ∈ V ∗ such that f(α) = qα ∈ Q++, for all α ∈ A. Multiplying f with the productof the denominators of the qα yields the desired positive functional.


1.5. The category SF, closed and strictly positive sets. We introduce thecategory SF of sets in free abelian groups whose objects are pairs (R,X) consistingof a free abelian group X and a subset R ⊂ X which spans X and satisfies 0 ∈ R.The morphisms f : (R,X)→ (S, Y ) of SF are the group homomorphisms f : X → Ysatisfying f(R) ⊂ S. The elements of

R× = R 0

will often be referred to as roots. More generally, for any subset A of R, we putA× = A 0.

Let (R,X) ∈ SF. Generalizing a concept of [63, 10.2], a subset C ⊂ R is calledadditively closed in R (or simply closed if there is no ambiguity) if C = R∩N+[C],i.e., if for all α1, . . . , αn ∈ C with β := α1 + · · · + αn ∈ R, we have β ∈ C.The additive closure Ac of a subset A of R is the smallest additively closed subsetcontaining A; it is given by

Ac = R ∩ N+[A]. (1)

In the special case A = α, β, we write[[[[α, β

]]]]:= α, βc = R ∩ mα+ nβ : m,n ∈ N, m+ n > 0 (2)

and call it the closed root interval between α and β. If f : (R,X) → (R′, X ′) is amorphism of SF, then

f(Ac) ⊂ f(A)c. (3)

This is immediate from the definitions.

A subset A of R is called strictly positive if it is additively closed and N-free.We remark that

A is strictly positive ⇐⇒ A is closed and 0 /∈ A. (4)

Indeed, the implication from left to right is clear because an N-free set does notcontain 0. Conversely, let A be closed and 0 /∈ A. If

∑nα · α = 0 then 0 ∈ A since

A is closed in R and 0 ∈ R, contradiction.

1.6. Commutator sets. Let (R,X) ∈ SF. For arbitrary subsets A,B of Rwe define the commutator set(((((((

A,B)))))))

:= R ∩(N+[A] + N+[B]

). (1)

Thus γ ∈(((((((A,B

)))))))if and only if γ belongs to R and has the form

γ = α1 + · · ·+ αp + β1 + · · ·+ βq (2)

where p, q > 1, αi ∈ A, and βj ∈ B.If A = α consists of a single element, we simply write

(((((((α,B

)))))))instead of(((((((

α, B)))))))

, and similarly(((((((α, β

))))))):=(((((((α, β

)))))))= R ∩

(N+α+ N+β

), (3)


called the open root interval from α to β. The following properties follow easilyfrom the definition: (((((((

A, ∅)))))))

= ∅, A ∪(((((((A,A

)))))))= Ac =

(((((((A, 0

))))))), (4)

A is closed ⇐⇒(((((((A,A

)))))))⊂ A, (5)

0 ∈ Bc =⇒ Ac ⊂(((((((A,B

))))))), (6)(((((((

A,B)))))))

=(((((((B,A

)))))))=(((((((Ac, B

)))))))=(((((((Ac, Bc

)))))))=(((((((A,B

)))))))c, (7)

A′ ⊂ A, B′ ⊂ B =⇒(((((((A′, B′

)))))))⊂(((((((A,B

))))))), (8)(

A ∪B)c

= Ac ∪(((((((A,B

)))))))∪ Bc, (9)(((((((

A,(((((((A,B

))))))))))))))⊂(((((((A,B

))))))). (10)

If f : (R,X)→ (R′, X ′) is a morphism of SF then for A,B ⊂ X,

f(((((((A,B

)))))))⊂(((((((f(A), f(B)

)))))))∩ f(R). (11)

Let A ⊂ R be additively closed. A subset B of A is called normal (in A) if(((((((A,B

)))))))⊂ B. We remark that in [94, p. 24], the terminology “B is an ideal in A”

is employed. By (5) and (8), a normal subset is in particular closed. Moreover, by(4) and (5), ∅ and A are always normal subsets of A, and by (6) any proper normalsubset B of A has 0 /∈ Bc.

1.7. The lower central series. Let (R,X) ∈ SF and let A ⊂ R be anarbitrary subset. The lower central series of A is defined inductively by

C 1(A) = Ac, C n+1(A) =(((((((A,C n(A)

))))))). (1)

From (1.6.7) and (1.6.8) it follows by induction that

C n(A) = C n(Ac) = C n(A)c, (2)

C n(A) ⊃ C n+1(A), (3)

and (1.6.6) and (1.6.4) yield

0 ∈ Ac =⇒ C n(A) = Ac, (4)

for all n > 1. Thus the lower central series is mainly of interest for closed subsetsnot containing 0, i.e., for strictly positive subsets, cf. (1.5.4). We note also that allC n(A) are normal subsets of A if A is closed. The lower central series behaves wellwith respect to inclusions and morphisms:

B ⊂ A =⇒ C n(B) ⊂ C n(A), (5)

f(C n(A)

)⊂ C n

(f(A)

). (6)

Indeed, (5) is a consequence of (1.6.8) while (6) follows from (1.5.3) and (1.6.11).


1.8. The upper central series. Let (R,X) ∈ SF and let A ⊂ R be a closedsubset. We define the upper central series of A inductively by

Z0(A) = ∅, Zn(A) = γ ∈ A :(((((((γ,A

)))))))⊂ Zn−1(A), (1)

and the centre of A by

Z (A) := Z1(A) = γ ∈ A :(((((((γ,A

)))))))= ∅. (2)

From the definition, it is clear that

∅ = Z0(A) ⊂ Z1(A) ⊂ Z2(A) ⊂ · · · ⊂ A, (3)

and that (((((((A,Zn(A)

)))))))⊂ Zn−1(A), (4)

in particular, the Zn(A) are normal in A.

As for the lower central series, only the case 0 /∈ A is of interest, because 0 ∈ Aimplies γ = γ + 0 ∈

(((((((γ,A

)))))))for all γ ∈ A, so Z (A) and therefore also all the other

Zn(A) are empty.

1.9. Prenilpotent and nilpotent subsets. Let (R,X) ∈ SF. A subset A ofR is said to be prenilpotent if C n(A) = ∅ for sufficiently large n, and it is callednilpotent if it is closed and prenilpotent. From 1.5, 1.6 and 1.7 it is immediate thata prenilpotent set cannot contain 0 and that the following implications hold:

A prenilpotent =⇒ 0 /∈ Ac, (1)

B ⊂ A and A prenilpotent =⇒ B prenilpotent, (2)

f(A) prenilpotent =⇒ A prenilpotent, (3)

A prenilpotent =⇒ Ac nilpotent, (4)

A nilpotent =⇒ A strictly positive. (5)

The class of a nilpotent A is the smallest k such that C k+1(A) = ∅. Thus

k 6 1 ⇐⇒ A = Z (A) ⇐⇒(((((((A,A

)))))))= ∅,

k 6 2 ⇐⇒(((((((A,A

)))))))⊂ Z (A) ⇐⇒

(((((((A,(((((((A,A

))))))))))))))= ∅,

and we will call an A of class 6 1 resp. 6 2 abelian resp. 2-step nilpotent.

As in the case of groups, nilpotence can also be characterized by the uppercentral series. More generally, let A be a strictly positive subset of R. A chain ofsubsets A ⊃ A1 ⊃ A2 ⊃ · · · is called a central chain if

(((((((A,An

)))))))⊂ An+1 for all n>1.

For example, the lower central series is a central chain, and so is Ai := Zm+1−i forsome fixed m, provided we define Zj(A) = ∅ for j < 0.

Clearly the terms An of a central chain are normal in A. From (1.7.1) and(1.8.1) it follows easily that

A1 = A =⇒ Ai ⊃ C i(A), (6)

An+1 = ∅ =⇒ Ai ⊂ Zn+1−i(A). (7)


Now (6) shows

A is nilpotent of class 6 n ⇐⇒ there exists a central chainwith A1 = A and An+1 = ∅, (8)

and (7) implies

A is nilpotent of class 6 n ⇐⇒ Zn(A) = A. (9)

Let us also note that the length of the upper central series of a nilpotent A of classk is exactly k. Indeed, Zk(A) = A holds by (9). Assuming Zk−1(A) = A wouldyield a central chain Ai := Zk−i(A) with A1 = A and Ak = Z0(A) = ∅, so A wouldhave class 6 k − 1, contradiction.

1.10. Lemma. Let A ⊂ X and let h: Z[A] → Z be a positive functional asdefined in 1.3.

(a) Then h(α)> n for all α ∈ C n(A).

(b) If h∣∣A is bounded by k then C k+1(A) = ∅, so A is prenilpotent of class at

most k.

Proof. (a) The proof is by induction on n. For n = 1 this is clear since apositive functional takes positive values on A and C 1(A) = Ac. For the inductionstep, we have C n+1(A) =

(((((((A,C n(A)

))))))). By (1.6.2), an element of C n+1(A) has the

form γ = α1 + · · ·+αp + β1 + · · ·+ βq where αi ∈ A, βj ∈ C n(A), and p> 1, q> 1.Hence by induction, h(γ)> p+ nq > 1 + n.

(b) This is immediate from (a).

1.11. Lemma. Let (R,X) ∈ SF and let A ⊂ R be a finite strictly positivesubset of cardinality n.

(a) There exist total orders > on A compatible with the partial order <A definedby A, cf. (1.1.1), in the sense that α<A β implies α> β.

(b) Let > be as in (a), and enumerate A = α1, . . . , αn in such a way thatα1 < · · · < αn. Then Ai := αi, . . . , αn for i = 1, . . . , n, and Ai := ∅ for i > n, isa central chain of A. In particular, A is nilpotent of class 6 n.

Proof. (a) This follows from the Szpilrajn-Marczewski Lemma [36, Chapter 8,Section 8.6].

(b) We show(((((((A,Ai

)))))))⊂ Ai+1. By (1.6.2), an element γ ∈

(((((((A,Ai

)))))))has the form

γ = αi1 +· · ·+αip+αj1 +· · ·+αjq where p, q>1, iλ ∈ 1, . . . , n and jµ ∈ i, . . . , n;in particular, γ A αj1 . On the other hand, γ ∈ A because A is closed, say, γ = αk.Hence k > j1 > i so αk ∈ Ak ⊂ Ai+1.

The statement about the nilpotence of A now follows from (1.9.8).

1.12. Lemma. Let F be a finite set and let NF , the set of functions F → N,be equipped with the partial order

v 6 w ⇐⇒ v(α)6 w(α) for all α ∈ F .

Then every infinite subset S of NF contains a strictly increasing sequence v1 <v2 < · · ·.


Proof. The proof is by induction on the cardinality of F , the case F = ∅being trivial. If S has no maximal element then the assertion is clear. Otherwise,let m be a maximal element of S. Then v > m holds for no v ∈ S, i.e., forevery v ∈ S there exists an element α ∈ F such that v(α) 6 m(α). LettingSα := v ∈ S : v(α)6m(α), we thus have S =

⋃α∈F Sα. Since S is infinite, there

must be a β ∈ F such that Sβ is infinite. Consider the evaluation map Sβ → N,v 7→ v(β), whose image is contained in the finite interval I := 0, 1, . . . ,m(β) of N.Since Sβ is infinite, there exists i ∈ I such that the fibre Siβ := v ∈ Sβ : v(β) = iis infinite. Let F ′ := F β, denote by res: NF → NF ′ the restriction map inducedby the inclusion F ′ → F , and put S′ := res(Siβ) ⊂ NF ′ . Clearly, res: Siβ → S′ is

bijective, with inverse ext: S′ → Siβ given by extending an element v′ ∈ S′ (whichafter all is a map F ′ → N) to a map F → N via β 7→ i. By induction, there exists astrictly increasing sequence v′1 < v′2 < · · · in S′. Then vk := ext(v′k) is the desiredsequence in S.

1.13. Proposition. Let (R,X) ∈ SF. For a subset F ⊂ R with closureF c = A, the following conditions are equivalent:

(i) F is finite and prenilpotent,

(ii) A is finite and nilpotent,

(iii) A is finite and strictly positive,

(iv) A is finite and 0 /∈ A.

Proof. (i) ⇐⇒ (ii): F is prenilpotent if and only if A is nilpotent by (1.9.4), soit remains to show that F finite implies A is finite. Assume, by way of contradiction,that A is infinite. Then by definition of the closure of a set in (1.5.1) we haveS := κ−1(A) ⊂ NF infinite, where κ is defined in 1.1. Choose a sequence (vk)k>1

in S as in Lemma 1.12 and put γk = κ(vk). We will show by induction thatγk ∈ C k(F ) for all k > 1, contradicting the fact that C k(F ) = ∅ for sufficiently bigk, by nilpotence of A. Obviously, γ1 ∈ A = C 1(F ). Suppose we have γk ∈ C k(F ).Then γk+1−γk =

∑α∈F nαα where all nα := vk+1(α)−vk(α) ∈ N, and at least one

nα is positive because vk+1 > vk. Hence γk+1 ∈(((((((F, γk

)))))))⊂(((((((F,C k(F )

)))))))= C k+1(F ).

(ii) =⇒ (iii) is (1.9.5), and the implication (iii) =⇒ (ii) is a consequence ofLemma 1.11(b). The equivalence of (iii) and (iv) follows from (1.5.4).

1.14. Corollary. A finite prenilpotent subset is N-free and admits a positivefunctional.

Proof. Let F be finite and prenilpotent with closure A. By Proposition 1.13(iii),A is finite and strictly positive, hence in particular N-free, see 1.5. By Proposi-tion 1.4, A admits a positive functional and hence so does F .

1.15. Corollary. The following conditions on a subset A of R are equivalent:

(i) A is closed in R and every finite subset of A is prenilpotent,

(ii) A is strictly positive and every finite subset of A has finite closure.

§2] Reflection systems and root systems 9

This follows easily from Proposition 1.13. A subset satisfying these conditionsis called locally nilpotent. In particular, if R is a locally finite root system, cf. 2.8,then a subset of R is locally nilpotent if and only if it is strictly positive.

§2. Reflection systems and root systems

2.1. Reflections. Let X be a free abelian group with dual X∗ = Hom(X,Z).By a (hyperplane) reflection we mean a linear map s: X → X of the form

s(x) = sv,f (x) = x− f(x) v

where v ∈ X and f ∈ X∗ satisfy f(v) = 2. Then s2 = IdX , the fixed point set of sis the hyperplane

X+(s) := x ∈ X : s(x) = x = Ker(f),

and we put

v ∈ X−(s) := x ∈ X : s(x) = −x.

Recall [17, VII, §3.1, Corollary 2] that a subgroup of a free group is free. More-over, since X is torsion-free it follows from [15, II, §7.10, Proposition 26] that thecanonical map X → XQ := X ⊗ Q is injective, so that we can identify X with asubgroup of XQ. Extending scalars from Z to Q, it is clear that X−(s)Q is one-dimensional and X+(s)Q is a hyperplane. Hence X−(s) is free of rank one, but v isnot necessarily a basis of X−(s).

We put X ′ = X+(s) + X−(s). Since X is torsion-free, it is clear that the sumis direct. In general, X ′ 6= X, depending on the divisibility properties of v and f .

First, if v or f are divisible in X or X∗ then they are divisible at most by 2, andthey cannot be both divisible by 2. This follows easily from the fact that f(v) = 2.Thus there are two cases:

(a) Suppose either v or f is divisible by 2, and put h(x) := vf(x). Thenh ∈ End(X) is divisible by 2, say, h = 2p, and p2 = p is a projection. Hence Xdecomposes X = Im(p)⊕Ker(p). Since s = Id− 2p, it follows that X−(s) = Im(p)and X+(x) = Ker(p), so X = X ′.

(b) If neither f nor v are divisible by 2 then X ′ has index 2 in X. Indeed,f(v) = 2 shows that 2Z ⊂ f(X) ⊂ Z, and since f is not divisible by 2, we musthave f(X) = Z. As X−(s) is free of rank one and v ∈ X−(s) is indivisible, v is abasis of X−(s).

Choose u ∈ X with f(u) = 1. Then X = Zu ⊕ Ker(f) = Zu + X ′. We havef(2u − v) = 2 · 1 − 2 = 0, so 2u − v ∈ Ker(f) ⊂ X ′ and therefore 2u ∈ X ′. Butu /∈ X ′, otherwise u = mv + y where m ∈ Z and y ∈ X+ = Ker(f), which wouldimply 1 = f(u) = mf(v) = 2m, contradiction. Thus X ′ has index 2 in X.

We now discuss to what extent the reflection s determines v and f . Let s = sv,fand s′ = sv′,f ′ . Then

X−(s)Q = X−(s′)Q =⇒ v = ±v′ or v = ±2v′ or v′ = ±2v. (1)


Indeed, since v and v′ are Q-bases of X−(s) and X−(s′), respectively, we have pv =qv′ for p, q ∈ Z 0. Put a = f ′(v) and b = f(v′). Then f ′(pv) = pa = f ′(qv′) = 2qand f(qv′) = qb = f(pv) = 2p. Hence 2pa = 4q = qba so ab = 4. Let σ = sgn(pq).Since sgn(a) = sgn(b) = σ, there are six cases, namely, (σa, σb) = (2, 2), (1, 4) and(4, 1), corresponding to the cases p = σq, p = σ2q and q = σ2p, corresponding tothe cases listed on the right hand side of (1).

Similarly, one can show that X+(s)Q = X+(s′)Q implies f ′ = ±f or f ′ = ±2for f = ±2f ′, and hence even X+(x) = X+(s′).

In the sequel, we will use the notation

f(x) = 〈x, f〉

for f ∈ X∗ and x ∈ X, following the usual practice for root systems.

2.2. Definition. Let (R,X) ∈ SF (see 1.5), let ∨: R → X∗ be a map, anddefine

Rim = α ∈ R : α∨ = 0, Rre = R Rim,

called the imaginary and reflective roots, respectively. The triple (R,X, ∨) is calleda reflection system over Z if the following axioms hold:

(ReS1) If α ∈ Rre then 〈α, α∨〉 = 2 (and hence α 6= 0). We then abbreviatesα := sα,α∨ so that

sα(x) = x− 〈x, α∨〉α (1)

for all x ∈ X.

(ReS2) sα(R) = R for all α ∈ Rre.

(ReS3) If α and nα in Rre for n ∈ Z then α∨ = n · (nα)∨.

(ReS4) (sαβ)∨ = β∨ − 〈α, β∨〉α∨ for all α ∈ Rre and β ∈ R.

This is a straightforward modification of the notion of an integral reflection systemover a field of characteristic zero given in [65, §2], see 2.5 for details. In this book,the unqualified term “reflection system” will always mean a reflection system overZ. By abuse of notation we will often refer to a reflection system simply by R or(R,X) instead of (R,X, ∨). It will be convenient to put sα = IdX for α ∈ Rim.

An equivalent form of (ReS4) is

(sαβ)∨ = β∨ sα. (2)

Indeed, 〈x, β∨ sα〉 = 〈sα(x), β∨〉 =⟨x−〈x, α∨〉α, β∨

⟩=⟨x, β∨−〈α, β∨〉α∨

⟩for all

x ∈ X. Since R spans X, (2) is in turn equivalent to

〈γ,(sα(β)

)∨〉 = 〈sα(γ), β∨〉 (3)

for α ∈ Rre and β, γ ∈ R. It also follows that β∨ = 0 implies (sαβ)∨ = 0. Hencesα(R Rre) = R Rre and therefore also

sα(Rre) = Rre (4)


for all α ∈ Rre.

The axioms (ReS1) and (ReS2) imply 0 6∈ Rre and Rre = −Rre. We say R issymmetric if also R = −R.

Let α, β ∈ Rre and p, q ∈ Z 0. Then

pα = qβ =⇒ α = σβ or α = σ2β or β = σ2α, (5)

where σ = sgn(pq). This follows from (2.1.1) since α and β are Q-bases of X−(sα)Qand X−(sβ)Q, respectively. We say R is reduced if α ∈ Rre and nα ∈ Rre for n ∈ Zimply n = ±1.

For α, β ∈ Rre we have

sα = sβ ⇐⇒ pα = qβ for 0 6= p, q ∈ Z. (6)

Indeed, by (1),

sα = sβ ⇐⇒ 〈x, α∨〉α = 〈x, β∨〉β for all x ∈ X.

Putting x = α shows 2α = 〈α, β∨〉β. Conversely, pα = qβ implies pβ∨ = qα∨ by (5)and (ReS3), and therefore pq〈x, α∨〉α = qp〈x, β∨〉β. Since pq 6= 0, we have sα = sβ .In particular, (6) implies

sα = s−α. (7)

2.3. Morphisms of reflection systems, the category ReS. Let (R,X, ∨)and (S, Y, ∨) be reflection systems. Unless this might lead to confusion, we will usethe same symbol ∨ for the maps R→ X∗ and S → Y ∗. A morphism f : (R,X, ∨)→(S, Y, ∨) of reflection systems is a linear map f : X → Y such that f(R) ⊂ S and

f(sα(β)

)= sf(α)

(f(β)

), (1)

for all α, β ∈ R. We denote by ReS the category of reflection systems. There is anobvious forgetful functor from ReS to SF.

We will give different characterizations for a morphism f : (R,X) → (S, Y ) ofSF to be a morphism of ReS. First, since R spans X, (1) is equivalent to

f sα = sf(α) f (2)

for all α ∈ R. For a morphism f : (R,X)→ (S, Y ) of reflection systems we have

f(Rre) ⊂ Sre ∪ 0, f(Rim) ⊂ Sim. (3)

Indeed, if sα 6= IdX but sf(α) = IdY then −f(α) = f(−α) = f(sα(α)) =sf(α)(f(α)) = f(α) shows 2f(α) = 0. Likewise, sα = Id implies f(α) = f(sα(α)) =sf(α)f(α) and hence sf(α) = Id. Thus, (3) holds. It is now obvious that for a linearmap f : X → Y with f(R) ⊂ S we have

f is a morphism of reflection systems

⇐⇒ f(Rim) ⊂ Sim and f sα = sf(α) f for all α ∈ Rre. (4)


The automorphism group of (R,X, ∨) is denoted by Aut(R,X, ∨) or simply byAut(R). We claim

sα ∈ Aut(R) for all α ∈ R.

To show this, we can assume α ∈ Rre. Then sα is a morphism in SF by (ReS2), andsatisfies the first condition in (4) by (2.2.4). Because of s2

α = IdX it thus remainsto verify

ssα(β) = sαsβsα (5)

for β ∈ R. This is clear for β ∈ R Rre where sβ = Id, since then sα(β) ∈ R Rre

by (2.2.4). For β ∈ Rre a straightforward calculation shows

(sα sβ sα)(x) = x− 〈x, β∨ sα〉sα(β)

for x ∈ X, so that (2.2.2) proves (5) for β ∈ Rre.

The subgroup of Aut(R) generated by all sα, α ∈ R, is called the Weyl groupof R and denoted W(R). It is a normal subgroup of Aut(R).

2.4. Direct sums and subsystems. A family (Ri, Xi)i∈I in SF has thecoproduct

(R,X) =∐i∈I

(Ri, Xi) =(⋃i∈I

Ri,⊕i∈I

Xi

)see [63, 1.2] or [65, 2.4]. Following tradition, we also write R =

⊕i∈I Ri and call

R the direct sum of the Ri.If each Ri is a reflection system so is R. Indeed, we extend each α∨i , αi ∈ Ri,

to a linear form on X by setting it to zero on all Xj , j 6= i. It is immediate thatRre =

⋃i∈I R

rei , and W(R) ∼=

⊕i∈I W(Ri), the restricted direct product of the

W(Ri).Let (R,X, ∨) be a reflection system, let R′ be a subset of R containing 0 and let

X ′ = spanZ(R′). We say R′ is a subsystem of R if sα(β) ∈ R′ for all α, β ∈ R′. Forα ∈ R′ let α∨′ be the restriction of α∨ to X ′. Then (R′, X ′, ∨′) is a reflection systemand the inclusion (R′, X ′, ∨′) → (R,X, ∨) is a morphism of reflection systems. Forexample, it follows from (ReS2) that

Re(R) := Rre ∪ 0 (1)

is always a subsystem of any reflection system R.

2.5. Examples of reflection systems. Let X be a free abelian group. Asmentioned in 2.1, we can identify X with a subgroup of XQ. Moreover, if K is afield of characteristic zero then the canonical map XQ → XK = X⊗QK is injectiveas well, so we also identify X with a subset of XK .

Let now (R,X) be a reflection system over Z. For α ∈ R ⊂ X ⊂ XK denotethe K-linear extension of α∨: X → Z to a linear form XK → K again by α∨. Then(R,XK) is an integral pre-reflection system over K in the sense of [65, 2.1], and itis a reflection system as soon as the following stronger version of (ReS3) holds:

(ReS3)K If α and cα in Rre for c ∈ K 0 then α∨ = c · (cα)∨.

Conversely, let (R,X) be an integral reflection system over K. The restriction ofα∨ ∈ X∗ to spanZ(R) is a linear form of the Z-module spanZ(R), again denoted


α∨. If spanZ(R) is free, e.g. if R is finite, then (R, spanZR) is a reflection systemover Z.

The relation between reflection systems over Z and integral reflection systemsover K discussed above provides us with many examples of reflection systems. Themost important examples for this book are the finite or locally finite root systems,see 2.8. Other important examples of reflection systems are:

— the roots of a Kac-Moody Lie algebra with Rre being the real and Rim beingthe imaginary roots,

— the extended affine root systems occurring in extended affine Lie algebras,

— the roots of classical Lie superalgebras.

Many more examples are to be found in [65, 2.12, 3.1, 4.3].

Let again X be a free abelian group. If X ′ is a subgroup of X then X ′Q iscanonically identified with a sub-vector space of XQ. A family (xi)i∈I in X isfree (linearly independent over Z) if and only if the family (xi ⊗ 1)i∈I is linearlyindependent in XQ.

2.6. Lemma and Definition. Let X be a free abelian group. A subset R ofX is called locally finite if it satisfies the following equivalent conditions where Kdenotes a field of characteristic zero:

(i) R ∩ Y is finite, for every finitely generated subgroup Y of X,

(ii) R ∩ U is finite, for every finite-dimensional sub-vector space U of XQ,

(iii) R ∩ V is finite, for every finite-dimensional sub-vector space V of XK .

Proof. (i) =⇒ (ii): Let b1, . . . , bd be a Q-basis of U . Then each bi is a finite linearcombination with rational coefficients of elements of X. By clearing denominators,we may assume that these coefficients are integers. Hence there exist finitely manyx1, . . . , xn ∈ X such that, putting M = spanZ(x1, . . . , xn), we have U ⊂ MQ.Therefore, it suffices to show that R ∩ MQ is finite. Since M is free, being asubgroup of the free abelian group X, it is no restriction to assume that x1, . . . , xnis a Z-basis of M . Then the xi ⊗ 1 are a Q-basis of MQ.

Now let Y = X ∩MQ. Then Y ⊂ X is a subgroup containing M , and M ⊂Y ⊂ MQ implies MQ = YQ. Since X is a free abelian group so is Y . If (yi)i∈I is aZ-basis of Y then (yi⊗ 1)i∈I is a Q-basis of YQ = MQ. Hence Card(I) = n is finite,so Y is a finitely generated subgroup of X. By (i), R∩Y = R∩X ∩MQ = R∩MQis finite.

(ii) =⇒ (iii): As before we can assume that there exists a finite-dimensionalsubspace U of XQ such that V ⊂ UK , and it suffices to show that R ∩UK is finite.The K-vector space XK has an obvious Q-structure with respect to which thesubspace UK is rational. Hence [15, II, §8 No. 2, Proposition 2] shows U = XQ∩UK .But then R ∩ UK = R ∩XQ ∩ UK = R ∩ U is finite.

(iii) =⇒ (i): The inclusion Y ⊂ X induces injections Y → YK → XK and hencean injection R ∩ Y → R ∩ YK . Putting V = YK , we have (i).


2.7. Lemma (Uniqueness of reflections). Let X be a free abelian groupand let R be a locally finite subset of X generating X. Then for any α ∈ R× thereexists at most one reflection s satisfying s(α) = −α and s(R) = R.

Proof. This is a well-known result [18, VI, §1.1, Lemma 1] or [63, Lemma 3.2].For the convenience of the reader, we include the proof. Let s = sα,f and s′ = sα,f ′

be reflections as in 2.1 with the stated properties. Then t = ss′ is given byt(x) = x+αd(x) where d = f ′− f , and t(α) = α because f(α) = f ′(α). Assumingd 6= 0, we can find β ∈ R such that 〈β, d〉 6= 0, because R spans X. Then the vectorstn(β) = β + nd(β)α (n ∈ N) form an infinite set in R ∩ (Zα + Zβ), contradictinglocal finiteness of R.

2.8. Locally finite root systems. In this book we mean by a locally finiteroot system a pair (R,X) consisting of a free abelian group X and a subset R of Xsatisfying the following conditions:

(i) R generates X and contains 0,

(ii) R is locally finite,

(iii) for every α ∈ R× = R 0 there exists a reflection s such that s(α) = −αand s(R) = R.

By Lemma 2.7, there is at most one such reflection. By 2.1, it has the forms = sv,f where v = α and f ∈ X∗ is uniquely determined by X+(s) = Ker(f) andf(α) = 2. We then put α∨ := f .

As in [63] the term root system will be an abbreviation for “locally finite rootsystem”, and a finite root system will be a root system (R,X) with Card(R) <∞,equivalently, with X finitely generated. The rank of a root system (R,X) is bydefinition the rank of X as an abelian group, i.e., the dimension of XQ.

We denote by RS the category whose objects are root systems and for whicha morphism f : (R,X) → (S, Y ) is a group homomorphism f : X → Y satisfyingf(R) ⊂ S. Thus RS is the full subcategory of SF whose objects are root systems.

The definition above is an obvious analogue of the definition of a locally finiteroot system (over R) in [63, Definition 3.3]. In fact, as the following result shows,the two concepts are equivalent. We denote by RSR the category of root systemsover R as defined in [63, 3.6] and denoted RS there: Its morphisms f : (R,X) →(S, Y ) are the R-linear maps satisfying f(R) ⊂ S.

2.9. Proposition. (a) If (R,X) is a root system over Z, then (R,XR) is aroot system over R in the sense of [63]. A morphism f : (R,X) → (S, Y ) in RSuniquely extends to a morphism fR: (R,XR)→ (S, YR) in RSR.

(b) If (R,E) is a root system over R, then (R, spanZ(R)) is a root system overZ. The restriction of a morphism f : (R,E) → (S, F ) in RSR to spanZ(R) andspanZ(S) is a morphism in RS.

(c) The constructions in (a) and (b) induce an equivalence between the cate-gories RS and RSR.

Proof. (a) Let (R,X) be a root system over Z. By Lemma 2.6, the subsetR ⊂ XR is locally finite. Moreover, the reflection sα for α ∈ R× uniquely extends


to a reflection of XR having the the properties required in [63, 3.3]. Hence (R,XR)is a root system over R. The second part is clear.

(b) Let R be a locally finite root system in a real vector space E as in [63], andlet X = spanZ(R) be the abelian group generated by R (in E), also called the rootlattice. By [63, Theorem 7.5(a)], X is a free abelian group, and by Lemma 2.6, Ris locally finite in X. The reflections of the real root system (R,E) restrict to Xand have the properties required in (iii) of Definition 2.8. Thus (R,X) is a rootsystem over Z. The second part of (b) and part (c) are obvious.

2.10. Remarks. (a) Mutatis mutandis, the above proposition holds for rootsystems over fields of characteristic zero as defined in [63, 4.14] as well.

(b) The categorical equivalence between RS and RSR allows us to transferresults of [63] to root systems over Z. For example, for α, β ∈ R, a root systemover Z, we have

|〈α, β∨〉| 6 4, (1)

〈α, β∨〉 = 〈β, α∨〉 = 2 ⇐⇒ α = β 6= 0. (2)

We call B ⊂ R a root basis (as opposed to a grid basis defined in 18.3) if B is Z-freeand every root in R is a linear combination of B with coefficients of the same sign.Then B is a root basis of (R,X) if and only if B is a root basis of (R,XR) in thesense of [63, Def. 6.1]. Hence, by [63, Proposition 6.7], ever countable R has a rootbasis.

2.11. Orthogonality and irreducible components. Let (R,X) be a rootsystem and let α, β ∈ R×. Then the following conditions are equivalent:

(i) 〈α, β∨〉 = 0,

(ii) sαsβ = sβsα, and α and β are Z-free,

(iii) 〈β, α∨〉 = 0.

This follows from [63, 3.5] and the transfer principle 2.10(b). We include a prooffor the convenience of the reader. Indeed, (i) implies sβ(α) = α − 〈α, β∨〉β = αand hence ssβ(α) = sβsαsβ = sα (by (2.3.5)). Assuming mα + nβ = 0 for nonzerom,n ∈ Z, it follows that 0 = m〈α, β∨〉 = 〈mα, β∨〉 = −〈nβ, β∨〉 = −2n, whencemα = 0 and α = 0, contradiction. Conversely, assume (ii) holds. A simplecomputation shows that sαsβ = sβsα if and only if

〈x, α∨〉〈α, β∨〉β = 〈x, β∨〉〈β, α∨〉α

for all x ∈ X. Putting x = α yields 〈α, β∨〉(2β − 〈β, α∨〉α

)= 0. Since α and β

are Z-free, (i) follows. Finally, (ii) is symmetric in α and β, so (ii) and (iii) areequivalent as well.

Two roots α, β are called orthogonal, written α ⊥ β, if they satisfy the conditions(i) – (iii) above. They are said to be connected if there exist αi ∈ R× such thatα = α0 6⊥ α1 6⊥ · · · 6⊥ αn = β. A direct sum of root systems is again a rootsystem. A nonzero root system is called irreducible if it is not isomorphic to adirect sum of two nonzero root systems. Any root system decomposes uniquely into


a direct sum of irreducible root subsystems, called its irreducible components. Theirreducible components of a root system R are precisely the subsets C = 0 ∪C×where C× is an equivalence class with respect to the equivalence relation definedby connectedness. The proof is a straightforward generalization of the one given in[63, 3.13], or can be deduced by transfer from root systems over R. In fact, sincetwo roots α, β ∈ R are orthogonal in (R,X) if and only if they are so in (R,XR),the same holds for connected roots. It follows that the connected components of(R,X) coincide with the connected components of (R,XR). In particular, (R,X)is irreducible if and only if (R,XR) is so.

2.12. The normalized inner product. By [63, 4.6], a root system (R,X)over Z admits a unique quadratic form Φ: X → Z with the property that theassociated bilinear form (x | y) = Φ(x + y) − Φ(x) − Φ(y) is positive definite, thesmallest non-zero value of Φ on each connected component of R is 1, and the linearforms α∨ are given by

〈x, α∨〉 =(x |α)

Φ(α)= 2

(x |α)

(α |α). (1)

We call ( | ) the normalized inner product . Then α ⊥ β in the sense of 2.11 if andonly if (α |β) = 0. The attributes “long” and “short” for roots refer to this innerproduct. From (1) it is clear that the reflection sα is the orthogonal reflection withrespect to ( | ).

We show next that locally finite root systems are in particular reflection systems.

2.13. Lemma. A locally finite root system is a reflection system.

Proof. Let (R,X) be a locally finite root system. By 2.8 we have a well-definedmap ∨: R× → X∗, which we extend to all of R by putting 0∨ = 0. Clearly,R× = Rre, and the axioms (ReS1) and (ReS2) hold.

Let α and nα be in R×, so that in particular n 6= 0. Then −nα = snα(nα) =−nsnα(α), whence snα(α) = −α. By Lemma 2.7, snα = sα, so sα(x) = x −〈x, α∨〉α = x − 〈x, (nα)∨〉(nα) for all x ∈ X. This shows n(nα)∨ = α∨ and proves(ReS3).

Let s = sv,f as in 2.1 and let g ∈ GL(X). Then it is immediate that gsg−1 =sgv,fg−1 . We apply this to s = sβ and g = sα where α, β ∈ R×, and obtain, puttingγ = sα(β), that sγ,β∨sα is a reflection stabilizing R and mapping γ to −γ. Sinceγ ∈ R×, Lemma 2.7 shows that sγ,β∨sα = sγ = sγ,γ∨ , so γ∨ = sα(β)∨ = β∨ sα,proving (2.2.2) and hence (ReS4).

2.14. Proposition (coroot system) Let (R,X) be a root system. Put R∨ =α∨ : α ∈ R and X∨ = spanZ(R∨) ⊂ X∗. Then (R∨, X∨) is a root system, calledthe coroot system of (R,X), with reflections given by

sα∨(β∨) :=(sα(β)

)∨= β∨ − 〈α, β∨〉α∨. (1)

Let (R,XR) be the real root system associated with (R,X) in 2.9(a). Then the realroot system associated with (R∨, X∨) is the coroot system of (R,XR) as defined in[63, Theorem 4.9].

Proof. The set B of integer-valued bounded functions on R× is a free abeliangroup with respect to the standard addition [11, Corollary 1.2]. By (2.10.1), X∨

can be identified with a subgroup of B, so it is free abelian, too.


Condition (i) of Definition 2.8 holds by construction. We show that also thesecond condition is fulfilled, i.e., that R∨ is locally finite in X∨. First, there is acanonical injective homomorphism of real vector spaces

θ: (X∗)R → (XR)∗,(ϕ(f)⊗ r

)(x⊗ t) = f(x)rt (2)

for f ∈ X∗, x ∈ X, and r, t ∈ R. This will be proved in two steps.

(a) There is an injective homomorphism

θ1: X∗ ⊗Z Q→ HomZ(X,Q) ∼= HomQ(XQ,Q) = (XQ)∗, (3)

given by f ⊗ r 7→ (x 7→ f(x)r), for f ∈ X∗, r ∈ Q, x ∈ X.Indeed, let r1, . . . , rn ∈ Q, f1, . . . , fn ∈ X∗ and assume θ1(

∑fi ⊗ ri) = 0. We

may assume that ri = pi/q for pi, q ∈ Z. Then∑fi(x)(pi/q) = 0 for all x ∈ X,

whence 0 =∑fi(x)pi =

(∑fipi

)(x), showing

∑fipi = 0 in X∗. Hence

∑fi⊗ri =∑

fi ⊗ (pi/q) =∑

(fipi)⊗ (1/q) = 0. The isomorphism HomZ(X,Q) ∼= (XQ)∗ is aspecial case of [15, II, §4, Proposition 1].

(b) Let E be a vector space over Q. Then there is an injective homomorphism

θ2: E∗ ⊗Q R→ HomQ(E,R) ∼= HomR(ER,R) = (ER)∗, (4)

given by g ⊗ t 7→ (y 7→ g(y)t), for g ∈ E∗, t ∈ R, y ∈ E. Indeed, since R is aQ-vector space, hence free over Q, this follows from [15, II, §4, No. 2, Corollary ofProposition 2]. Again, the second isomorphism is a special case of loc. cit.

To prove (2), we use the canonical isomorphism X∗ ⊗Z R ∼= (X∗ ⊗Z Q) ⊗Q R.Then θ factors as follows, where we put E = XQ:

θ: (X∗ ⊗Q)⊗ R θ1⊗R−→ E∗ ⊗ R θ2−→ (ER)∗ = [(XQ)R]∗ = (XR)∗.

Here θ1 ⊗ R is injective by (3) and since R is free (hence flat) over Q, and θ2 isinjective by (4).

Next, we show that θ induces an isomorphism

θ: X∨ ⊗Z R = (X∨)R∼=−→ (XR)∨. (5)

Let α ∈ R. Then α∨ ∈ R∨ and α∨ ⊗ 1R ∈ (X∨)R. One checks that

θ(α∨ ⊗ 1R) = (α∨)R,

the R-linear extension of the linear form α∨: X → Z to a linear form (α∨)R: XR →R. Since X∨ is spanned by all α∨, α ∈ R, and (XR)∨ is spanned by all (α∨)R, theassertion follows.

Now let V = XR and S = RR. Then (S, V ) is a root system over R byProposition 2.9, hence S∨ is locally finite in V ∨ by [63, Theorem 4.9]. By (5),we have R∨ ⊗ 1R locally finite in X∨ ⊗ R, and therefore R∨ is locally finite in X∨

by Lemma 2.6.

The equality in (1) is (ReS4) for the reflection system (R,X). It is immediatefrom that equation that the sα∨ , α∨ ∈ R∨, are reflections satisfying condition (iii)of the Definition 2.8 of a root system.


2.15. Embeddings of root systems. Let f : (R,X)→ (S, Y ) be a morphismof root systems as defined in 2.8. In general, f will not be a morphism of theassociated reflection systems, see [63, 3.6]. We say f is an embedding of rootsystems if it satisfies the following equivalent conditions:

(i) f is injective and f(R) is a subsystem of S,

(ii) f is injective and f sα = sf(α) f holds for all α ∈ R×,

(iii) f is an injective morphism of reflection systems,

(iv) 〈f(x), f(α)∨〉 = 〈x, α∨〉 for all x ∈ X, α ∈ R,

(v) 〈f(β), f(α)∨〉 = 〈β, α∨〉 for all α, β ∈ R.

Proof. (i) =⇒ (ii): Let g: f(X) → X be the inverse of f : X → f(X), and fixα ∈ R×. Since f(R) is a subsystem of S and f(α) ∈ S, sf(α) stabilizes f(R), hencealso spanZ f(R) = f(X). Therefore, the map s := g sf(α) f is well-defined. It isa reflection, determined by α and the linear form x 7→ 〈f(x), f(α)∨〉, and satisfiess(α) = −α and s(R) = R. From Lemma 2.7 we conclude s = sα, whence (ii).Conversely, the condition f sα = sf(α) f implies that f(R) is a subsystem of S,proving (ii) =⇒ (i).

Since R R× = 0 and f(0) = 0, the equivalence (ii) ⇐⇒ (iii) follows from(2.3.4), which also shows that

(iii) ⇐⇒ f injective and f sα = sf(α) f for all α ∈ R×

⇐⇒ f injective and 〈x, α∨〉f(α) = 〈f(x), f(α)∨〉f(α) for all x ∈ X, α ∈ R×

⇐⇒ f injective and 〈x, α∨〉 = 〈f(x), f(α)∨〉 for all x ∈ X, α ∈ R×

=⇒ (iv).

Thus, to prove that, conversely, (iv) =⇒ (iii), it suffices to establish that (iv) impliesthat f injective. But, by (iv), any x ∈ Ker(f) lies in

⋂α∈R Ker(α∨) which vanishes

by (2.12.1) and nondegeneracy of the normalized inner product ( | ). Finally, theequivalence (iv) ⇐⇒ (v) holds since R spans X.

The proof above also shows that an embedding f : (R,X) → (S, Y ) of rootsystems satisfies

(vi) f(sα(β)

)= sf(α)f(β) for all α, β ∈ R,

(vii) f(sα(x)

)= sf(α)

(f(x)

)for all x ∈ X and α ∈ R.

Remark. We take the opportunity to correct an error in [63, Lemma 3.7].There it was asserted that the conditions (i) – (vii) stated above are all equivalent.The equivalent conditions (vi) and (vii) in fact do not imply (i) – (v), as can beseen from the example of a non-zero root system R and f = 0.

2.16. Classification of root systems. Under the correspondence betweenroot systems over Z and over R described in 2.9, the notions of direct sums,connectedness and irreducibility are preserved. Hence the classification of rootsystems over Z can be easily deduced from [63, §8] as follows.


Let I be a non-empty set, let L (I) = Z(I) =⊕

i∈I Zεi be the free abelian groupon the set I, and let t: L (I) → Z be the linear form determined by t(εi) = 1 forall i, called the trace form. We put

Ln(I) = t−1(nZ) = x ∈ L (I) : t(x) ∈ nZ,

and define further

AI = εi − εj : i, j ∈ I, (1)

DI = AI ∪ ±(εi + εj) : i 6= j, (2)

BI = DI ∪ ±εi : i ∈ I, (3)

CI = DI ∪ ±2εi : i ∈ I = ±εi ± εj : i, j ∈ I, (4)

BCI = BI ∪ CI = ±εi : i ∈ I ∪ ±εi ± εj : i, j ∈ I. (5)

Then AI is a root system in L0(I), BI and BCI are root systems in L (I), and CIand DI are root systems in L2(I), with the exception of DI for Card(I) = 1 whereDI = 0 is a root system in L0(I). The rank of AI is Card(I)− 1 while the rankin the other cases is Card(I). The notation A (instead of the traditional A) servesto indicate this fact. For a finite I, say |I| = n, we will use the standard notationBn = BI , Cn = CI , Dn = DI and BCn = BCI , while the usual notation for theroot systems of type A is linked to our notation by

An−1 = A0,...,n−1 = An. (6)

A root system R is called classical if it is isomorphic to one of the root systems (1)– (5) for a suitable, possibly infinite, set I. For infinite I, these root systems areirreducible and pairwise not isomorphic. For small ranks, there are the well-knownisomorphisms A1

∼= B1∼= C1, B2

∼= C2, D2∼= A1 ⊕ A1, D3

∼= A3. Because of thefirst isomorphism we will sometimes identify A1 with the set −1, 0, 1 ⊂ Z.

The exceptional root systems are the well-known finite irreducible root systemsof type E6,E7,E8,F4 and G2, see for example [18]. An irreducible root system iseither classical or isomorphic to an exceptional root system [63, Theorem 8.4].

Recall the notion of prenilpotent subset introduced in 1.9. The following lemmagives a detailed description of the prenilpotent two-element subsets of locally finiteroot systems.

2.17. Lemma. Let (R,X) be a locally finite root system, and let α, β ∈ R×.Then

α, β is prenilpotent ⇐⇒ α+ β 6= 0, 2α+ β 6= 0, α+ 2β 6= 0. (1)

Assume this to be the case and put Rαβ := R ∩(Zα+ Zβ

)and C :=

(((((((α, β

))))))). Then[[[[

α, β]]]]

= αc ∪ C ∪ βc (2)

is nilpotent of class k65 and of cardinality 66. Moreover, CardC64, Card(((((((C,C

)))))))6 1, and C 6= ∅ if and only if α+ β ∈ R.


Proof. By (1.5.2) and 1.13, α, β is not prenilpotent if and only if there existp, q ∈ N, p + q > 0, such that pα + qβ = 0. Since α, β 6= 0, this is equivalent topα + qβ = 0 for some p, q ∈ N+. Hence (1) follows from (2.2.5). Formula (2) is aconsequence of (1.6.9). The remaining assertions follow easily from the classificationof root systems of rank 6 2 in [18]. The details are left to the reader.

Note that (2) easily implies

C 2([[[[α, β

]]]])=(αc α

)∪ C ∪

(βc β

).

Also, αc = α, 2α or αc = α depending on whether 2α does or does notbelong to R.

We now list the cases where C 6= ∅ in more detail. It is no restriction to assumethat ‖α‖6 ‖β‖ with respect to the normalized inner product.

Case 〈α, β∨〉〈β, α∨〉 C =(((((((α, β

)))))))k

∣∣[[[[α, β]]]]∣∣ Rαβ

1 2 2 2α 2 2 BC1

2 1 1 α+ β 2 3 G2

3 0 0 α+ β 2 3 or 5 B2 or BC2

4 −1 −1 α+ β 2 3 A2

5 −1 −1 α+ β, 2α+ β, α+ 2β 3 5 G2

6 −1 −2 α+ β, 2α+ β 3 4 B2

7 −1 −2 α+ β, 2α+ β, 2α+ 2β 4 6 BC2

8 −1 −3α+ β, 2α+ β, 3α+ β,

3α+ 2β5 6 G2

Remarks. We put B := α, β.

Case 1: Here α = β.

Case 2: α and β are two short roots of G2 whose sum is a long root.

Case 3: α and β are weakly orthogonal short roots.

Case 4: B is a root basis of A2.

Case 5: B is a root basis for the subsystem of short roots of G2.

Case 6: Rαβ = B2 and B is a root basis of B2.

Case 7: B is a root basis of BC2.

Case 8: B is a root basis of G2.

§3] Groups with commutator relations 21

§3. Groups with commutator relations

3.1. Nilpotent pairs and division of roots. Let (R,X) ∈ SF and α, β ∈ R.Recall from (1.5.2) and (1.6.9) that the closed root interval from α to β is[[[[

α, β]]]]

= α, βc = αc ∪(((((((α, β

)))))))∪ βc

= R ∩ pα+ qβ : p, q ∈ N, p+ q > 1. (1)

We will call (α, β) a nilpotent pair if[[[[α, β

]]]]is a nilpotent subset of R, in other

words, if the subset α, β of R is prenilpotent, see 1.9. Clearly, if (α, β) is anilpotent pair then so is (β, α). By Proposition 1.13,

(α, β) is a nilpotent pair ⇐⇒[[[[α, β

]]]]is finite and 0 6∈

[[[[α, β

]]]]. (2)

On the other hand, one shows easily that, for α, β ∈ R×,

0 ∈[[[[α, β

]]]]⇐⇒ 0 ∈

(((((((α, β

)))))))⇐⇒ α ∈

(((((((α, β

)))))))⇐⇒ β ∈

(((((((α, β

))))))). (3)

Let α, β ∈ R. We say α divides β, written α∣∣β, if β ∈ N · α, i.e., β = nα for some

n ∈ N. Note that

α∣∣ 0 always holds, while 0

∣∣β implies β = 0. (4)

Divisibility is a partial order, in particular it is transitive:

α∣∣β and β

∣∣ γ =⇒ α∣∣ γ. (5)

3.2. Definition. Let G be a group. The inner automorphism of G defined bya ∈ G is

Int(a) · b = aba−1.

The commutator of a, b ∈ G is (((((((a, b)))))))

= aba−1b−1.

If S is a subset of G, we denote by⟨S⟩

the subgroup of G generated by S, andextend this notation to a family (Si)i∈I of subsets by⟨

Si : i ∈ I⟩

:=⟨⋃i∈I

Si

⟩.

For subsets S and T of G, we use the notation(((((((S, T

)))))))=⟨(((((((a, b)))))))

: a ∈ S, b ∈ T⟩

for the subgroup of G generated by all(((((((a, b)))))))

, a ∈ S, b ∈ T .Let (R,X) ∈ SF and let (Uα)α∈R be a family of subgroups of G. For a subset

A of R we put

UA =⟨Uα : α ∈ A

⟩;


thus in particular U∅ = 1. We say G has R-commutator relations with rootgroups (Uα)α∈R if G is generated by the Uα and the following conditions hold forall α, β ∈ R:

U0 = 1, (1)

α divides β =⇒ Uβ ⊂ Uα, (2)

(α, β) nilpotent =⇒(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))). (3)

By (3.1.4) and (1), the relation (2) holds automatically if α or β is zero. Wewill usually refer to the subgroups Uα as root groups. Because of (1) only the Uα,α 6= 0, are of interest. In particular, when considering examples it is sufficient tospecify Uα, α ∈ R×.

The condition that the family of root groups generate G is not serious. Itsimplifies the presentation in the general setting and is fulfilled for all groups thatwe will consider in the following chapters. Of course, if G is a group with a family(Uα)α∈R satisfying (1)–(3) then the subgroup

r(G, (Uα)) = UR (4)

generated by all root groups is a group which has R-commutator relations.

3.3. Examples. (a) The case R = A1. The reader will find many examplesof groups with R-commutator relations throughout this book. A very simple caseis R = −1, 0, 1 ⊂ Z, the root system of type A1. Then a group G has A1-commutator relations if and only if U1 = U+ and U−1 = U− are two abeliansubgroups generating G. In particular, the projective elementary group PE(V )of a Jordan pair V as in 7.5 has A1-commutator relations with root subgroupsU± = exp±(V ±).

(b) Commutator relations are inherited by homomorphic images: if ϕ: G→ His a surjective group homomorphism then H has R-commutator relations with rootgroups ϕ(Uα).

(c) Algebraic groups over fields. Let G be a connected reductive algebraic groupover an algebraically closed field in the sense of [13, 90], and let Φ = Φ(G,T ) bethe root system of G with respect to a maximal torus T . For α ∈ Φ let Uα be theroot group defined for example in [13, Theorem 13.18(4.d)]. Then R = Φ ∪ 0 isa reduced finite root system and the family (Uα)α∈R satisfies the R-commutatorrelations (3.2.1) – (3.2.3), [13, 14.5 (*)]. Hence the subgroup r(G) of G is a groupwith R-commutator relations. In particular, G itself is a group with R-commutatorrelations if G is semisimple [90, Theorem 8.1.5(i)].

More generally, let G be a connected reductive algebraic group defined over anarbitrary field k in the sense of [13, 90], and let Φ′ be the set of k-roots of Gwith respect to a maximal k-split torus T ′ of G. One knows [13, Theorem 21.6]that R′ = Φ′ ∪ 0 is a finite but not necessarily reduced root system. Moreover,for every α′ ∈ Φ′ there exists a unique closed connected unipotent k-subgroup U ′α′normalized by the centralizer of T ′ and with Lie algebra gα′ ⊕ g2α′ , where g2α′ = 0if 2α′ 6∈ Φ′ [13, Proposition 21.9]. The construction of the U ′α′ in [13] is a specialcase of the construction given in Lemma 4.18. Even if G is generated by the Uα,this will in general no longer hold for the U ′α′ .


The analogous statement also holds for the groups G(k) and U ′α′(k) of k-rationalpoints of G and the U ′α: the group r

(G(k)

)has R′-commutator relations with

respect to the family(U ′α′(k)

). However, G(k) is in general not generated by the

U ′α′(k), even when G is semisimple and simply connected, see for example [32,Ch. 2.2.E] for a discussion of this question for groups of type A (the Tannaka-Artinproblem).

(d) Split reductive group schemes. Let G be a split reductive group scheme overa scheme S [23, Exp. XXII, Def. 1.13]. Recall [23, Exp. XXII, Proposition 1.14]that the root system R of G is reduced. Let Uα, α ∈ R×, be the root subgroupsof G. Let S′ → S be a morphism of schemes. Then the root groups Uα = Uα(S′)satisfy the relations (3.2.2) – (3.2.3) [23, Exp. XXII, Corollary 5.5.2]. If G is simplyconnected and hence in particular semisimple, by [23, Exp. XXII, Proposition 4.3.4],and S′ is a local scheme, the group G = G(S′) is generated by the root subgroups[23, Exp. XXII, Corollary 5.7.6] and thus has R-commutator relations with respectto the family (Uα)α∈R.

Let in particular S = Spec(Z) and let S′ = Spec(k) where k is any field. ThenUR =

⟨Uα : α ∈ R

⟩is a Chevalley group in the sense of [94, §3] and hence has

R-commutator relations.

(e) Groups associated with Moufang buildings. Let B be a thick irreduciblespherical Moufang building over I = 1, . . . , l with l > 2 and different from anoctagon (we use the notation of [95, II, §5] and [104]). Let Φ be the set of roots ofan apartment A of B. It is known [95, p. 126] that Φ∪ 0 can be identified witha finite irreducible reduced root system. Moreover, there exist an irreducible finiteroot system R ⊃ Φ and a subgroup G ⊂ Aut(B) with root groups Uα, α ∈ R, suchthat G = UR has R-commutator relations. We have R = Φ∪0 if Φ is not of typeB or C, and R ∈ Bl,Cl,BCl otherwise depending on B.

The construction of these groups is for example given in [95, II, §5]. It isimmediate from this construction that the relation (3.2.2) holds. We remark thatit has to be verified only in case α and 2α ∈ R. The relation (3.2.3) for thenilpotent pair (α, α) is also clear from the construction. To verify (3.2.3) in theremaining cases, we may therefore, in view of 2.17, assume that α and β are Z-linearly independent. For l = 2 the commutator relation (3.2.3) then follows bycomparing the list in 2.17 with the one in [103] or [107, §5.4]. The case l > 3 canbe reduced to the case l = 2, see, e.g., [95, II, (5.7)]. For example, for a Moufangquadrangle of type F4 in the sense of [104, (16.7)], the root system R is of typeBC2, see [104, (40.59)].

(f) The reader can find more examples of groups with commutator relationsin 3.14 (nilpotent groups), 3.16 (groups with unique factorization, in particularelementary linear groups), 5.22 (rank one groups), 9.3 (elementary groups of specialJordan pairs), 9.18 (Steinberg groups Stn(A) for a ring A). In particular, wewill see in 3.14 that nilpotent groups provide natural examples of groups withR-commutator relations where αc can have any finite cardinality.

3.4. Remarks. Let G have R-commutator relations with root groups Uα.

(a) Suppose (α, β) ∈ R× × R× is not nilpotent, so that either 0 ∈[[[[α, β

]]]]or[[[[

α, β]]]]

is infinite. In the first case, (3.1.3) implies that even

24 GROUPS WITH COMMUTATOR RELATIONS [Ch. I⟨Uα ∪ Uβ

⟩⊂ U(((((α,β))))) (1)

holds. In the second case we do not have nor do we require any relations.

(b) In many interesting examples, for example, if R is a root system or, moregenerally, if R = Re(S) where S is an extended affine root system, R satisfies thefiniteness condition

(F1): αc is finite, for every α ∈ R. (2)

Then a pair (α, β) ∈ R××R× is nilpotent if and only if the commutator set(((((((α, β

)))))))is finite and does not contain zero.

(c) Suppose (R,X) satisfies the stronger finiteness condition

(F2):[[[[α, β

]]]]is finite, for all α, β ∈ R. (3)

Then also (F1) holds, and for α, β ∈ R× either (α, β) is nilpotent or 0 ∈[[[[α, β

]]]].

Hence by (1), a group with R-commutator relations actually satisfies(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) for all α, β ∈ R. (4)

This condition (F2) is always fulfilled if (R,X) is a locally finite root system, see2.8.

(d) Suppose(((((((α, α

)))))))⊂ 2α for all α ∈ R×, which holds if R is a root system.

Then (3.2.3) implies (((((((Uα, Uα

)))))))⊂ U2α,

(((((((Uα, U2α

)))))))= 1. (5)

Hence Uα is 2-step nilpotent (the derived group is central), and even abelian if Ris reduced.

3.5. Lemma. Let G be a group with R-commutator relations and let (α, β) bea nilpotent pair.

(a) For all γ, δ ∈[[[[α, β

]]]], the pair (γ, δ) is nilpotent and satisfies(((((((

γ, δ)))))))⊂(((((((α, β

))))))). (1)

(b) Uα and Uβ normalize U(((((α,β))))), and

U[[[α,β]]] = Uα · U(((((α,β))))) · Uβ . (2)

Proof. (a) Clearly[[[[γ, δ]]]]⊂[[[[α, β

]]]], and

(((((((γ, δ)))))))⊂(((((((α, β

)))))))holds by (1.6.8). Since

subsets of prenilpotent sets are prenilpotent (cf. (1.9.2)), we have (a).

(b) Let γ, δ ∈[[[[α, β

]]]]. By (a) and the commutator relation (3.2.3), we have

Int(Uγ) · Uδ ⊂(((((((Uγ , Uδ

)))))))· Uδ ⊂ U(((((γ,δ))))) · Uδ. (3)

In particular, for γ ∈ α, β and δ ∈(((((((α, β

))))))), (3) and (1) imply Int(Uγ)·Uδ ⊂ U(((((α,β))))),

from which it follows that Uα and Uβ normalize U(((((α,β))))). Hence Uα · U(((((α,β))))) =U(((((α,β))))) · Uα and K = Uβ · U(((((α,β))))) = U(((((α,β))))) · Uβ are subgroups of G. The inclusionfrom right to left in (2) is clear from (3.1.1). For the reverse inclusion, let H denotethe right hand side of (2). Note first that, because of (3.1.1) and the relation (3.2.2),H contains all Uγ , γ ∈

[[[[α, β

]]]]. Hence it suffices to show that H is a subgroup of

G. Now, for γ, δ = α, β, (3) shows that Int(Uα) · Uβ ⊂ U(((((α,β))))) · Uβ , andInt(Uβ) · Uα ⊂ U(((((α,β))))) · Uα. Thus U(((((α,β))))) · Uβ is a subgroup of G which is itselfnormalized by Uα. This proves that H is a subgroup, and hence that (2) holds.


3.6. Commutator formulas. Let G be a group. We denote by Z (G) thecentre of G. Then the following formulas hold:(((((((

a, b)))))))−1 =

(((((((b, a))))))), (1)(((((((

ab, c)))))))

=(

Int(a) ·(((((((b, c))))))))·(((((((a, c)))))))

=(((((((a,(((((((b, c))))))))))))))·(((((((b, c)))))))·(((((((a, c))))))), (2)(((((((

a, bc)))))))

=(((((((a, b)))))))·(

Int(b) ·(((((((a, c))))))))

=(((((((a, b)))))))·(((((((a, c)))))))·((((((((((((((c, a))))))), b))))))), (3)(((((((

a−1, b)))))))

= a−1(((((((a, b)))))))−1a, (4)(((((((

a, b−1)))))))

= b−1(((((((a, b)))))))−1b. (5)(((((((

a,(((((((b, c))))))))))))))

=(((((((ab, c

)))))))·(((((((c, a)))))))·(((((((c, b))))))), (6)((((((((((((((

a, b))))))), c)))))))

=(((((((a, b)))))))·(((((((c, b)))))))·(((((((b, ca

))))))), (7)

a ≡ a′ and b ≡ b′ mod Z (G) =⇒(((((((a, b)))))))

=(((((((a′, b′

))))))). (8)

If(((((((a, c)))))))∈ Z (G) then(((((((a,(((((((b, c))))))))))))))

=((((((( (((((((

a, b))))))),(((((((b, c))))))) )))))))·((((((( (((((((

b, c))))))),((((((((((((((a, b))))))), c))))))) )))))))·((((((((((((((a, b))))))), c))))))). (9)

We also have the following relations, see [15, I, §6.2]. Formula (12) is a group-theoretic analogue of the Jacobi identity in Lie algebras.(((((((

a, bc)))))))·(((((((b, ca

)))))))·(((((((c, ab

)))))))= 1, (10)(((((((

ab, c)))))))·(((((((ca, b

)))))))·(((((((bc, a

)))))))= 1, (11)(((((((

Int(b) · a,(((((((c, b))))))))))))))·(((((((

Int(c) · b,(((((((a, c))))))))))))))·(((((((

Int(a) · c,(((((((b, a))))))))))))))

= 1. (12)

The proofs are straightforward verifications.

3.7. Lemma. Let G be a group, H a subgroup and let X1, X2 be subsets of Gnormalizing H and satisfying

(((((((X1, X2

)))))))⊂ H. Then the subgroups Gi generated by

Xi normalize H and(((((((G1, G2

)))))))⊂ H.

Proof. Since the normalizer of any subset is a subgroup, it is clear that the Ginormalize H. For the proof of the second claim, we first note that

(((((((x±1

1 , x±12

)))))))∈ H

by (3.6.4) and (3.6.5). We may therefore assume Xi = X−1i , so that Gi is the

submonoid generated by Xi. Then(((((((g1, g2

)))))))∈ H for all gi ∈ Gi by a straightforward

induction, using (3.6.2) and (3.6.3), and this in turn implies(((((((G1, G2

)))))))⊂ H.

3.8. Lemma. Let (R,X) ∈ SF and let G be a group with a family of subgroupsUα indexed by α ∈ R, which generate G and satisfy (3.2.1) and (3.2.2). For eachα ∈ R let Xα = X−1

α ⊂ Uα be a symmetric set of generators of Uα, and supposethat (((((((

Xα, Xβ

)))))))⊂ U(((((α,β))))) (1)

holds for all nilpotent pairs (α, β). Then G has R-commutator relations with rootgroups Uα.

Proof. It remains to verify the commutator relation (3.2.3) for every nilpotentpair (α, β). To do so, we apply Lemma 3.7 to X1 = Xα, X2 = Xβ and H = U(((((α,β))))),so we must show that Xα and Xβ normalize H. Now H is generated by all Uγ ,γ ∈

(((((((α, β

))))))), and (α, γ) is a nilpotent pair, by 3.5(a). Thus our hypothesis (1) yields


xαyγx−1α ∈ U(((((α,γ))))) · yγ ⊂ H for all xα ∈ Xα, yγ ∈ Xγ . As Uγ is generated by Xγ ,

this implies xαUγx−1α ⊂ H. By definition, H is generated by all Uγ , γ ∈

(((((((α, β

))))))).

Conjugation with xα is an automorphism so xαHx−1α ⊂ H. As also x−1

α ∈ Xα bysymmetry of Xα, we have xαHx

−1α = H, so xα does normalize H. In the same way,

one shows that Xβ normalizes H. Now Lemma 3.7 yields(((((((Uα, Uβ

)))))))⊂ H.

Recall [84, p. 117] that H = Z1 ⊃ Z2 ⊃ · · · ⊃ Zh+1 = 1 is a central chainof a group H if the Zi are normal in H and Zi/Zi+1 ⊂ Z (H/Zi+1). In particular,the lower and upper central series of H are defined inductively by C 1(H) = H,C n+1(H) =

(((((((H, C n(H)

)))))))and Z0(H) = 1 and Zn(H) = a ∈ H :

(((((((a,H

)))))))⊂

Zn−1(H), respectively.

3.9. Lemma. Let G be a group with R-commutator relations.

(a) Let A and B be subsets of R with the property that, for all α ∈ A, β ∈ B andγ ∈

(((((((A,B

))))))), the pairs (α, β), (α, γ) and (β, γ) are nilpotent. Then the subgroups

UA and UB normalize U(((((A,B))))), and the generalized commutator relations(((((((UA, UB

)))))))⊂ U(((((A,B))))) (1)

hold.

(b) Let A be a strictly positive subset (cf. 1.5) of R with the property that (α, β)is a nilpotent pair, for all α, β ∈ A. Then a central chain A ⊃ A1 ⊃ A2 ⊃ · · · in Agives rise to a central chain UA ⊃ UA1

⊃ UA2⊃ · · · in UA. The lower and upper

central series of A and UA are related by

C n(UA) ⊂ UCn(A), UZn(A) ⊂ Zn(UA). (2)

If D ⊂ A is a normal subset then UD is a normal subgroup of UA.

Proof. (a) Let C :=(((((((A,B

)))))))and put X1 :=

⋃α∈A Uα, X2 :=

⋃β∈B Uβ and

H := UC . Since UA =⟨X1

⟩and UB =

⟨X2

⟩, our claim will follow from Lemma 3.7

once we verify the assumptions of that lemma. First, since the pair (α, β) isnilpotent for all α ∈ A, β ∈ B and satisfies

(((((((α, β

)))))))⊂ C by (1.6.8), we obtain(((((((

X1, X2

)))))))⊂ UC = H. Thus, in order to apply 3.7, it remains to show that X1

and X2 normalize H. By symmetry, it is enough to do so for X1. Let α ∈ A andγ ∈ C. Then (α, γ) is a nilpotent pair by assumption, and

(((((((α, γ

)))))))⊂(((((((A,C

)))))))⊂ C

by (1.6.10). Hence, for all xα ∈ Uα,

xαUγx−1α ⊂

(((((((Uα, Uγ

)))))))· Uγ ⊂ U(((((α,γ))))) · Uγ ⊂ UC = H.

This implies xαHx−1α ⊂ H because the Uγ generate H, and even xαHx

−1α = H

because Uα, being a subgroup, is closed under inversion. Thus X1 normalizes H,as required.

(b) Formula (1) applies to any subset B of A. Indeed,(((((((A,B

)))))))⊂ A because A

is closed, so our assumption on A shows that the property required in (a) holds. By1.9, a central chain in A satisfies

(((((((A,Ai

)))))))⊂ Ai+1. Hence

(((((((UA, UAi

)))))))⊂ U(((((A,Ai))))) ⊂

UAi+1 , showing the UAi form a central chain in UA.For n = 1 the first formula of (2) is clear. The induction step follows by putting

B = C n(A), whence C n+1(UA) =(((((((UA,C n(UA)

)))))))⊂(((((((UA, UB

)))))))(by induction)

⊂ U(((((A,B))))) = UCn+1(A) (by (1.7.1)).


The second formula of (2) obviously holds for n = 0. Assume it is true for n− 1and let α ∈ Zn(A). Then

(((((((Uα, UA

)))))))⊂ U(((((α,A))))) ⊂ UZn−1(A) (by (1.8.1)) ⊂ Zn−1(UA)

(by induction). A normal subset D of A satisfies(((((((A,D

)))))))⊂ D. Hence the last

statement follows from Int(UA) · UD ⊂(((((((UA, UD

)))))))UD ⊂ U(((((A,D)))))UD ⊂ UD.

The following lemma is due to J. Tits [102, 4.7, Lemma 2], for (b) see also [94,Lemma 18].

3.10. Lemma. Let H be a group generated by subgroups H1, . . . ,Hn. Supposethat H has a central chain H = Z1 ⊃ Z2 ⊃ · · · ⊃ Zh ⊃ Zh+1 = 1 such that, forall i ∈ 1, . . . , h, there exists j ∈ 1, . . . , n for which the inclusion Zi ⊂ Hj ·Zi+1

holds. Then:

(a) For every permutation σ ∈ Sn, the product map Hσ(1) × · · · ×Hσ(n) → His surjective.

(b) If that map is injective for one permutation σ, it is injective for all σ.

3.11. Indivisibility. Let A ⊂ R. A root β ∈ A is said to be indivisible in Aif β 6= 0 and, for all α ∈ A, the relation α

∣∣β implies α = β. We denote the set ofindivisible roots in A by Aind. Indivisibility of a root is not an absolute propertybut depends on the set A containing α:

B ⊂ A =⇒ B ∩Aind ⊂ Bind, (1)

and this is in general a proper inclusion. For example, if α and 2α belong to R×

then 2α is indivisible in B = 2α while it is divisible in A = α, 2α.

3.12. Proposition. Let G be a group with R-commutator relations and rootgroups Uα.

(a) If A ⊂ R is a nilpotent subset of class at most k then UA is a nilpotentsubgroup of G of class at most k.

(b) Let A be a finite nilpotent subset of R. Then UA is nilpotent of class atmost Card(A), and for any ordering Aind = β1, . . . , βn,

UA = UAind= Uβ1

· · · Uβn . (1)

If the product map Uβ1× · · · ×Uβn → UA is injective for one ordering of Aind then

it is so for all orderings.

Proof. (a) A is strictly positive by (1.9.5), and (α, β) is a nilpotent pair for allα, β ∈ A. Thus Lemma 3.9(b) is applicable to A, and the assertion follows fromthe definition of nilpotence of a given class in (1.9.8) which is analogous to thedefinition for groups.

(b) Let h = Card(A), order A = α1, . . . , αh as in Lemma 1.11(a), andconsider the central chain Ai = αi, . . . , αh of A as in Lemma 1.11(b). By 3.9(b),the Zi := UAi form a central chain in UA. Since Z1 = UA and Zh+1 = 1, UA isnilpotent of class at most h. Now let Aind = β1, . . . , βn and put Xj := Uβj , forj = 1, . . . , n. By finiteness of A, for every α ∈ A there exists a β ∈ Aind dividingα. Hence for every i ∈ 1, . . . , h there exists some j = j(i) ∈ 1, . . . , n such that


βj∣∣αi, and therefore Xj = Uβj ⊃ Uαi (by (3.2.2)). This shows that the Xj generate

UA. The members of a central chain are normal subgroups. Hence Uαi · Zi+1 is asubgroup of UA, and therefore

Zi =⟨Uαi ∪ · · · ∪ Uαn

⟩=⟨Uαi ∪ Zi+1

⟩= Uαi · Zi+1 ⊂ Xj(i) · Zi+1.

Now the assertion follows from Tits’ Lemma 3.10.

3.13. Corollary. Let G be a group having R-commutator relations with rootgroups Uα.

(a) If αc is finite then Uα is nilpotent.

(b) If A ⊂ R is locally nilpotent in the sense of 1.15 then UA is a locallynilpotent group.

(c) Let R be a locally finite root system and let (α, β) be a nilpotent pair whichdoes not fall under the cases 7 or 8 of the table in 2.17. Then U(((((α,β))))) is abelian. In

particular, this is so if Card(((((((α, β

)))))))6 2.

Proof. (a) We may assume α 6= 0. The set A := αc is finite by assumption,and obviously strictly positive, hence nilpotent by Proposition 1.13. As Aind = α,the assertion follows from Proposition 3.12.

(b) We must show that every finite subset E of UA is contained in a nilpotentsubgroup. Now E ⊂ UF where F is a suitable finite subset of A. By Corollary 1.15,B := F c is nilpotent, and therefore so is UB by Proposition 3.12(a). Since UF ⊂ UBwe are done.

(c) It follows from 2.17 that in the cases 1 – 6, the set A =(((((((α, β

)))))))is abelian.

Hence by Proposition 3.12(a), UA is abelian as well.

We now show how nilpotent groups fit into our framework.

3.14. Corollary. For k ∈ N+ let R = 0, 1, . . . , k ⊂ X = Z. Then any groupG with R-commutator relations and root groups (Ui)i∈R is nilpotent of class at mostk. Conversely, if G is a nilpotent group of class at most k then G has R-commutatorrelations with root groups U0 = 1 and Ui = C i(G) for i = 1, . . . , k.

Proof. Let A = R× = 1, . . . , k. One shows easily by induction that C i(A) =i, . . . , n, so A is nilpotent of class k. By Proposition 3.12(a), G = UA is nilpotentof class 6 k.

Conversely, let G be nilpotent of class 6 k and put Ui = C i(G) for i = 1, . . . , k(and of course U0 = 1). Then the R-commutator relations hold. Indeed, thecondition (3.2.2) follows from the fact that Ui ⊃ Uj for i 6 j. To verify (3.2.3)observe that the nilpotent pairs of R are the pairs (i, j) with 1 6 i, j 6 k. Theysatisfy i+ j ∈

(((((((i, j)))))))⊂ l ∈ N : i+ j6 l6k if i+ j6k, while

(((((((i, j)))))))

= ∅ if i+ j > k.It now follows that(((((((

Ui, Uj)))))))

=(((((((C i(G),C j(G)

)))))))⊂ C i+j(G) = Ui+j ⊂ U(((((i,j)))))

in either case.


Remark. There is a similar result for a group G generated by two nilpotentsubgroups, say U+ of class at most k and U− of class at most l, respectively. Indeed,put R = −l, . . . ,−1, 0, 1, . . . , k ⊂ Z. Then

G has R-commutator relations with root groups U±i = C i(U±). (1)

Conversely, ifG hasR-commutator relations with root groups Ui then the subgroupsU1 and U−1 are nilpotent of class 6 k and of class 6 l respectively. Details are leftto the reader.

3.15. Groups with unique factorization. Let G have R-commutator rela-tions and let A ⊂ R be a finite nilpotent subset. We say G has unique factorizationfor A if there exists an enumeration Aind = γ1, . . . , γn such that the product map

µ: Uγ1 × · · · × Uγn → UA

is injective. Recall from Proposition 3.12(b) that µ is surjective and that theinjectivity of µ is independent of the choice of enumeration. A group having uniquefactorization for all finite nilpotent subsets is said to have (unqualified) uniquefactorization.

It is convenient to introduce the following weaker form, called unique factoriza-tion for nilpotent pairs: For all nilpotent pairs (α, β), unique factorization holdsfor the sets

[[[[α, β

]]]]and

(((((((α, β

))))))).

In general, unique factorization for[[[[α, β

]]]]will not imply this property for(((((((

α, β)))))))

. For example, let α ∈ R× and assume that αc = α, 2α, 3α. Then[[[[α, α

]]]]ind = α so unique factorization for

[[[[α, α

]]]]holds trivially, while

(((((((α, α

)))))))=

2α, 3α =(((((((α, α

)))))))ind, and so unique factorization for

(((((((α, α

)))))))means that U2α∩U3α =

1. On the other hand,

α and β linearly independent =⇒(((((((α, β

)))))))ind ⊂

[[[[α, β

]]]]ind,

and hence unique factorization for[[[[α, β

]]]]implies that for

(((((((α, β

))))))).

For the proof, let γ ∈(((((((α, β

)))))))ind and assume that γ is divisible by some δ 6= γ in[[[[

α, β]]]]. Thus δ = pα+ qβ where p, q ∈ N, and γ = npα+nqβ for some n> 2. Since

α and β are linearly independent,(((((((α, β

)))))))contains no multiple of α or β, so we have

np> 1 and nq> 1. But then also p> 1 and q> 1, whence δ ∈(((((((α, β

))))))), contradicting

the fact that γ is indivisible in(((((((α, β

))))))).

As a consequence, we see: if R has Card(((((((α, β

)))))))6 1 for all linearly dependent

nilpotent pairs, which is for instance the case when R is a locally finite root systemby 2.17, then unique factorization for nilpotent pairs follows from that for all

[[[[α, β

]]]].

3.16. Examples. (a) The condition that the map µ of 3.15 be injective istrivially fulfilled for any finite nilpotent A with Card(Aind) = 1. For example, letR = A1 = −1, 0, 1, let V be a Jordan pair, and let G = PE(V ). As noted in3.3(a), G then has R-commutator relations with root groups U±1 = U±. Since 1and −1 are the only nilpotent subsets of R, PE(V ) has unique factorization.

(b) Here is an example which shows that not every group has unique fac-torization for nilpotent pairs and that unique factorization for nilpotent pairs is


weaker than unqualified unique factorization. Let X = Zn with standard basisB = ε1, . . . , εn and let R = 0 ∪ B. Then B is an abelian subset of R in thesense of 1.9. A group G with R-commutator relations is a group with a family ofabelian subgroups Ui = Uεi which commute pairwise. We may replace G by thesubgroup generated by the Ui. Then G is commutative, and in additive notation,we have G =

∑ni=1 Ui. The nilpotent pairs (α, β) are the pairs (εi, εj), and since(((((((

εi, εj)))))))

= ∅, we have[[[[εi, εj

]]]]= εi, εj, so that U[[[εj ,εj ]]] = Ui + Uj . Hence G has

unique factorization for all nilpotent pairs if and only if Ui∩Uj = 0 for i 6= j. Onthe other hand, G has (unqualified) unique factorization if and only if G =

⊕ni=1 Ui.

(c) Elementary linear groups. Let I be an index set, let A be a unital associativering and M = A(I) the free right A-module with standard basis (ei)i∈I . LetEI(A) ⊂ GL(M) be the elementary linear group, that is, the subgroup of GL(M)generated by all transvections

eij(a) = Id + Eij(a) (a ∈ A, i 6= j),

where the Eij(a) are the usual matrix units mapping ek to δjkeia. Let R = AI =εi − εj : i, j ∈ I be the locally finite root system as in (2.16.1). Then it is

well-known and easy to see (cf. [32]) that G = EI(A) has AI -commutator relationsand root groups Uεi−εj = eij(A) for i 6= j. Moreover, it is well-known that G hasunqualified unique factorization. But we will give now an elementary proof for thespecial case of unique factorization for nilpotent pairs.

Thus, let (α, β) be a nilpotent pair in R. In the present situation, this meansα = εi − εj , β = εk − εl where i 6= j, k 6= l, and α + β 6= 0. It suffices to treatthe case α 6= β. By 2.17, either α + β /∈ R× and then

(((((((α, β

)))))))= ∅, or α + β ∈ R×

and then(((((((α, β

)))))))= α + β (all this holds for any simply laced locally finite root

system).

Case 1: α+β /∈ R. This is equivalent to j 6= k and i 6= l. We must show that themultiplication map Uα×Uβ → G is injective, which is equivalent to Uα∩Uβ = 1.Thus assume eij(a) = ekl(b) ∈ Uα ∩ Uβ . Applying this to ej and el yields

eij(a) · ej = ej + eia = ekl(b) · ej = ej + δjlekb,

eij(a) · el = el + δjleia = ekl(b) · el = el + ekb.

If i 6= k then these equations imply a = b = 0, as required. If i = k then necessarilyj 6= l, otherwise α = β. Hence these equations again show a = b = 0.

Case 2: γ = α + β ∈ R×. Possibly after exchanging α and β we may assumej = k and have i 6= l, so γ = εi − εl. We show that the multiplication map Uα ×Uβ × Uγ → G is injective. Let g = eij(a)ejl(b)eil(c). Then a simple computationshows that

g · ej = ej + eia, g · el = el + ejb+ ei(c+ ab).

This shows that a, b, c are uniquely determined by g and proves our claim.

(d) All the groups considered in Examples 3.3(c) – (e) have unique factorization.Indeed, if G is a connected semisimple algebraic group defined over a field k and sohas commutator relations with respect to some finite root system R, it follows from[65, Lemma 3.4] that any nilpotent subset A ⊂ R lies in a positive system of R.


Then [13, Prop. 14.5] shows that the product map∏γ∈A Uγ → UA is bijective if k

is algebraically closed. The case of an arbitrary base field k then follows from thealgebraically closed case. For split reductive group schemes as in Example 3.3(d)unique factorization is a consequence of [23, Exp. XXII, Prop. 5.5.1] while forChevalley groups this is proved in [94, p. 24, Lemma 17].

For the groups in Example 3.3(e) one can argue as above: any nilpotent subsetlies in a positive system so that it suffices to know that the product map

∏γ∈P Uγ →

UP is bijective. This is for example proved in [104, 8.10] for l = 2 or [110,Proposition 11.11] in general. More examples are discussed in 4.15. Theorem 21.19furnishes another class of examples. The following lemma will be useful in the proofof that theorem.

3.17. Lemma. If a subset A of R is the disjoint union of two closed subsetsB and C, then

Aind = Bind ∪ Cind. (1)

Proof. Since B and C are disjoint so is the union on the right hand side of (1).By (3.11.1) we have B∩Aind ⊂ Bind and C∩Aind ⊂ Cind. This proves the inclusionfrom left to right. Conversely, let β ∈ Bind and assume β /∈ Aind. Then β = nα forsome α ∈ A and n> 2. We cannot have α ∈ B because β is indivisible in B. Henceα ∈ C. But then also nα ∈ C because C is closed, whence β ∈ Bind ∩ C = ∅, acontradiction which proves β ∈ Aind. By symmetry, we have Cind ⊂ Aind, so theinclusion from right to left in (1) holds as well.

3.18. Corollary. Let α and β be linearly independent. Then[[[[α, β

]]]]ind = α ∪

(((((((α, β

)))))))ind ∪ β.

If moreover (α, β) is a nilpotent pair and G is a group with R-commutator relationsand root groups Uα and unique factorization for nilpotent pairs, then Uβ ∩U(((((α,β))))) =1.

Proof. From linear independence it follows that the union A :=[[[[α, β

]]]]=

αc ∪(((((((α, β

)))))))∪ βc is disjoint. Moreover, B := αc is closed by definition, and

C :=(((((((α, β

)))))))∪βc is easily seen to be closed. By Lemma 3.17, Aind = Bind ∪Cind,

and obviously Bind = α. Repeating this argument for the disjoint decompositionC =

(((((((α, β

)))))))∪ βc into two closed subsets yields the first assertion. The second is

then immediate from the definitions.

3.19. Lemma. Let G be a group with R-commutator relations and assume Rand G have the following property: for every finite non-empty nilpotent subset A ofR there exists α0 ∈ A such that

(i) B := A α0c is closed, and

(ii) Uα0 ∩ UB = 1.

Then G has unique factorization.

Proof. Let A be a finite nilpotent subset of R and assume α0 ∈ A satisfiescondition (i). Then


Aind = α0 ∪Bind (1)

by Lemma 3.17. We show by induction on n = Card(Aind): there exists anenumeration Aind = α1, . . . , αn such that the multiplication map Uα1

× · · · ×Uαn → G is injective.

This is trivial for Card(Aind) = 1. Let Card(Aind) = n+ 1 and let α0 ∈ A andB be as in (i). Observe that B is nilpotent, being a closed subset of a nilpotent set,see (1.9.4) and (1.9.2). By (1), Bind has cardinality n, so by induction hypothesis,there exists an enumeration Bind = α1, . . . , αn such that the multiplication mapUα1× · · · × Uαn → G is injective. Let gi, hi ∈ Uαi for i = 0, . . . , n, and assume

g0g1 · · · gn = h0h1 · · ·hn. Then h−10 g0 = h1 · · ·hng−1

n · · · g−11 ∈ Uα0

∩ UB = 1.This implies g0 = h0, hence also g1 · · · gn = h1 · · ·hn, and therefore, by induction,gi = hi for i = 1, . . . , n. Hence the multiplication map Uα0 × · · · × Uαn → G isinjective.

Remark. Condition (i) is always fulfilled if R is reduced in the sense thatR ∩ N+α = α for every α ∈ R×. Indeed, let h be a positive functional as inCorollary 1.14, and choose α0 ∈ A with h(α0) minimal. If B = A α0 were notclosed there would exist β1, . . . βp ∈ B such that β1 + · · ·+ βp = α0. Applying h tothis equation yields p = 1 and α0 = β1 ∈ B, contradiction.

§4. Categories of groups with commutator relations

4.1. Definition. Fix (R,X) ∈ SF. We define a category gcR as follows: itsobjects are pairs (G,U) consisting of a group G having R-commutator relationswith respect to the family U = (Uα)α∈R of subgroups, called root groups, see 3.2.Let also (H,V) = (H, (Vα)α∈R) ∈ gcR. A morphism

ϕ: (G,U)→ (H,V)

is a group homomorphisms ϕ: G → H preserving root groups: ϕ(Uα) ⊂ Vα for allα ∈ R.

There is an obvious forgetful functor Φ: gcR → grp, the category of groups,which sends an object (G,U) to G, and a morphism ϕ: (G,U) → (H,V) toϕ: G → H. Clearly, Φ is faithful (injective on morphisms). Recall that monomor-phisms resp. epimorphisms in any category are defined by the property of beingleft resp. right cancellable: f g = f h resp. g f = h f implies g = h. In thecategory of groups, the monomorphisms and epimorphisms are just the injectiveand surjective group homomorphisms. If ϕ is a morphism of gcR and Φ(ϕ) is amonomorphism resp. epimorphism of groups then it has the same property in gcR.This follows immediately from the fact that Φ is faithful. The converse is false: wewill see later that a monomorphism in gcR need not be an injective homomorphismof the underlying groups. To simplify notation, we often write an object (G,U) ofgcR simply as G, as long as no confusion appears possible.

Let ϕ: (G,U)→ (H,V) be a morphism of gcR, and define

ϕα = ϕ∣∣Uα: Uα → Vα,

for all α ∈ R. Then the assignments (G,U) 7→ Uα on objects and ϕ 7→ ϕα onmorphisms define a functor Φα: gcR → grp.

§4] Categories of groups with commutator relations 33

4.2. Definition. A morphism ϕ: (G,U) → (G′,U′) of gcR is called injective(surjective, bijective) on root groups if ϕα: Uα → U ′α has the respective propertyfor all α ∈ R. Morphisms of this type are stable under composition, and hencedefine (non-full) subcategories of gcR.

Since G′ is generated by its root subgroups, a morphism of gcR which is surjec-tive on root groups is actually surjective (and hence an epimorphism). An analogousresult does not hold for morphisms which are injective on root groups. However,such morphisms are still monomorphisms of gcR. Indeed, assume ϕ: G → G′ isinjective on root groups and we have morphisms ψ1, ψ2: (H,V) → (G,U) of gcRsuch that ϕ ψ1 = ϕ ψ2. As Φα is a functor, ϕα ψ1,α = ϕα ψ2,α for all α, henceψ1,α = ψ2,α for all α by injectivity of ϕα. Since the Vα generate H, it follows thatψ1 = ψ2.

Below there is a stronger positive result on injectivity under suitable assump-tions on unique factorization. Recall the notation UA =

⟨Uα : α ∈ A

⟩of 3.2 for a

subset A of R.

4.3. Lemma. Let ϕ: (G,U)→ (G′,U′) be a morphism of gcR.

(a) If ϕ is surjective on root groups then ϕ(UA) = U ′A for any subset A of R.

(b) Suppose ϕ is injective on root groups and G′ has unique factorization for afinite nilpotent subset A of R, see 3.15. Then G has unique factorization for A aswell, and ϕ: UA → U ′A is injective.

(c) Let again ϕ be injective on root groups. If G′ has unique factorization forall finite nilpotent subsets then so does G, and ϕ: UA → U ′A is injective, for all(possibly infinite) nilpotent subsets A of R.

(d) If ϕ is bijective on root groups and G′ has unique factorization then G hasunique factorization, and ϕ: UA → U ′A is an isomorphism, for all nilpotent A ⊂ R.

Proof. (a) This is evident from the definitions.

(b) Enumerate Aind = γ1, . . . , γn. Then the diagram

Uγ1 × · · · × Uγnµ //

UA

U ′γ1 × · · · × U

′γn µ′

// U ′A

commutes, where the horizontal maps are the product maps of G and G′, respec-tively, and the vertical maps are induced by ϕ. By (3.12.1) the horizontal maps aresurjective. Since ϕ is injective on root groups, the left hand map is injective, andsince G′ has unique factorization for A, the bottom map µ′ is injective. Hence µ andthe right hand map are injective as well. In particular, G has unique factorizationfor A.

(c) By (b), G has unique factorization for all finite nilpotent subsets. Now letA be an arbitrary nilpotent subset, and let u ∈ UA with ϕ(u) = 1. Then thereexists a finite subset F ⊂ A such that u ∈ UF . By Proposition 1.13, the closure F c

is finite and nilpotent, so after replacing F by F c we may assume F nilpotent. Bywhat we proved in the finite case, u = 1, as desired.

(d) This follows immediately from (a) – (c).


4.4. Definition. A morphism ϕ: (G,U)→ (G′,U′) of gcR is called a coveringif

(i) ϕ is bijective on root groups, and

(ii) ϕ: U[[[α,β]]] → U ′[[[α,β]]] is bijective, for all nilpotent pairs (α, β).

The reader may wonder about the relation between (i) and (ii). The followingexamples show that these conditions are in general independent.

(i) does not imply (ii): Let ε1, ε2 be the standard basis of X = Z2 and letR = 0, ε1, ε2 ⊂ X. Let e1, e2 be the standard basis of G := Z2 and put Uεi = Z·ei.Define G′ = Z with U ′εi = Z and let π: G → G′ be defined by π(ei) = 1. All pairs(εi, εj) for i, j ∈ 1, 2 are nilpotent pairs. Then π is bijective on all Uα, butπ∣∣U[[[ε1,ε2]]] is not.

(ii) does not imply (i): Let R = N ⊂ X = Z, let G = Z (additive group), andput U1 = G and Un = 0 for n 6= 1. Let G′ = 1 and let π: G→ G′ be the onlypossible map. There are no nilpotent pairs, because

[[[[α, β

]]]]either contains 0 (in

case α = 0 or β = 0), or it is infinite. Hence (ii) holds trivially. On the other hand,π: U1 → U ′1 = G′ = 1 is not injective.

However, if R satisfies the finiteness condition (F1) of (3.4.2), that is, if αc isfinite for all α, then (ii) implies (i). Indeed, then any pair (α, α) (for α ∈ R×) isnilpotent, and

[[[[α, α

]]]]= αc as well as U[[[α,α]]] = Uα because of the relation (3.2.2).

In particular, this is so in the important case where R is a locally finite root system.

4.5. Lemma. (a) Coverings are closed under composition. They are bothmonomorphisms and epimorphisms of the category gcR, but in general not isomor-phisms.

(b) Consider a commutative diagram

G′ϕ //

ϕ′ AAAAAAAA G′′

ϕ′′~~||||||||

G

of gcR. If ϕ′ is a covering and ϕ is surjective on root groups then both ϕ and ϕ′′

are coverings. Moreover, if two of the morphisms are coverings then so is the third,and they determine the third uniquely.

(c) Assume G′ ∈ gcR has unique factorization for all nilpotent sets of the form[[[[α, β

]]]]where (α, β) is a nilpotent pair. Then a morphism ϕ: G → G′ of gcR is a

covering if and only if it is bijective on root groups.

Proof. (a) follows easily from the remarks made in 4.2 and the definitions. For(b), consider the induced commutative diagrams

U ′αϕα //

ϕ′α BBBBBBBU ′′α

ϕ′′α~~||||||||

Uα

U ′[[[α,β]]]

ϕαβ //

ϕ′αβ ##GGGGGGGGU ′′[[[α,β]]]

ϕ′′αβwwwwwwww

U[[[α,β]]]


Assume first that ϕ′ is a covering. Then ϕ′α and ϕ′α,β are isomorphisms and thediagrams show that ϕα and ϕαβ are injective, and ϕ′′α and ϕ′′αβ are surjective. If ϕ issurjective on root groups then the maps ϕα and ϕαβ are surjective, hence bijective,so ϕ is a covering, and then so is ϕ′′. If ϕ′′ is injective on root groups, then the mapsϕ′′α and ϕ′′αβ are injective, hence bijective, whence ϕ′′ and then ϕ are coverings. Ifboth ϕ and ϕ′′ are coverings, the commutative diagrams above show that ϕ′ is acovering too. Finally, (c) is an immediate consequence of Lemma 4.3.

4.6. Definition. An object G = (G,U) of gcR is called simply connected ifevery covering ϕ: H → G is an isomorphism. Thus G is simply connected if it doesnot admit proper coverings.

Let sgcR be the full subcategory of gcR whose objects are the simply connectedones. Our aim in the sequel is to show that sgcR is a coreflective subcategoryof gcR. As a consequence, we will show that every G ∈ gcR admits a simply

connected covering group G which is essentially unique, and that G is in fact theuniversal covering group of G in an obvious sense. Before entering into the details,we quickly recall some standard facts about coreflective subcategories and refer to[1, Chapter I, Section 4] or [66, IV.3] for details and proofs.

4.7. Coreflective subcategories. Let S be a full subcategory of a categoryC, and let G ∈ C. A morphism π: G → G is called an S-coreflection for G ifG ∈ S and G has the following universal property: for all H ∈ S and all morphismsϕ: H → G there exists a unique ψ: H → G such that the diagram

G

π

H

∃!ψ99

ϕ// G

(1)

is commutative. Such π is essentially unique: if also π′: G′ → G has the sameproperty then there exists a unique isomorphism ψ: G→ G′ such that

G

π????????∃!ψ∼=

// G′

π′~~~~~~~~

G

(2)

commutes. This follows by applying the universal property (1) with (ϕ, π) replacedby (π, π′). The subcategory S is called coreflective if every G ∈ C admits anS-coreflection.

Assume this to be the case and choose an S-coreflection πG: G → G for everyG ∈ C. Then there exists a unique functor Σ: C → S, called a coreflector, suchthat

(i) Σ(G) = G for all G ∈ C, and

(ii) for every morphism ϕ: G→ H of C the diagram


Σ(G)

πG

Σ(ϕ) // Σ(H)

πH

G

ϕ// H

(3)

commutes.

The functor Σ is right adjoint to the inclusion functor I: S→ C:

MorC(I(H), G

) ∼= MorS(H, Σ(G)

),

for all H ∈ S and G ∈ C. Conversely, a subcategory S whose inclusion functor hasa right adjoint is coreflective.

We return to the situation of 4.6. The construction of the simply connectedcovering group G from G is based on the following lemma, inspired by [102, 3.6].

4.8. Lemma. Let (R,X) ∈ SF. For all α ∈ R× and all nilpotent pairs (α, β) ∈R × R, let Lα and L(α,β) be groups. For simpler notation, we write Lαβ = L(α,β).Let

iβα: Lβ → Lα whenever α∣∣β, and (1)

iγαβ : Lγ → Lαβ for all γ ∈[[[[α, β

]]]], (2)

be group homomorphisms satisfying

Lαβ =⟨iγαβ(Lγ) : γ ∈

[[[[α, β

]]]]⟩, (3)(((((((

iααβ(Lα), iβαβ(Lβ))))))))⊂⟨iγαβ(Lγ) : γ ∈

(((((((α, β

)))))))⟩. (4)

LetL = lim

−→(Lα, Lαβ)

be the colimit of the groups Lα and Lαβ with respect to the maps iβα and iγαβ in thecategory of groups. Denote by jα: Lα → L and jαβ : Lαβ → L the canonical mapsinto the colimit. Then L has R-commutator relations with root groups

Y0 = 1, Yα = jα(Lα) (α ∈ R×), (5)

and is generated by the Yα; that is,(L, (Yα)α∈R

)∈ gcR.

Proof. By [87, I.1.1] the colimit exists in the category of groups and has thefollowing properties which characterize it uniquely up to unique isomorphism: thehomomorphisms jα and jαβ , for all α ∈ R× and all nilpotent pairs (α, β), make theinner left hand triangles of the diagrams

Lβjβ

''OOOOOOOOO

iβα

ϕβ

$$L

∃!ϕ // M

Lαjα

77oooooooooϕα

::

Lγjγ

''OOOOOOOOO

iγαβ

ϕγ

$$L

∃!ϕ // M

Lαβjαβ

77oooooooooϕαβ

:: (6)


commutative, for all α∣∣β and all γ ∈

[[[[α, β

]]]], (α, β) nilpotent. Furthermore, given

any group M and given homomorphisms

ϕα: Lα →M, ϕαβ : Lαβ →M

making the outer triangles of (6) commute, there exists a unique homomorphismϕ: L→M making the entire diagrams commute.

We first observe thatY[[[α,β]]] = jαβ(Lαβ), (7)

for all nilpotent pairs (α, β). Recall here that, by the definition given in 3.2, for asubset A of R, YA is the subgroup of L generated by all Yγ , γ ∈ A. This is appliedhere to the subsets

[[[[α, β

]]]]of R. Indeed,

Y[[[α,β]]] =⟨Yγ : γ ∈

[[[[α, β

]]]]⟩=⟨jγ(Lγ) : γ ∈

[[[[α, β

]]]]⟩=⟨jαβ iγαβ(Lγ) : γ ∈

[[[[α, β

]]]]⟩(by (6))

= jαβ

⟨iγαβ(Lγ) : γ ∈

[[[[α, β

]]]]⟩(since jαβ is a homomorphism)

= jαβ(Lαβ) (by (3)).

By uniqueness of ϕ in (6), L is generated by the subgroups jα(Lα) = Yα, α ∈ R,and jαβ(Lαβ), (α, β) nilpotent. Now (7) shows that L is already generated by theYα.

Next, the relation (3.2.2) for L, i.e., Yβ ⊂ Yα if α∣∣β, follows immediately from

the first diagram of (6) and the definition of the root groups, so it remains to verifythat L satisfies the commutator relations (3.2.3). Let (α, β) be a nilpotent pair.Then, since α and β belong to

[[[[α, β

]]]], by (6) and by (4),(((((((

Yα, Yβ)))))))

=(((((((jα(Lα), jβ(Lβ)

)))))))=(((((((

(jαβ iααβ)(Lα), (jαβ iβαβ)(Lβ))))))))

= jαβ((((((((iααβ(Lα), iβαβ(Lβ)

))))))))⊂ jαβ

⟨iγαβ(Lγ) : γ ∈

(((((((α, β

)))))))⟩=⟨

(jαβ iγαβ)(Lγ) : γ ∈(((((((α, β

)))))))⟩=⟨jγ(Lγ) : γ ∈

(((((((α, β

)))))))⟩=⟨Yγ : γ ∈

(((((((α, β

)))))))⟩= Y(((((α,β))))).

Remark. For this computation to work it is essential that iααβ : Lα → Lαβ and

iβαβ : Lβ → Lαβ be defined. This explains why we have to allow γ ∈[[[[α, β

]]]]in (2);

it would not be sufficient to require (2) only for γ ∈(((((((α, β

))))))).

4.9. Theorem. With the terminology and notation introduced in 4.6, sgcR isa coreflective subcategory of gcR. In more detail, let (G,U) ∈ gcR, and let

G = (G, U) = lim−→

(Uα, U[[[α,β]]]

)∈ gcR

be the colimit of the Uα and U[[[α,β]]] as in Lemma 4.8. Let jα: Uα → Uα be as in(4.8.5).


(a) There exists a unique morphism πG: G → G of gcR such that πG jα =inc: Uα → G for all α ∈ R, and πG is a covering.

(b) πG is an sgcR-coreflection for G.

Proof. Lemma 4.8 is applicable to Lα = Uα and Lαβ = U[[[α,β]]] with the obvious

inclusion maps Uβ ⊂ Uα for α∣∣β and Uγ ⊂ U[[[α,β]]] because (4.8.3) holds by definition

of U[[[α,β]]], and (4.8.4) follows from (3.2.3).

(a) We use the universal property of G (cf. (4.8.6)) in the case where M = Gand the ϕα: Lα → G and ϕαβ : Lαβ → G are the inclusion maps. Then the outertriangles of the diagrams (4.8.6) obviously commute, proving the existence of πG.From the first diagram of (4.8.6) we see that πG jα = IdUα . By definition, the

root groups of G are Uα = Yα = jα(Lα) = jα(Uα). Hence jα: Uα → Uα is anisomorphism, so πG is bijective on root groups. This also shows the uniquenessof πG. In the same way, the second diagram of (4.8.6) shows that πG jαβ isthe identity on U[[[α,β]]]. Hence πG satisfies condition (ii) of 4.4 as well, so πG is acovering.

The proof of (b) is subdivided in the following steps.

(b1) For every morphism ϕ: (G,U) → (H,V) of gcR there exists a unique

morphism ϕ: (G, U)→ (H, V) of gcR making the diagram

G

πG

ϕ // H

πH

G

ϕ// H

commutative.

We apply Lemma 4.8 to the case where L = G and M = H. Thus we needhomomorphisms ϕα and ϕα,β such that the outer triangles of the diagrams

Uβjβ

''NNNNNNNNN

iβα

ϕβ

##G

∃! ϕ // H

Uαjα

77ppppppppp

ϕα

;;

Uγjγ

''OOOOOOOOO

iγαβ

ϕγ

$$G

∃! ϕ // H

Uαβjαβ

77ooooooooo

ϕαβ

:: (1)

for α∣∣β in the first case, and (α, β) nilpotent and γ ∈

[[[[α, β

]]]]in the second case,

commute. Define ϕα and ϕαβ by the commutativity of the diagrams

Vα

πH∼=

Uα

ϕα

99rrrrrrrrrrrrϕ

// Vα

V[[[α,β]]]

∼= πH

Uαβ

ϕαβ88qqqqqqqqqqqq

ϕ// V[[[α,β]]]

(2)


where we use that, by (a), πH : H → H is a covering, so that the vertical maps of(2) are isomorphisms. Then the required commutativity of the diagrams in (1) isobtained as follows. Let β = nα. We must show ϕβ = ϕα iβα. By the first diagramin (2) it suffices to show that πH ϕβ = πH ϕα iβα, and this holds because ϕ mapsUα to Vα. The second case is done similarly. Now it follows that ϕ is a morphismof gcR. Finally, uniqueness of ϕ follows easily from the fact that πH is a covering,hence a monomorphism.

(b2) Let (H,V) ∈ gcR and let η: H → G be a covering. Then there exists a

unique ϕ: G→ H such that

G∃!ϕ

yyπG

H

η// G

is commutative.

We use again the universal property of the colimit. Define ϕα: Uα → H andϕαβ : U[[[α,β]]] → H to be the inverses of η

∣∣Vα and η∣∣V[[[α,β]]]. Since η is a covering,

this makes sense. Then the outer triangles of (4.8.6) are clearly commutative, so

we have a unique ϕ: G → H making the diagrams (4.8.6) commute. From thesediagrams one sees that ϕ preserves root groups, and uniqueness of ϕ follows fromLemma 4.5(b). Finally, by definition of ϕα, ηα ϕα = (πG)α for all α, whenceη ϕ = πG. Uniqueness of ϕ follows from the fact that η is a monomorphism.

(b3) G is simply connected.

Let ψ: H → G be a covering. By 4.6, we must show that ψ is an isomorphism.By Lemma 4.5(a), η := πG ψ: H → G → G is a covering as well. By (b2), there

exists a unique ϕ: G→ H such that π = η ϕ:

G

ϕyyssssssssssss

πG

H

ψ

99ssssssssssssη

// G

We claim that ϕ and ψ are inverses of each other. Indeed, from the diagram wehave η = πG ψ and πG = η ϕ. Hence πG = η ϕ = (πG ψ) ϕ = πG (ψ ϕ),and in the same way, η = πG ψ = (η ϕ) ψ = η (ϕ ψ). Since πG and η aremonomorphisms by Lemma 4.5(a), this implies ψ ϕ = Id

Gand ϕ ψ = IdH .

(b4) πG: G→ G is an sgcR-coreflection for G.

Let H ∈ sgcR and let ϕ: H → G be a morphism of gcR. By (b1), there exists

a unique ϕ: H → G making the diagram

H

πH

ϕ // G

πG

H

ϕ//

ψ

99ssssssssssssG


commutative. By (b3), H ∈ sgcR, so πH is an isomorphism. Put ψ = ϕ π−1H .

Then πG ψ = πG ϕ π−1H = ϕ πH π−1

H = ϕ. Assume also ψ′: H → G satisfiesϕ = πG ψ′. Then ψ = ψ′ because πG is a covering and therefore a monomorphismby Lemma 4.5(a).

4.10. Definition. If πG: G→ G is an sgcR-coreflection for G then G is calleda Steinberg group of G and denoted St(G). For a morphism ϕ: G → H of gcR,

we write ϕ = St(ϕ): G → H. A justification for this choice of terminology will begiven later.

4.11. Corollary. Let ϕ: H → G be a a covering. Then

ϕ: St(H) = H → St(G) = G

is an isomorphism. There exists a unique morphism η: G→ H such that πG = ϕη,and η is a covering.

Proof. By (4.7.3) we have a commutative diagram

H

πH

ϕ // G

πG

η

yyH

ϕ// G

Here πG and ϕ πH are coverings. Hence ϕ is a covering as well, by Lemma 4.5(b).

Since G is simply connected, ϕ is an isomorphism. Finally, put η = πH (ϕ)−1.

4.12. Corollary. Let G ∈ gcR and let st(G) be the category whose objects

are the coverings ϕ: H → G and whose morphisms η: (Hϕ→ G) → (K

ψ→ G) arecommutative triangles

Hη //

ϕ @@@@@@@@ K

ψ~~~~~~~~~~

G

of the category gcR. Then st(G) is a preordered category, i.e., there is at most one

morphism between any two objects in st(G), and πG: G→ G is an initial object ofst(G).

Proof. Suppose η1, η2 are morphisms from ϕ: H → G to ψ: K → G. Thenϕ = ψ η1 = ψ η2 and therefore η1 = η2, since ψ is a covering and therefore amonomorphism. Finally, if ϕ: H → G is a covering then by Corollary 4.11 thereexists η: G → H such that ϕ η = πG. Hence πG: G → G is an initial object ofst(G).

Remark. The category st(G) is a full subcategory of the comma category(gcR ↓ G), see [66, II.6], of gcR overG. There is a forgetful functor Υ : st(G)→ gcRsending an object H → G to H and a morphism η: (H → G) → (K → G) toη: H → K. The functor Υ is faithful but it is not an embedding. For example,


let ϕ: H → G be an object of st(G) and let η be a non-trivial automorphism of Hstabilizing each root group of H. Then ϕη: H → G is an object in st(G), differentfrom ϕ: H → G, but with the same image under Υ . A terminal object of st(G) isIdG: G→ G.

4.13. Another construction of the Steinberg group. Generalizing again[102, 3.6], we now give a more concrete (but less canonical) description of St(G) incase G ∈ gcR has unique factorization for nilpotent pairs as defined in 3.15.

For every nilpotent pair (α, β) choose an ordering(((((((α, β

)))))))ind = γ1, . . . , γn

(where n = nαβ will of course depend on α, β). Then there are well-definedfunctions f iαβ : Uα × Uβ → Uγi such that

(((((((a, b)))))))

=

n∏i=1

f iαβ(a, b), (1)

for all a ∈ Uα, b ∈ Uβ , the product being taken in the sense of the ordering. (Thef iαβ will in general depend on the chosen ordering). Let

F =∐α∈R

Uα

be the free product of the Uα in the category of groups, and let

hα: Uα → F

be the canonical injections, with image Fα = hα(Uα). Let N be the normalsubgroup of F generated by all

hβ(b)−1 · hα(b), (2)(((((((hα(a), hβ(b)

)))))))−1 ·n∏i=1

hγi(f iαβ(a, b)

), (3)

where α∣∣β and b ∈ Uβ in the first formula, and (α, β) is nilpotent and a ∈ Uα,

b ∈ Uβ in the second. Finally, let

L := F/N (4)

and denote by can: F → L the canonical map. We define

kα := can hα: Uα → L, Yα = can(Fα) = kα(Uα) ⊂ L, Y = (Lα)α∈R. (5)

4.14. Theorem. Suppose (G,U) ∈ gcR has unique factorization for nilpotentpairs. Then, with the notations of 4.13, (L,Y) belongs to gcR, and there exists aunique morphism π: L→ G of gcR making L an initial object of st(G) (and henceL “is” the Steinberg group St(G)).

Proof. First, we show that (L,Y) ∈ gcR, that is, L has R-commutator relationswith root groups Yα and is generated by the Yα. Since F is generated by the Fα,


it follows from (4.13.4) that L is generated by the Yα = can(Fα). If α∣∣β, we have

Yβ ⊂ Yα by applying can to the relations (4.13.2), and the commutator relations(3.2.3) follow in the same way by applying can to (4.13.3).

Since G satisfies the commutator relations, it is clear from the universal propertyof L that there is a unique homomorphism π: L→ G of groups in gcR such that

πα kα = IdUα (1)

for all α ∈ R. The maps kα: Uα → Yα are surjective by definition, and injectiveby (1). Hence πα is bijective, so π is bijective on root groups, and therefore π isa covering, by Lemma 4.5(c) and our assumption that G have unique factorizationfor nilpotent pairs. This shows that L ∈ st(G).

Now let ϕ: (H,V) → (G,U) be a covering. We must show that there exists amorphism ψ: L → H such that ϕ ψ = π. Since F is the free product of the Uαand ϕ is bijective on root groups, there is a unique homomorphism κ: F → H suchthat the restriction κα of κ to Fα is given by

κα := ϕ−1α h−1

α : Fα → Uα → Vα. (2)

We claim that π can = ϕ κ: F → G. Indeed, since F is generated by the Fα, itsuffices to show that the squares

Fαcan //

κα

Yα

πα

Vα ϕα

// Uα

(3)

are commutative, for all α ∈ R. Now Fα = hα(Uα), and

ϕα κα = ϕα (ϕ−1α h−1

α ) = h−1α , (4)

while πα canhα = πα kα = IdUα by (1) and therefore πα can = h−1α = ϕα κα,

as desired.Now we show that κ factors via L = F/N , that is, that all elements of type

(4.13.2) and (4.13.3) belong to the kernel of κ.Let α

∣∣β and b = ϕ(b′) ∈ Uβ where b′ ∈ Vβ . Then Uβ ⊂ Uα and (4) imply

ϕ(κ(hβ(b))

)= b = ϕ

(κ(hα(b))

).

Since ϕ∣∣Vα is injective, we conclude κ(hβ(b)) = κ(hα(b)), so κ vanishes on elements

of type (4.13.2).Next, let (α, β) be a nilpotent pair, and let a ∈ Uα, b ∈ Uβ . We must show that

κ(((((((hα(a), hβ(b)

)))))))= κ

( n∏i=1

hγi(f iαβ(a, b)

)). (5)

By (2), κα hα = ϕ−1α : Uα → Vα, so that κ(hα(a)) ∈ Vα. Hence the left hand side

of (5) is, since κ is a homomorphism and H has R-commutator relations,


κ(((((((hα(a), hβ(b)

)))))))=(((((((κ(hα(a)), κ(hα(b))

)))))))∈(((((((Vα, Vβ

)))))))⊂ V(((((α,β))))).

In the same way, one sees that the right hand side belongs to Vγ1 · · ·Vγn =V(((((α,β))))), see Proposition 3.12(b). By assumption, ϕ is a covering, so it satisfies (ii)of 4.4. In particular ϕ: V(((((α,β))))) → U(((((α,β))))) is injective. Hence it suffices that (5) holdafter applying ϕ. This follows now immediately from (4) and (4.13.1): the left handside is

ϕ(κ((((((((hα(a), hβ(b)

)))))))))

=(((((((a, b)))))))

=

n∏i=1

f iαβ(ab),

while the right hand side is

n∏i=1

ϕ(κ(hγi(f

iαβ(a, b))

))=

n∏i=1

f iαβ(a, b).

Hence also the generators of type (4.13.3) belong to the kernel of κ.Thus we have a unique homomorphism ψ: F/N = L → H satisfying κ =

ψ can: F → L→ H, and making the upper triangle in

Fαcan //

κα

Yαψα

xxrrrrrrrrrrrrπα

Vα ϕα

// Uα

commutative, for all α ∈ R. Since the outer square is commutative by (3), thisimplies ϕα ψα can = ϕα κα = πα can, and since can: Fα → Yα is surjective, itfollows that πα = ϕα ψα. Now L is generated by the Yα, so we conclude π = ϕψ,as desired.

4.15. Examples. (a) Let R = −1, 0, 1 be the root system A1 and letG ∈ gcR. Since R contains only the nilpotent pairs (1, 1) and (−1,−1), G issimply a group generated by two abelian subgroups U1 and U−1. It is immediatelyseen that the free product of U1 and U−1 is an initial object of st(G). This appliesin particular to the projective elementary group of a Jordan pair, see §7.

(b) We take up Example (b) of 3.16 and let G be a group with R-commutatorrelations and root groups Ui = Uεi . Suppose that G has unique factorization forall nilpotent pairs, i.e., that Ui ∩ Uj = 0 for i 6= j. Then it follows easily fromTheorem 4.14 that L =

⊕ni=1 Ui, with π(x1 ⊕ · · · ⊕ xn) = x1 + · · · + xn, is the

Steinberg group of G. As shown in 3.16, G has unique factorization if and only ifG =

⊕ni=1 Ui, i.e., if and only if L = G is its own Steinberg group.

(c) The linear elementary group EI(A) has unique factorization for nilpotentpairs by Example (c) of 3.16. Hence it follows easily from Theorem 4.14 thatSt(EI(A)) is the usual Steinberg group, at least when I is finite or countably infinite:

St(En(A)) = Stn(A), St(EN(A)) = St(A)

in the notation of [32, 1.4], see also 9.18 where we will relate Stn(A) to the Steinberggroup of the Jordan pair V = (Matpq(A),Matqp(A)).


(d) Similarly, the usual elementary unitary group EU2n(A, J, ε, Λ) (n> 3) of aform ring (A, J, ε, Λ) in the sense of [32] has Cn-commutator relations, and

St(EU2n(A, J, ε, Λ)) = StU2n(A, J, ε, Λ)

is the usual unitary Steinberg group.

(e) Let A be a generalized Cartan matrix, let g(A) be the Kac-Moody Liealgebra associated with A, and let Rre be the set of real roots of g(A). ThenR = 0 ∪ Rre ∈ SF. In fact, R is a reduced reflection system as defined in 2.2,even a partial root system in the sense of [65].

Let G be the Kac-Moody group functor associated with a (in our contextunimportant) root datum D(A), constructed by J. Tits in [102], see also [81,Chapter 8 and 9]. By definition, G is an amalgam of a torus and the so-calledSteinberg group functor StA depending only on A. This is an example of theconstruction of Lemma 4.8 with R as above. Indeed, for every commutative ringS one puts StA(S) = lim

−→(Uα(S),Uαβ(S)) where Uα, α ∈ Rre, are certain Z-group

schemes and where Uα,β =∏γ∈[[[α,β]]] Uγ for any nilpotent pair (α, β) in R. While

the definitions of a nilpotent pair here and in [102] are not the same, we have shownin [65, Corollary 3.8] that they are equivalent.

4.16. Definition. We now define a category gc encompassing all gcR intro-duced in 4.1. To keep notation manageable, we will refer to an object (R,X) ∈ SFsimply by R and thus omit the free abelian group X generated by R, see 1.5.

The objects of gc are pairs (R, (G,U)) where R ∈ SF and (G,U) ∈ gcR. Amorphism from (R, (G,U)) to (S, (H,V)) in gc is a pair (f, ϕ), where f : R→ S isa morphism of SF and ϕ: G→ H is a group homomorphism, such that

ϕ(Uα) ⊂ Vf(α) for all α ∈ R. (1)

By our convention that U0 = 1 (cf. (3.2.1)), this means in particular that

f(α) = 0 =⇒ Uα ⊂ Ker(ϕ). (2)

It is easily verified that, together with the natural composition of morphisms, thisdefines indeed a category. Thus gc is the category of all groups with R-commutatorrelations, for all possible R ∈ SF.

The projections onto the first factor, (R, (G,U)) 7→ R on objects and (f, ϕ) 7→ fon morphisms, define a functor

Π : gc→ SF. (3)

For fixed R ∈ SF we may identify gcR with the fibre of Π at R, defined below.

We will show in Proposition 4.19 that Π is an opfibration. For the convenienceof the reader, we first review this concept.

4.17. Opfibrations. Let Π: X→ B be a covariant functor between categoriesX and B. For a ∈ B, denote by Xa the (non-full) subcategory of X with objectsall X ∈ X such that Π(X) = a, and morphisms those morphisms t: X → X ′ of X


satisfying Π(t) = Ida, called the fibre over a. Objects of Xa are also called objectsover a.

A morphism u: X → Y of X, say with Π(u) = f : a = Π(X) → b = Π(Y ), iscalled a morphism over f : a → b. We say u is opcartesian if for every morphismw: X → Z of X over h: a → c = Π(Z), every factorization h = g f in B can belifted uniquely to a factorization of w, i.e., there exists a unique morphism v: Y → Zof X such that w = v u:

Xw //

u $$IIIIII Z

Yv

∃!55

a

f $$JJJJJJh // c

bg

55kkkkkkkkkkk

(1)

Here a vertical line such asX∣∣a

indicates that Π(X) = a, i.e., that X is an object

over a.

Then Π (or by abuse of terminology X if Π is clear from the context) is calledan opfibration if for every morphism f : a→ b of B and every X ∈ Xa, there existsY ∈ Xb and an opcartesian morphism u: X → Y over f . Such u and Y are uniqueup to unique isomorphism.

Let X be an opfibration over B. For every X ∈ X and every morphismf : a → b of B choose an opcartesian morphism u: X → Y , and put f∗(X) := Yand ω(f,X) := u. Then for every morphism t: X → X ′ in the category Xa thereexists a unique morphism f∗(t): f∗(X)→ f∗(X

′) in Xb such that the diagram

X

t

ω(f,X) // f∗(X)

f∗(t)

X ′

ω(f,X′)

// f∗(X ′)

(2)

commutes. This follows by applying the diagram (1) to the case where w =ω(f,X ′) t, Z = f∗(X

′) and g = Idb. In this way, we obtain a functor f∗: Xa →Xb. A choice of ω(f,X) for all f and X is called a cleavage. There are uniqueisomorphisms (Ida)∗ ∼= IdXa

and g∗ f∗ ∼= (g f)∗, but these are in general notequalities, so the assignment a 7→ Xa, f 7→ f∗ is merely a pseudofunctor (or lax2-functor) from B to Cat, the category of categories.

An opfibration is called split if it possesses a cleavage satisfying (Ida)∗ = IdXa

and g∗ f∗ = (g f)∗, in which case the above pseudofunctor is a functor.

4.18. Lemma. Let (G,U) ∈ gcR be a group with R-commutator relations andlet f : R→ S be a morphism of SF. For all ξ ∈ S define subsets R[ξ] of R by

R[ξ] := R[ξ, f ] := α ∈ R : f(α) 6= 0 and ξ∣∣ f(α), (1)

in particular R[0] = ∅, and define subgroups U ′ξ and G′ of G by


U ′ξ := UR[ξ], G′ =⟨U ′ξ : ξ ∈ S

⟩. (2)

ThenG′ =

⟨Uα : f(α) 6= 0

⟩, (3)

and putting U′ = (U ′ξ)ξ∈S, we have

(G′,U′) ∈ gcS . (4)

Proof. If α ∈ R[ξ] then in particular f(α) 6= 0. Conversely, f(α) 6= 0 impliesα ∈ R[f(α)], so (3) holds.

Let us show that G′ has S-commutator relations with root groups U ′ξ. We have

R[0] = ∅ by (3.1.4), so U ′0 = U∅ = 1. Next, η∣∣ ξ implies R[η] ⊃ R[ξ] by (3.1.5),

from which the relation U ′η ⊃ U ′ξ follows. It remains to verify the commutator

relation (3.2.3). Let Xξ =⋃Uα : α ∈ R[ξ], and note Xξ = X−1

ξ and⟨Xξ

⟩= U ′ξ.

Hence, by Lemma 3.8 applied to the family of subgroups (U ′ξ)ξ∈S , it suffices toprove that (((((((

Uα, Uβ)))))))⊂ U ′(((((ξ,η))))), (5)

whenever (ξ, η) is a nilpotent pair in S and α ∈ R[ξ] and β ∈ R[η]. By (1.9.3), α, βis prenilpotent, hence the R-commutator relations for G yield

(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))),

and U(((((α,β))))) ⊂ U ′(((((ξ,η))))) follows from f((((((((α, β

))))))))⊂(((((((ξ, η)))))))

and Uγ ⊂ U ′f(γ) for f(γ) 6= 0.

This establishes (5) and completes the proof.

4.19. Proposition. The functor Π: gc → SF of (4.16.3) is an opfibration.A cleavage of Π is given as follows. For a morphism f : R → S of SF and anobject (R, (G,U)) of gc, let Nf be the normal subgroup of G generated by Uα :f(α) = 0. Let H = G/Nf and denote can: G → H the canonical map. Forξ ∈ S define Vξ := can(U ′ξ) where the U ′ξ are as in (4.18.2), and put V = (Vξ)ξ∈S.Then f∗((R, (G,U)) := (S, (H,V)) ∈ gc, and ω(f, (R, (G,U))) = (f, can) defines acleavage.

Proof. We apply 4.17 in the case where B = SF, X = gc, and Π is theprojection onto the first factor. By (4.18.4), (G′,U′) is an object of gcS . By 3.3,homomorphic images inherit commutator relations. Hence to show that (H,V) ∈gcS , it suffices to prove can(G′) = H. This follows from (4.18.3) and the fact thatthe Uα with f(α) = 0 lie in Nf . It remains to show that (f, can) is opcartesian.Explicitly, this means:

(∗) Let K be an object of gc over T , let (h, χ): G → K be a morphism of gcover h: R → T , and let h = g f be factored via a morphism g: S → T of SF.Then there exists a unique morphism (g, ψ): f∗(G) → K in gc over g such thatχ = ψ can.

First, (f, can) is a morphism of gc over f . Indeed, can: G → G/Nf is a grouphomomorphism, and Uα ⊂ U ′f(α) obviously holds by the definition (4.18.2), whence

can(Uα) ⊂ can(U ′f(α)) = Vf(α) for all α ∈ R. Now let K = (T, (K,W = (Wτ )τ∈T )),

let (h, χ): G → K and h = g f as in (∗). For α ∈ R with f(α) = 0 we haveh(α) = g(f(α)) = 0 and therefore Uα ⊂ Ker(χ) by (4.16.2), which implies that alsoNf ⊂ Ker(χ). Hence there exists a unique group homomorphism ψ: H → K such


that χ = ψ can, and it remains to show that ψ is a homomorphism of gc over g,i.e., that ψ(Vξ) ⊂Wg(ξ) for all ξ ∈ S. Now Vξ = can(U ′ξ) is generated by all can(Uα)

where f(α) 6= 0 and ξ∣∣ f(α). For such α, we have ψ(can(Uα)) = χ(Uα) ⊂ Wh(α),

since χ: G → K is a morphism over h. But ξ∣∣ f(α) implies g(ξ)

∣∣h(α) (becauseh = g f), and hence Wg(ξ) ⊃ Wh(α) by (3.2.2). This shows ψ(Vξ) ⊂ Wg(ξ), asdesired.

4.20. Remarks. (a) Let ϑ: G→ G′ be a morphism of groups in gcR, and letN ′f be the normal subgroup of G′ defined analogously to Nf . Then it easy to seethat ϑ(Nf ) ⊂ N ′f , and that the homomorphism f∗(ϑ): G/Nf → G′/N ′f of (4.17.2)is the one induced from ϑ by passing to the quotient groups.

(b) Let g: S → T be a morphism of SF. The isomorphisms of functorsIdgcR

∼= (IdR)∗ and (g f)∗ ∼= g∗ f∗ mentioned in 4.17 are given on objectsas follows. For f = IdR, we have Nf = 1, and hence (IdR)∗(G) = G/1 ∼= Gin the obvious way. Let Ngf be the normal subgroup of G generated by all Uαwith g(f(α)) = 0 and let Ng the normal subgroup of G := G/Nf generated by allUξ with g(ξ) = 0. Then Ngf/Nf = Ng, and we have an isomorphism of groups(g f)∗(G) = G/Ngf ∼= G/Ng = g∗(f∗(G)) by the first isomorphism theorem. Onechecks easily that this is compatible with the respective root subgroups.

(c) If f : R → S is an immersion in the sense that R ∩ Ker(f) = 0 thenNf = 1 and therefore the underlying group of f∗(G) is the same as that of G,although of course the root groups differ.

Our next aim is to show that there is a second way of embedding the categoriesgcR as the fibres of an opfibration which is even split. This is based on the followingconstruction.

4.21. The Grothendieck construction. We review the Grothendieck con-struction in the special case of functors (as opposed to pseudofunctors) with valuesin the category Cat of categories and refer to [38, B.1.3.1] and [108, 3.1.3] fordetails and proofs.

Let B be a category, and let Φ: B→ Cat be a (covariant) functor. The objectsof B will be written as a, b, c, and morphisms are f : a → b, g: b → c, etc. Theobjects of the category Φ(a) are denoted A,A′ etc. For a morphism f : a→ b of B,we write

Φ(f) = f•: Φ(a)→ Φ(b).

The Grothendieck construction associates with these data the following split op-fibration Π =

∫Φ: X → B. The objects of X are pairs (a,A) where a ∈ B and

A ∈ Φ(a). A morphism of X from (a,A) to (b, B) is a pair (f, ϕ) where

f : a→ b and ϕ: f•(A)→ B

are morphisms of B and Φ(b), respectively.The composition of two morphisms (f, ϕ): (a,A) → (b, B) and (g, ψ): (b, B)

→ (c, C) of X is defined by

(g, ψ) (f, ϕ) =(g f, ψ g•(ϕ)

).


Since ϕ: f•(A)→ B we have

g•(ϕ): g•(f•(A)) = (g f)•(A)→ g•(B),

a morphism of the category Φ(b), and the composite

ψ g•(ϕ): (g f)•(A)→ g•(B)→ C

is a morphism from (gf)•(A) to C, as it should be. From the functorial propertiesof Φ it follows that the composition of morphisms is associative, so X is a category.

The projection onto the first factor, Π(a,A) = a on objects and Π(f, ϕ) = fon morphisms, makes X a split opfibration over B. A splitting is given by

f∗(a,A) = (b, f•(A)) and ω(f, (a,A)) = (f, Idf•(A)),

for all morphism f : a→ b of B and A ∈ Φ(a).The fibre of X over a ∈ B in the sense of 4.17 is the subcategory Xa with

objects all (a,A), A ∈ Φ(a), and morphisms all (f, ϕ): (a,A) → (a,B) of X withΠ(f, ϕ) = f = Ida. Clearly, Xa is canonically isomorphic with the category Φ(a).

4.22. Proposition. Let (G,U) ∈ gcR be a group with R-commutator relations.

(a) Let f : R → S and g: S → T be morphisms of SF. Let (G′,U′) ∈ gcS bedefined as in Lemma 4.18, and for τ ∈ T , define subgroups U ′′τ of G by

U ′′τ :=⟨Uα : g(f(α)) 6= 0, τ

∣∣ (g f)(α)⟩

= UR[τ,gf ]. (1)

ThenU ′′τ =

⟨U ′ξ : g(ξ) 6= 0, τ

∣∣ g(ξ)⟩ (

= U ′S[τ,g]

). (2)

(b) The assignments (G,U) 7→ (G′,U′) on objects and ϕ 7→ ϕ∣∣G′ on morphisms

define a covariant functor f•: gcR → gcS.

(c) f• itself depends functorially on f ; i.e., (Id)• = Id and (g f)• = g• f•.Hence the assignments R 7→ gcR on objects and f 7→ f• on morphisms define afunctor Φ: SF→ Cat.

Proof. (a) For the inclusion from left to right in (2), let α ∈ R[τ, g f ] andput ξ = f(α). Then g(ξ) = g(f(α)) 6= 0 and τ

∣∣ g(ξ), so ξ ∈ S[τ, g]. Obviously,α ∈ R[ξ, f ], and hence Uα ⊂ U ′ξ.

To prove the inclusion from right to left in (2), let ξ ∈ S[τ, g]. We must showU ′ξ ⊂ U ′′τ . By (4.18.2), U ′ξ is generated by all Uα, α ∈ R[ξ, f ], so it suffices to showthat α ∈ R[ξ, f ] and ξ ∈ S[τ, g] imply α ∈ R[τ, g f ], because then Uα ⊂ U ′′τ willfollow. Now g is a homomorphism of the abelian groups generated by S and T ,respectively, so f(α) = nξ 6= 0 and g(ξ) = pτ 6= 0 for suitable n, p ∈ N+ implyg(f(α)) = ng(ξ) = npτ 6= 0, whence α ∈ R[τ, g f ], as desired.

(b) By (4.18.4), we have f•(G,U) = (G′,U′) ∈ gcS . Now let also (H,V) ∈ gcRand let ϕ: G → H be a morphism of gcR, so that ϕ(Uα) ⊂ Vα for all α ∈ R. Weuse the notations of (4.18.1) and (4.18.2) for H as well. Then ϕ(U ′ξ) ⊂ V ′ξ is clear

from the definition (4.18.2), so f•(ϕ) = ϕ∣∣G′: G′ → H ′ is a morphism of gcS . It

§5] Weyl elements 49

is immediately verified that f•(IdG) = Idf•(G) and f•(ψ ϕ) = f•(ψ) f•(ϕ) for amorphism ψ: H → K of gcR, so f• is indeed a covariant functor.

(c) From (4.18.2) and the fact that Uβ ⊂ Uα for α∣∣β (by (3.2.2)) it follows

that Id•(G) = G and Id•(ϕ) = ϕ for a morphism ϕ of gcR. Now let g: S → T be amorphism of SF. We must show that the functors (gf)• and g• f• have the sameeffect on objects and morphisms of gcR. We have (g f)•(G) = (G′′,U′′) ∈ gcTwhere U′′ = (U ′′τ )τ∈T is given by (1) and G′′ is the subgroup of G generated by theU ′′τ . Hence g•(f•(G)) = (g f)•(G) follows from (2). Thus g• f• and (g f)• agreeon objects of gcR. That also g•(f•(ϕ)) = (g f)•(ϕ) for morphisms of gcR, is thenan easy consequence.

4.23. Proposition. Let Π ′ =∫Φ: gc′ → SF be the split opfibration obtained

from the functor Φ: SF → Cat of Proposition 4.22(c) by the Grothendieck con-struction 4.21. Then there is an embedding E: gc → gc′, compatible with theprojections, given by the identity on objects and by E(f, ϕ) = (f, ϕ

∣∣f•(G,U)) onmorphisms (f, ϕ): (R, (G,U))→ (S, (H,V)) of gc.

Proof. Here the base category is B = SF, so the objects of X = gc′ arethe pairs (R, (G,U)) where R ∈ SF and (G,U) ∈ Φ(R). Given a morphism(f, ϕ): (R, (G,U)) → (S, (H,V)) of gc, we must show that ϕ′ := ϕ

∣∣G′: (G′,U′) →(H,V) is a morphism of gcS , i.e., ϕ(U ′ξ) ⊂ Vξ for all ξ ∈ S. Let α ∈ R[ξ]. By(4.18.1), (4.18.2) and (4.16.1), f(α) = nξ for some positive integer n, hence by(3.2.2),

ϕ(Uα) ⊂ Vf(α) = Vnξ ⊂ Vξ.

This implies ϕ(U ′ξ) ⊂ Vξ, so E(f, ϕ) is a morphism of gc′. We leave it to the readerto show that E is a functor, and show that E is faithful. Suppose ϕ and ψ aremorphisms from (R, (G,U)) to (S, (H,V)) in gc for which ϕ

∣∣G′ = ψ∣∣G′. We claim

that ϕ = ψ. Since G is generated by the root groups Uα, it suffices to show that ϕand ψ agree on all Uα. If f(α) 6= 0 then α ∈ R[f(α)], so Uα ⊂ G′. If f(α) = 0 thenUα ⊂ Ker(ϕ) and Uα ⊂ Ker(ψ) by (4.16.2).

§5. Weyl elements

5.1. Weyl elements and Weyl triples. Let R = (R,X, ∨) be a reflectionsystem, see 2.2, and let G = (G,U) ∈ gcR be a group with R-commutator relationsand root groups U = (Uα)α∈R as in 4.1. Let α ∈ Re(R) = Rre ∪ 0 and let sα bethe reflection associated with α. An element w ∈ G is called a Weyl element forα if w ∈ U−α Uα U−α and conjugation by w realizes the reflection sα on the rootgroups in the sense that

wUβ w−1 = Usα(β) for all β ∈ R. (1)

This follows Faulkner [28], except that he requires w ∈ Uα U−α Uα. Thus a Weylelement for α has a representation w = x−1x0x1 where x±1 ∈ U−α and x0 ∈ Uα.In general, w does not determine the triple (x−1, x0, x1) uniquely. Therefore, wedefine: a Weyl triple for α is a triple x = (x−1, x0, x1) ∈ U−α×Uα×U−α such thatµ(x) = x−1x0x1 is a Weyl element for α were µ denotes the multiplication map.

We denote by Wα(G) or simply Wα, as long as G is fixed, the set of Weylelements for α, and by Tα or Tα(G) the set of Weyl triples for α. Strictly speaking,


we should write Wα(G,U) since the notions of Weyl element and Weyl triple dependof course on the family of root groups U. But we will use the simplified notation,hoping that the reader will keep this dependence in mind. Clearly, µ: Tα →Wα issurjective. By our conventions 0 ∈ Re(R), s0 = Id and U0 = 1. Hence, W0 = 1and T0 = (1, 1, 1). It is convenient to consider also the following sets:

Θα(G) = U−α × Uα × U−α, Θ(G) =∐α∈R

Θα(G) (2)

as well asT(G) =

∐α∈Re(R)

Tα(G). (3)

Clearly, Θα(G) and Θ(G) depend functorially on G: if ϕ: G→ H is a morphism ofgcR (thus mapping root groups to root groups) then Θα(ϕ): Θα(G) → Θα(H) isdefined by (x−1, x0, x1) 7→ (ϕ(x−1), ϕ(x0), ϕ(x1)). In general, the sets Wα(G) andTα(G) do not depend functorially on G. However, if ϕ is surjective on root groups,then ϕ

(Wα(G)

)⊂ Wα(H) which is seen by applying ϕ to (1). This easily implies

Tα(ϕ)(Tα(G)

)⊂ Tα(H) as well, where we define Tα(ϕ) = Θα(ϕ)

∣∣Tα(G).

5.2. Example: linear elementary groups. Let A be a unital associativering and G = E2(A) the elementary group of 2×2-matrices over A as in 3.16(c). Weview G as a group with commutator relations with root system R = A1 = −1, 0, 1and root groups U±1 = U± = e±(A), where the maps e±: A→ G are defined by

e+(x) =

(1 x0 1

), e−(y) =

(1 0−y 1

). (1)

Thus e+(x) = e12(x) and e−(y) = e21(−y), in the notation of 3.16(c). Then theWeyl elements for α = 1 are precisely the elements

wu =

(0 u−u−1 0

)= e−(u−1) e+(u) e−(u−1),

where u ∈ A×, the set of units of A.

Indeed, let w =

(a bc d

)∈W1. Then for all x ∈ A there exists y ∈ A such that

(a bc d

)(1 x0 1

)=

(1 0y 1

)(a bc d

).

By working out the matrix products, this is equivalent to(a ax+ bc cx+ d

)=

(a b

ya+ c yb+ d

),

in particular ax = 0 for all x ∈ A, so a = 0. Similarly, for all y ∈ A there existsx ∈ A such that (

a bc d

)(1 0y 1

)=

(1 x0 1

)(a bc d

),


equivalently, (a+ by bc+ dy d

)=

(a+ xc b+ xdc d

),

whence dy = 0 for all y ∈ A, so d = 0. It follows that w =

(0 bc 0

)with inverse

w−1 =

(0 c−1

b−1 0

).

We next exploit the fact that w ∈ U−U+U−, so

w =

(1 0−t 1

)(1 u0 1

)(1 0−y 1

)∈ U−U+U−.

By working out the product on the right, we obtain

w =

(0 bc 0

)=

(1− uy u

−t− (1− tu)y 1− tu

).

This shows that u = b ∈ A× with inverse u−1 = t = −c, so

w =

(0 u−u−1 0

)= wu. (2)

Conversely, it is easily seen that these elements are indeed Weyl elements for α = 1.By interchanging the roles of 1 and −1, one sees that the Weyl elements for theroot −1 are the elements(

1 u−1

0 1

)(1 0−u 1

)(1 u−1

0 1

)=

(0 u−1

−u 0

)= wu−1 , u ∈ A×. (3)

Hence, the Weyl triples for α = σ1 (σ ∈ +,−) are the elements

tσ(u) =(e−σ(u−1), eσ(u), e−σ(u−1)

), u ∈ A×. (4)

We also see that here the multiplication maps µ: Tα →Wα are bijective.All this can be generalized to the group En(A), viewed as group with An−1-

commutator relations, see for example [32, 1.4E] (but note the different normal-ization in loc. cit.), or even to the group EI(A) of 3.16(c) for I any index set: theWeyl triples and Weyl elements for the root α = εi − εj ∈ R = AI are

tα,u = (Id− u−1Eji, Id + uEij , Id− u−1Eji),

wα,u = (Id− Eii − Ejj) + uEij − u−1Eji.(5)

for u ∈ A×. Indeed, the subgroup of EI(A) generated by Id + aEij , Id + aEji :a ∈ A is a group with A1-commutator relations isomorphic to E2(A). Hence,the calculations above show that the Weyl triples and Weyl elements of EI(A) arenecessarily of the form given in (5). It then suffices to check that wα,u not onlysatisfies (5.1.1) for β = ±α but in fact for all β ∈ R. This is an easy matrixcalculation.

Another generalization of the example E2(A) is presented in 9.14.


5.3. Example: Moufang polygons. Let G be the group associated in 3.3(e)with a Moufang building. There we have seen that G has R-commutator relationswith respect to root groups Uα, α ∈ R, where R is a finite irreducible root systemof rank l > 2. A theorem of Tits (see for example [110, Proposition 11.22]) saysthat for any α ∈ R× and 1 6= uα ∈ Uα there exist u′−α and u′′−α ∈ U−α such thatu′−αuαu

′′−α ∈Wα. In particular, Wα 6= ∅.

More examples of groups with Weyl elements will be given later in 5.14, 5.17,5.19 and 5.22.

5.4. Proposition. Let R be a reflection system and let G be a group with R-commutator relations.

(a) The sets Wα and Tα, α ∈ Re(R), satisfy the following relations:

Wα = W−1α = W−α, (1)

Wnα ⊂Wα, Tnα ⊂ Tα if n ∈ N+ and nα ∈ Re(R), (2)

In particular, if w is a Weyl element for α then so is w−1, and both w and w−1

are Weyl elements for −α.

(b) Let N =⋂β∈R NormG(Uβ). Then for all α, β ∈ Re(R) and wα ∈Wα,

WαWα ⊂ N, (3)

wαWβw−1α = Wsαβ , (4)

wαTβw−1α = Tsαβ , (5)

where conjugation by wα in (5) is understood componentwise.

Proof. (a) Let w ∈ Wα. Since s2α = Id we have w−1Uβw = w−1Us2αβw =

w−1w2Uβw−2w = wUβw

−1 = Usαβ for all β ∈ R, and obviously w−1 ∈ U−αUαU−α.This proves Wα = W−1

α .Next, sα = s−α since sα = sβ whenever pα = qβ for some 0 6= p, q ∈ Z, by

(2.2.6). Hence wUβw−1 = Us−α(β) for all β. Moreover, sα(±α) = ∓α implies

w = www−1 ∈ w · U−αUαU−α · w−1

= wU−αw−1 · wUαw−1 · wU−αw−1 = UαU−αUα,

so we have w ∈W−α. Thus Wα ⊂W−α and then also W−α ⊂W−(−α) = Wα.If nα ∈ R then (2.2.6) shows snα = sα, and (3.2.2) implies Unα ⊂ Uα. This

easily implies Wnα ⊂Wα and Tnα ⊂ Tα.

(b) If w,w′ ∈ Wα then ww′Uβw′−1w−1 = wUsαβw

−1 = Us2αβ = Uβ for allβ ∈ R, so ww′ ∈ N . If wα ∈Wα and wβ ∈Wβ (where now α, β ∈ Re(R)) then, forall γ ∈ R,

wαwβw−1α Uγwαw

−1β w−1

α = Usαsβsα(γ) = Ussα(β)γ

by (2.3.5). Moreover,

wαwβw−1α ∈ wαU−βUβU−βw−1

α = wαU−βw−1α · wαUβw−1

α · wαU−βw−1α

= U−sα(β)Usα(β)U−sα(β),

which shows wαWβw−1α ⊂Wsαβ , and in fact we have equality because from s2

α = Idand w−1

α ∈ Wα we see w−1α Wsαβwα ⊂ Ws2αβ

= Wβ . Now (5) is an immediateconsequence.


5.5. An algebraic structure on the set of Weyl triples. Let G be a groupwith R-commutator relations and root groups U = (Uα)α∈R and let Tα = Tα(G) bethe set of Weyl triples for α ∈ Re(R). Let T = T(G) be as in (5.1.3). We define thefollowing operations on T. First, let x ∈ Tα, say, x = (x−1, x0, x1) ∈ U−α×Uα×U−α,and let w = µ(x) = x−1x0x1 be the corresponding Weyl element. By (5.4.1),w−1 = x−1

1 x−10 x−1

−1 is again a Weyl element for α. Hence,

x−1 := (x−11 , x−1

0 , x−1−1) ∈ Tα. (1)

This yields a unary operation ( )−1: T → T which maps each Tα to itself and isobviously involutive and compatible with multiplication:

(x−1)−1 = x and µ(x−1) = µ(x)−1. (2)

Next, consider the triples

x] := (wx1w−1, x−1, x0), x[ := (x0, x1, w

−1x−1w). (3)

Since wU−α w−1 = Uα and xi ∈ U(−1)iα, we see that x] and x[ are in Uα×U−α×Uα.

Moreover,

µ(x]) = wx1(x−11 x−1

0 x−1−1)x−1x0 = w, µ(x[) = x0x1(x−1

1 x−10 x−1

−1)x−1w = w, (4)

so x] and x[ are Weyl triples for the root −α. This yields two more unary operations] and [ on T.

Let x and w be as before, and let y = (y−1, y0, y1) ∈ Tβ . Then

x • y := Int(µ(x)

)· y = (wy−1w

−1, wy0w−1, wy1w

−1) ∈ Tsα(β) (5)

by (5.4.5). This defines a binary operation • on T. Finally, we let

1 = (1, 1, 1) ∈ T0

and define a projection p: T→ Re(R) by mapping the elements of Tα to α.

Example. Let G = E2(A) as in 5.2 and use the notation for the Weyl triplesintroduced in (5.2.4). Then the unary operations are

tσ(u)−1 = tσ(−u), tσ(u)] = tσ(u)[ = t−σ(u−1). (6)

This follows easily from the action of wu on the root groups. For the products •one computes

t+(u) • t−(v) = t+(uvu). (7)

This yields formulas of type tσ(u) • tτ (v) by using (5.6.2) below. For example,t+(u)• t+(v) = t+(u)• t−(v−1)] = (t+(u)• t−(v−1))] = t+(uv−1u)] = t−(u−1vu−1).


5.6. Lemma. (a) The operations just introduced satisfy the following rules.

x • (x−1 • y) = x−1 • (x • y) = y, (1)

(x • y)−1 = x • y−1, (x • y)] = x • y], (x • y)[ = x • y[, (2)

x] • y = x[ • y = x • y, (3)

x • (y • z) = (x • y) • (x • z), (4)

1 • x = x, x • 1 = 1, (5)

p(x • y) = sp(x)p(y), p(x−1) = p(x), p(x]) = p(x[) = −p(x), (6)

(x])[ = (x[)] = x, (7)

(x])−1 = (x−1)[, (x[)−1 = (x−1)], (8)

x]]] = x • x, x[[[ = x−1 • x. (9)

(b) Let ϕ: G→ H be a morphism of gcR which is surjective on root groups, anddefine T(ϕ): T(G) → T(H) by T(ϕ)

∣∣Tα = Tα(ϕ) as in 5.1. Then T(ϕ) preservesthe algebraic operations of 5.5.

Proof. (a) (1) is immediate from the definition, since µ(x−1) = x−11 x−1

0 x−1−1 =

µ(x)−1. To prove the first formula of (2), put w = µ(x) and observe that

(x • y)−1 =(

Intw · (y−1, y0, y1))−1

= (wy−1w−1, wy0w

−1, wy1w−1)−1

= (wy−11 w−1, wy−1

0 w−1, wy−1−1w

−1) = Int(w) · y−1 = x • y−1.

The proof of the second and third formula is similar. Formula (3) follows from(5.5.4) and (5.5.5). For (4), we compute

x • (y • z) = Intµ(x) ·(

Intµ(y) · z)

= Int(µ(x)µ(y)) · z= Int(µ(x)µ(y)µ(x)−1) Int(µ(x) · z).

On the other hand,

µ(x)µ(y)µ(x)−1 = µ(

Int(µ(x)) · y)

= µ(x • y),

from which the assertion follows. It is obvious that (5) and (6) hold.For (7), observe (x])[ = (wx1w

−1, x−1, x0)[ = (x−1, x0, w−1wx1w

−1w) = x, andsimilarly for the second formula. The first formula of (8) follows from (x])−1 =(wx1w

−1, x−1, x0)−1 = (x−10 , x−1

−1, wx−11 w−1) = (x−1)[, and the second formula is

proved similarly. Finally, since µ(x]) = µ(x[) = µ(x) = w, we have

x]]] = (wx1w−1, x−1, x0)]] = (wx0w

−1, wx1w−1, x−1)]

= (wx−1w−1, wx0w

−1, wx1w−1) = x • x,

and similarly for the second formula.(b) This follows immediately from the definitions.

We remark that the left multiplications Lx: y 7→ x • y are bijective, with L−1x =

Lx−1 , by (1). Formulas (2) and (4) say that Lx is an automorphism of T, equippedwith the algebraic structures of multiplication, inversion, ] and [. By (7), [ is justthe inverse map of ], and by (8), [ can also be defined in terms of ] and inversionas

x[ = ((x−1)])−1 and x] = ((x−1)[)−1. (10)

For example, this can be used to prove the formulas for [ in (2) and (9), once thecorresponding formula for ] has been established.


5.7. Subsystems. We say a subset S of T is closed or a subsystem if it containsthe element 1 ∈ T0 and is closed under the operations of multiplication •, inversion,and ] or, equivalently, [. If X ⊂ T is an arbitrary subset, the closure of X (or thesubsystem generated by X), denoted

⟨X⟩, is defined as the smallest subsystem of

T containing X. Its existence and uniqueness is clear: just take the intersection ofall subsystems containing X, this set being non-empty because T belongs to it. Wenow give a more explicit description of the closure of a subset X.

Example. Let G = E2(A) as in 5.2 and let X = T1 = t+(u) : u ∈ A× as in(5.2.4). Then

⟨X⟩

= T follows from (5.5.6).

5.8. Lemma. Let X ⊂ T be an arbitrary subset.

(a) The set⟨X⟩

is obtained as follows. Let Y = X ∪ X−1 ∪ 1 and put

Y :=⋃n∈Z

Yn],

where xn] = x]···] (n times) for n> 0 and xn] = x[···[ (n times) for n < 0. Then⟨X⟩

is the set of all finite products, with arbitrary parentheses, of elements taken fromY.

(b) If ϕ: G→ H is a morphism of gcR which is surjective on root groups, thenT(ϕ)

(⟨X⟩)

=⟨T(ϕ)(X)

⟩.

Proof. (a) Clearly, Y ⊂⟨X⟩. It follows from (5.6.7) and (5.6.8) that Y is

stable under the unary operations ( )−1, ] and [. Let P be the set of all products of

elements from Y. Then clearly P ⊂ 〈X〉 so it suffices to show that P is a subsystem.Evidently, P contains 1 and is closed under products. To show it is closed underthe unary operations, we use induction on the length of a product. Products oflength 1 are just the elements of Y. A product of length n > 1 is of the form a • bwhere a and b are products of length < n. Then (a • b)−1 = a • b−1 by (5.6.2), andby induction b−1 ∈ P. Hence (a • b)−1 ∈ P as well. Similarly, one shows that P isstable under ] and [, using the second and third formula of (5.6.2).

(b) This follows from Lemma 5.6(b).

5.9. Categories defined by sets of Weyl triples. Let G = (G, U) ∈ gcRand let st(G) be the category of coverings of G defined in Corollary 4.12. We definefull subcategories of st(G) depending on a set X of Weyl triples of G as follows.

Let π: G → G be an object of st(G), i.e., a covering, and define Θ(G) andΘ(G) as in (5.1.2). Since π is in particular bijective on root groups, the inducedmaps Θ(π): Θ(G)→ Θ(G) sending a triple x = (x−1, x0, x1) ∈ U−α × Uα × U−α toπ(x) =

(π(x−1), π(x0), π(x1)

), are bijective as well. By abuse of notation, we will

often simply write π instead of Θ(π) or T(π). For an element t = (t−1, t0, t1) ∈ Θ(G)we call π−1(t) ∈ Θ(G) the lift of t to G. The lift of a Weyl triple for G will ingeneral no longer be a Weyl triple for G. Therefore, we define the full subcategoryst(G, X) of st(G) by(

π: G→ G)∈ st(G, X) ⇐⇒ π−1(X) ⊂ T(G).

This subcategory has the following property:


if π ∈ st(G, X) and ϕ: π → η is a morphism of st(G) then also η ∈ st(G, X).

Indeed, the morphism ϕ is a commutative triangle

G

π????????ϕ // H

η~~~~~~~~

G

(1)

of morphisms of gcR, so by Lemma 4.5(b) ϕ is also a covering, in particular, it isbijective on root groups. Hence it induces a commutative triangle of bijections

Θ(G)

π##GGGGGGGGϕ // Θ(H)

ηwwwwwwww

Θ(G)

which impliesϕ(π−1(t)) = η−1(t) (2)

for all t ∈ Θ(G). Since ϕ is surjective on root groups, the image of a Weyl tripleof G under ϕ is a Weyl triple of H, as noted in 5.1. Hence η−1(X) = ϕ(π−1(X)) ⊂ϕ(T(G)) ⊂ T(H).

5.10. Lemma. The subcategories st(G, X) have the following properties.

st(G, ∅) = st(G, 1) = st(G), (1)

X ⊂ Y =⇒ st(G, X) ⊃ st(G, Y), (2)

st(G, X) = st(G,⟨X⟩). (3)

Proof. (1) and (2) are evident from the definition. We prove (3). Since X ⊂⟨X⟩,

we have st(G, X) ⊃ st(G,⟨X⟩) by (2). Conversely, let π: G→ G belong to st(G, X),

so X := π−1(X) ⊂ T(G). We must show that π−1(⟨X⟩) ⊂ T(G) as well. By 5.5,⟨

X⟩⊂ T(G), so it suffices to show that π−1(

⟨X⟩) =

⟨X⟩. But this follows from

Lemma 5.8 and bijectivity of π on Θ(G): π(⟨X⟩)

=⟨π(X)

⟩=⟨X⟩.

5.11. Reflective subcategories. Our next aim is to show that st(G, X) is areflective subcategory of st(G). For the convenience of the reader, we recall from[1, Chapter I, 4.16] or [66, IV.3] the following notions, dual to those reviewed in4.7. Let B be a full subcategory of a category A. A B-reflection for an objectA ∈ A is an object B ∈ B and a morphism r: A→ B with the following universalproperty: for all f : A → B′ (where B′ ∈ B) there exists a unique g: B → B′ suchthat

Af //

r

B′

B

∃! g

88(1)


is commutative. As in 4.7 one sees that a B-reflection is unique up to a uniqueisomorphism. The subcategory B is called reflective if every A ∈ A admits aB-reflection.

An equivalent condition is that the inclusion functor I: B ⊂ A admit a leftadjoint P : for all A ∈ A and B ∈ B there exist natural isomorphisms

MorB(P (A), B) ∼= MorA(A, I(B)). (2)

5.12. Theorem. Let R be a reflection system, let G ∈ gcR be a group withR-commutator relations and let X be a set of Weyl triples for G. Then the categoryst(G, X) is a reflective subcategory of st(G).

Proof. We put Xα = X∩Tα(G), so that X =∐α∈Re(R) Xα. Let π: G→ G be an

object of st(G), let t ∈ Xα be a Weyl triple, and let wt = µ(t) be the correspondingWeyl element. Also let x = π−1(t) ∈ Θ(G) be the lift of t to G. Since π is bijectiveon root groups and wt is a Weyl element for α in G, there exists, for every β ∈ R,a unique isomorphism fαβ(t): Uβ → Usα(β) making the diagram

Uβfαβ(t) //

π ∼=

Usα(β)

∼= π

Uβ

Int(wt)// Usα(β)

(1)

commutative. It is clear that wt := µ(x) is a Weyl element for α in G if and only if

Int(wt) · u = fαβ(t) · u, (2)

for all u ∈ Uβ and all β ∈ R. The desired reflection r: A→ B is obtained as follows.

Let G be the largest quotient of G for which the relations (2) hold. Then π inducesa morphism π: G → G which is an object B of st(G, X), and the reflection r isinduced by the canonical map can: G→ G.

We now work out the details. Let K be the normal subgroup of G generatedby all elements

Z(t, u, α, β) :=(

Int(wt) · u)−1(

fαβ(t) · u)

where α ∈ Rre, β ∈ R, t ∈ Xα and u ∈ Uβ , and observe that K ⊂ Ker(π). Let

G = G/K and let can: G→ G be the canonical map. Then π factors π = π can.Lemma 4.5(b) shows that G with root groups Uα = can(Uα) and projection πbelongs to st(G), and we have a commutative triangle

Gcan //

π>>>>>>>> G

π

G

It follows from the definition of K that can(wt) is a Weyl element (and hencecan(π−1(t)) = π−1(t) is a Weyl triple) for α in G, for all t ∈ Xα and all α ∈ Rre.


Hence π belongs to st(G, X), and we claim that r = can: G→ G is a reflection forG, thus has the universal property of (5.11.1).

Indeed, let η: (G′,U′)→ (G, U) be in st(G, X) and let ϕ: π → η be a morphismof st(G), thus a commutative triangle

Gϕ //

π???????? G′

η~~~~~~~~~~

G

By Lemma 4.5(b), ϕ is then also a covering, in particular, ϕβ : Uβ → U ′β is an

isomorphism for all β ∈ R. To have an induced homomorphism ϕ: G → G′ itsuffices to show that K ⊂ Ker(ϕ). Let t ∈ Xα and let w′t = µ(η−1(t)) ∈ G′. SinceG′ ∈ st(G, X), this is a Weyl element for G′, and since ϕ is a group homomorphismwe have, using (5.9.2),

ϕ(wt) = ϕ(µ(π−1(t))) = µ(ϕ(π−1(t))) = µ(η−1(t)) = w′t. (3)

From (1) and the analogous diagram for G′ and the fact that η is bijective on rootgroups it follows that

ϕ(fαβ(t) · u

)= f ′αβ(t) · ϕ(u), (4)

for all u ∈ Uα. Indeed, let γ = sα(β). Then both ϕ(fαβ(t) · u) and f ′αβ(t) · ϕ(u)

belong to U ′γ , and η: U ′γ → Uγ is bijective. Hence it suffices to show that theirimages under η are equal. Now compute, using π = η ϕ and (1) and the analogousdiagram for G′:

η(ϕ(fαβ(t) · u)) = π(fαβ(t) · u) = Int(wt) · π(u)

= Int(wt) · η(ϕ(u)) = η(f ′αβ(t) · ϕ(u)).

Now (3) and (4) imply

ϕ(Z(t, u, α, β)

)=(

Int(w′t) · ϕ(u))−1(

f ′αβ(t) · ϕ(u))

= (∗).

We have G′ ∈ st(G, X), so w′t is a Weyl element for G′. Hence (2), applied toG′, shows that (∗) = 1. This implies K ⊂ Ker(ϕ), so there is a unique inducedmorphism ϕ: G→ G′ of st(G) making the diagram

Gϕ //

can

G′

G

∃! ϕ

99

commutative. From π can = π = η ϕ = η ϕ can we infer π = η ϕ, provingthat ϕ is indeed a morphism from π to η in st(G).


5.13. Corollary and Definition. Let G ∈ gcR be a group with R-commutatorrelations and let X be a set of Weyl triples for G. Let St(G) = π: G → G be itsSteinberg group in st(G) as in 4.10, and let r: G → G be a st(G, X)-reflection forG. Then G is an initial object in st(G, X), called the Steinberg group of (G, X) anddenoted St(G, X). This group does not change when replacing X by its closure:

St(G, X) = St(G,⟨X⟩). (1)

Proof. This is immediate from Theorem 5.12 and the fact that St(G) is an initialobject in st(G). The last statement follows from (5.10.3).

5.14. Example: St2(A) of a ring. Let A be a unital associative ring. Thelinear Steinberg group St2(A) is the group presented by generators and relations asfollows. The generators are all symbols xσ(a), a ∈ A, σ ∈ +,−. For u ∈ A× put

wσ(u) = x−σ(u−1) xσ(u) x−σ(u−1).

Then the relations are

xσ(a+ b) = xσ(a)xσ(b), (1)

wσ(u)x−σ(a)wσ(u)−1 = xσ(uau), (2)

for all σ ∈ +,−, a, b ∈ A, u ∈ A×. The reader will easily verify that this definitionagrees with the one in [74, 10.4], see also [32, p. 57], by setting x+(a) = x12(a)and x−(a) = x21(−a). Let us show that this group is the Steinberg group of anappropriately defined (G, X) in the sense of 5.13. First observe that (1) and (2)imply

wσ(u)xσ(a)wσ(u)−1 = x−σ(u−1au−1) (3)

for u ∈ A× and a ∈ A. Indeed, wσ(u)−1 = wσ(−u) by (1), whence x−σ(a) =wσ(−u)xσ(uau)wσ(−u)−1 by (2), so that (3) follows by replacing u by −u and aby u−1au−1.

Let R = A1 = 0,±1. We have already noted in 3.3(a) that the objects of thecategory gcR are the groups G generated by two abelian subgroups U± = U±1.

In particular, this is so for the group St2(A), generated by the two abeliansubgroups x±(A). The relations (2) and (3) say that

tσ(u) =(x−σ(u−1), xσ(u), x−σ(u−1)

), σ = ±, u ∈ A×,

is a Weyl triple for the root σ1 ∈ R.Let now G = E2(A) be the group of Example 5.2 which, by definition, is

generated by the two abelian subgroups e±(A) for e± as in (5.2.1). Hence alsoE2(A) ∈ gcR. Moreover, we have seen in loc. cit. that the elements

tσ(u) =(e−σ(u−1), eσ(u), e−σ(u−1)

), σ = ±, u ∈ A×,

of (5.2.4) are the Weyl triples of G ∈ gcR. Thus there exists a unique grouphomomorphism π: St2(A) → G mapping xσ(a), a ∈ A, onto eσ(a), hence alsoπ(tσ(u)

)= tσ(u) ∈ E2(A). It follows that π is an object of the category st(G, X)

for X = tσ(u) : σ = ±, u ∈ A×. Since any other object π: G → G in st(G, X)satisfies the relations (1) and (2), the map π: St2(A) → E2(A) is an initial objectin st(G, X). Thus, by the usual abuse of notation, St2(A) is canonically isomorphicto the Steinberg group St(G, X).

In view of (5.13.1) and the example in 5.7, one obtains the same group by takingfor X only the set T1(G) of all Weyl triples for the root α = 1.


5.15. Balanced Weyl triples. We return to a group G with R-commutatorrelations and root groups Uα where R is a reflection system. Recall from (5.5.3)the unary operations ] and [ on the set T of Weyl triples. A Weyl triple x is calledbalanced if x] = x[.

Here are some properties of balanced Weyl triples. First, the following condi-tions for a Weyl triple x = (x−1, x0, x1) with w = µ(x) = x−1x0x1 are equivalent:

(i) x is balanced,

(ii) x−1 = x1 and wx1w−1 = x0,

(iii) x−1 = x1 and wx0w−1 = x1,

(iv) x−1 = x1 and x1x0x1 = x0x1x0,

In this case, x] = x[ = (x0, x1, x0).

Indeed, let x] = x[. By (5.5.3) this means

(wx1w−1, x−1, x0) = (x0, x1, w

−1x−1w),

equivalently, x−1 = x1 and x0 = wx1w−1 = w−1x1w. This proves (i) =⇒ (ii)

and (i) =⇒ (iii). Now suppose (ii) holds. Since w = x1x0x1, it follows thatx1x0x1x1 = wx1 = x0w = x0x1x0x1 whence x0x1x0 = x1x0x1, proving (iv).In the same way, one shows (iii) =⇒ (iv). Finally, suppose (iv) holds. Thenw = x1x0x1 = x0x1x0, which implies wx1w

−1 = x0x1x0 ·x1 ·x−11 x−1

0 x−11 = x0, and

w−1x1w = x−11 x−1

0 x−11 · x1 · x0x1x0 = x0. But this says x] = x[ = (x0, x1, x0) by

(5.5.3).As a consequence of these characterizations, we note:

Let x and y be balanced Weyl triples having one componentin common and satisfying w = µ(x) = µ(y). Then x = y.

(1)

Indeed, if x−1 = y−1 then x0 = wx−1w−1 = y0 by (ii), and if x0 = y0 then

x−1 = y−1 follows from (iii). See also Proposition 5.20 for a similar result.

From Lemma 5.6(a), one sees immediately that the set of balanced Weyl triplesis stable under the unary operations ( )−1, ] and [, as well as under all left multi-plications by elements of T. In particular,

the set of balanced Weyl triples is a subsystem of T, (2)

if X ⊂ T is balanced, then so is⟨X⟩. (3)

Example. Let G = E2(A). Then by (5.5.6), all Weyl triples are balanced, soa Weyl triple is uniquely determined by any one of its components (which is alsoevident from the explicit formulas (5.2.2) and (5.2.4)). Moreover,

wu = µ(t+(u)) = µ(t−(u)) = e+(u)e−(u−1)e+(u).

By (5.2.5) the analogous statements hold for EI(A) for I an index set with |I|> 2.We will discuss other examples of groups in which all Weyl triples are balanced

in 5.17, 5.18 and 5.19. However, the reader should not get the impression that thisis always true. For example, by 9.12(b), the projective elementary group PE(V ) ofa Jordan pair V , viewed as group with A1-commutator relations, has non-balancedWeyl triples as soon as the extreme radical Extr(V −) is non-trivial.


5.16. Lemma. Let ϕ: G → H be a morphism of gcR which is surjective onroot groups. Then T(ϕ) preserves balanced Weyl triples, and if ϕ is bijective onroot groups, a Weyl triple x ∈ T(G) is balanced if and only if T(ϕ)(x) is balanced.

Proof. The first statement follows immediately from Lemma 5.6(b). For thesecond, it suffices to remark that T(ϕ) is injective if ϕ is bijective on root groups.

5.17. Example: projective elementary linear groups. Let A be a unitalassociative ring, and let G = E2(A) be the elementary linear group of 5.2. We haveseen there that G has A1-commutator relations with respect to the root groupsU±1. The centre of G is Z (G) = z · 12 ∈ G : z ∈ Z (A)× where Z (A)× is theset of invertible elements of the centre Z (A) of A. Let

π: G→ G := G/Z (G)

be the canonical map. Then G has A1-commutator relations with respect to theroot groups U±1 = π(U±1). Since π is surjective on root groups, it sends Weylelements of G to Weyl elements of G. In fact, standard matrix calculations shows

π(W±1(G)

)= W±1(G). (1)

The induced map T(π) on the Weyl triples is well-defined by 5.1, injective sinceZ (G) ∩ U± = 1 and surjective by (1), whence

T(π): T(G)→ T(G) is a bijection.

In particular it follows from 5.16 and the example in 5.15 that all Weyl triples inG are balanced.

5.18. Example: semisimple algebraic groups. Let G be a connected semi-simple algebraic group defined over an algebraically closed field k (in the sense of[13, 90]) and let R be the root system of G with respect to a maximal torus T ⊂ G.We have seen in Example (c) of 3.3 that G has commutator relations with respect tothe family (Uα)α∈R of root groups. By [90, Lemma 8.1.4] it also has Weyl elementsfor all roots. We claim that all Weyl triples are balanced.

For the proof, let Gα be the centralizer of the subtorus (Kerα)0 ⊂ T in G. By[13, Theorem 13.18], Gα is a connected reductive algebraic group of semisimplerank 1. In particular, it contains root subgroups U±α with respect to T suchthat the subgroup G′α of Gα generated by Uα ∪U−α has A1-commutator relations.By [13, Proposition 13.13] there exists a surjective morphism ϕ from Gα ontoPGL2(k) = GL2(k)/k× · 12. One knows that Ker(ϕ) = Z (Gα) ⊂ T and that ϕmaps the root groups U±α isomorphically onto root groups of PGL2(k), which inthe setting of this example equals the group G of Example 5.17. By conjugacy ofmaximal tori and hence of the associated root groups, we can assume that ϕ(U±α)are the root groups of 5.17.

Let now x ∈ Tα(G) be a Weyl triple. Then x is in particular a Weyl triple of G′αand so, by 5.1, T(ϕ)(x) is a Weyl triple of PGL2(k), hence balanced by 5.17. Butthen x is balanced by Lemma 5.16, applied to ϕ: G′α → PGL2(k).


5.19. Example: Split reductive group schemes. Let G be a split reductivegroup scheme over a scheme S, and let S′ → S be a morphism of schemes. Wehave seen in Example 3.3(d) that the subgroup G of G(S′) generated by the rootgroups Uα = Uα(S′) is a group with R-commutator relations, where R is the rootsystem of G.

The elements wα of [23, XX, 3.1] are Weyl elements (for the root −α) ([23,XXII, Proposition 1.4]) and are obtained from balanced Weyl triples. In particular,if G is a Chevalley group, the existence of balanced Weyl triples for all α ∈ Rfollows from [94, Chapter 3, Lemma 19].

We have seen in 5.15 that a balanced Weyl triple is uniquely determined byany one of its components. We now investigate this property for not necessarilybalanced Weyl triples.

5.20. Proposition. Let R be a symmetric reflection system, see 2.2, and let(G,U) be a group with R-commutator relations and unique factorization for nilpo-tent pairs, see 3.15. Let α ∈ R and suppose that there exists a root β ∈ Rre withthe following properties:

(i) α and β are Z-linearly independent, and both (α, β) and (α,−β)are nilpotent pairs,

(ii)(((((((α, β

)))))))= ∅,

(iii) sβ(α) 6= α,

(iv) Wβ 6= ∅.

(a) Then Uα ∩ U−α = 1.

(b) If x and y are Weyl triples for the root α having one component in commonthen x = y.

Remarks. (1) If β has properties (i) – (iv) for α then −β has these propertiesfor −α. This follows from symmetry of R and Wβ = W−β (by (5.4.1)).

(2) Suppose R satisfies the condition (F2) of 3.4 and α and β are Z-linearlyindependent. Then (α, β) and (α,−β) are nilpotent pairs.

(3) Let R be locally finite root system without irreducible components of rankone. Then for every α ∈ R× there exists β ∈ R× satisfying (i) – (iii).

Indeed, since R has no irreducible components of rank 1, there exists a linearlyindependent β not orthogonal to α. Possibly after replacing β by its negative, wemay assume 〈α, β∨〉 > 0. Since sβ(α) = α− 〈α, β∨〉β, it is clear that (iii) holds forβ. Now we distinguish two cases. First suppose that α+β /∈ R. From the structureof the commutator set of two roots given in 2.17, it follows that

(((((((α, β

)))))))= ∅. Hence

β has the required property (ii) as well, and condition (i) holds by (b).Now suppose that γ := α + β ∈ R. Then we modify β as follows. First, note

that 〈γ, β∨〉 = 3. Indeed, by standard facts [63, A.2], 〈α, β∨〉 ∈ 1, 2, 3. Hence〈γ, β∨〉 = 〈α + β, β∨〉 = 〈α, β∨〉 + 2 ∈ 3, 4, 5. Assuming 〈γ, β∨〉 = 4 yields, byloc. cit., γ = 2β which implies α = β, contradicting linear independence of α andβ. The case 〈γ, β∨〉 = 5 is impossible, again by loc. cit. Now it follows from 2.17,case 8, that B = −β, γ is a root basis of a subsystem of type G2, with −β the


short root. From the well-known structure of such root systems, one sees easilythat β′ = −γ = −β − α has the required properties.

Proof. We first draw some consequences from our assumptions. From (i) and(ii) and symmetry of R follows(((((((

Uα, Uβ)))))))⊂ U(((((α,β))))) = U∅ = 1 and

(((((((U−α, U−β

)))))))= 1. (1)

By unique factorization and Z-linear independence of α and β, it follows fromCorollary 3.18 that

Uα ∩ U(((((α,−β))))) = U(((((α,−β))))) ∩ U−β = 1. (2)

By (2.3.5), sβ(α) = α − 〈α, β∨〉β, where 0 6= 〈α, β∨〉 ∈ Z by (iii) and integralityof R. Moreover, (ii) implies 〈α, β∨〉 > 0, else sβ(α) ∈

(((((((α, β

))))))). Hence we have

sβ(α) ∈(((((((α,−β

))))))). Thus if wβ ∈Wβ then

wβUαw−1β = Usβ(α) ⊂ U(((((α,−β))))). (3)

As a consequence,

uα ∈ Uα and(((((((uα, U−β

)))))))= 1 =⇒ uα = 1. (4)

Indeed, the assumptions on uα and (1) show that uα commutes with wβ (whichbelongs to U−βUβU−β). Hence uα = wβuαw

−1β ∈ Uα∩Usβ(α) ⊂ Uα∩U(((((α,−β))))) = 1

by (3) and (2). By symmetry of R and Wβ = W−β , see (5.4.1), we also have

v−α ∈ U−α and(((((((v−α, Uβ

)))))))= 1 =⇒ v−α = 1. (5)

Proof of (a). If u ∈ Uα∩U−α then (1) implies(((((((u, U±β

)))))))= 1, so u = 1 by (4).

Proof of (b). Let x = (x−1, x0, x1) and y = (y−1, y0, y1) be Weyl triples forα, and let wα = x−1x0x1 and w′α = y−1y0y1 be the Weyl elements determined bythem. We put v := w−1

α w′α ∈WαWα so v normalizes all root groups.

Case 1 : Suppose x−1 = y−1. Then v = x−αuαz−α where x−α = x−11 ∈ U−α,

uα = x−10 y0 ∈ Uα, and z−α = y1 ∈ U−α.

We show first that uα = 1, i.e., that the second components of x and y agree.Since x−α and z−α commute with U−β by (1) and v normalizes all root groups,it follows that uα = x−1

−αvz−1−α normalizes U−β , too, so we have

(((((((uα, U−β

)))))))⊂ U−β .

But also(((((((uα, U−β

)))))))⊂ U(((((α,−β))))) by the commutator relations. Hence,(((((((

uα, U−β)))))))⊂ U(((((α,−β))))) ∩ U−β = 1

by (2), and therefore uα = 1 by (4). This says x0 = y0, so the second componentsof x and y agree.

Now v = x−αz−α ∈ U−α normalizes all root groups, so(((((((v, Uβ

)))))))⊂ U(((((−α,β)))))∩Uβ =

1 by (2). By (5), v = 1, so x1 = y1, and the third components of x and y agree.

Case 2 : Suppose that x0 = y0. Then x[ = (x0, x1, w−1α x−1wα) and y[ =

(y0, y1, w′α−1y−1w

′α) are two Weyl triples for −α with the same first component.

By Remark (1) above, −β satisfies the assumptions of the proposition for −α.Hence x[ = y[ by Case 1, and therefore x = y by (5.6.7).

Case 3 : Finally, suppose x1 = y1. Then x−1 and y−1 are two Weyl triples forα with the same first component x−1

1 = y−11 , so x−1 = y−1 by Case 1 and therefore

x = y by (5.5.2).


5.21. Rank one groups. A rank one group is a triple (G,U+, U−) consistingof a non-trivial group G and two subgroups U+ and U− satisfying the followingconditions:

(R0) G is generated by U+ and U−,

(R1) U+ ∩ U− = 1,(R2) for all σ ∈ +,− and all x ∈ Uσ := Uσ 1 there exists y ∈ U−σ

with the property that

xU−σx−1 = y Uσy−1. (1)

This generalizes Timmesfeld’s definition [95, I, (1.1)] because we do not assumeU± nilpotent.

By abuse of language, we often do not specify the subgroups U± when speakingof a rank one group. Here are some standard properties of rank one groups, see[95, I, §1, §2].

(a) The normalizer of Uσ in U−σ is trivial. Assume x ∈ Uσ normalizes U−σ

and choose y ∈ U−σ as in (1). Then U−σ = xU−σx−1 = y Uσy−1 and henceUσ = y−1 U−σy = U−σ, so 1 6= x ∈ U+ ∩ U−, contradicting (R1).

(b) For a given x ∈ Uσ, the element y of (1) is uniquely determined and is6= 1. Indeed, assume xU−σx−1 = y Uσy−1 = v Uσv−1 for y, v ∈ U−σ. Then v−1ynormalizes Uσ and belongs to U−σ, so v−1y = 1 by (a). We also have y 6= 1 elsexU−σx−1 = Uσ which implies U−σ = x−1 Uσx = Uσ and 1 6= x ∈ U+ ∩ U−,contradicting (R1).

We write y = jσ(x) and then have the defining relation

xU−σx−1 = jσ(x)Uσjσ(x)−1. (2)

(c) The map jσ: Uσ → U−σ is bijective, with inverse map j−1σ = j−σ. Indeed,

put y = jσ(x). By (2), with σ replaced by −σ and x replaced by y, we havey Uσy−1 = j−σ(y)U−σj−σ(y)−1. By comparing this with (2) and using (b), itfollows that x = j−σ(y) = j−σ(jσ(x)). Since σ can be + and −, this shows that jσis bijective with inverse map j−σ.

(d) Let sbgr(G) be the set of all subgroups of G and let

Ξ = U+ ∪ xU−x−1 : x ∈ U+ ⊂ sbgr(G). (3)

It follows easily from (1) that also

Ξ = U− ∪ yU+y−1 : y ∈ U−. (4)

We have U+ 6= U−, else G = 1 by (R0) and (R1). It follows that the unions in(3) and (4) are disjoint. The group G acts on sbgr(G) by conjugation. From (3) itis clear that Ξ is stable under the action of U+, and (4) shows that it is invariantunder the action of U− as well. Since G is generated by U+ and U−, it follows thatΞ is precisely the orbit of U+ (or U−) under the action of G on sbgr(G). Moreover,


G acts doubly transitively on Ξ.

Indeed, by transitivity of G on Ξ, it suffices to show that, for some ξ ∈ Ξ, theisotropy group Gξ of ξ in G acts transitively on Ξ ξ. Let ξ = U+. ThenU+ ⊂ Gξ and (3) shows that U+ is transitive on Ξ ξ.

(e) If ξ, η ∈ Ξ with ξ ∩ η 6= 1 then ξ = η. Indeed, assume to the contrarythat ξ 6= η. By the double transitivity of G on Ξ, there exists g ∈ G such thatξ = gU+g−1 and η = gU−g−1. Hence ξ ∩ η 6= 1 implies U+ ∩ U− 6= 1,contradicting (R1).

We now characterize rank one groups in terms of Weyl triples. For x ∈ Uσ, weintroduce the notation

∨x = jσ(x)−1, x∨ = jσ(x−1), (5)

and note that the maps x 7→ ∨x and x 7→ x∨ are inverses of each other:

(∨x)∨ = x = ∨(x∨). (6)

Indeed,

(∨x)∨ =(jσ(x)−1

)∨= j−σ

((jσ(x)−1

)−1)

= j−σ(jσ(x)

)= x,

by (c), and similarly the second formula.

5.22. Proposition. Let G be a group generated by two non-trivial subgroupsU+ and U−. Let R = X = Z and define Rre = 1,−1 and sα(x) = −x for α ∈ Rre

and x ∈ X. It is easily seen that this defines a reflection system and that G hasR-commutator relations with root groups U±n = U± for n > 0. Denote by Tα theset of Weyl triples for the root α. Then the following conditions are equivalent:

(i) G is a rank one group,

(ii) for all x ∈ U± there exist unique elements v, t ∈ U∓ such that vxt is aWeyl element for the root ±1,

(iii) the projection onto the second factor pr2: T±1 → U± maps T±1 bijectivelyonto U±.

If these conditions hold, then the elements v, t of (ii) are v = ∨x and t = x∨, andthe inverse of pr2 is given by x 7→ t(x) := (∨x, x, x∨).

Remark. Since the maps x 7→ ∨x and x 7→ x∨ are bijective, it follows thatthe first and third projections pr1: T±1 → U∓ and pr3: T±1 → U∓ are bijective aswell.

Proof. (i) =⇒ (ii): We first prove uniqueness. Let σ ∈ +,−. Supposex ∈ Uσ and v, t ∈ U−σ are such that w = vxt is a Weyl element for α = σ1.We show that v = ∨x. Since w is a Weyl element for α, wU−σ = Uσw whichimplies vxtU−σ = vxU−σ = Uσvxt and, by multiplying with t−1 on the right,vxU−σ = Uσvx. Hence xU−σx−1 = v−1Uσv, which by 5.21(b) shows v−1 = jσ(x),so v = jσ(x)−1 = ∨x.


Similarly, wUσ = U−σw says vxtUσ = U−σvxt = U−σxt, and implies xtUσ =U−σxt, whence x−1Uσx = tUσt−1, and shows t = jσ(x−1) = x∨.

To prove existence, suppose x ∈ Uσ and put v = ∨x, t = x∨ and w = vxt. Thenw is a Weyl element for the root α = σ1. Indeed, tUσt−1 = jσ(x−1)Uσjσ(x−1)−1 =x−1U−σx by (5.21.2), hence

wUσw−1 = (vx)[tUσt−1](x−1v−1)

= (vx)[x−1U−σx](x−1v−1) = vU−σv−1 = U−σ

since v = ∨x ∈ U−σ. Similarly, since also t = x∨ ∈ U−σ,

wU−σw−1 = vx(tU−σt−1)x−1v−1 = v(xU−σx−1)v−1

= vjσ(x)Uσjσ(x)−1v−1 = Uσ,

by (5.21.2) and since vjσ(x) = 1 by (5.21.5). Thus, the relation (5.1.1) holds forα = σ1 and β = ±σ1. Since U±i = U± for i> 1, (5.1.1) follows for all β ∈ R, so wis a Weyl element for α = σ1.

(ii) =⇒ (iii): We show first that pr2 maps Tσ1 to Uσ. Assume to the contrarythat there exists x = (v, x, t) ∈ Tσ1 with x = 1. Then w = vt ∈ U−σ is a Weylelement, which implies Uσ = wU−σw−1 = U−σ, so we have U+ = U−. But thenevery element of G = U+ = U− is trivially a Weyl element. Hence for 1 6= x ∈ U+

we have x = 1 · x · 1 = x · x · x−1, so both v = 1, t = 1 and v = x, t = x−1 satisfy(ii), contradicting the uniqueness property. Now it follows immediately from (ii)that pr2: Tσ1 → Uσ is bijective.

(iii) =⇒ (i): We first show that U+ ∩ U− = 1. Assume to the contrary that1 6= x ∈ U+∩U−. Then there exists a Weyl triple x = (v, x, t) ∈ T1, with v, t ∈ U−.Since also x ∈ U−, the corresponding Weyl element w = vxt ∈ U−. This impliesU+ = wU−w−1 = U−. But then any x ∈ U+ is a Weyl element for α = 1 . Hence(1, x, 1) and (x, x, x−1) are Weyl triples in T1 having the same middle component.By (ii), the middle projection is injective, so we have x = 1, contradiction.

It remains to verify (5.21.1). Let x ∈ Uσ. Since pr2: Tσ1 → Uσ is surjective,there exists a Weyl triple x = (v, x, t) ∈ Tσ1. Let w = vxt be the correspondingWeyl element. Then

wU−σ = vxtU−σ = Uσw = Uσvxt.

Since t ∈ U−σ, this implies vxU−σ = Uσvx or xU−σx−1 = v−1Uσv, so condition(5.21.1) holds for y = v−1.

Example. The group G = E2(A) as in 5.2 is a rank one group if and only ifA is a division ring. In this case ∨eσ(u) = eσ(u)∨ = e−σ(u−1) for all 0 6= u ∈ A.

Indeed, if G is a rank one group then, by the proposition, every eσ(u) 6= 1, i.e.,u 6= 0, is the middle component of a Weyl triple. By (5.2.4), u is invertible and∨eσ(u) = eσ(u)∨ = e−σ(u−1). Conversely, if A is a division ring then, again by(5.2.4), every 1 6= eσ(u) is the middle component of a Weyl triple, so that G is arank one group. We will generalize this example in 9.13 by replacing (A,A) by anyJordan division pair.


5.23. Corollary. Let (G,U+, U−) be a rank one group. For an element x ∈ Uσlet t(x) = (∨x, x, x∨) be the associated Weyl triple as in Proposition 5.22. Recallthe operations ] and [ on Weyl triples from 5.5.

(a) Then t(x)] = t(∨x) and t(x)[ = t(x∨).

(b) The following conditions are equivalent:

(i) t(x) is balanced,

(ii) ∨x = x∨,

(iii) jσ(x−1) = jσ(x)−1,

(iv) x · x∨ · x = x∨ · x · x∨,

(v) x · ∨x · x = ∨x · x · ∨x.

Proof. (a) By (5.21.6) we have

t(x∨) = (∨(x∨), x∨, x∨∨) = (x, x∨, x∨∨), t(∨x) = (∨∨x, ∨x, (∨x)∨) = (∨∨x, ∨x, x).

On the other hand, by the definition of ] and [, t(x)] = (w · x∨ · w−1, ∨x, x) andt(x)[ = (x, x∨, w−1 · ∨x · w) where w is the Weyl element defined by t(x), t(x)]

and t(x)[. Since t(x∨) and t(x)[ have the same middle component, it follows fromProposition 5.22 that they are equal. The second statement is proved in the sameway.

(b) Recall from 5.15 that a Weyl triple x is balanced if x] = x[. Hence theequivalence of (i) and (ii) follows from (a) and the fact that x 7→ t(x) is bijective,by Proposition 5.22, and the equivalence of (ii) and (iii) is clear from (5.21.5).

Now suppose (i) and (ii) hold. Then t(x)[ = (x, x∨, x) = t(x)] = (x, ∨x, x). By(5.5.4), these Weyl triples determine the same Weyl element w as t(x) = (∨x, x, x∨).It follows that w = ∨x · x · x∨ = x · x∨ · x = x · ∨x · x, so we have (iv) and (v).

(iv) =⇒ (ii): Put t = x∨ for simpler notation. Then xtx = txt impliesxtx−1 = t−1xt ∈ t−1Uσt. By (5.21.2), xtx−1 ∈ xU−σx−1 = jσ(x)Uσjσ(x)−1,so 1 6= xtx−1 ∈ t−1Uσt ∩ jσ(x)Uσjσ(x)−1. By 5.21(e), these subgroups are equal.Hence tjσ(x) ∈ U−σ normalizes Uσ. By 5.21(a), we have tjσ(x) = 1 or t =jσ(x)−1 = ∨x.

(v) =⇒ (ii): Let v = ∨x. Then xvx = vxv implies vxv−1 = x−1vx ∈x−1U−σx = jσ(x−1)Uσjσ(x−1)−1 by (5.21.2). Since also vxv−1 ∈ vUσv−1, wesee again that the groups vUσv−1 and jσ(x−1)Uσjσ(x−1)−1 have non-trivial in-tersection and hence are equal. As before, this implies that v−1jσ(x−1) ∈ U−σ

normalizes Uσ, so v = jσ(x−1) = x∨, as desired.

Notes

Most of §1 is contained in [65, §1], except for the replacement of the category of torsion-freeabelian groups used there by the category SF. The notion of N-free subsets is due to N. Bardy

[8]. Lemma 1.3 and Proposition 1.4 are new, the latter generalizes [65, Lemma 1.1].

Nilpotent pairs in generalizations of locally finite root systems (partial root systems) are

investigated in [65, §3] where we show that our notion of a prenilpotent pair agrees with the one

defined by J. Tits for Kac-Moody root systems in terms of positive systems and the Weyl group.


§2. The material of this section is essentially taken from [63] and [65], which contain anelaborate theory of locally finite root systems and reflection systems, at least over fields of char-

acteristic zero. A generalization of finite root systems over Z in our sense are the root data of [23,

Exp. XXI].

The reader may have noticed that the equivalences (i) – (iii) in 2.11 are in fact true forelements of Rre where R is a reflection system. As in [65, 2.6] it is then possible to define the

concepts of orthogonality, connected subsets of R, connected components of R and show that

Re(R) is the direct sum of its connected components. We will not use these results for arbitraryreflection systems.

The structure of prenilpotent pairs, elaborated in 2.17 in the locally finite case, can become

quite complicated in root systems related to Kac-Moody algebras, see for example the recent paper

[2] by Allcock where the number of prenilpotent pairs in the set of real roots of the Kac-Moodyroot system E10 is estimated.

§3. The idea to study groups via commutator relations came to prominence in Chevalley’sseminal paper [22]. Chevalley’s work was later expanded in Steinberg’s lecture notes [94]. Nowa-

days, groups with commutator relations are abundant in the theory of algebraic groups, spherical

buildings [101, 104] and groups with abstract root subgroups, see for example [20], [95, 96] or[25, 26]. Also, Tits’ construction of Kac-Moody groups [102, 81] is based on this concept.

The definition of a group with commutator relations is inspired by Faulkner’s definition of

groups with Steinberg relations [27, Chapter 1]. In our terminology, Faulkner considers groups

with R-commutator relations where R is a finite root systems, and requires the existence of aWeyl element for each root. In the simply laced case, Faulkner’s results were later also obtained

by Shi [88].

We remark that, unless R is a reduced root system, the assumption α 6= −β in [27, (1.1)]

is not sufficient to guarantee that(((((((α, β

)))))))not contain both a root and its negative, a condition

required in [27, (1.4)]. A counterexample is(((((((α,−2α

)))))))= 0,±α,±2α in BC1.

§4. The theory of Steinberg groups originates with Steinberg’s paper [92], see also [94, §6],which in our setting studies st(G) for G a Chevalley group over a field, cf. Example 3.3(d). An

account of Steinberg groups associated with classical groups is given in the book [32] by Hahn and

O’Meara. These references represent only a small portion of the many papers devoted to variousaspects of Steinberg groups.

A special case of Lemma 4.18 is the construction of root groups in reductive algebraic groups

over arbitrary fields [13, Proposition 21.9], see Example 3.3(c).

§5. The terminology “Weyl element” goes back to [28]. These elements have of course

been considered before in many types of groups (algebraic groups, Chevalley groups, elementary

groups).

The question of when µ: Tα →Wα is injective seems to be rather delicate. Injectivity holds inthe example of GL2(A) treated below, but fails for PGL2(A). The group NormU−α (Uα) operates

freely on Tα by

x · n = (x−1, x0, x1) · n = (x−1n, n−1x0n, n

−1x1)

and obviously x and x · n determine the same Weyl element. Examples where this group is non-trivial (and hence µ is not injective) occur for Jordan pairs with invertible elements and non-trivialextreme radical, see 7.11 and 8.7.

A definition of St2(A) different from the one in 5.14 is given in [46]. For a division algebraA the group St2(A) is isomorphic to the group defined in [59, 1.4] for the division Jordan pairV = (A,A). This follows from [59, Corollary 1.8]. An example of a Steinberg group in the spiritof 5.13 is considered in [29].

Our definition of a rank one group, a slight generalization of Timmesfeld’s, is taken from [60].The properties of rank one groups established in 5.21 are proved in [95, I, §1, §2] and in [60,Lemmata 2.2 and 2.3] in general. The equivalence of (iii) and (iv) in Corollary 5.23 is [95, I,

Lemma (2.2)].

The Steinberg group St(V ) associated with a (not necessarily Jordan) division pair V in thesense of [60, 1.4] is an example of a Steinberg group St(G, X). The group G can be taken as the

little projective group of the based Moufang set M associated with V [60, 3.4]. For this group, X

is the set of all Weyl elements defined in [60, (3.4.5)].


Following [95, I, Definition (1.1)], a rank one group is called special if all elements of Uσ satisfy

the equivalent conditions of Corollary 5.23(b). Not all rank one groups are special. An example

is the 1-dimensional affine group over a field with more than four elements [95, I, Example (1.7)].

CHAPTER II

GROUPS ASSOCIATED WITH JORDAN PAIRS

Summary. The algebraic foundation of the book is the theory of Jordan pairs. In order

to accommodate readers unfamiliar with this little known area of algebra, we give a leisurelyintroduction to Jordan pairs in §6. In particular, we present the most important examples, and

introduce fundamental notions, such as quasi-invertible pairs and the inner automorphisms definedby them, idempotents and their Peirce decompositions.

The prototype of a Jordan pair is the algebraic structure given by the two off-diagonal blocks

of a matrix over some ring, decomposed into four blocks. The basic observation is this: the upperright-hand and the lower left-hand entries, say of size p×q and q×p, are naturally abelian groups,

say V + and V +. Given x ∈ V + and y ∈ V −, the products of the corresponding matrices satisfy(0 x

0 0

)(0 0

y 0

)(0 x

0 0

)=

(0 xyx

0 0

),

(0 0

y 0

)(0 x

0 0

)(0 0

y 0

)=

(0 0

yxy 0

).

Thus we obtain quadratic-linear compositions (x, y) 7→ Qxy := xyx from V + × V − to V +, and(y, x) 7→ Qyx := yxy from V − × V + to V −. They make sense not only in the rectangular case,

but also for hermitian or alternating matrices. Abstract Jordan pairs are then defined as pairs of

abelian groups (more generally, modules over commutative rings) with such “crosswise” quadratic-linear compositions, satisfying identities derived from these concrete examples. The non-linear

character of the compositions causes many technical difficulties but is absolutely essential if one

wishes for a theory valid over arbitrary rings of scalars, in particular, over the integers.

With every Jordan pair V we associate in §7 its projective elementary group PE(V ) which,

for the purpose of defining Steinberg groups, plays the role of the elementary group of a ring orthe elementary unitary groups of a form ring. This is based on the Tits-Kantor-Koecher algebra

L(V ) =⊕i∈Z Li(V ),

a Z-graded Lie algebra with Li(V ) = 0 for |i| > 1, L±1(V ) = V ± and L0(V ) isomorphic to a

suitable subalgebra of the derivation algebra of V . From the Z-graded structure it is clear that(adx)3 = 0 for x ∈ V ±, and due to the definition of Jordan pairs involving a quadratic-linear

composition, (adx)2 is divisible by 2. Thus

exp±(x) = exp(adx) = Id + adx+1

2(adx)2

is well-defined, and turns out to be an automorphisms of L(V ). The maps exp± are injective grouphomomorphisms. The projective elementary group PE(V ) is then defined as the subgroup of the

automorphism group of L(V ) generated by the subgroups U± = exp±(V ±). These subgroups areabelian, so that PE(V ) is a group with A1-commutator relations and root groups U+ and U−.

The next section §8 deals with more specialized topics: the projective elementary group of a

subpair and of a direct sum, the centre of PE(V ) (Theorem 8.7), and a comparison with Faulkner’sprojective elementary group [28].

In the final §9 we introduce the category st(V ) of groups over V , consisting of all coverings(in the sense of §4) of G = PE(V ) with root groups (U+, U−). Thus an object of st(V ) canbe considered as a quadruple (G,U±, π) consisting of a group G, two abelian subgroups U+ and

U− generating G, and a group homomorphism π: G → G which induces an isomorphism from

U± ⊂ G onto the subgroups U± of G. The inner automorphisms defined by quasi-invertible pairsembed in PE(V ) and act on U± by conjugation. They have canonical lifts to any group G over V

which in general do not normalize the U±. Imposing this condition gives rise to relations whichare used later to define subcategories of st(V ). A similar mechanism is applied to idempotents ofV which yield candidates for Weyl elements in G. Finally, we show that the classical Steinberg

group of a ring lies in st(V ) for appropriate V (9.18).

70

§6] Introduction to Jordan pairs 71

§6. Introduction to Jordan pairs

6.1. The elementary group of a Morita context. Let us start with some-thing very simple, namely 2× 2 matrices(

a bc d

)with coefficients in a ring A. (By “ring” we always mean an associative ring withunit element.) Recall from 3.16(c) that the elementary group E2(A) is the subgroupof GL2(A) generated by the elementary matrices

e+(x) =

(1 x0 1

), e−(y) =

(1 0−y 1

)(x, y ∈ A).

More generally, the elementary group En(A) ⊂ GLn(A) is generated by all eij(x) :=1n + xEij , i 6= j, x ∈ A. This can also be done with (formal) 2 × 2 matrices bysubdividing an n× n matrix in 4 blocks, say of size p× p, p× q, q × p, q × q, withp+ q = n. Then En(A) equals the group G generated by the matrices

e+(x) =

(1p x0 1q

), e−(y) =

(1p 0−y 1q

) (x ∈ Matpq(A), y ∈ Matqp(A)

).

Indeed, if x = (xij) ∈ Matpq(A) and y = (yji) ∈ Matqp(A) then

(1p x0 1q

)=

p∏i=1

q∏j=1

ei,p+j(xij),

(1p 0y 1q

)=

p∏i=1

q∏j=1

ep+j,i(yji),

the product being taken in any order, so G ⊂ En(A). To prove equality, it remainsto show that the generators eij(a) where 16 i 6= j 6 p and p+ 16 i 6= j 6 n lie inG. For 16 i 6= j 6 p we have

eij(a) =(((((((ein(a), enj(1)

)))))))∈ G,

and the remaining generators of En(A) are recovered similarly.This suggests to consider right away the following situation: replace Matn(A) by

a ring A with a formal block matrix decomposition with respect to an idempotente = e1 ∈ A. Putting e2 = 1− e, the ring A is isomorphic to the formal matrix ring

A ∼=(

R M+

M− S

)(1)

whereR = e1Ae1, M+ = e1Ae2, M− = e1Ae1, S = e2Ae2.

Thus R and S are rings with unit elements e1 and e2, and M+ and M− are (R,S)-and (S,R)-bimodules, respectively, and M = (R,M+,M−, S) is a Morita context.One defines the elementary group of M by

72 GROUPS ASSOCIATED WITH JORDAN PAIRS [Ch. II

E(M) =⟨(

1 M+

0 1

)∪(

1 0M− 1

)⟩⊂ A×.

(By abuse of notation, we simply write 1 for the unit elements of R and S, respec-tively).

Let us now always work over an arbitrary commutative base ring k. All objectsfor which this makes sense are modules over k, rings are k-algebras, and so on. IfA and M are as above, then M+ and M− are in particular k-modules and R andS are k-algebras.

The associative algebra A gives rise to a Lie algebra A− having the same under-lying k-module and the Lie bracket [a, b] = ab− ba. This Lie algebra has a naturalZ-grading A− =

⊕i∈Z Ai, where Ai = 0 for i /∈ −1, 0, 1 and

A−1 =

(0 0M− 0

), A0 =

(R 00 S

), A1 =

(0 M+

0 0

).

A Z-graded Lie algebra concentrated in degrees −1, 0, 1 is also called 3-graded.

6.2. Generalized elementary groups. We keep the notation introduced be-fore. Let V ± ⊂M± be k-submodules and let V be the pair (V +, V −). We considerthe subgroup

E(M, V ) =⟨(

1 V +

0 1

)∪(

1 0V − 1

)⟩of E(M), called the elementary group of V . Since the V ± are in particular additive

subgroups of M±, it is clear that(

1 V +

0 1

)and

(1 0V − 1

)are multiplicative subgroups

of A×, isomorphic to the additive groups V + and V − under the maps x 7→ e+(x) =(1 x0 1

)and y 7→ e−(y) =

(1 0−y 1

).

Examples. Let A = Mat2n(k), subdivided into 4 blocks of size n× n, and letM be the corresponding Morita context. Then E(M) = E2n(k) is the elementarygroup as in 6.1. Choosing V ± = Hn(k), the n × n symmetric matrices, yields forE(M, V ) the elementary symplectic group ESp2n(k), and choosing V ± = Altn(k),the alternating n × n matrices, i.e., skew-symmetric with zeros on the diagonal,E(M, V ) is the elementary orthogonal group EO2n(k), see [32, 5.3A, 5.3B].

Returning to the general situation, we define k-submodules ei of Ai by

e−1 =

(0 0V − 0

), e1 =

(0 V +

0 0

),

e0 = k ·(

1 00 0

)+ k ·

(0 00 1

)+ [e1, e−1],

ei = 0 for i /∈ −1, 0, 1,

and put

e(M, V ) =⊕i∈Z

ei = e−1 ⊕ e0 ⊕ e1.

Let us consider the following closure conditions for V :


x, z ∈ V σ, y ∈ V −σ =⇒ xyz + zyx ∈ V σ, (1)

x ∈ V σ, y ∈ V −σ =⇒ xyx ∈ V σ. (2)

Here and in the sequel the index σ always takes values in +,− and −σ hasthe obvious meaning. Note that (2) implies (1) by linearization, since V ± is inparticular an abelian subgroup of M±:

xyz + zyx = (x+ z)y(x+ z)− xyx− zyz.

But (1) does in general not imply (2): by specializing x = z in (1) we only get2xyx ∈ V σ. In the examples treated so far, the conditions (2) and hence (1) aresatisfied. Their significance is shown by the following lemma.

6.3. Lemma. Let V = (V +, V −) be a pair of submodules of (M+,M−).

(a) V satisfies (6.2.1) ⇐⇒ e(M, V ) is a graded Lie subalgebra of the Liealgebra A−.

(b) V satisfies (6.2.2) ⇐⇒ e(M, V ) is a graded subalgebra of A− stable underconjugation by elements of E(M, V ).

Proof. (a) “=⇒”: This is shown by direct computation. For example, the rule[e0, e1] ⊂ e1 follows from the relations[(

1 00 0

),

(0 x0 0

)]=

(0 x0 0

)= −

[(0 00 1

),

(0 x0 0

)], (1)[(

0 x0 0

),

(0 0−y 0

)]=

(−xy 0

0 yx

), (2)[(

xy 00 −yx

),

(0 z0 0

)]=

(0 xyz + zyx0 0

). (3)

Similarly, the fact that e0 is a subalgebra of A− follows from the formula

[xy, uv] = xyuv − uvxy = (xyu+ uyx)v − u(yxv + vxy).

The remaining details are left to the reader.

“⇐=”: We know [e1, e−1] ⊂ e0 and [e0, e1] ⊂ e1, so (2) and (3) show that (6.2.1)holds for σ = +, and the case σ = − is proved similarly.

(b) “=⇒”: Since (6.2.2) implies (6.2.1), e(M, V ) is a 3-graded Lie algebra by(a). It follows easily from the formula(

1 x0 1

)(0 0−y 0

)(1 −x0 1

)=

(−xy xyx−y yx

)(4)

that e(M, V ) is stable under conjugation with(

1 V +

0 1

), and a similar computation

shows stability under the remaining generators of E(M, V ).

“⇐=”: From (4) we see that xyx ∈ V + and similarly one has yxy ∈ V −, for allx ∈ V +, y ∈ V −. Hence V satisfies (6.2.2).

For any pair V satisfying (6.2.2) we call e(M, V ) the elementary Lie algebra of(M, V ). (In fact, we will never consider pairs V satisfying (6.2.1) but not (6.2.2).)


6.4. Concrete Jordan pairs. We will first define “concrete” Jordan pairsas pairs of off-diagonal submodules of a Morita context satisfying (6.2.2), andthen, by abstracting from their properties, Jordan pairs in general. This followsa well-established procedure in algebra. For example, concrete Lie algebras aresubmodules of associative algebras closed under the commutator product [a, b] =ab − ba, abstract Lie algebras are modules equipped with an alternating product[a, b] satisfying “the same” identities (in this case, the Jacobi identity) as concreteLie algebras.

Let M be a Morita context as above. A Jordan subpair of M is a pair ofsubmodules V = (V +, V −) of (M+,M−) satisfying condition (6.2.2). Thus Vcomes equipped with the following somewhat unusual algebraic structure: a pair ofmaps Q+: V + × V − → V + and Q−: V − × V + → V −, given by

Q+(x; y) = xyx and Q−(y;x) = yxy. (1)

Clearly, these maps are quadratic in the first and linear in the second variable.They can also be considered as quadratic maps Q+: V + → Homk(V −, V +) andQ−: V − → Homk(V +, V −), by defining

Q+(x) · y = Q+(x; y), Q−(y) · x = Q−(y;x).

For the definition of abstract Jordan pairs we have to find the relevant identitiesholding for the quadratic-linear compositions of a Jordan subpair V as above. Thisturns out to be fairly complicated. To avoid a proliferation of parentheses andindices ±, we introduce the following conventions: for x ∈ V σ and y ∈ V −σ (whereσ ∈ +,−), we simply write

Qσ(x)y = Qxy (= xyx in the concrete situation).

This notation does not lead to confusion as long as care is taken to ensure that inan expression Qxy, the elements x and y come from different spaces: x ∈ V + andy ∈ V −, or vice versa.

We will also need efficient notation for the linearizations of the quadratic-linearexpression Qxy. First, we denote the linearization of Qxy with respect to x in thedirection of z by

Qx,zy = Q(x, z)y = Qx+zy −Qxy −Qzy,

and then have Qx,z = Qz,x and Qx,x = 2Qx. Next, define trilinear compositions−, −, −: V σ × V −σ × V σ → V σ by

xyz = zyx = Qx,zy, (2)

referred to as the Jordan triple product. Again, the entries in the trilinear productxyz have to be taken alternatingly in V + and V −. In the concrete situation, wehave

xyz = xyz + zyx. (3)

Among all the identities satisfied by the compositions of a Jordan subpair V asabove, the following three have turned out to be the essential ones:


x, y,Qxv = Qxy, x, v, (JP1)

Qxy, y, z = x,Qyx, z, (JP2)

QQxyv = QxQyQxv, (JP3)

for all x, z ∈ V σ, y, v ∈ V −σ, and σ ∈ +,−. The identity (JP3) is also known asthe fundamental formula. Here and in the sequel, the numbering of the identities(JPx) follows the one in [52]. In the concrete situation of a Jordan subpair of aMorita context M, (JP1) amounts to the following computation, valid because ofthe associativity of A:

x, y,Qxv = xy(xvx) + (xvx)yx = x(yxv + vxy)x = Qxy, v, x.

Similarly, (JP2) and (JP3) say concretely

Qxy, y, z = (xyx)yz + zy(xyx) = x(yxy)z + z(yxy)x = x,Qyx, z,QQxyv = (xyx)v(xyx) = x(y(xvx)y)x = QxQyQxv.

Thus, the identities (JP1) – (JP3) should not be regarded as saying that V is a non-associative algebraic system but rather as an expression of the essential associativityof the non-linear composition xyx. Inspection shows that (JP1) is of degree 3 in xand (JP3) is of degree 4 in x. In turns out that one needs all (formal) linearizationsof these identities to hold as well. (For (JP2) this is automatically the case becauseit is only of degree 2 in x and y). A more concise way of expressing this fact isas follows. Suppose K ∈ k-alg is an arbitrary commutative associative unital k-algebra, and let VK = (V +

K , V−K ) be the corresponding base ring extension. Since

the maps Qσ: V σ × V −σ → V σ of (1) are of bi-degree (2, 1), they have naturalextensions to maps QσK : V σK × V

−σK → V σK of K-modules, again of bi-degree (2, 1).

We refer to [82, Proposition II.1] or [35, 1.2, Lemma] for the fact that quadraticmaps extend naturally to base ring extensions. For a Jordan subpair V of M theselinearizations hold because one sees easily that VK is a Jordan subpair of MK , theMorita context obtained by base ring extension from M.

6.5. Abstract Jordan pairs. The formal definition of an (abstract) Jordanpair is now as follows: a Jordan pair over the commutative ring k is a pair ofk-modules V = (V +, V −) equipped with a pair Qσ: V σ × V −σ → V σ of maps,bi-homogeneous of bi-degree (2, 1), such that, using the notations introduced in6.4, the identities (JP1) – (JP3) hold in all base ring extensions. By taking inparticular as base ring extension a polynomial ring in sufficiently many variables,one sees that it is equivalent to require the validity of (JP1) – (JP3) and all theirlinearizations in V .

From the definition, it is evident that Jordan pairs admit arbitrary base change:if V is a Jordan pair over k then VK is a Jordan pair over K, for all K ∈ k-alg.Similarly, if k ∈ k′-alg then k′V , obtained from V by restricting the scalars to k′,is a Jordan pair over k′. This is evident from the fact that the Jordan identitiesand their linearizations remain trivially true in k′V .

As expected, a homomorphism h: V →W of Jordan pairs is a pair (h+, h−) ofk-linear maps hσ: V σ →Wσ satisfying hσ(Qxy) = Qhσ(x)h−σ(y) for all x ∈ V σ, y ∈V −σ. Jordan pairs then form a category admitting arbitrary base ring extensions.


The definition of isomorphisms and automorphisms is clear. For example, any unitµ ∈ k× gives rise to an automorphism (µIdV + , µ−1IdV −) of V .

Unlike the case of rings, it makes no sense to define the opposite of a Jordanpair by reversing the order of the factors in a product. However, it is possible tointerchange the roles of V + and V −, so we define: the opposite of V is the Jordanpair V op = (V −, V +) with quadratic maps Qop

x y = Qxy for x ∈ (V op)σ = V −σ andy ∈ (V op)−σ = V σ. If V = (M+,M−) is the Jordan pair of a Morita context withA and an idempotent e = e1 as in 6.1, then V op is the Jordan pair associated withthe opposite algebra Aop and the same idempotent since e1A

ope2 = e2Ae1 = M−

and e2Aope1 = e1Ae2 = M+.

The reader will not be surprised to learn that a subpair of a Jordan pair V isa pair S = (S+, S−) of submodules of V = (V +, V −) satisfying Q(Sσ)S−σ ⊂ Sσ

for σ = ±, while an ideal of V is a pair I = (I+, I−) of submodules such thatQ(Iσ)V −σ + Q(V σ)I−σ + V σ, V −σ, Iσ ⊂ Iσ holds for σ = ±. Then V/I =(V +/I+, V −/I−) is a Jordan pair with the obvious operations. The role of one-sided ideals in ring theory is played in Jordan theory by the inner ideals: these arethe k-submodules M ⊂ V σ satisfying QMV

−σ ⊂M .

6.6. Examples, special and exceptional Jordan pairs. A natural questionarises here: is every abstract Jordan pair a subpair of some Morita context? ForLie algebras, a positive answer to the analogous question is, at least over fields,furnished by the Poincare-Birkhoff-Witt theorem. For Jordan pairs, the answer isno: there are Jordan pairs even over the complex numbers, called exceptional, whichcannot be embedded into any Morita context. This leads to the following definition:an (abstract) Jordan pair V is called special if it can be embedded into some Moritacontext (this may be possible in many different ways). In other words, there existsa (non-unique) Morita context M = (R,M+,M−, S) and a pair U = (U+, U−) ofsubmodules Uσ ⊂ Mσ satisfying xyx ∈ Uσ for x ∈ Uσ, y ∈ U−σ such that V isisomorphic to the concrete Jordan pair U . The most important examples of specialJordan pairs are the following.

(a) Rectangular matrices Mpq(A). Let A be an arbitrary associative unital (notnecessarily commutative) k-algebra and put V + = Matpq(A), V − = Matqp(A).This Jordan pair embeds in the Morita context(

Matpp(A),Matpq(A),Matqp(A),Matqq(A)),

and is even closed under the associative triple product 〈xyz〉 = xyz for x, z ∈ V σand y ∈ V −σ. This is not true for the examples (b) and (c) below, which aresubpairs of Mnn(k).

(b) Alternating matrices An(k). Here V + = V − = Altn(k), alternating n × nmatrices over k.

(c) Symmetric matrices Hn(k). Here V + = V − = Hn(k), symmetric n × nmatrices over k.

In all three cases, we obtain special Jordan pairs with composition Qxy = xyx(matrix product), cf. the examples in 6.2.

(d) Hermitian matrices over a form ring. The following example puts (b) and(c) in a more general context. Let (A, J, ε, Λ) be a form ring in the sense of [32,


5.1C]. Thus A is an associative unital k-algebra, J is an anti-automorphism of A,ε ∈ A× is a unit of A with the property that εJ = ε−1 and aJJ = εaε−1 for alla ∈ A, and Λ is a form parameter; i.e., a k-submodule of A with the property that

a− aJε : a ∈ A ⊂ Λ ⊂ a ∈ A : a = −aJε and aJλa ∈ Λ

for all a ∈ A and λ ∈ Λ. We extend J to an anti-automorphism of Matn(A) bydefining xJ = (xJji) for an n × n-matrix x = (xij) with entries from A. Now put

Λ+ = ε−1Λ, Λ− = Λ and define

V + = x ∈ Matn(A) : xJ = −εx and xii ∈ Λ+ for all i,V − = y ∈ Matn(A) : yJ = −yε−1 and yii ∈ Λ− for all i.

Then V = (V +, V −) is a Jordan subpair of the Jordan pair Mnn(A) of Example(a), hence special. We denote this Jordan pair by

Hn(A, J, ε, Λ).

Observe that V − = Λn and V + = ΛJn in the notation of [32, 5.1C].For example, if Λ = 0 then necessarily ε = 1, J = IdA and A is commutative,

so we obtainHn(A, Id, 1, 0) = An(A) = (Altn(A),Altn(A))

the Jordan pair of alternating matrices over A.Let ε = −1. Then J is an involution of A and a form parameter Λ is a k-

submodule satisfying

a+ aJ : a ∈ A ⊂ Λ ⊂ a ∈ A : aJ = a =: H(A, J) and aJΛa ⊂ Λ

for all a ∈ A. In this case, Λ+ = Λ = Λ− and V + = V − consists of the hermitiann× n matrices over A with diagonal entries in Λ. We write this simply as

Hn(A, J,−1, Λ) = Hn(A, J, Λ).

In particular, for A = k, J = Id and Λ = k we get

Hn(k, Idk, k) = Hn(k),

the example (c) of symmetric matrices above.Important examples of form parameters Λ are the ones with 1 ∈ Λ, traditionally

called ample subspaces in Jordan theory. For A = Z and J = Id, Λ = 2Z is a formparameter which is not ample. The submodule H(A, J) is always an ample subspace,but in general Λ & H(A, J). For example, for k a field of characteristic 2 and Aa quaternion algebra over k with standard involution J , the subspace k · 1A is anample subspace different from H(A, J).

(e) Examples of exceptional Jordan pairs are obtained by taking in (a) analternative instead of an associative coordinate algebra, but only for small sizes ofthe respective matrices. Thus let A be an alternative k-algebra. Then M12(A) =(Mat12(A),Mat21(A)) is still a Jordan pair with quadratic operators


Qxy = x(yx), Qyx = (yx)y,

for x ∈ Mat12(A), y ∈ Mat21(A) [52, 8.15]. If A is an octonion algebra this Jordanpair is exceptional [61, Theorem 3.3].

(f) Let q: M → k be a quadratic form on a k-module M and denote byb(x, y) = q(x+ y)− q(x)− q(y) the polar form of q. Then V = (M,M) is a Jordanpair, called the Jordan pair of q and denoted J(M, q), with quadratic operatorsQxy = b(x, y)x− q(x)y.

(g) In some of the examples above, the Jordan pairs V = (V +, V −) had theproperty that V + = V − and Q+ = Q−. These types of Jordan pairs are essentiallythe same as Jordan triple systems, see [52, 1.13] for details.

(h) A special case of (g) is the Jordan pair (J, J) associated with a unitalquadratic Jordan algebra J [37], [71, App. C], see 6.13 for more details. Themost famous example of this is the exceptional Jordan algebra H3(C) of hermitian3×3-matrices with scalar diagonal entries over an octonion algebra C [68, 80]. Wedenote the corresponding Jordan pair by H3(C) = (H3(C),H3(C)).

Other examples of non-special Jordan algebras are obtained by starting froman associative algebra and dividing by a Jordan ideal which is not an associativeideal. See 8.10 for a concrete example.

(i) Since in previous sections we have used algebraic groups as examples, itis appropriate to point out the following connection between algebraic groups andJordan pairs.

Let G be a reductive group scheme over a ring k, and suppose that the mul-tiplicative group acts on G such that the induced action on the Lie algebra g ofG has only the weights ±1 and 0. Then the pair of weight spaces (g1, g−1) car-ries a canonical Jordan pair structure which can be defined in terms of the groupstructure of G. Details and generalizations are given in [55].

For example, let G be a (connected) quasi-simple algebraic group over an al-gebraically closed field k in the sense of [90, 13]. Then G admits such an actionif and only if the root system of G has a minuscule coweight in the sense of 14.6,i.e., the dual root system has a minuscule weight as defined in [16, VIII, §7.3]. TheJordan pairs V obtained in this way are:

(1) V = Mpq(k) for G of type Ap+q−1,

(2) V = An(k) for G of type Dn, n> 4,

(3) V = Hn(k) for G of type Cn, n> 2,

(4) V = J(km, q) with q non-degenerate and m = 2n − 1 for G of type Bn,n> 2, and m = 2n− 2 for G of type Dn, n> 4,

(5) the exceptional Jordan pairs M12(C) for G of type E6 and H3(C) for G oftype E7. Here C is an octonion algebra over k.

Here a quadratic form q with polar form b as in (f) is called non-degenerate ifq(x) = b(x,M) = 0 implies x = 0.

(j) Let (Vi)i∈I be a family of Jordan pairs. Then the product∏i∈I Vi of k-

modules is a Jordan pair with componentwise operations. The direct sum⊕

i∈I Viis a subpair of

∏i∈I Vi.


6.7. Identities. Jordan theory requires a large amount of sometimes non-trivial identities, all of which are consequences of the defining identities (JP1) –(JP3). We derive some of them here and refer to [52] for a more complete list.

Let us define bilinear maps Dσ: V σ × V −σ → V σ by

Dσ(x, y) · z = Qx,zy.

We follow the same convention as for Q and drop the index σ at D. To saveparentheses, we will often write Dx,y instead of D(x, y).

Since the right hand side of (JP1) is symmetric in y and v so must be the lefthand side. This yields x, y,Qxv = x, v,Qxy = Qxyxv, or in operator form:

Dx,yQx = QxDy,x = Q(x,Qxy). (JP4)

Linearizing (JP2) with respect to x in the direction of u resp. with respect to y inthe direction of v yields

xyu, y, z = x,Qyu, z+ u,Qyx, z,x, yxv, z = Qxy, v, z+ Qxv, y, z.

Written in operator form, this becomes

D(xyu, y) = D(x,Qyu) +D(u,Qyx), (JP7)

D(x, yxv) = D(Qxy, v) +D(Qxv, y), (JP8)

Dz,yDx,y = Qx,zQy +D(z,Qyx), (JP9)

Qx,zDy,x = Q(Qxy, z) +Dz,yQx. (JP10)

Similarly, linearize (JP1) with respect to x in direction z:

x, y, xvz+ z, y,Qxv = Qxyzv+Qx,zyxv.

Reading this as a function of y yields

Q(x, xvz) +Q(z,Qxv) = QxDv,z +Qx,zDv,x.

Replace here v by y and add the result to (JP10). After switching x and z, weobtain

Dx,yQz +QzDy,x = Q(z, xyz). (JP12)

Applying this to v and reading the result as a function of x yields

D(Qzv, y) +QzQy,v = Dz,vDz,y. (JP13)

Linearizing (JP13) with respect to z in the direction u and applying the result toan element x shows

zvu, y, x+ z, yxv, u = z, v, uyx+ u, v, zyx.

By reading this as a function of z, we see


[Dx,y, Du,v] = D(xyu, v)−D(u, yxv). (JP15)

Since the left hand side of (JP15) changes sign when we interchange (x, y) and(u, v) so does the right hand side. This yields

D(xyu, v)−D(u, yxv) = D(x, yuv)−D(uvx, y). (JP16)

The identities derived so far are all consequences of (JP1) and (JP2). The followingtwo identities require (JP3). For the proof, we refer to [52, 2.10].

Q(xyz) +Q(Qxy,Qzy) = QxQyQz +QzQyQx +Qx,zQyQx,z, (JP20)

Q(xyz) +Q(QxQyz, z) = QxQyQz +QzQyQx +Dx,yQzDy,x. (JP21)

For some applications it is useful to know that under suitable conditions on Vthe identity (JP15) implies (JP1)–(JP3). For example, let 2 be a unit in k, let V =(V +, V −) be a pair of k-modules without 3-torsion and suppose V σ×V −σ×V σ →V σ, (x, y, z) → xyz =: D(x, y)z are trilinear maps which are symmetric in theouter variables and satisfy (JP15). Then V becomes a Jordan pair with respect toQ(x)y = 1

2xyx [52, Proposition 2.2]. This can be used in the following situation.Let L = L−1⊕L0⊕L1 be a 3-graded Lie algebra over a ring k in which 6 is a unit.Then V = (L1, L−1) becomes a Jordan pair with respect to Qxy = 1

2 [x, [x, y]]. Ofcourse, V is also a Jordan pair with respect to Q′(x)y = s[[x, [x, y]] for any scalars ∈ k. The normalization s = 1

2 recovers the original Jordan pair V in case L is theTits-Kantor-Koecher algebra of V , cf. 7.1.

6.8. Derivations and inner derivations. Derivations of Jordan pairs aredefined by the usual mechanism: a pair ∆ = (∆+, ∆−) of linear maps ∆σ ∈End(V σ) is called a derivation if Id + ε∆ is an automorphism of the base ringextension V ⊗k(ε) where k(ε) is the algebra of dual numbers. A simple computationshows that this is equivalent to the conditions

∆σ(Qzv) = ∆σ(z), v, z+Qz∆−σ(v), (1)

for all z ∈ V σ, v ∈ V −σ. With componentwise operations, the derivations of Vform a Lie subalgebra Der(V ) of End(V +)× End(V −).

Identity (JP12) says precisely that, for any pair (x, y) ∈ V + × V −, the pair

δ(x, y) := (Dx,y,−Dy,x) (2)

is a derivation of V . We call this the inner derivation determined by (x, y). From(JP15) it follows that the k-linear span

Inder(V ) = spanδ(x, y) : (x, y) ∈ V (3)

is a subalgebra of Der(V ), called the inner derivation algebra of V .In any Jordan pair, we have the derivation

ζV = (IdV + ,−IdV −) (4)

which obviously belongs to the centre of Der(V ).Linearizing (1) one sees that any derivation ∆ = (∆+, ∆−) satisfies

[∆, δ(x, y)] = δ(∆+(x), y

)+ δ(x,∆−(y)

)(5)

for all (x, y) ∈ V . In particular, this shows that Inder(V ) is an ideal of Der(V ).Conversely, if V ± has no 2-torsion then (5) is sufficient for a pair (∆+, ∆−) ∈End(V +)× End(V −) to be a derivation.


Example. Let M = (R,M+,M−, S) be a Morita context and consider thespecial Jordan pair M = (M+,M−). A pair (a, d) ∈ R×S gives rise to a derivation∆(a, d) = (∆(a, d)+, ∆(a, d)−) of M by

∆(a, d)+(u) = au− ud, ∆(a, d)−(v) = dv − va. (6)

In particular, for (x, y) ∈M we know (xy,−yx) ∈ R × S by (6.3.2), and it followsfrom (6.4.3) that

∆(xy,−yx) = δ(x, y). (7)

Also∆(1, 0) = ∆(0,−1) = ζV . (8)

Let now V ⊂M be a Jordan subpair, and let e = e(M, V ) be the elementary Liealgebra of 6.3. Since the 0-component e0 of the 3-graded Lie algebra e is spannedby the matrices

e1 =

(1 00 0

), e2 =

(0 00 1

)and

(xy 00 −yx

)((x, y) ∈ V ),

cf. (6.3.2), it follows from (6.3.3) and the corresponding formula for e−1 that the

derivation ∆(a, d) for(a 00 d

)∈ e0 leaves V ⊂ S invariant and thus induces a

derivation ∆V (a, d) of V .

6.9. The Bergmann operators and the quasi-inverse. For a pair (x, y) ∈V σ × V −σ we define the Bergmann operator B(x, y) = Bx,y ∈ EndV σ by

Bx,y = IdV σ −Dx,y +QxQy.

The name “Bergmann” comes from the fact that Jordan pairs over the complexnumbers equipped with positive hermitian involutions are in correspondence withbounded symmetric domains. Then the determinant of Bx,y is related to theBergmann kernel of the domain, see [53].

The Bergmann operators play a fundamental role in the theory of Jordan pairs.Clearly,

B(λx, y) = B(x, λy) (1)

for all λ ∈ k. Of the many identities satisfied by them, we list only the followingtwo and refer to [52, 2.11] for more.

B(x, y)2 = B(2x−Qxy, y) = B(x, 2y −Qyx), (JP25)

Q(Bx,yz) = Bx,yQzBy,x. (JP26)

A pair (x, y) ∈ V σ × V −σ is called quasi-invertible if the Bergmann operatorB(x, y) is invertible (as an endomorphism of V σ). Then, the quasi-inverse of (x, y)is defined by

xy := B(x, y)−1(x−Qxy) ∈ V σ. (2)

By [52, Proposition 3.2], (x, y) is quasi-invertible if and only if there exists z ∈ V σsuch that

Bx,yz = x−Qxy and Bx,yQzy = Qxy, (3)

and then z = xy is the quasi-inverse.


6.10. Example. Let V be a special Jordan pair, V ⊂ (M+,M−), embeddedinto a Morita context M = (R,M+,M−, S) as in 6.1. Then

B(x, y)z = z − xyz+QxQyz = z − xyz − zyx+ xyzyx

= (1− xy)z(1− yx). (1)

For (x, y) ∈ V + × V − the following conditions are equivalent:

(i) (x, y) is quasi-invertible in V ,

(ii) (y, x) is quasi-invertible in V op,

(iii) 1− xy ∈ R× and (1− xy)−1x ∈ V +.

(iv) 1− yx ∈ S× and (1− yx)−1y ∈ V −.

In this case, the quasi-inverses are given by

xy = (1− xy)−1x = x(1− yx)−1, yx = (1− yx)−1y = y(1− yx)−1. (2)

Proof. (i) ⇐⇒ (ii): This is a general fact, see (6.11.1).

(i) and (ii) =⇒ (iii): Since Bx,y and By,x are bijective, there exist u ∈ V + andv ∈ V − such that x = Bx,yu = (1−xy)u(1−yx) and y = By,xv = (1−yx)v(1−xy).Put b = −u(1− yx) ∈ B and c = (1− yx)v and compute in the elementary groupE(M): (

1 0c 1

)(1 x0 1

)(1 0−y 1

)(1 b0 1

)=

(1− xy 0

0 1 + cx

).

This proves 1− xy ∈ R×.A straightforward calculation now shows that 1− yx ∈ S× with inverse

(1− yx)−1 = 1 + y(1− xy)−1x.

FromV + = Bx,yV

+ = (1− xy)V +(1− yx)

it now follows that also (1 − xy)−1V +(1 − yx)−1 = V + and that B−1x,ys = (1 −

xy)−1s(1 − yx)−1 for all s ∈ V +. Since V is a Jordan subpair of (M+,M−), wehave x−Qxy ∈ V +. Hence

z = (1− xy)−1x = (1− xy)−1x(1− yx)(1− yx)−1

= (1− xy)−1(x−Qxy)(1− yx)−1 ∈ V +,

and z = B−1x,y(x−Qxy) = xy is the quasi-inverse of (x, y). Moreover,

x(1− yx)−1 = (1− xy)−1(x−Qxy)(1− yx)−1 = B−1x,y(x−Qxy) = xy.

The implication (ii) =⇒ (iv) and the formulas for yx are proved similarly.

(iii) =⇒ (i): Put z = (1−xy)−1x ∈ V +. We show that the conditions of (6.9.3)hold:


Bx,yz = (1− xy)(1− xy)−1x(1− yx) = x(1− yx) = x−Qxy,Bx,yQzy = (1− xy)zyz(1− yx) = xy(1− xy)−1x(1− yx)

= xy(1− xy)−1(1− xy)x = xyx = Qxy.

Hence (x, y) is quasi-invertible. Again, (iv) =⇒ (ii) is proved similarly and left tothe reader.

In particular, if M is the Morita context of 2×2 matrices over a ring R, then thepair (x, 1R) is quasi-invertible (in the Jordan pair sense) if and only if x is quasi-invertible in the sense of ring theory [85, p. 180]. The group-theoretic significanceof quasi-invertibility will be shown in Theorem 7.7.

6.11. Properties of the quasi-inverse. We return to an arbitrary Jordanpair. Proofs of the following facts can be found in [52, §3]. The following conditionson a pair (x, y) ∈ V are equivalent:

(i) (x, y) is quasi-invertible,

(ii) B(x, y) is surjective,

(iii) 2x−Qxy belongs to the image of B(x, y),

(iv) x belongs to the image of B(x, y).

For the equivalence of (i) – (iii) see [52, Proposition 3.2]. Clearly, (ii) implies(iv). Now suppose (iv) holds, so there exists z ∈ V + such that x = B(x, y)z. From(JP26) we get Qxy = Q(B(x, y)z)y = B(x, y)QzB(y, x)y and hence 2x − Qxy =B(x, y)

(2z −QzB(y, x)y

)belongs to the image of B(x, y).

In ring theory, invertibility in a ring and the opposite ring are equivalent. Theanalogue for Jordan pairs is the “symmetry principle” [52, Proposition 3.3]:

(x, y) is quasi-invertible in V ⇐⇒ (y, x) is quasi-invertible in V op, (1)

and in this case, the quasi-inverses are related by the formula

xy = x+Qxyx. (2)

Let h: V → W be a homomorphism of Jordan pairs. Condition (iii) (or (iv))together with (6.9.2) immediately imply: If (x, y) is quasi-invertible in V then(h+(x), h−(y)) is quasi-invertible in W , and then

h+

(xy)

= h+(x)h−(y), h−(yx) = h−(y)h+(x). (3)

An important property of the quasi-inverse, and the reason for the exponentialnotation xy, is the following. Suppose (x, y) is quasi-invertible and let v ∈ V −.Then (x, y + v) is quasi-invertible if and only if (xy, v) is quasi-invertible, in whichcase

(xy)v = xy+v, (y + v)x = yx +B(y, x)−1 · v(xy). (4)

We refer to [52, 3.7] for the proof.There are numerous identities relating the quasi-inverse and the Bergmann

operators; refer to [52, 3.6] for the proof. It is always assumed that (x, y) isquasi-invertible, while z ∈ V + and v ∈ V − can be arbitrary.


B(x, y)Q(xy) = Q(xy)B(y, x) = Q(x), (JP28)

B(x, y)Q(xy, z) +QxD(y, z) = Q(xy, z)B(y, x) +D(z, y)Qx = Q(x, z), (JP29)

B(x, y)D(xy, v) = D(x, v)−QxQ(y, v), (JP30)

D(v, xy)B(y, x) = D(v, x)−Q(y, v)Qx, (JP31)

D(xy, y −Qyx) = D(x−Qxy, yx) = D(x, y), (JP32)

B(x, y)B(xy, v) = B(x, y + v), (JP33)

B(z, yx)B(x, y) = B(x+ z, y), (JP34)

B(x, y)−1 = B(xy,−y) = B(−x, yx). (JP35)

Generalizations of Bergmann operators and quasi-inverses will be introduced in 8.4and 8.5.

6.12. Structural transformations. Besides homomorphisms, the followingtypes of maps between Jordan pairs play an important role. Let V = (V +, V −)and W = (W+,W−) be Jordan pairs. A structural transformation from V to Wis a pair of k-linear maps f : V + → W+ and g: W− → V − (note the change ofdirection!) such that

Qf(x) = f Qx g and Qg(y) = g Qy f,

for all x ∈ V + and y ∈W−. We write this as

(f, g): V W

and note that

(f, g): V W ⇐⇒ (g, f): W op V op.

The basic examples are given by the quadratic operators and the Bergmann oper-ators: for all x ∈ V +, y ∈ V −, we have the structural transformations

(Qx, Qx): V op V, (Qy, Qy): V V op, (Bx,y, By,x): V V.

This is just another way of expressing the identities (JP3) and (JP26). For anyscalar λ ∈ k, the homotheties f(x) = λx and g(y) = λy define a structuraltransformation from V to itself.

An invertible structural transformation is essentially an isomorphism; moreprecisely, the following conditions are equivalent:

(i) (f, g): V W is a structural transformation with f and g invertible,

(ii) (f, g−1): V →W is an isomorphism.

The proof is immediate from the definitions.In particular, let (x, y) be quasi-invertible. Then both Bx,y and By,x are invert-

ible, so

β(x, y) := (Bx,y, B−1y,x)

is an automorphism of V , called the inner automorphism determined by (x, y).From (JP33) – (JP35) we get the formulas


β(x, y)β(xy, v) = β(x, y + v), (1)

β(z, yx)β(x, y) = β(x+ z, y), (2)

β(x, y)−1 = β(xy,−y) = β(−x, yx). (3)

The inner automorphism group Inn(V ) is the subgroup of Aut(V ) generated by allβ(x, y), (x, y) ∈ V quasi-invertible.

Structural transformations relate well to the quasi-inverse. Let (f, g): V Wbe structural and let x ∈ V + and y ∈W−. Then (f(x), y) is quasi-invertible in Wif and only if (x, g(y)) is quasi-invertible in V , in which case the formula

f(x)y = f(xg(y)

)(4)

holds, known as the “shifting principle”.Indeed, by linearizing the defining equations of a structural transformation, one

obtains the formulas

D(f(x), y) f = f D(x, g(y)), D(g(y), x) = g D(y, f(x))

and then also

B(f(x), y) f = f B(x, g(y)), B(g(y), x) = g B(y, f(x)),

for all x ∈ V +, y ∈ V −. Now suppose (x, g(y)) quasi-invertible. By the equivalentconditions of 6.11 there exists z ∈ V + such that x = B(x, g(y))z. Applying f tothis equation and using the above formula yields f(x) = B(f(x), y)f(z), so (f(x), y)is quasi-invertible.

Next, let (f(x), y) be quasi-invertible. By the symmetry principle, (y, f(x))is quasi-invertible, and since (g, f) is structural, it follows that (g(y), x) is quasi-invertible, which implies (x, g(y)) quasi-invertible, again by symmetry. Finally, wehave B(x, g(y))xg(y) = x−Qxg(y). Applying f to this and using the above formulasshows

f(B(x, g(y))xg(y)

)= B(f(x), y)f

(xg(y)

)= f(x)− fQxg(y) = f(x)−Qf(x)y.

By applying B(f(x), y)−1 to this we obtain (4).

6.13. Inverses and Jordan algebras. An element u ∈ V σ is called invertibleif Qu: V −σ → V σ is invertible (as a linear map). If u ∈ V σ is invertible then theinverse of u is defined by

u−1 = Q−1u u ∈ V −σ

Recall here that Qu maps V −σ to V σ, so Q−1u : V σ → V −σ. It follows easily from

(JP3) thatQ−1u = Qu−1 and (u−1)−1 = u (1)

for an invertible u ∈ V σ.In the example M11(A) = (A,A) of a ring A, it is immediate that an element

is invertible in M11(A) if and only if it is invertible in A. Care has to be taken todistinguish between x−1A , the quasi-inverse of (x,−1A), and the inverse x−1. Moregenerally, invertibility in the Jordan pair of rectangular matrices is described asfollows.


Example. Let A be a ring and let V = Mp,q(A) be the Jordan pair of rectan-gular matrices as in 6.6(a). Then u ∈ V + = Matp,q(A) is invertible if and only ifu: Aq → Ap (left multiplication with the matrix u acting on the column space Aq)is an A-module isomorphism.

Proof. If u: Aq → Ap is invertible as an A-module homomorphism with inversev: Ap → Aq then it is immediate that Qu: V − → V + is invertible with inverse Qv.Conversely, let u ∈ V + be invertible in the Jordan pair sense. To see that u isinjective, let y ∈ Aq with uy = 0. Extend y to an q× p-matrix y = ( y 0 · · · 0 )by adding p − 1 columns of zeros. Then uy = 0, hence uyu = Quy = 0 whichimplies y = 0 and therefore y = 0. For surjectivity, let x ∈ Ap and extend it tox = (x 0 · · · 0 ) ∈ Matp,q(A) by adding q − 1 columns of zeros. Since Qu issurjective, there exists v ∈ Matq,p(A) with x = uvu. Here vu ∈ Matq(A), say, withcolumn vectors v1, . . . , vq ∈ Aq. Then

(x 0 · · · 0 ) = x = u ( v1 · · · vq ) = (uv1 · · · uvq ) ,

so x = uv1.

If p = q then clearly V contains invertible elements. The converse holds if A isa ring with invariant basis number, for instance, if A is commutative, local, or hasstable rank 1, see, for example, [47, §1.5].

We say V is a Jordan division pair if V 6= (0, 0), and if every non-zero elementof V σ is invertible. For example, the Jordan pair M11(A) = (A,A) of an associativek-algebra is a Jordan division pair if and only if A is a division algebra.

Jordan pairs containing invertible elements are related to (unital quadratic)Jordan algebras as follows. First, we recall from [37] that a unital quadratic Jordanalgebra over k is a k-module J equipped with a distinguished element 1J ∈ J anda quadratic map U : J → Endk(J) such that the identities

U1J = IdJ , (QJ1)

UUxy = UxUyUx, (QJ2)

UxVy,x = Vx,yUx (QJ3)

hold in all scalar extensions. Here Vx,y ∈ Endk(J) is defined (similarly to Dx,y forJordan pairs) by

Vx,yz = Ux+zy − Uxy − Uzy.

Then any Jordan algebra J determines a Jordan pair (V +, V −) = (J, J) withquadratic operators Qx = Ux (x ∈ V ±). Indeed, (JP1) is (QJ3), (JP3) is (QJ2),and (JP2) is the identity QJ21 of [37, p. 3.10].

Conversely, let V be a Jordan pair containing an invertible element v ∈ V −.Then the k-module V + becomes a unital quadratic Jordan algebra J by defining1J = v−1 and Ux = QxQv for all x ∈ V +, and the Jordan pair (J, J) is isomorphicto V under the pair of maps (IdV + , Qv): (J, J) → (V +, V −). The Jordan alge-bras arising from a different choice of invertible element in V − are not necessarilyisomorphic, but they are isotopic. For details, we refer to [52, §1].

There is the following relation between the inverse and the quasi-inverse. Sup-pose u ∈ V + (resp. v ∈ V −) is an invertible element of the Jordan pair V , and lety ∈ V − (resp. x ∈ V +) be arbitrary. Then


B(u, y) = QuQ(u−1 − y), B(x, v) = Q(x− v−1)Qv. (2)

Moreover, (u, y) is quasi-invertible if and only if u−1 − y is invertible in V −, andthen the formula

uy = (u−1 − y)−1 (3)

holds; see [52, 2.12, 3.13] for a proof.

6.14. Idempotents and Peirce decomposition. Let V be a Jordan pair.A pair e = (e+, e−) ∈ V + × V − is called an idempotent if

Qe+e− = e+ and Qe−e+ = e−.

If V is a special Jordan pair embedded in a Morita context as in 6.4 then this meanse+ = e+e−e+ and e− = e−e+e−, so e+ and e− are in particular von Neumannregular. Also, e+e− and e−e+ are ring idempotents in R and S, respectively.

Clearly, idempotents are mapped to idempotents under Jordan pair homomor-phisms. Since Qx is a quadratic function of x, it is immediate that

(e+, e−) idempotent =⇒ −e = (−e+,−e−) idempotent. (1)

An idempotent e of V gives rise to an action of the multiplicative group on V byautomorphisms and hence to the important Peirce decomposition as follows. Forany K ∈ k-alg it is clear that e ⊗ 1K is again an idempotent in VK . By abuse ofnotation, we denote this again simply by e. Let t ∈ K×, the group of units of K.Then the pair (e+, (1− t)e−) ∈ VK is quasi-invertible with quasi-inverse t−1e+, andthe formula

%e(t) := β(e+, (1− t)e−) (2)

defines a homomorphism %e: K× → Inn(VK) [52, Lemma 5.2].

For σ ∈ +,− and i ∈ 0, 1, 2 define endomorphisms Eσi of V σ by

Eσ2 = QeσQe−σ , Eσ1 = D(eσ, e−σ)− 2Eσ2 , Eσ0 = B(eσ, e−σ). (3)

Then%e(t) = (t2E+

2 + tE+1 + E+

0 , t−2E−2 + t−1E−1 + E−0 ) (4)

and the Eσi are orthogonal projections whose sum is the identity on V σ, so that

V σ = V σ2 ⊕ V σ1 ⊕ V σ0 where V σi = Im(Eσi ).

This is the Peirce decomposition of V with respect to e, the Vi = V σi (e) are calledthe Peirce spaces of V with respect to e. From (3) it is clear that e and −e havethe same Peirce spaces:

Vi(−e) = Vi(e). (5)

From (4) we see that %e has weights 0,±1,±2 on V ± and the V ±i are the corre-sponding weight spaces. In particular,

se := %e(−1) = β(e+, 2e−) (6)


is an involutorial inner automorphism, called the Peirce reflection with respect toe, and given by xi 7→ (−1)ixi for xi ∈ V σi .

We often put Vi = (V +i , V

−i ) and write the Peirce decomposition as V =

V2 ⊕ V1 ⊕ V0, to be read componentwise. The Peirce spaces can also be describedby

V σ2 = x ∈ V σ : QeσQe−σx = x = ImQeσV σ1 = x ∈ V σ : eσ, e−σ, x = xV σ0 = x ∈ V σ : Qe−σx = eσ, e−σ, x = 0

(7)

In particular, for x ∈ V σ2 we have x = Eσ2 (x) and 0 = Eσ1 (x) = eσ, e−σ, x −2Eσ2 (x), whence eσ, e−σ, x = 2x.

Idempotents and Peirce decompositions behave well with respect to homomor-phisms: suppose h: V → W is a homomorphism of Jordan pairs. Then the imageh(e) = (h+(e+), h−(e−)) of an idempotent of V is an idempotent of W , and itfollows from the definition of the Peirce spaces that h

(V σi (e)

)⊂Wσ

i

(h(e)

).

The Peirce spaces satisfy the following multiplication rules, where we put V σj = 0for j /∈ 0, 1, 2:

Q(V σi )V −σj ⊂ V σ2i−jV σi , V

−σj , V σl ⊂ V σi−j+l

V σ2 , V −σ0 , V σ = V σ0 , V −σ2 , V σ = 0

(8)

Let in particular u ∈ V + be invertible with inverse u−1 ∈ V −. Then e = (u, u−1) isan idempotent with the property that V = V2(e). Conversely, if e is an idempotentwith V = V2(e) then e+ is invertible with inverse e−. In general, the (+)-componente+ of an idempotent e is always invertible in the subpair V2(e), and its inverse ise−. We refer to [52, Theorem 5.4] for proofs.

6.15. Compatible idempotents. Two idempotents e and f are said to becompatible if their Peirce projections Eσi and Fσj commute:

[Eσi , Fσj ] = 0 for all i, j ∈ 0, 1, 2 and σ ∈ +,−. (1)

It is easily seen that this holds if and only if %e and %f commute in the sense that

%e(s)%f (t) = %f (t)%e(s) for all s, t ∈ K× and all K ∈ k-alg. (2)

(Choose K = k[S, T, S−1, T−1], the Laurent ring in two variables, to prove that (2)implies (1).) In this case, V admits the joint Peirce decomposition

V =⊕

i,j∈0,1,2

V(ij), where V(ij) = Vi(e) ∩ Vj(f). (3)

The necessary and sufficient conditions for compatibility in terms of V alone arefairly complicated, they were given by K. McCrimmon [70, 1.6]. For our purposes,the following simple criterion will suffice:

If f ∈ Vi(e) or e ∈ Vi(f) for some i ∈ 0, 1, 2 then e and f are compatible. (4)

Indeed, suppose f ∈ Vi(e). We show (2). Since the map %e(s) of (6.14.2) is anautomorphism of VK for any K ∈ k-alg, we have


%e(s) %f (t) %e(s)−1 = %e(s)β(f+, (1− t)f−) %e(s)

−1 = β(%e(s)f+, %e(s)(1− t)f−)

= β(sif+, s−i(1− t)f−) = β(f+, (1− t)f−) = %f (t).

Here we used the fact that scalars may be shifted from left to right in innerautomorphisms: β(λx, y) = β(x, λy), see (6.9.1). By interchanging the roles ofe and f , one sees in the same way that e ∈ Vi(f) implies e and f compatible.

The following special cases will play a distinguished role in the sequel:

(i) e and f are orthogonal, written e ⊥ f , if e ∈ V0(f) and f ∈ V0(e),

(ii) e and f are associated, written e ≈ f , if e ∈ V2(f) and f ∈ V2(e),

(iii) e and f are collinear, written e > f , if e ∈ V1(f) and f ∈ V1(e),

(iv) f governs e or e is governed by e, written f ` e or e a f , if f ∈ V1(e) ande ∈ V2(f).

In the next subsection, we discuss orthogonality in more detail. For associationsee 6.17 and 6.18.

6.16. Orthogonal idempotents. Two idempotents e and f ∈ V are orthog-onal as soon as one of the two conditions defining orthogonality is satisfied:

e ∈ V0(f) ⇐⇒ e ⊥ f. (1)

Indeed, if e ∈ V0(f) then eσ e−σ fσ = 0 and Qe−σfσ ∈ V σ4 (e) = 0 by (6.14.8),

whence f ∈ V0(e) by (6.14.7).If e and f are orthogonal idempotents then it is easily seen that e+ f (defined

componentwise) is again an idempotent.Suppose e1, . . . , en is a finite family of pairwise orthogonal idempotents, and

define V σij ⊂ V σ for i, j ∈ 0, 1, . . . , n by

Vii = V2(ei), Vij = Vji = V1(ei) ∩ V1(ej) (i 6= j),

V00 =

n⋂i=1

V0(ei), Vi0 = V0i = V1(ei) ∩⋂j 6=i

V0(ej).(2)

Then V decomposes as

V =⊕

06i6j6n

Vij . (3)

To formulate the multiplication rules which these spaces satisfy, we consider triplesof unordered pairs of indices (ij, lm, pq) taken from 0, 1, . . . , n, and furthermoreidentify this with (pq, lm, ij). We call such a triple connected if it is of the form(ij, jm,mp). Then the following composition rules hold:

V σij , V −σjm , V σmp ⊂ V σip. (4)

If (ij, jl, ij) is connected and ij = lm then

Q(V σij )V−σjl ⊂ V

σim. (5)

If (ij, lm, pq) resp. (ij, lm, ij) is not connected then


V σij , V −σlm , V σpq = Q(V σij )V−σlm = 0. (6)

If I ⊂ 1, . . . , n then eI =∑i∈I ei is an idempotent of V whose Peirce spaces are

given by

V2(eI) =∑i,j∈I

Vij , V1(eI) =∑

i∈I,j 6∈I

Vij , V0(eI) =∑i,j 6∈I

Vij , (7)

and the maps %e of (6.14.2) satisfy

%eI (t) =∏i∈I

%ei(t) for all t ∈ K× and all K ∈ k-alg. (8)

Proofs can be found in [52, Lemma 5.13 and Theorem 5.14].

Let I be an arbitrary index set, possibly infinite. A family O = (ei)i∈I ofidempotents is said to be an orthogonal system if ei ⊥ ej for i 6= j. The Peircespaces with respect to O are defined as in (2), and they satisfy the multiplicationrules listed above. However, (3) need not hold if I is infinite.

Example. Let X and Y be k-modules and let V =(

Hom(Y,X),Hom(X,Y ))

be the special Jordan pair of the Morita context(

End(X),Hom(Y,X),Hom(X,Y ),

End(Y )). Associate to e = (e+, e−) ∈ V the endomorphisms p+ = e+e− ∈ End(X)

and p− = e−e+ ∈ End(Y ). Then e is an idempotent if and only if p2σ = pσ for

σ = ±, eσ(

Ker(pσ))

= 0 and eσ| Im(p−σ) is an isomorphism onto Im(pσ) withinverse e−σ| Im(pσ). Assume this to be the case. Let X = X1 ⊕X0, X1 = Im(p+),and X0 = Ker(p+) be the eigenspace decomposition of X with respect to p+, anddefine Y = Y1 ⊕ Y0 analogously. Then, using the obvious identifications,

V2(e) =(

Hom(Y1, X1),Hom(X1, Y1)),

V1(e) =(

Hom(Y1, X0),Hom(X0, Y1))⊕(

Hom(Y0, X1),Hom(X1, Y0)),

V0(e) =(

Hom(Y0, X0),Hom(X0, Y0)).

In particular, V1(e) is a direct sum of two subpairs. Two idempotents e and f withassociated projections pσ(e) and pσ(f) are orthogonal if and only if pσ(e)pσ(f) =0 = pσ(f)pσ(e) for σ = ±. In this case, denoting X0(e) = Ker(p+(e)) andX0(f) = Ker(p+(f)), we have X = X1(e)⊕X1(f)⊕

(X0(e)∩X0(f)

)with X1(e) =

Im(p+(e)) ⊂ Ker(p+(f)) and X1(f) = Im(p+(f)) ⊂ X0(e). Analogously, Y =Y1(e) ⊕ Y1(f) ⊕

(Y0(e) ∩ Y0(f)

). The joint Peirce spaces of e and f are easily

described in terms of the two decompositions of X and Y . For example,

V +1 (e) ∩ V +

1 (f) = Hom(Y1(e), X1(f)

)⊕Hom

(Y1(f), X1(e)

),

V +0 (e) ∩ V +

0 (f) = Hom(Y0(e) ∩ Y0(f), X0(e) ∩X0(f)

).

Decompositions X =⊕

i∈NXi and Y =⊕

i∈N Yi with Xi∼= Yi as k-modules give

rise to an obvious infinite family (ei)i∈N of orthogonal idempotents. In order fora homomorphism ϕ: Y → X to lie in the sum of the Peirce spaces Vij of theorthogonal family (ei) it must vanish on almost all of the Yi’s and its image mustlie in the sum of finitely many of the Xi’s, showing that (3) need not hold in general.

§7] The projective elementary group I 91

6.17. Associated idempotents. Recall from 6.15 that two idempotents e andf of V are associated, written e ≈ f , if e ∈ V2(f) and f ∈ V2(e). The followingconditions are equivalent [76, I.2.3]:

(i) e ≈ f ,

(ii) V2(e) = V2(f),

(iii) Vi(e) = Vi(f), for all i ∈ 0, 1, 2,(iv) f ∈ V2(e) and f± is invertible in V2(e),

(v) %e(t) = %f (t) for all t ∈ K× and K ∈ k-alg.

To prove these equivalences we will use the multiplication rules and propertiesof Peirce decompositions stated in 6.14 without further comment. The implica-tions (iii) =⇒ (ii) =⇒ (i) being obvious, let us assume (i). It then follows thatV σ2 (f) = Q(fσ)V −σ = Q(fσ)V −σ2 (e) ⊂ V σ2 (e) for σ = ±, whence (ii) by symme-try. Condition (ii) implies B(fσ, f−σ)u0 = u0 for u0 ∈ V σ0 (e), thus V0(e) ⊂ V0(f)and then V0(e) = V0(f) by symmetry again. But then the Peirce projectionssatisfy Eσ1 (e) = Id − Eσ2 (e) − Eσ0 (e) = Id − Eσ2 (f) − Eσ0 (f) = Eσ1 (f), thus (iii)holds. We have now established that (i), (ii) and (iii) are equivalent. Sincef± is invertible in V2(f), it is clear that (ii)=⇒ (iv). Assuming (iv), we haveV σ2 (f) = Q(fσ)V = Q(fσ)V σ2 (e) = V σ2 (e) using invertibility of f± in V2(e), i.e.,(ii). If (iii) holds then (v) follows from (6.14.4). Conversely, evaluating (v) for theLaurent polynomial ring k[T, T−1] shows Eσi (e) = Eσi (f), and thus (iii).

It is clear that associated idempotents satisfy

D(eσ, e−σ) = D(fσ, f−σ). (1)

Conversely, this equality implies e ≈ f if 2 ∈ k×. We also note that wheneverx ∈ V +

2 (e) is invertible then (x, x−1) ∈ V2(e) is an idempotent associated with e.

Example. Let A be a unital associative k-algebra and let V = Mpq(A) be theJordan pair of rectangular matrices over A as in 6.6(a). For i, j satisfying 16 i6 pand 16 j6 q the pair eij = (Eij , Eji) is an idempotent of V , where Eij and Eji arethe usual matrix units in the corresponding module of matrices. For any u ∈ A×the pair (uEij , u

−1Eji) is an idempotent associated with eij and, conversely, thecondition (iv) above together with the Example in 6.13 shows that every idempotentassociated with eij has this form.

6.18. Lemma. Let e ∈ V be an idempotent and let x2 ∈ V +2 (e). Then (x2, e−)

is quasi-invertible if and only if there exists an idempotent f ∈ V2(e) such thatf ≈ e and x2 = e+ + f+. In this case,

(e−)e++f+ = −f−, (e+ + f+)e− = −e+ −Q(e+)f−, (1)

B(e+ + f+, e−)u2 = Qf+Qe−u2, B(e−, e+ + f+)v2 = Qe−Qf+v2 (2)

holds for all (u2, v2) ∈ V2(e).

Proof. Since e− is invertible in V2(e) with inverse e+, it follows from (6.13.2)that

B(x2, e−)u2 = Qx2−e+Qe−u2, B(e−, x2)v2 = Qe−Qe+−x2v2 (3)


for (u2, v2) ∈ V2(e). By condition (iii) of 6.11, quasi-invertibility of (x2, e−) inV2(e) and in V is the same. Hence, by (3), (x2, e−) is quasi-invertible if and onlyif f+ = x2 − e+ is invertible in V2(e). As observed in 6.17, it then follows thatf = (f+, f−) with f− = (f+)−1 is an idempotent associated with e.

§7. The projective elementary group I

7.1. The Tits-Kantor-Koecher algebra. Let V be a Jordan pair over k.We use the notation introduced in 6.8 and note that

L0(V ) := k · ζV + Inder(V )

is a subalgebra of Der(V ) and ζV is central in L0(V ) (indeed, in all of Der(V )).When V is clear from the context, we often write simply ζ instead of ζV .

In this book, the Tits-Kantor-Koecher algebra of V (or TKK-algebra for short)is the Lie algebra

L(V ) = V + ⊕ L0(V )⊕ V − (1)

The multiplication is determined by the requirements that it be alternating, thatL0(V ) be a subalgebra, and furthermore

[V σ, V σ] = 0, [D, z] = Dσ(z), [x, y] = −δ(x, y) (2)

for D = (D+, D−) ∈ L0(V ), z ∈ V ± and (x, y) ∈ V . In particular, this says

[[x, y], u] = −δ(x, y) · u = −D(x, y)u = −xyu (3)

for x, u ∈ V + and y ∈ V −, and similarly

[[x, y], v] = −δ(x, y) · v = D(y, x)v = yxv (4)

for x ∈ V + and y, v ∈ V −. This definition differs from the one used elsewhere, butit is the most appropriate for our purposes.

Let us indicate why L(V ) is a Lie algebra. We put g0 = L0(V ), g1 = V +,g−1 = V − and gi = 0 for i ∈ Z 0,±1. Then

L(V ) = g =⊕i∈Z

gi = g−1 ⊕ g0 ⊕ g1 (5)

is an alternating and Z-graded algebra: [gi, gj ] ⊂ gi+j for all i, j ∈ Z which is3-graded in the sense that gi = 0 if i /∈ 0,±1. Therefore, it suffices to verify theJacobi identity [[a, b], c] + [[b, c], a] + [[c, a], b] = 0 for homogeneous elements. SinceL0(V ) is a subalgebra, it suffices to check it in the following cases: (a, b, c) ∈ V σ ×V −σ×V σ where it follows from the symmetry (6.4.2); (a, b, c) ∈ L0(V )×V σ×V −σin which case it holds by (6.8.5); (a, b, c) ∈ L0(V ) × L0(V ) × V σ, where one usesthe fact that L0(V ) acts on V .

Moreover, ad ζ is the grading derivation, i.e., [ζ,X] = iX for X ∈ gi. From thedefinition, it follows easily that the derived algebra of L(V ) is

DL(V ) = [L(V ),L(V )] = V + ⊕ Inder(V )⊕ V −. (6)


The Lie algebra L(V ) has trivial centre. Indeed, if Z = x⊕∆⊕ y ∈ V + ⊕ g0 ⊕ V −is central in g then 0 = [ζ, Z] = x ⊕ 0 ⊕ (−y) shows Z = ∆ ∈ g0, and since theadjoint representation of g0 on g1 ⊕ g−1 is faithful, it follows that Z = 0.

The TKK-algebra depends functorially on V with respect to surjective homo-morphisms. In more detail, let f = (f+, f−): V → W be a surjective homomor-phism of Jordan pairs. Then it follows easily from the definitions that f induces asurjective homomorphism of Lie algebras

f0: L0(V )→ L0(W ), δ(x, y) 7→ δ(f+(x), f−(y)), ζV 7→ ζW

mapping Inder(V ) onto Inder(W ), and a surjective homomorphism of the TKK-algebras

L(f): L(V )→ L(W ), x⊕∆⊕ y 7→ f+(x)⊕ f0(∆)⊕ f−(y), (7)

which maps DL(V ) onto DL(W ).In particular, this shows that every automorphism h = (h+, h−) of V induces

an automorphism, again denoted by h, of L(V ) by

h(x⊕∆⊕ y) = h+(x)⊕ (h∆h−1)⊕ h−(y). (8)

In this way, Aut(V ) embeds into the group of automorphisms of L(V ) preservingthe grading. We finally note that L(V op) is canonically isomorphic to L(V ) asungraded Lie algebras, but has the inverse grading L(V op)i = L(V )−i.

7.2. Example: TKK-algebras of special Jordan pairs. Let V be a specialJordan pair, embedded in the off-diagonal part (M+,M−) of a Morita contextM = (R,M+,M−, S). Let g = e(M, V ) be the elementary Lie algebra of 6.3 andwrite accordingly gi = ei. Also let g = L(V ). We have seen in 6.8 that any(a 0

0 d ) ∈ g0 gives rise to a derivation ∆V (a, d) of V . Hence there is a well-definedlinear map

Ψ : g→ g,

(a bc d

)7→ b⊕∆V (a, d)⊕ (−c). (1)

By a straightforward computation Ψ is a Lie algebra homomorphism. The genera-tors ζ = Ψ(e1) = Ψ(−e2) and V ± = Ψ(g±1) of g belong to the image of Ψ so Ψ issurjective. We claim that

z(g) = KerΨ ⊂ g0. (2)

Indeed, by (1),

γ =

(a bc d

)∈ KerΨ ⇐⇒ b = c = ∆V (a, d) = 0. (3)

In particular, this shows KerΨ ⊂ g0. On the other hand, by 6.2 and 6.3 g is

generated as a Lie algebra by e1, e2 and all(

0 xy 0

)where (x, y) ∈ V . Hence

γ ∈ z(g) ⇐⇒ [γ, ei] =[γ,

(0 xy 0

)]= 0,


for i = 1, 2 and all (x, y) ∈ V . A simple matrix calculation shows that [γ, ei] = 0 ifand only if γ = diag(a, d) is diagonal, and then[

γ,

(0 xy 0

)]=

(0 ax− xd

dy − ya 0

)=

(0 ∆V (a, d)+(x)

∆V (a, d)−(y) 0

).

Comparing this with (3) shows KerΨ = z(g). Summarizing, we have shown for aspecial Jordan pair V that

L(V ) ∼= g/z(g), g = e(M, V ). (4)

7.3. Example. Let k be an algebraically closed field of characteristic zero.The finite-dimensional simple Jordan pairs over k are classified in [52, 17.12] (fork of arbitrary characteristic). The table below lists these Jordan pairs, using thenotation of 6.6 in the first and the notation of [52, 17.12] in the second column. IfV is such a pair, ζV is an inner derivation of V , and L(V ) is a finite-dimensionalsimple Lie algebra over k of the following type:

V type of V type of L(V )

Mpq(k), p+ q > 2 Ipq Ap+q = Ap+q−1

J(k2n−1, q), n> 2, q non-degenerate IV2n−1 Bn

Hn(k), n> 2 IIIn Cn

An(k), n> 4 IIn Dn

J(k2n−2, q), n> 4, q non-degenerate IV2n−2 Dn

M12(C), C an octonion algebra V E6

H3(C), C an octonion algebra VI E7

Thus L(V ) is a finite-dimensional simple Lie algebra not of type E8, F4 or G2. Anyfinite-dimensional simple Lie algebra not of type E8, F4 or G2 is the TKK-algebra ofsome finite-dimensional simple Jordan pair, but non-isomorphic such Jordan pairsmay have isomorphic Lie algebras.

7.4. Lemma. Let g = L(V ) be the Tits-Kantor-Koecher algebra of a Jordanpair V . For all z ∈ V σ and σ ∈ +,− define an endomorphism expσ(z) of thek-module g by

expσ(z)u = u, expσ(z)∆ = ∆+ [z,∆], expσ(z)v = v + [z, v] +Qzv, (1)

where u ∈ V σ, ∆ ∈ g0 and v ∈ V −σ. Then expσ(z) is an automorphism of the Liealgebra g, and the maps expσ are injective group homomorphisms from the additivegroups V σ to Aut(g).


Remark. With respect to the decomposition g = V + ⊕ g0 ⊕ V −, the mapsexpσ are given explicitly by the following formal 3×3 matrices of homomorphisms:

exp+(x) =

1 adx Qx0 1 adx0 0 1

, exp−(y) =

1 0 0ad y 1 0Qy ad y 1

, (2)

for x ∈ V + and y ∈ V −.

Proof. We start by showing that expσ(x) is a Lie algebra endomorphism ofg. By symmetry it is enough to do so for σ = +. Write exp+(x) in the formexp+(x) = Id + adx + E2 where E2 ∈ End g is defined by E2(g0 ⊕ g1) = 0 andE2(y) = Qxy ∈ V + ⊂ g for y ∈ V −. Evaluating the condition exp+(x)[u, v] =[exp+(x)u, exp+(x)v] for u, v ∈ g and keeping in mind the Z-grading (7.1.5) of g,one sees that it suffices to show

[E2(u), v] +[[x, u], [x, v]

]+ [u,E2(v)] = E2

([u, v]

), (3)[

[x, u], E2(v)]

+[E2(u), [x, v]

]= 0. (4)

Again because of the Z-grading of g, (3) holds for u ∈ V + and for u, v ∈ g0. Incase u = ∆ = (∆+, ∆−) ∈ g0, (3) becomes

[[x,∆], [x, v]] + [∆,Qxv] = Qx[∆, v].

which is equivalent to

−∆+(x), v, x+∆+Qxv = Qx∆−(v),

and therefore follows from (6.8.1). By antisymmetry of (3) in u, v, it now remainsto consider the case u, v ∈ g−1. Then the equation becomes, in view of (JP15),

0 = −δ(Qxu, v) + [δ(x, u), δ(x, v)] + δ(Qxv, u)

= δ(Qxu, v) + δ(Qxv, u)− δ(x, uxv)

which is a linearized version of (JP2).We next verify (4). The left hand side of (4) is an alternating function of u and

v, and it follows from the definition of E2 and the Lie product in g that it vanishesif either u or v belongs to g0 ⊕ g1. Hence it suffices to verify (4) for u, v ∈ g−1.Then the left hand side becomes, using (JP4) in the third equation,[

[x, u], Qxv]

+[Qxu, [x, v]

]= −[δ(x, u), Qxv]− [Qxu, δ(x, v)]

= −D(x, u)Qxv +D(x, v)Qxu = −QxD(u, x)v +QxD(v, x)u

= Qx(uxv − vxu

)= 0.

This finishes the proof that exp+(x) is an endomorphism of the Lie algebra g.It is easily seen that exp+ is a homomorphism from the additive group V + intoAut(g). To see that it is injective, assume exp+(x) = Id. Then (1) impliesζ = exp+(x)ζ = ζ + [x, ζ] = ζ − x, so x = 0.


7.5. The projective elementary group. Let V be a Jordan pair and letUσ := expσ(V σ). The projective elementary group of V is the subgroup of Aut(g)generated by U+ and U−:

G = PE(V ) =⟨U+, U−

⟩.

Since the groups U± are abelian, PE(V ) has A1-commutator relations with rootgroups U±1 = U±, cf. 3.3(a). It is evident from (7.4.2) that

U+ ∩ U− = 1.

The diagonal subgroup of G is the group

H = PE0(V ) = G ∩Aut(V )

where Aut(V ) is diagonally embedded in Aut(g) as in (7.1.8). The big cell of G isdefined as

Ω = U−HU+.

Recall from 6.5 the notion of the opposite of a Jordan pair. As in 7.1 we identifyL(V ) and L(V op) as Lie algebras (disregarding the grading). Then it is immediatefrom the definitions that PE(V ) = PE(V op) as groups, but that the roles of exp+

and exp−, and therefore of U+ and U−, are interchanged:

expopσ (x) = exp−σ(x), (1)

for all x ∈ (V op)σ = V −σ. From the functoriality of L(V ) (see 7.1) it followsthat PE(V ) depends functorially on V with respect to surjective homomorphisms.Namely, let f : V → V ′ be a surjective homomorphism of Jordan pairs, givingrise to the surjective homomorphism L(f): L(V ) → L(V ′) of TKK-algebras as in(7.1.7). Let expσ, G, H be as in 7.5, and denote the corresponding objects for V ′

by exp′σ, G′, and H ′. It is immediate that L(f) expσ(x) = exp′σ(fσ(x)) L(f)holds for x ∈ V σ. Hence expσ(x) leaves the kernel k = Ker

(L(f)

)invariant,

and exp′σ(fσ(x)) is the automorphism of L(V ′) ∼= L(V )/k induced by expσ(x).Therefore, any element of G leaves k invariant and induces an automorphism ofL(V ′). We thus get a unique surjective homomorphism

ϕ = PE(f): G→ G′, expσ(x) 7→ exp′σ(fσ(x)), (2)

which maps H to H ′ and satisfies

ϕ(β(x, y)

)= β

(f+(x), f−(y)

)(3)

for quasi-invertible pairs (x, y) ∈ V . For the details, we refer to [58, Proposi-tion 1.6].

7.6. Examples. (a) Let M = (R,M+,M−, S) be a Morita context and letV ⊂ (M+,M−) be a special Jordan pair, as in 6.4. Let G = E(M, V ) be theelementary group as in 6.2 and let Ψ : g = e(M, V ) → g = L(V ) be the surjectiveLie algebra homomorphism with kernel z(g) of 7.2. Let g ∈ G. By Lemma 6.3(b),


Ad g leaves g invariant and therefore also its centre z(g). Since g ∼= g/z(g) by (7.2.4),it follows that Ad g induces an automorphism π(g) of g, making the diagram

gAd g //

Ψ

g

Ψ

g

π(g)// g

(1)

commutative.We claim that π: G→ G = PE(V ) is a surjective homomorphism and the kernel

of π is precisely the diagonal part of the centre of G:

Kerπ = Z0(G) := Z (G) ∩(R× 00 S×

). (2)

HenceG ∼= G/Z0(G) (3)

is a central quotient of the elementary group G = E(M, V ).Indeed, one checks, using (7.2.1), that for x ∈ V σ, the diagram

gAd eσ(x) //

Ψ

g

Ψ

g

expσ(x)// g

is commutative, so that π(eσ(x)) = expσ(x). Since G and G are generated by thesubgroups Uσ = eσ(V σ) and Uσ = expσ(V σ), respectively, it follows that π: G→ Gis a surjective homomorphism, which furthermore induces isomorphisms Uσ ∼= Uσ.

It remains to show (2). Let g = diag(a, d) ∈ Z0(G). Then Ad g(ei) = ei and

ge+(x)g−1 = e+(x) =(

1 axd−1

0 1

)=(

1 x0 1

)implies Ad g ·

(0 x0 0

)=(

0 x0 0

)for all

x ∈ V +. In the same way, Ad g is the identity on g−1. As g is generated by the eiand g±1, it follows that Ad g = Id, so by (1), we have π(g) = 1. Thus

Z0(G) ⊂ Ker Ad ⊂ Kerπ.

For the proof of the inclusion Kerπ ⊂ Z0(G) we refer to [58, Theorem 2.8].

(b) In (a), it is not clear whether the centre of G is always diagonal. Hereis a sufficient condition for this to happen. Suppose that 1R ∈ V +V −, i.e., thatthere exist xi ∈ V + ⊂ M+ and yi ∈ V − ⊂ M− such that 1R =

∑xiyi. Then

Z (G) is diagonal. Indeed, let g =(a bc d

)∈ Z (G). Since g commutes with all

e+(x) and e−(y), (x, y) ∈ V , a straightforward matrix computation shows thatxc = cx = yb = by = 0. Hence b = 1R b =

∑xiyib = 0 and c = c 1R =

∑cxiyi = 0,

as desired.This condition holds in particular when V = Mpq(A) = (Matpq(A),Matqp(A))

is the Jordan pair of rectangular matrices over a ring A, see 6.1. Here G = En(A) is


the elementary group as in [32, 1.2C]. Since R = Matp(A) and 1R =∑pi=1EinEni ∈

V +V −, the condition above is satisfied, so that we have, putting p+ q = n,

PE(Mpq(A)) ∼= En(A)/Z (En(A)) = PEn(A), (4)

in the notation of [32, 2.2.13]. It is known [32, 1.2.14] that the centralizer of En(A)in GLn(A) consists of all a · 1n where a is an invertible element of the centre of A.

(c) Let K be a commutative k-algebra. Since the generators of En(K) havedeterminant 1, it is clear that

En(K) ⊂ SLn(K). (5)

If K is a field then En(K) = SLn(K) by [32, 2.2.6], so we have

PE(Mpq(K)) ∼= SLn(K)/Z(SLn(K)

) ∼= SLn(K)/µn(K) · 1n, (6)

where µn(K) denotes the group of nth roots of unity in K.

(d) The group PE(V ) is commutative if and only if V is trivial, i.e., Qu = 0 forall u ∈ V σ. Indeed, by Theorem 7.7(b),(c), commutativity of PE(V ) is equivalentto β(x, y) = Id and xy = x, yx = y for all (x, y) ∈ V . By the definition of thequasi-inverse (6.9.2) this is in turn equivalent to V being trivial. In this case,PE(V ) = U+ ×U− ∼= V + × V −, the direct product of the additive groups V + andV −.

7.7. Theorem. Let V be a Jordan pair. We keep the notations of 7.5.

(a) The subgroups U± are normalized by H; more precisely, for h = (h+, h−) ∈H and x ∈ V σ,

h expσ(x)h−1 = expσ(hσ(x)). (1)

(b) The map V − ×H × V + → Ω, (y, h, x) 7→ exp−(y)h exp+(x), is bijective.

(c) A pair (x, y) ∈ V is quasi-invertible if and only if exp+(x) exp−(y) ∈ Ω. Inthis case,

exp+(x) exp−(y) = exp−(yx)β(x, y) exp+(xy), (2)

exp−(y) exp+(x) = exp+(xy)β(x, y)−1 exp−(yx). (3)

Hence, the inner automorphism group Inn(V ) is contained in H.

Proof. (a) We show that h expσ(x) = expσ(hσ(x))h. Since both sides areautomorphisms of g it suffices to show that they agree on the generators ζ, V + andV − of g. Indeed, h(ζ) = h(IdV + ,−IdV −)h−1 = (IdV + ,−IdV −) = ζ, hence

h expσ(x)ζ = h(ζ + [x, ζ]) = ζ + [hσ(x), h(ζ)] = ζ + [hσ(x), ζ],

while expσ(hσ(x))ζ = ζ + [hσ(x), ζ], by (7.4.1).Next, let u ∈ V σ. Then h expσ(x)u = hσ(u) while expσ(hσ(x))h(u) =

expσ(hσ(x))hσ(u) = hσ(u) as well. Finally, for v ∈ V −σ we have


h expσ(x)v = h(v + [x, v] +Qxv

)= h−σ(v) + (h[x, v]h−1) + hσ(Qxv)

= h−σ(v) + [hσ(x), h−σ(v)] +Qhσ(x)h−σ(v)

since h is an automorphism of V , while

expσ(hσ(x))h(v) = expσ(hσ(x))h−σ(v)

= h−σ(v) + [hσ(x), h−σ(v)] +Qhσ(x)h−σ(v),

by (7.4.1).

(b) Since the exponential maps are injective by Lemma 7.4, it suffices to showthat the multiplication map U− ×H × U+ → G is injective. As H normalizes U±

it is enough that u−hu+ = 1 imply u± = 1 = h. But this follows from (7.4.2) andthe definition of H ⊂ Aut(g) in (7.1.8) by a simple matrix calculation.

(c) First, let exp+(x) exp−(y) ∈ Ω, say,

exp+(x) exp−(y) = exp−(v)h exp+(u),

and apply both sides to an element z ∈ V +. On the left, we obtain,

exp+(x) exp−(y)z = exp+(x)(z + [y, z] +Qyz

)= z + [y, z] + [x, [y, z]] +Qyz + [x,Qyz] +QxQyz

≡ B(x, y)z (mod g0 ⊕ g−1),

while on the right,

exp−(v)h exp+(u)z = exp−(v)h · z = exp−(v)h+(z)

≡ h+(z) (mod g0 ⊕ g−1)).

This shows B(x, y) = h+ invertible, so (x, y) is quasi-invertible.Now suppose (x, y) quasi-invertible. We then prove

exp+(−x) exp−(yx)β(x, y) = exp−(y) exp+(−xy) (4)

which is equivalent to (2) but easier to work with. Since both sides of (4) areautomorphisms of g it suffices to show that they agree on the generators V ± and ζof g. Thus let z ∈ V + and put z′ = B(x, y)z. Then Q(yx)z′ = Qyz by the identity(JP28). Hence the left hand side of (5) applied to z is

exp+(−x) exp−(yx)β(x, y)z = exp+(−x) exp−(yx)z′

= exp+(−x)(z′ ⊕ δ(z′, yx

)⊕Qyz)

=(z′ +QxQyz +D(x, yx)z′

)⊕(δ(z′, yx) + δ(x,Qyz)

)⊕Qyz, (5)

while the right hand side yields

exp−(y) exp+(−xy)z = z ⊕ δ(z, y)⊕Qyz. (6)

Now D(x, yx)z′ = D(x, y)z−2QxQyz (by (JP31)) = z−B(x, y)z−QxQy and hencez′ + QxQyz + D(x, yx)z′ = z so the components in V + of (5) and (6) agree. For


the components in g0 we must show δ(z′, yx) + δ(x,Qyz) = δ(z, y). By the dualityprinciple [52, 2.5] it suffices to prove that D(z′, yx) +D(x,Qyz) = D(z, y). Applythis to an arbitrary element u ∈ V + and read the result as a linear map in z. Thenit is equivalent to D(u, yx)B(x, y) +D(u,Qyz) = D(u, y) which is just the identity(JP31).

Next, we show that the two sides of (4) agree on v ∈ V −. Since B(y, x) isinvertible on V −, we can write v = B(y, x)v′. Then β(x, y)v = B(y, x)−1v = v′

(remember the identification of Aut(V ) with a subgroup of Aut(g)). Hence thedesired equality is

exp+(−x) exp−(yx)v′ = exp−(y) exp+(−xy)v.

This follows from the already proved formula

exp−(y) exp+(−xy)z = exp+(−x) exp−(yx)z′

by passing to V op and the substitution x→ −y, z → v′, together with the fact that(x, y) is quasi-invertible if and only (−x,−y) is, in which case −(xy) = (−x)−y,since (−Id,−Id) is an automorphism of V .

Finally we show that the two sides of (4) agree on ζ. Indeed,

exp+(−x) exp−(yx)β(x, y)ζ = exp+(−x) exp−(yx)ζ

= exp+(−x)(ζ + yx)

=(x+Qxy

x)⊕(ζ + δ(x, yx)

)⊕ yx

and similarly

exp−(y) exp+(−xy)ζ = xy ⊕(ζ + δ(xy, y)

)⊕(y +Qyx

y).

Here the components in g±1 agree by the symmetry principle (6.11.1), and for thecomponents in g0 it suffices to prove the identity

D(xy, y) = D(x, yx). (7)

Indeed, by (JP28), Q(xy)Q(y) = B(x, y)−1QxQy = QxQyB(x, y)−1 since QxQyand B(x, y) commute by (JP1). Also, QxQ(yx) = QxQyB(x, y)−1 (by (JP28))= Q(xy)Qy. Now (7) follows from the identity (JP35) by comparing the D-terms.This finishes the proof that (2) holds for quasi-invertible (x, y) and at the sametime shows Inn(V ) ⊂ H since β(x, y) ∈ U−U+U−U+ by (2).

Finally, (3) follows from (2) by inversion and −(xy)

= (−x)−y, as noted before.

7.8. The Weyl element defined by an idempotent. Let e = (e+, e−) bean idempotent of V , see 6.14. We introduce the notations

θe =(

exp−(e−), exp+(e+), exp−(e−))∈ U− × U+ × U−,

ωe = µ(θe) = exp−(e−) exp+(e+) exp−(e−) ∈ G = PE(V ).

As noted in 7.5, G has A1-commutator relations with root groups U1 = U+ andU−1 = U−, so θe ∈ Θ1(G) in the notation of (5.1.2). If e is an invertible idempotent


we will see in 7.11 that ωe is a Weyl element and hence θe is a Weyl triple for G. Ingeneral, this is not the case. However, it will be shown later in Section 12 that thePeirce decomposition of V with respect to e gives rise to C2-commutator relationsfor G (for suitably defined root groups) and then ωe is indeed a Weyl element forone of the long roots. By abuse of language, we will therefore often refer to ωe(resp. θe) as the Weyl element (resp. Weyl triple) defined by e without furtherspecification.

Let V op = (V −, V +) be the opposite Jordan pair, see 6.5. We define eop =(e−, e+) and correspondingly

θeop =(

exp+(e+), exp−(e−), exp+(e+))∈ U+ × U− × U+,

ωeop = exp+(e+) exp−(e−) exp+(e+).

By (6.14.1), −e is an idempotent along with e. Since expσ is a group homomorphismit is evident that

ω−1e = ω−e. (1)

7.9. Lemma. Let e be an idempotent of V and let V = V2 ⊕ V1 ⊕ V0 be theassociated Peirce decomposition as in 6.14. Then the action of ωe on the generatorsof L(V ) is given by

ωe · x =

x if x ∈ V σ0[e−σ, x] if x ∈ V σ1Qe−σ · x if x ∈ V σ2x+ [e+, e−] if x = ζ

. (1)

Moreover, ω2e = se = %e(−1) is the Peirce reflection defined in (6.14.6), and

therefore ω4e = Id.

Proof. For the first formula, let x = xσ0 ∈ V σ0 . Then expσ(eσ) · x = x holds bydefinition of exp in (7.4.1), while

exp−σ(e−σ) · xσ0 = x0 + [e−σ, x0] +Qe−σ · x0 = x0

follows from the multiplication rules for the Peirce spaces in 6.14. Next, let x1 ∈ V +1 .

Then exp−(e−) · x1 = x1 + [e−, x1] since Qe−x1 = 0 by the Peirce relations, hence

ωe · x1 = exp−(e−) exp+(e+) ·(x1 + [e−, x1]

)= exp−(e−) ·

(x1 + [e−, x1] + [e+, [e−, x1]]

)Since [e+, [e−, x1]] = −e+, e−, x1 = −x1 we get

ωe · x1 = exp−(e−) · [e−, x1] = [e−, x1] + [e−, [e−, x1]] = [e−, x1]

because [e−, [e−, x1]] = e−, x1, e− = 0 by the Peirce multiplication rules. Nowlet y1 ∈ V −1 . Then similarly

ωe · y1 = exp−(e−) exp+(e+) · y1 = exp−(e−) ·(y1 + [e+, y1]

)= y1 + [e+, y1] + [e−, [e+, y1]] = [e+, y1].


Next, consider x2 ∈ V +2 . Then

ωe · x2 = exp−(e−) exp+(e+) · (x2 + [e−, x2] +Qe−x2)

= exp−(e−) ·(x2 +

([e−, x2] + [e+, [e−, x2]]

)+(Qe−x2 + [e+, Qe−x2] +Qe+Qe−x2

)).

The map Qe− : V +2 → V −2 is bijective with inverse Qe+ : V −2 → V +

2 . More-over, [e+, [e−, x2]] = −e+, e−, x2 = −2x2 and it will be shown in (10.8.1) that[Qe−x2, e+] = δ(e+, Qe−x2) = δ(x2, e−). Hence ωe · x2 = exp−(e−) · (Qe−x2) =Qe−x2, proving the case σ = + of the third formula of (1). The case σ = − canbe done by a similar computation, or by arguing as follows: recall that ω−e = ω−1

e

and that e and −e have the same Peirce spaces, while Qx is a quadratic functionof x. Hence Qe−x2 = ω−e · x2 = ω−1

e · x2 or x2 = ωe · Qe−x2 for all x2 ∈ V +2 . By

putting y2 = Qe−x2 we see that ωe · y2 = Qe+y2 for all y2 ∈ V −2 .

It remains to compute the action of ωe on ζ. Since ad ζ is the grading derivation,we have [e+, ζ] = −e+ and [e−, ζ] = e−. Hence

ωe · ζ = exp−(e−) exp+(e+) ·(ζ + [e−, ζ]

)= exp−(e−) ·

((ζ + [e+, ζ]) + (e− + [e+, e−] +Qe+e−)

)= exp−(e−) ·

(e− + (ζ + [e+, e−])

)= e− + ζ + [e−, ζ] + [e+, e−] + [e−, [e+, e−]]

= ζ + [e+, e−],

because Qe+e− = e+ and [e−, [e+, e−]] = −2Qe−e+ = −2e−.From (1) one deduces easily that ω2

e · xi = (−1)ixi for xi ∈ V σi , which provesthe last statement.

7.10. Proposition. Let e ∈ V be an idempotent. Then for all ui ∈ V +i (e),

vi ∈ V −i (e) and z0 ∈ V σ0 (e), the element ωe satisfies the conjugation formulas

ωe exp+(u2)ω−1e = exp−

(Qe−u2

), (1)

ωe exp−(v2)ω−1e = exp+

(Qe+v2

), (2)

ωe exp+(u1)ω−1e = β(u1,−e−), (3)

ωe exp−(v1)ω−1e = β(e+, v1), (4)

ωe expσ(z0)ω−1e = expσ(z0). (5)

Moreover, θe is balanced and hence ωe satisfies

ωe = ωeop . (6)

Proof. It is easier to prove (1) in the form

ωe exp+(u2) = exp−(Qe−u2)ωe, (7)

for all u2 ∈ V +2 . Since both sides of (7) are automorphisms of the TKK-algebra

g = L(V ) which is generated by V + = g1, V − = g−1 and ζ, it suffices to show


that both sides agree when applied to the generators x = xσi ∈ V σi (e) (σ ∈ +,−,i ∈ 0, 1, 2) and x = ζ. This amounts to seven cases. We do the case x = x−2 =y2 ∈ V −2 , using (7.9.1), and leave the others to the reader:

(ωe exp+(u2)

)· y2 = ωe ·

(y2 + [u2, y2] +Qu2

y2

)= Qe+y2 + [ωe(u2), ωe(y2)] +Qe−Qu2

y2.

In the second step, we have used the fact that ωe is an automorphism of g. On theother hand,(

exp−(Qe−u2)ωe)· y2 = exp−(Qe−u2) ·Qe+y2

= Qe+y2 + [Qe−u2, Qe+y2] +Q(Qe−u2)Qe+y2.

The second terms agree by (7.9.1), and so do the third terms by the Jordan identity(JP3) and the fact that Qe−Qe+ is the identity on V −2 (e): Q(Qe−u2)Qe+y2 =Qe−Qu2

Qe−Qe+y2 = Qe−Qu2y2.

The relation (2) is now a consequence of the one just proved and the followingobservation. Since (1) holds for all idempotents, it does so in particular for −e.Now ω−e = ω−1

e and Q is a quadratic map, so putting u2 = Qe+v2, we haveω−1e exp+(Qe+v2)ωe = exp−(Q−e−Qe+v2) = exp−(v2) which is (2).

The remaining formulas (3) – (5) can be proved in the same way. For a differentmethod see the proof of Lemma 12.2.

Finally, put v2 = e− in (2) and use Qe+e− = e+. Then ωe exp−(e−)ω−1e =

exp+(e+), so θe is balanced by 5.15(ii) which implies (6) by 5.15(iv).

7.11. Corollary. Let e be an invertible idempotent of V , i.e., e+ ∈ V + isinvertible with inverse e−. Consider G = PE(V ) as a group with A1-commutatorrelations as in 7.5. Then θe is a balanced Weyl triple and ωe is a Weyl elementfor the root α = 1 in the sense of 5.1: ωeU

σω−1e = U−σ for σ = ±. Moreover, ωe

normalizes H.

Proof. We have seen in 7.10 that θe is a balanced Weyl triple. By assumption,V = V2(e), so Qe− : V + → V − is an isomorphism with inverse Qe+ : V − → V +.Now the first assertion of the corollary follows from (7.10.1) and (7.10.2). To showthat ωe normalizes H, we argue as follows. From the fundamental formula (JP3)it follows that f = (Qe− , Qe+): V → V op is an isomorphism. Hence by (7.5.2) and(7.5.1), f induces an isomorphism ϕ = PE(f): PE(V )→ PE(V op) satisfying

ϕ(expσ(x)) = expopσ (fσ(x)) = exp−σ(Qe−σ (x)),

and mapping H isomorphically to H ′ = PE(V op) ∩ Aut(V op). Comparison with(7.10.1) and (7.10.2) shows that ϕ = Int(ωe). Up to a switch of V + and V −,we identify the automorphism groups, the Tits-Kantor-Koecher algebras and theelementary projective groups of V and V op. Then H = H ′, so ωe normalizes H.

In general, however, not all Weyl elements of PE(V ) are of the form ωe, seeProposition 9.12 for details.


7.12. Example. In the following example, we work out in some detail thegroups E2(A) and PE2(A) = PE(A,A) for the following ring A. Let k = F2 bethe field with two elements, let T be an indeterminate, and let A = k[T ]/(Tn+1), acommutative associative ring. Putting ε = can(T ), we have A = k·1⊕k·ε⊕· · ·⊕k·εnand εn+1 = 0. Thus A = k for n = 0, and for n = 1 we have the usual algebra ofdual numbers over k. Let J be the Jordan algebra determined by A, and V = (J, J)the corresponding Jordan pair. Equivalently, V is the Jordan pair determined by

the Morita context(A AA A

).

(a) Let H be the subgroup of G := E2(A) consisting of diagonal matrices. Weclaim that

H = diag(u, u−1) : u ∈ A×. (1)

Indeed, E2(A) ⊂ SL2(A) by (7.6.5). Hence the inclusion from left to right holds.For the converse, observe that the units u of A are precisely the elements u = 1 +xwhere x ∈

∑ni=1 k · εi. Now the matrix identity(

1 x0 1

)(1 01 1

)=

(1 0u−1 1

)(u 00 u−1

)(1 u−1x0 1

)shows that H contains all diag(u, u−1), u ∈ A×.

We claim that G and G := PE(V ) = PE2(A) (cf. (7.6.4)) are solvable groups oforders ∣∣E2(A)

∣∣ = 6 · 23n,∣∣PE2(A)

∣∣ = 6 · 22n+[n/2], (2)

respectively.

Proof. We prove the first formula by induction on n. For n = 0, by [32, 2.2.6]and [24, §8], E2(k) = SL2(k) ∼= S3, the symmetric group on three letters, of order6.

Now let n > 1, let A′ = k[T ]/(Tn), and let ϕ: A = k(ε) → A′ = k(δ) (whereεn+1 = 0 and δn = 0) be the homomorphism sending ε to δ, with kernel k ·εn. Thisinduces a surjective group homomorphism Φ: E2(A)→ E2(A′); let K be its kernel.We claim that

K = e−(k · εn) · (K ∩ H) · e+(k · εn), (3)

The inclusion from right to left is clear. Conversely, let g =(a bc d

)∈ K, so

Φ(g) =(ϕ(a) ϕ(b)ϕ(c) ϕ(d)

)=(

1 00 1

). In particular, ϕ(a) = 1, and since the kernel of ϕ is a

nil ideal, it follows that a ∈ A×. Since E2(A) ⊂ SL2(A), the easily verified matrixidentity (

a bc d

)=

(1 0

ca−1 1

)(a 00 a−1

)(1 a−1b0 1

)shows that g ∈ e−(A) · H · e+(A), and since the kernel of ϕ is k · εn, we have (3).We show next that

K ∩ H =

diag(1, 1), diag(1 + εn, 1 + εn) ∼= Z/2Z. (4)

Indeed, by (1), we have to determine which units of A belong to the kernel of ϕ.

Let u = 1 +∑ni=1 λiε

i ∈ A×. Then ϕ(u) = 1 +∑n−1i=1 λiδ

i = 1 if and only if


λ1 = · · · = λn−1 = 0. This proves (4). From this and (3) it follows easily thatK ∼= (Z/2Z)3 has order 8. Now the first formula of (2) follows by induction. It alsofollows by induction (and the fact that S3 is solvable) that E2(A) is solvable. Forthe second formula of (2), we use the fact that, by (7.6.2), PE2(A) is the quotientof E2(A) by the diagonal part of its centre. By [32, 1.2.14], the centralizer of En(A)in GLn(A) consists of all r · 1n where r is an invertible element of the centre of A.Hence

Z (G) = diag(r, r) : r2 = 1 ∼= µ2(A), (5)

the second roots of unity in A. To determine this in more detail, let u = 1 +∑ni=1 λiε

i ∈ A× and put p =[n2

]. Then u2 = 1 +

∑pi=1 λiε

2i, so u ∈ µ2(A) if andonly if λ1 = · · · = λp = 0. This shows that µ2(A) consists of all 1 +

∑ni=p+1 λiε

i

where the λi ∈ k = F2. It follows that

Z (G) ∼= µ2(A) ∼=

(Z/2Z)p if n = 2p is even(Z/2Z)p+1 if n = 2p+ 1 is odd

. (6)

Now the second formula of (2) follows from the first one by a simple computation.

(b) Next, we determine the centre of G. Since G = G/Z (G), we have

Z (G) = Z2(G)/Z (G),

where Z2(G) = g ∈ G :(((((((g, G

)))))))⊂ Z (G) is the second term in the ascending

central series of G. We claim that

Z (G) ∼=Z/2Z if n = 2p is even1 if n = 2p+ 1 is odd

. (7)

In the first case, a representative of the non-trivial element of Z (G) is the matrixr

m :=

(1 + εp ε2p

ε2p (1 + εp)−1

).

Proof. Let g =(a bc d

)∈ Z2(G). Then a straightforward computation shows

that (((((((g, e+(x)

)))))))=

(1 + acx (1 + a2)x+ acx2

c2x 1 + acx+ c2x2

)∈ Z (G),

for all x ∈ A. From (5) it follows that

c2x = 0, (8)

(1 + a2)x+ acx2 = 0, (9)

for all x ∈ A. In particular, putting x = 1 in (8) shows c2 = 0. Also ad = 1 + bc, so

(ad)2 = 1 + b2c2 = 1, (10)

in particular, a and d are invertible. Let x = 1 in (9) and multiply with a−1. Thenwe obtain


ac = 1 + a2, equivalently, c = a+ a−1, (11)

and by squaring 0 = c2 = a2 + a−2, so a4 = 1. Now (10) implies

d2 = a−2 = a2 and d4 = 1. (12)

Analogously, one computes

(((((((g, e−(y)

)))))))=

(1 + bdy + b2y2 by2

(1 + d2)y + bdy2 1 + bd

)for all y ∈ A, from which one concludes as above

b2 = 0 and b = d+ d−1. (13)

Hence 1 = det g = ad+ (d+ d−1)(a+ a−1), which implies by multiplying with withad and observing (12) that

ad = a2d2 + (a2 + 1)(d2 + 1) = 2a2d2 + a2 + d2 + 1 = 1 + 2a2 = 1. (14)

Thus d = a−1, and (11) and (13) show

b = c = a+ a−1 = d+ d−1. (15)

Altogether, g has the form

g =

(a a+ a−1

a+ a−1 a−1

), a ∈ µ4(A). (16)

We now determine a in more detail. From (9) and (11) we obtain, for x = ε, that(1 + a2)ε(ε+ 1) = 0. Since ε is nilpotent, 1 + ε is a unit, so this implies

(1 + a2)ε = 0. (17)

Write a =∑ni=0 λiε

i with λi ∈ k. Since a is a unit and ε is nilpotent, λ0 = 1. Hence1+a2 =

∑pi=1 λiε

2i, where p = [n2 ]. Then (17) implies (1+a2)ε =∑pi=1 λiε

2i+1 = 0,from which we conclude λ1 = · · · = λp = 0 if n = 2p + 1 is odd, while λ1 = · · · =λp−1 = 0 if n = 2p is even. This shows

1 + a2 =

0 if n = 2p+ 1λpε

2p if n = 2p

. (18)

Now we distinguish two cases.

Case 1: a2 = 1. Then (16) shows g = diag(a, a) ∈ Z (G). This always holds ifn = 2p+ 1 is odd. Thus it remains to consider

Case 2: n = 2p even and a2 6= 1. Then a2 = 1 + ε2p by (18).Let us put r = a + εp. Then r2 = a2 + ε2p = 1, so r is a second root of unity,

anda = r + εp = r(1 + rεp).

§8] The projective elementary group II 107

Since a is a fourth root of unity, we have

a−1 = a3 = (r + εp)3 = r3 + 3r2εp + 3rε2p + ε3p

= r + εp + rε2p = a+ rε2p,

and therefore a+ a−1 = rε2p. Now (16) shows

g =

(r(1 + εp) rε2p

rε2p r(1 + εp)−1

)=

(r 00 r

)(1 + εp ε2p

ε2p (1 + εp)−1

)= diag(r, r) ·m.

Conversely, a direct matrix computation shows that m ∈ Z2(G) and that m2 =diag(1 + εp, 1 + εp) ∈ Z (G). This proves (7).

(c) Finally, we consider the special case n = 1, thus A = k[T ]/(T 2) = k(ε),ε2 = 0 the ring of dual numbers, and show that

PE2(k(ε)) ∼= S4. (19)

For the proof, consider the matrices g1 =(

1 10 1

), g2 =

(1 01 1

), g3 =

(1 1+ε0 1

)in

G = E2(A) and let w = g1g2g1 =(

0 11 0

). Then w

(1 ε0 1

)w−1 =

(1 0ε 1

). This

implies that G is generated by g1, g2 and g3. Moreover, straightforward matrixcomputation shows the relations

w2 = (g1g2)3 = 1, (g1g3)2 = 1, (g2g3)3 = diag(1 + ε, 1 + ε) ∈ Z (G).

Hence the images of the gi generate G and satisfy the well-known relations for thegenerators of S4. Since G has, by (2), order 24, the assertion follows.

§8. The projective elementary group II

8.1. The projective elementary group of a subpair. Consider a subpairV ′ = (V ′+, V ′−) of V , let g = L(V ) and define g′ ⊂ g by

g′ = V ′+ ⊕(k · ζV + [V ′+, V ′−]

)⊕ V ′− = g′1 ⊕ g′0 ⊕ g′−1.

Since V ′ is a subpair, it follows from the definition of the multiplication in g that[[V ′+, V ′−], V ′σ] ⊂ V ′σ and this implies that g′ is a (graded) subalgebra of g.Let L(V ′) = V ′+ ⊕

(k · ζV ′ + Inder(V ′)

)⊕ V ′− be the TKK-algebra of V ′. It is

easily verified that there is a surjective homomorphism ϕ: g′ → L(V ′) of gradedLie algebras given by ϕ(x′) = x′ for x′ ∈ g′±1, and on g′0 by restricting an elementD′ ∈ g′0 to V ′±. Thus we have an exact sequence of Lie algebras

0 // k // g′ϕ // L(V ′) // 0

with Ker(ϕ) = k ⊂ g′0, and X ∈ k if and only if [X,V ′±] = 0. This implies [k, g′] = 0,so k is central in g′.


Now consider the projective elementary groups PE(V ) of V and PE(V ′) of V ′

with exponential maps exp± and exp′±, respectively. Let G′ ⊂ PE(V ) be thesubgroup generated by exp+(V ′+) ∪ exp−(V ′−). We claim that there is a uniquesurjective group homomorphism

ψ: G′ → PE(V ′) (1)

with the property thatψ(

expσ(x′))

= exp′σ(x′), (2)

for all x′ ∈ V ′σ and σ ∈ +,−.Indeed, it follows from (7.4.1) and the fact that V ′ is a subpair of V that expσ(x′)

stabilizes g′. Moreover, for X ∈ k we have expσ(x′) ·X = X + [x′, X] (by (7.4.1))= X, so the generators of G′ fix the elements of k. Hence G′ stabilizes g′ and k,showing there is a well-defined homomorphism ψ: G′ → GL

(L(V ′)

). By applying

ϕ to (7.4.1) (with x and y now belonging to V ′) we see that the diagrams

g′

ϕ

expσ(x′) // g′

ϕ

L(V ′)

exp′σ(x′)

// L(V ′)

are commutative. Hence (2) holds so ψ maps G′ onto PE(V ′), as asserted.

8.2. The projective elementary group of a direct sum. Let V =⊕

i∈I Vibe a direct sum of ideals. Our aim is to show that there is a natural isomorphism

PE(V ) ∼=⊕i∈I

PE(Vi) (1)

where the direct sum symbol on the right denotes the restricted direct productof groups, i.e., the subgroup of the full direct product whose elements have onlyfinitely many components different from 1.

Clearly, the inner derivation algebra of V commutes with direct sums:

Inder(V ) ∼=⊕i

Inder(Vi). (2)

Recall from 7.1 that DL(V ) = V − ⊕ Inder(V ) ⊕ V + is the derived algebra of theTKK-algebra L(V ) of V . Then (2) immediately implies that the functor DL alsocommutes with direct sums:

DL(V ) ∼=⊕i∈I

DL(Vi). (3)

The relation between the full TKK-algebra L(V ) = k · ζV + DL(V ) of V and thatof the Vi is more complicated.

Let pi: V → Vi be the projection onto the i-th factor. By (7.1.7), there areinduced homomorphisms fi = L(pi): L(V )→ L(Vi), so we have a homomorphism


f : L(V )→∏i∈I

L(Vi)

with components fi. Explicitly, fi and f are given as follows. Let

X = x⊕ (λ · ζV + d)⊕ y ∈ L(V )

where x =∑xi ∈ V + =

⊕V +i , y =

∑yi ∈

⊕V −i , λ ∈ k, and d =

∑di ∈⊕

Inder(Vi). Then

fi(X) = xi ⊕ (λζVi + di)⊕ yi, f(X) = (fi(X))i∈I = x⊕ (λξ + d)⊕ y, (4)

where we identify DL(V ) with the subalgebra of∏i L(Vi) determined by (3) and

the embeddings DL(Vi) ⊂ L(Vi), and ξ := (ζVi)i∈I ∈∏

L(Vi). From (4) it is clearthat f is an isomorphism of L(V ) onto the subalgebra g := k ·ξ+DL(V ) of

∏L(Vi).

We will therefore identify the TKK-algebra of V with the subalgebra g of∏

L(Vi).Let Gi := PE(Vi) and G′ =

⊕iGi ⊂

∏iGi. The latter group acts on

∏L(Vi)

diagonally (i.e., componentwise) by automorphisms. Then the assertion (1) is aconsequence of the following result.

8.3. Proposition. With the notations introduced above, G′ stabilizes the subal-gebra g ∼= L(V ), and the induced homomorphism ψ: G′ → Aut(g) is an isomorphismof G′ onto PE(V ). The inverse ψ−1: PE(V ) → G′ is given by g 7→

(PE(pi)

)i∈I

where PE(pi) is the surjective group homomorphism of (7.5.2) associated with thesurjective homomorphism pi: V → Vi.

Proof. We start with the following remark. Let W be an arbitrary Jordanpair with TKK-algebra L(W ) = k · ζW + DL(W ) and let h ∈ PE(W ). Then hstabilizes DL(W ) (since h is an automorphism of L(W ) and the derived algebrais a characteristic ideal). Moreover, h(ζW ) ≡ ζW ( mod DL(W )). Indeed, itsuffices to check this for the generators exp±(x) of PE(W ), where exp±(x) · ζW =ζW + [x, ζW ] = ζW ∓ x.

Now let g = (gi)i∈I ∈ G′. Then clearly g stabilizes DL(V ). Moreover, applyingthe remark above to W = Vi and h = gi, we have gi(ζVi) = ζVi + Xi whereXi ∈ DL(Vi), and since gi 6= 1 for only finitely many i, only finitely many Xi aredifferent from 0. Hence g(ξ) = ξ + (Xi)i∈I ∈ ξ + DL(V ) ⊂ g. This shows that g isindeed stable under G′ and proves the existence of a group homomorphism ψ.

Next, we show that ψ is injective. If g = (gi)i∈I ∈ Ker(ψ) then gi acts like theidentity on DL(Vi), and g(ξ) = ξ implies gi(ζVi) = ζVi , for all i ∈ I, whence g = 1.

Finally, we show ψ(G′) = PE(V ). Denote the exponential maps of L(Vi) (i ∈ I)

and of L(V ) by exp(i)σ : V σi → Gi and expσ: V σ → Aut(g), respectively. Then one

easily verifies that the diagram

V σiinc //

exp(i)σ

V σ

expσ

Gi

ψ// Aut(g)

commutes. Since PE(V ) is generated by expσ(V σ) (σ ∈ +,−), and expσ(x) =

expσ(∑xi) =

∏i expσ(xi) =

∏i ψ(

exp(i)σ (xi)

), this shows that ψ(G′) = PE(V ).


8.4. Generalized Bergmann operators. For x = (x1, . . . , x2n) ∈ V n =(V + × V −)n define

exp(x) = exp+(x1) · · · exp−(x2n) ∈ G.

Observe that every element of G is of the form exp(x) for a suitable x and n,since one can always add trivial factors exp±(0). We put xop = (x2n, . . . , x1)and exp(xop) = exp−(x2n) · · · exp+(x1) ∈ G. Now define generalized Bergmannoperators B(x) ∈ End(V +) and B(xop) ∈ End(V −) by

B(x)u ≡ exp(x)(u) mod g0 ⊕ V −, B(xop)v ≡ exp(xop)(v) mod g0 ⊕ V +, (1)

for (u, v) ∈ V . Using the identification of exp± with 3 × 3 matrices of homomor-phisms as in (7.4.2) one calculates for (x, y) ∈ V that

exp+(x) exp−(y) =

1 + adx ad y +QxQy adx+Qx ad y Qxad y + adxQy 1 + adx ad y adx

Qy ad y 1

. (2)

Since adx ad y∣∣V + = −D(x, y) this shows that B(x) = B(x, y) for x = (x, y).

Similarly one sees B(xop) = B(y, x). Thus, the generalized Bergmann operatorsare indeed generalizations of the usual Bergmann operators. For another example,a lengthy but straightforward calculation shows that for n = 2,

B(x, y, z, v) = B(x, y)B(z, v)−D(x, v) +QxQv +QxQ(y, v)

+Q(x, z)Qv −QxD(y, z)Qv. (3)

There are also analogues of the Q-operators depending on 2n − 1 arguments. Forn = 2 and (x, y, z) ∈ V σ × V −σ × V σ, they are given by

T (x, y, z) = Qx+z −QxDy,z −Dx,yQz +QxQyQz.

The connection with the B(x, y, z, v) is given by

B(x, y, z, v) = T (x, y, z)Qv +B(x, y)(Id−D(z, v)) +QxQy,v −D(x, v)

= QxT (y, z, v) + (Id−D(x, y))B(z, v) +Qx,zQv −D(x, v).

If V is a special Jordan pair as in 6.6 then

T (x, y, z)v = (x+ z − xyz)v(z + x− zyx),

B(x, y, z, v)t =((1− xy)(1− zv)− xv

)t((1− vz)(1− yx)− vx

).

The generalized Bergmann operators share with the ordinary ones the property ofyielding structural transformations, see 6.12:

(B(x), B(xop)): V V is structural. (4)

Moreover,B(x) = B(−x), B(xop) = B(−xop). (5)


For the proof of (5) write exp(x) as a matrix(eij(x)

)16i,j63

of homomorphisms

with respect to the decomposition g = V + ⊕ g0 ⊕ V −. Since B(x) = e11(x) ourclaim will follow from

eij(−x) = (−1)i+jeij(x), 16 i, j 6 3. (6)

We prove (6) by induction on n where x ∈ V n. For n = 1, (6) follows from(2) by inspection. For n > 2, we write x = (x′,x′′) for x′ = (x1, x2). Then

eij(x) =∑3l=1 eil(x

′)elj(x′′) and each term satisfies

eil(−x′)elj(−x′′) = (−1)i+j+2leil(x′)elj(x

′′),

proving (6). One can prove B(xop) = B(−xop) similarly, or by interpreting B(x) =B(−x) for V op.

8.5. The generalized quasi-inverse. Generalizing from the case n = 1, wecall a 2n-tuple x quasi-invertible if both B(x) and B(xop) are invertible, and thendefine β(x) = (B(x), B(xop)−1). Since (B(x), B(xop)) is structural by (8.4.4), itfollows that

x quasi-invertible =⇒ β(x) ∈ Aut(V ). (1)

Generalizing Theorem 7.7(c) we have

x quasi-invertible ⇐⇒ exp(x) ∈ Ω, (2)

and in this case

exp(x) = exp−(q(xop)

)β(x) exp+

(q(x)

). (3)

The unique elements q(x) ∈ V + and q(xop) ∈ V − appearing in this formula arecalled the quasi-inverses of x and xop.

As an application of (3) we show that the subgroup H = G ∩Aut(V ) is in factthe subgroup of G consisting of “diagonal” maps with respect to the decompositiong = g1 ⊕ g0 ⊕ g−1:

β(x) : x quasi-invertible = H = f ∈ G : f(gi) = gi for i = ±1, 0. (4)

Indeed, by (1) and (3) we have β(x) : x quasi-invertible ⊂ H, while H ⊂ h ∈G : h(gi) ⊂ gi for i = ±1, 0 is immediate from the definition of H. Conversely, letg = diag(a1, a2, a3) ∈ G be diagonal and write g = exp(x) for a suitable x ∈ V n. Itfollows from (8.4.1) and (8.4.5) that B(x) = a1 and B(xop) = B(−xop) = a−1

3 areinvertible. Thus g = exp−(y)h exp+(x) ∈ Ω and therefore h = β(x) by (3). Wehave

g(ζ) = a2ζ = exp−(y)h(−x⊕ ζ) = exp−(y)(−h+(x)⊕ ζ)

= −h+(x)⊕ (ζ + [h+(x), y])⊕ (y −Qyh+(x)) ∈ g0.

This shows h+(x) = 0 = y and hence also x = 0, proving g = h = β(x).


8.6. The extreme radical. Recall from [52, 4.21] that the extreme radicalExtr(V ) = (E+, E−) of a Jordan pair V is

Eσ = z ∈ V σ : Qz = D(z, V −σ) = D(V −σ, z) = 0 (σ ∈ +,−). (1)

The extreme radical is a characteristic ideal. From the formulas for the Bergmannoperators and the quasi-inverse in 6.9 and 6.11 it is easy to see that E+ can alsobe characterized by

z ∈ E+ ⇐⇒ zy = z and β(z, y) = Id for all y ∈ V −, (2)

and similarly for E−. We note:

if V has invertible elements then 2z = 0 for any z ∈ Eσ. (3)

Indeed, suppose u ∈ V + is invertible. By 6.14, e = (u, u−1) is an idempotent withV = V2(e), so that D(z, V −σ) = 0 implies 0 = ze−σeσ = 2z. The extreme radicalof a Jordan division pair vanishes since z 6= 0 implies Qz invertible in a divisionpair.

We now describe the normalizer of U± and the centre Z (G) of the projectiveelementary group G = PE(V ) in terms of the extreme radical.

8.7. Theorem. (a) Let G = PE(V ) be the projective elementary group of aJordan pair V . Let N be the intersection of the normalizers of U+ and U−. ThenN is given by

N = exp−(E−) ·H · exp+(E+), (1)

where E = (E+, E−) is the extreme radical. In particular, the normalizer of U−σ

in Uσ is expσ(Eσ).

(b) An element g of G belongs to Z (G) if and only if g = exp−(v)h exp+(z)where (z, v) ∈ Extr(V ) and h = (h+, h−) ∈ H is determined by v and z by meansof the formulas

h+(x) = x+Qxv, h−(y) = y +Qyz (x ∈ V +, y ∈ V −). (2)

The maps Φ: Z (G)→ E+×E− and Φ0: Z (G)→ H sending g = exp−(v)h exp+(z)to (z, v) and h, respectively, are group homomorphisms with Φ injective, while

Ker(Φ0) = exp−(v) exp+(z) : (z, v) ∈ E+ × E−, QV −z = QV +v = 0.

In particular,

Z (G) ∩H = 1 and Extr(V ) = 0 =⇒ Z (G) = 1. (3)

Proof. (a) We first show that N ⊂ Ω. Let g ∈ N and t ∈ V −. Since gnormalizes U−, there exists t′ ∈ V − such that g exp−(t) = exp−(t′)g. With respectto the decomposition g = V + ⊕ g0 ⊕ V −, the automorphism g is given by a formal3× 3-matrix of homomorphisms, say g = (aij). With this identification, we obtain

g exp−(t)(ζ) = g(ζ + t) = (a12ζ + a13t)⊕ · · ·= exp−(t′)g(ζ) = exp−(t′)(a12ζ ⊕ a22ζ ⊕ a32ζ) = a12ζ ⊕ · · · .


Hence a13 = 0. An analogous computation yields, for s ∈ V +,

g exp−(t)(s) = g(s⊕ [t, s]⊕Qts) = (a11s+ a12[t, s])⊕ · · ·= exp−(t′)g(s) = exp−(t′)(a11s⊕ · · ·) = a11s⊕ · · · .

Since (s, t) ∈ V is arbitrary, this implies a12(Inder(V )) = 0. It is clear that everyautomorphism of g leaves the derived algebra L′(V ) (cf. (7.1.6)) invariant. Forg−1 = (bij), the g0-component of g−1(s) is b21s which, by the remark just made,belongs to Inder(V ). It follows that a12b21 = 0. The relation gg−1 = 1 now yieldsIdV + =

∑i a1ibi1 = a11b11. By switching the roles of g and g−1 we also have

IdV + = b11a11. Hence, a11 is invertible with inverse b11. Similarly, one shows thatb33 is invertible with inverse a33 by using the fact that g normalizes U+. Lettingg = exp(x) for a suitable x ∈ V n we have a11 = B(x), b33 = B(−xop) = B(xop)in view of (8.4.1) and (8.4.5). Thus, x is quasi-invertible so (8.5.2) shows g ∈ Ω =U−HU+.

By (7.7.1), H normalizes Uσ and also expσ(Eσ) since the extreme radical isstable under all automorphisms of V . Hence it remains to show that an elementg = exp−(v) exp+(z) belongs to N if and only if (z, v) ∈ Extr(V ). Now for anyy ∈ V −,

g exp−(y) g−1 = exp−(v) exp+(z) exp−(y) exp+(−z) exp−(−v) ∈ U−

if and only if exp+(z) exp−(y) ∈ U− exp+(z), which, by Theorem 7.7(c), is equiv-alent to (z, y) being quasi-invertible, z = zy and β(z, y) = 1. By (8.6.2), this isequivalent to z ∈ E+. Similarly, g normalizes U+ if and only if v ∈ E−.

(b) By (a), any g ∈ Z (G) has the form g = exp−(v)h exp+(z) with (z, v) ∈Extr(V ). Furthermore, for all x ∈ V +,

g exp+(x) g−1 = exp−(v)h exp+(x)h−1 exp−(−v)

= exp−(v) exp+(h+(x)) exp−(−v) = exp+(x)

if and only if exp+(x) exp−(v) = exp−(v) exp+(h+(x)). By (7.7.2) and v ∈ E− thisis equivalent to h+(x) = xv = x − Qxv = x + Qxv, since 2Qxv = xvx = 0. Bysymmetry, z ∈ E+ and h−(y) = y+Qyz for all y ∈ V −. This proves that g belongsto the centre if and only if it has the stated form.

For the second part we first observe(((((((exp+(E+), exp−(E−)

)))))))= 1, (4)

and, for g = exp−(v)h exp+(z) ∈ Z (G),(((((((h, expσ(Eσ)

)))))))= 1. (5)

Indeed, (z, v) ∈ Extr(V ) is quasi-invertible by 8.6 and satisfies zv = z, vz = vand β(z, v) = Id. Hence by (7.7.2), exp+(z) exp−(v) = exp−(vz)β(z, v) exp+(zv) =exp−(v) exp+(z). For (5), one uses (7.7.1), (2) and (8.6.1). Now let

gi = exp−(vi)hi exp+(zi) ∈ Z (G)

for i = 1, 2. Then, by (4) and (5), the factors of

g1g2 = exp−(v1)h1 exp+(z1) · exp−(v2)h2 exp+(z2)

commute pairwise, so that

g1g2 = exp−(v1 + v2) · h1h2 · exp+(z1 + z2).

Hence the maps Φ and Φ0 are group homomorphisms. The injectivity of Φ and thedescription of KerΦ0 follows from 8.7. Now (3) is immediate.


Remarks. (i) Let g = exp−(v)h exp+(z) ∈ Z (G). In view of (2), the mapsh± − IdV ± are both linear and quadratic over k. Therefore, they are zero (andhence h = 1) provided V satisfies the condition

(λ− λ2)u = 0 for all λ ∈ k =⇒ u = 0, (6)

for any u ∈ V σ. In particular, this is the case if V has no 2-torsion or if there existsan element λ ∈ k× with 1 − λ ∈ k×, for example, when k is a field with at least3 elements. On the other hand, here is an example where h 6= 1. Let k = F2, letA = k[T ]/(T 3) = k ·1⊕k ·ε⊕k ·ε2, ε3 = 0, and let V = (A,A) be the special Jordanpair determined by A. It is easily seen that the extreme radical is Eσ = k · ε2. Thepair (ε, 1) ∈ V is quasi-invertible because B(ε, 1)x = x− 2εx+ ε2x = (1 + ε2)x and(1 + ε2)−1 = 1 + ε2 ∈ A×. Hence Id 6= h = (h+, h−) = β(1, ε) ∈ H where hσ ismultiplication with 1 + ε2 in A. We claim that g = exp−(ε2)h exp+(ε2) ∈ Z (G).By (2) we have to show that hσ(x) = x + x2ε2 for all x ∈ V σ = A. Now

x =∑2i=0 λiε

i where λi ∈ k, whence (hσ − Id)(x) = ε2x = λ0ε2. On the other

hand, x2 = λ01 + λ1ε2, and therefore Qxε

2 = x2ε2 = λ0ε2, as desired.

(ii) The map Φ is in general not an isomorphism, the converse of (3) is not true.An example will be given in 8.10.

8.8. Faulkner’s projective elementary group. J. Faulkner [28, Sect. 3]introduced a group which is closely related to the projective elementary groupPE(V ) as defined in 7.5. We describe here the precise relation between these twogroups.

Faulkner’s Tits-Kantor-Koecher algebra is LFau(V ) = V +⊕Inder(V )⊕V − withthe multiplication

[V σ, V σ] = 0, [D, z] = Dσ(z), [x, y] = δ(x, y),

for D = (D+, D−) ∈ Inder(V ), z ∈ V σ and (x, y) ∈ V . Observe that LFau(V )differs from our Tits-Kantor-Koecher-algebra L(V ) of 7.1 in two respects: the 0-part of L(V ) is enlarged by the degree derivation ζV , and for (x, y) ∈ V we have[x, y] = −δ(x, y) in g. One easily verifies that the map

f : LFau(V )→ DL(V ), x⊕D ⊕ y 7→ x⊕D ⊕ (−y)

is an isomorphism.Faulkner’s group, which we denote FPE(V ), is the subgroup of Aut(LFau(V ))

generated by automorphisms xF (V σ) as defined in [28]. For u ∈ V σ one checksthat

f xF (u) f−1 = expσ(−u)∣∣DL(V ). (1)

In the sequel, we will identify LFau(V ) and DL(V ) by means of f . Then, letting%: PE(V )→ PE(V )

∣∣DL(V ) be the restriction map, we have

FPE(V ) = %(PE(V )

) ∼= PE(V )/Ker(%). (2)


8.9. Proposition. Let Extr(V ) = (E+, E−) be the extreme radical of V anddefine ϕ: E+ × E− → PE(V ) by

ϕ(u, v) = exp−(v) exp+(u).

Then the sequence

0 // E+ × E−ϕ // PE(V )

% // FPE(V ) // 1

is exact. In particular, Ker(%) is abelian, and PE(V ) ∼= FPE(V ) if and only ifExtr(V ) = 0.

Proof. By (8.8.2), % is surjective. Using (7.7.2), (8.6.2) and Theorem 7.7(b) it iseasy to see that ϕ is an injective group homomorphism into Ker(%). Any g ∈ PE(V )is of the form exp(x) for a suitable x ∈ V n. If g ∈ Ker(%) we have B(x) = Id by(8.4.1). Similarly, g−1 ∈ Ker(%) implies B(xop) = Id. Hence, by (8.5.3), x is quasi-invertible and g ∈ Ω, so that g = exp−(v)h exp+(u) for (u, v) ∈ V and h ∈ H.Now for all x ∈ V +,

x = g(x) = exp−(v)h+(x) = h+(x)⊕ [v, h+(x)]⊕Qvh+(x)

whence h+ = Id and Qv = D(v, V +) = D(V +, v) = 0, i.e., v ∈ E−. Similarly,h− = Id and u ∈ E+, proving that ϕ maps E+ × E− onto Ker %.

8.10. Example. Let k be a commutative ring satisfying 2k = 0. By [35,I.5, Example (3)], the polynomial algebra k[T ], considered as a quadratic Jordanalgebra, contains the Jordan ideal I = k · T 2 +

∑n>4 k · Tn. This is not an ideal

of the polynomial algebra, and the quotient J = k[T ]/I is a non-special Jordanalgebra, free of rank 3 as a k-module, with basis 1J , a = can(T ), b = can(T 3) andthe multiplication rules

J, J, J = 0, U1J = Id, Ua1J = a2 = 0, Uaa = b, Uab = 0, Ub = 0. (1)

From 2k = 0 it follows that k2 = λ2 : λ ∈ k is a subring of k. We claim that

A := UJ = Ux : x ∈ J

is a k2-subalgebra of Endk(J), free of rank 2 over k2 with basis IdJ and Ua, andU2a = 0. Indeed, U2

a = 0 is immediate from (1). Since the map x 7→ Ux ishomogeneous of degree 2, it is clear that A is stable under k2. From J, J, J = 0it follows that the map x 7→ Ux is additive, so A is additively closed. By (1),

x = λ1J + µa+ νb ∈ J =⇒ Ux = λ2IdJ + µ2Ua. (2)

If also y = α1J + βa+ γa3 ∈ J then it follows from U2a = 0 and 2k = 0 that

UxUy = (λ2IdJ + µ2Ua)(α2IdJ + β2Ua) = λ2α2IdJ + (λ2β2 + µ2α2)Ua

= U((λα)1J + (λβ + µα)a

)∈ A.


Thus A is a k2-algebra. Finally, assume that sIdJ + tUa = 0 for s, t ∈ k2. Applyingthis to 1J yields s1J = 0 so s = 0, and then 0 = tUaa = tb implies t = 0 since Jhas the k-basis 1J , a, b. Altogether, we have shown that

A ∼= k2(ε),

the algebra of dual numbers over k2.Now consider the Jordan pair V = (J, J) with quadratic operators Qx = Ux for

all x ∈ V σ, cf. 6.13. Since V σ, V −σ, V σ = 0, an element x ∈ V σ belongs to theextreme radical Eσ if and only if Ux = 0. Let n = λ ∈ k : λ2 = 0. Then (2)shows that

Eσ = n1J + na+ kb. (3)

We claim thatFPE(V ) = E2(A), (4)

the elementary group over A ∼= k2(ε), see 6.1.

Proof. Since J, J, J = 0 we have Inder(V ) = 0 so that g = L(V ) = V +⊕kζ⊕V − and the derived algebra is Dg = V +⊕V −, an abelian characteristic ideal. Anyg ∈ Aut(g) stabilizes Dg. Moreover, we have gζ ≡ ζ mod Dg. Indeed, gζ = λζ +Xfor some λ ∈ k, X ∈ Dg. Since 2 = 0 in k, we have [ζ, Y ] = Y for all Y ∈ Dg.Hence gY = g[ζ, Y ] = [gζ, gY ] = [λζ +X, gY ] = λgY shows λ = 1, since Dg is freeof positive rank and g induces an automorphism of the k-module Dg.

We identify g ∈ Aut(g) with a formal 3 × 3-matrix (gij) with respect to thedecomposition g = V + ⊕ kζ ⊕ V −. We also identify kζ, which is free of rank onewith basis ζ, with k. Then g has the form

g =

g++ g+0 g+−0 1 0g−+ g−0 g−−

, (5)

and the matrix of %(g), the restriction of g to Dg with respect to the decompositionV + ⊕ V −, is

%(g) =

(g++ g+−g−+ g−−

).

Here g±0 is an element of V ± = J while gστ : V τ = J → V σ = J (for σ, τ ∈ +,−)is a k-linear map.

Let x ∈ V + and y ∈ V −. Then (7.4.2) and Inder(V ) = 0 shows

exp+(x) =

IdJ x Ux0 1 00 0 IdJ

, exp−(y) =

IdJ 0 00 1 0Uy y IdJ

. (6)

Hence the generators of G belong to M =

(A J A0 1 0A J A

)⊂ Endk(g). It is easily seen

that M is multiplicatively closed, so that G ⊂M . Also,

%(exp+(x)) =

(1 Ux0 1

), %(exp−(y)) =

(1 0Uy 1

).

§9] Groups over Jordan pairs 117

Since A = UJ , this shows that FPE(V ) = %(G) is the group generated by(

1 A0 1

)and

(1 0A 1

)and proves (4).

Now we show that G has trivial centre,

Z (G) = 1. (7)

Indeed, let g = exp−(v)h exp+(u) ∈ Z (G) as in Theorem 8.7, with (u, v) ∈ Extr(V )and h ∈ H given by (8.7.2). By (3), v = α1J + βa+ γb where α, β ∈ n and γ ∈ k.Applying h+ to 1J and to b yields

h+(1J) = 1J + U1J v = 1J + v = (1 + α)1J + βa+ γb, (8)

h+(b) = b+ Ubv = b. (9)

On the other hand, h+ ∈ A, say, h+ = Uc for some c = λ1J + µa + νb ∈ J , andhence Uc = λ2IdJ + µ2Ua by (2). Applying this again to 1J and to b yields

h+(1J) = Uc1J = λ21J , (10)

h+(b) = Ucb = λ2b+ µ2Uab = λ2b. (11)

Comparing (8) and (10) shows 1 + α = λ2, β = γ = 0, and comparing (9) and (11)yields λ2 = 1. Hence α = 0 as well, so we have v = 0. In the same way, one showsu = 0, and therefore g = 1, proving (7).

§9. Groups over Jordan pairs

9.1. Groups over a Jordan pair. Let R = −1, 0, 1 ⊂ Z be the root systemof type A1. Recall from Example 3.3(a) that a group with A1 -commutator relationsis a group G generated by two abelian subgroups U1 = U+ and U−1 = U−.

Let us fix a Jordan pair V = (V +, V −). We modify the notations of 7.5 for theprojective elementary group and its subgroups by defining

G = PE(V ), U± = exp±(V ±), G0 = PE0(V ). (1)

By 7.5, G is a group with A1-commutator relations and root groups U±. Nowspecialize Corollary 4.12 to the present situation, see also Example (a) of 4.15. Thecategory st(G) will simply be denoted by st(V ) and its objects will be called groupsover V . In more detail, an object of st(V ) is a quadruple (G,U+, U−, π) where G isa group, U± are subgroups generating G, and π: G→ G is a homomorphism withthe property that π

∣∣Uσ: Uσ → Uσ is an isomorphism, for σ ∈ +,−; in particular,Uσ is abelian. A morphism ϕ: (G,U+, U−, π) → (G′, U ′+, U ′−, π′) of st(V ) is agroup homomorphism ϕ: G→ G′ making the diagrams

U±ϕ //

π !!DDDDDDDD U ′±

π′||yyyyyyyy

U±

(2)


commutative.We define isomorphisms x±: V ±

∼=−→ U± by the commutative diagrams

V ±x±

∼=//

exp±

∼=

!!DDDDDDDD U±

π

∼=

zzzzzzzz

U±

(3)

Then an object of st(V ) can also be identified with a quadruple (G, x+, x−, π) con-sisting of a group G and homomorphisms x±: V ± → G and π: G→ G satisfying (3)and such that the Uσ := xσ(V σ) generate G, and a morphism ϕ: (G, x+, x−, π)→(G′, x′+, x

′−, π

′) is the same as a group homomorphism ϕ: G→ G′ making the dia-grams

V ±

x±

~~||||||||x′±

!!CCCCCCCC

Gϕ //

π BBBBBBBB G′

π′

G

(4)

commutative. Since G is generated by U±, ϕ is automatically surjective, andthere is at most one morphism between two objects of st(V ), so st(V ) is a pre-ordered category, see Corollary 4.12. Also, ϕ induces isomorphisms Uσ ∼= U ′σ forσ ∈ +,−. To simplify notation, we will often denote an object of st(V ) simplyby G, and also use the same letters x± and π for different groups G in st(V ).

9.2. Lemma. Let G ∈ st(V ) with root groups Uσ = xσ(V σ). Using the nota-tion (9.1.1) we define subgroups G0 and N of G by

G0 := π−1(G0), N := NormG(U+) ∩NormG(U−).

Let Ω = U− · G0 · U+ be the big cell of G = PE(V ) as in 7.5 and define

Ω := π−1(Ω).

(a) The map Φ: V − ×G0 × V + → G, (y, h, x) 7→ x−(y) · h · x+(x), is injective(equivalently, the map µ: U− ×G0 × U+ → G given by multiplication is injective)with image Ω.

(b) Let h ∈ G0 and let π(h) = h = (h+, h−) ∈ PE0(V ). Then

h ∈ N ⇐⇒ Int(h) · xσ(v) = xσ(hσ(v)) for all v ∈ V σ, σ = ±. (1)

(c) Ker(π) ⊂ G0 and N ∩Ker(π) is central in G.

(d) Let ϕ: G→ G′ be a morphism of groups over V , with subgroups Uσ, G0, N ⊂G and U ′σ, G′0, N

′ ⊂ G′ as above. Then

(i) ϕ: Uσ → U ′σ is an isomorphism, and ϕ−1(U ′σ) = Uσ · Ker(ϕ)= Ker(ϕ) · Uσ.

(ii) ϕ−1(G′0) = G0 and ϕ(G0) = G′0.

(iii) ϕ(N) ⊂ N ′.


(e) Let K ⊂ Ker(π) be a normal subgroup of G and let can: G→ G := G/K bethe canonical quotient map. Then G, equipped with the root groups U± := can(U±)and the induced map π: G→ G is an object of st(V ).

Proof. (a) If x−(y)hx+(x) = x−(y′)h′x+(x′) then by applying π and Theo-rem 7.7(b) we obtain y = y′ and x = x′ whence also h = h′. Clearly Φ has rangecontained in π−1(Ω). Conversely, if g ∈ π−1(Ω) and, say, π(g) = exp−(y)h exp+(x)then x−(−y)gx+(−x) ∈ π−1(PE0(V )) = G0 so that g is in the range of Φ.

(b) By (7.7.1) we always have π(h xσ(v)h−1) = h expσ(v)h−1 = expσ(hσ(v))= π(xσ(hσ(v)). The asserted equivalence then follows from injectivity of π on Uσ.

(c) Clearly Ker(π) = π−1(1) ⊂ π−1(G0) = G0. Let h ∈ N∩Ker(π) ⊂ N∩G0.Then π(h) = h = 1 and therefore (1) shows that h centralizes the subgroups U±.

(d) For (i), we have ϕ(Uσ) = U ′σ by (9.1.4). Since π∣∣Uσ = (π′ ϕ)

∣∣Uσ is an

isomorphism, it follows that ϕ∣∣Uσ is injective, and since G and G′ are generated

by their respective root groups, ϕ is surjective. Next, let g ∈ ϕ−1(U ′σ), say,ϕ(g) = x′σ(v). Then ϕ(gxσ(−v)) = x′σ(v)x′σ(−v) = 1G′ , so g ∈ Ker(ϕ) · Uσ. Thusϕ−1(U ′σ) ⊂ Ker(ϕ) · Uσ, and the reverse inclusion is clear. The second formula isproved similarly.

(ii) We have

ϕ−1(G′0) = ϕ−1(π′−1(PE0(V ))) = (π′ ϕ)−1(PE0(V )) = π−1(PE0(V )) = G0.

For the second statement, let h ∈ G0. Then π′(ϕ(h)) = π(h) ∈ G0 whenceϕ(h) ∈ G′0. Let h′ ∈ G′0. Since ϕ is surjective, we have h′ = ϕ(g) for someg ∈ G. Then π(g) = π′(ϕ(g)) = π′(h′) ∈ G0, whence g ∈ G0.

Finally, (iii) as well as (e) are immediate from the definitions.

9.3. Example: Elementary groups of special Jordan pairs. Let M =(R,M+,M−, S) be a Morita context, let V ⊂ (M+,M−) be a special Jordan pairand let G = E(M, V ) be the elementary group of (M, V ) as in 6.2. By 7.6(a), wehave a surjective homomorphism π: G→ G = PE(V ) with kernel the diagonal part

of the centre of G, satisfying π(

1 x0 1

)= exp+(x) and π

(1 0−y 1

)= exp−(y), for all

x ∈ V +, y ∈ V −. Thus G is a group over V with subgroups U− =(

1 V +

0 1

)and

U− =(

1 0V − 1

), and homomorphisms

x+(x) =

(1 x0 1

), x−(y) =

(1 0−y 1

). (1)

By [58, Theorem 2.8], π−1(PE0(V )) = G0 is given by

G0 = g ∈ G : ge1 = e1g = G ∩(R× 00 S×

).


9.4. The lattice of groups over V . As remarked in 9.1, st(V ) is a preorderedcategory: there is at most one morphism ϕ: G → G′ between any two groupsG,G′ ∈ st(V ), and this morphism is necessarily surjective. We indicate this byG G′ or G′ ≺ G. If also ψ: G′ → G is a morphism then necessarily ϕ ψ = IdG′

and ψ ϕ = IdG because the only morphism from G to G is the identity, so G andG′ are isomorphic. Hence a skeleton K of st(V ) is a partially ordered set. It maybe realized as follows.

The free product Fr(V ) = V + ∗ V − of the additive groups V + and V − belongsto st(V ), with x± the canonical injections of V ± into Fr(V ), and this is in fact theSteinberg group of G = PE(V ), as remarked in 4.15(a). For any G ∈ st(V ) we havea unique surjective homomorphism

ηG: Fr(V )→ G

whose kernel K(G) is the set of relations defining the group G. The assignmentG 7→ K(G) induces a natural bijection between K and the set of normal subgroupsof Fr(V ) contained in K := K(G). Also, it is clear that G G′ if and only ifK(G) ⊂ K(G′) for the corresponding normal subgroups; in particular, K

(Fr(V )

)=

1. Hence K is a complete modular lattice, anti-isomorphic to the lattice ofnormal subgroups of Fr(V ) contained in K.

More generally, if G ∈ st(V ) then G G′ if and only if G′ ∼= G/K whereK = Ker(ϕ) ⊂ Ker(π). Hence the isomorphism classes of G′ ≺ G are in bijectionwith the normal subgroups of G contained in Ker(π).

It should be noted that the categorical setup employed here is not indispensableand merely serves as a convenient language. Certain concepts are more easilyexpressed in the category st(V ) than in the set of quotients of Fr(V ) lying overFr(V )/K. If the reader so desires she may of course adopt the latter point of view.

9.5. Lifting Jordan pair homomorphisms. Let V and V ′ be Jordan pairs,and let G and G′ be groups over V and V ′, respectively. Consider a homomorphismf : V → V ′ of Jordan pairs. We say a group homomorphism ϕf : G → G′ is a liftof f if the diagrams

Gϕf // G′

V ±

x±

OO

f±

// V ′±

x′±

OO

(1)

commute.This condition determines ϕf uniquely (and thus justifies the notation ϕf )

because G is generated by U+ and U−. If f ′: V ′ → V ′′ is a second Jordan pairhomomorphism with lift ϕf ′ : G

′ → G′′ then ϕf ′ ϕf is a lift of f ′ f . The lifts ofthe identity Id: V → V are just the morphisms of the category st(V ).

By the universal property of the free product, it is clear that every Jordan pairhomomorphisms f : V → V ′ lifts to a homomorphisms Fr(f): Fr(V )→ Fr(V ′).

Suppose G ∈ st(V ). An automorphism a of V is said to normalize G if thereexists an automorphism ϕa of G which lifts a. One sees easily that in this case,ϕ−1a is a lift of a−1.

Let A = Aut(V ) be the automorphism group of V and denote by NormA(G)the set of a ∈ A normalizing G. One shows easily that NormA(G) is a subgroup of


A, and the map ϕ: NormA(G)→ Aut(G), a 7→ ϕa, is an injective group homomor-phism.

With the notations of 9.2 let h ∈ N ∩G0. By (9.2.1),

a = π(h) ∈ NormA(G) ∩ G0, ϕa = Inth, (2)

the inner automorphism of G determined by h, is a lift of a. Thus the diagram

G0 ∩NInt //

π$$IIIIIIIII Aut(G)

G0 ∩NormA(G)

ϕ

::uuuuuuuuu(3)

is commutative.

9.6. Groups induced by subpairs. Let G be a group over V and let V ′ ⊂ Vbe a subpair of V . Define U ′σ = xσ(V ′σ) ⊂ Uσ and let

G′ =⟨U ′+ ∪ U ′−

⟩⊂ G.

Then x′σ := xσ∣∣V ′σ: V ′σ → U ′σ is an isomorphism. We claim that, with the fol-

lowing definition of the projection π′: G′ → PE(V ′), the quadruple (G′, x′+, x′−, π

′)is a group over V ′, and that the inclusion G′ ⊂ G is a lift (in the sense of 9.5) ofthe inclusion V ′ ⊂ V .

Let PE(V ′) be the projective elementary group of V ′ with exponential mapsexp′±, and let G′ ⊂ G = PE(V ) be the subgroup generated by exp+(V ′+) ∪exp−(V ′−) as in 8.1. From (9.1.3), it follows that π(G′) = G′. By (8.1.1) wehave a surjective homomorphism ψ: G′ → PE(V ′) satisfying (8.1.2). Now defineπ′: G′ → PE(V ′) by π′ = ψ (π

∣∣G′). The diagram below summarizes the variousdefinitions.

G′

π′

wwoooooooooooo

π|G′

inc // G

π

PE(V ′) G′

ψooinc

// G = PE(V )

To show that (G′, x′+, x′−, π

′) ∈ st(V ′), we need to verify (9.1.3). By the definitionof π′, we must show that the diagram

V ′σx′σ //

exp′σ

U ′σ

π′

uulllllllllllllll

π

exp′σ(V ′σ) expσ(V ′σ)

ψoo

is commutative and that π′ is an isomorphism. But this follows readily from (8.1.2).Finally, the fact that the inclusion G′ ⊂ G is a lift of the inclusion V ′ ⊂ V followsimmediately from the definition of x′σ.


9.7. The elements b(x, y). Let G be a group over V . For a quasi-invertiblepair (x, y) ∈ V we define the element b(x, y) ∈ G by the formula

x+(x) · x−(y) = x−(yx) · b(x, y) · x+(xy), (1)

equivalently, (((((((x−(−y), x+(x)

)))))))= x−(yx − y) · b(x, y) · x+(xy − x). (2)

Then (7.7.2) shows that

π(b(x, y)) = β(x, y); hence b(x, y) ∈ G0 = π−1(G0). (3)

One also sees immediately from the definition that

b(x, 0) = b(0, y) = 1. (4)

If there is no risk of confusion, we will use the same letter b for different groupsover V . If f : V → V ′ is a homomorphism of Jordan pairs and ϕf : G→ G′ is a liftof f as in 9.5 then

ϕf (b(x, y)) = b(f+(x), f−(y)). (5)

Indeed, if (x, y) is quasi-invertible then (f+(x), f−(y)) is quasi-invertible in V ′ andf+(xy) = f+(x)f−(y) as well as f−(yx) = f−(y)f+(x), by (6.11.3). Hence

ϕf(x+(x) · x−(y)

)= x+(f+(x)) · x−(f−(y))

= x−((f−(y))f+(x)

)· b(f+(x), f−(y)) · x+

((f+(x))f−(y)

)= x−(f−(yx)) · b(f+(x), f−(y)) · x+(f+(xy))

and also

ϕf(x+(x) · x−(y)

)= ϕf

(x−(yx) · b(x, y) · x+(xy)

)= x−(f−(yx)) · ϕf

(b(x, y)

)· x+(f+(xy)),

so that (5) follows by comparison. In particular, this applies to f = a = π(h) whereh ∈ N ∩G0, and then (9.5.2) yields the formula

h · b(x, y) · h−1 = b(a+(x), a−(y)) (a = π(h), h ∈ N ∩G0). (6)

Example. Let G = E(M, V ) as in 9.3 and let (x, y) ∈ V be quasi-invertible.Then a simple computation, using 6.10, shows

b(x, y) = x−(−yx) x+(x) x−(y) x+(−xy) =

(1 0yx 1

)(1− xy x−y 1

)(1 −xy0 1

)=

(1− xy x

0 1 + yxx

)(1 −xy0 1

)=

(1− xy 0

0 1 + yxx

)=

(1− xy 0

0 (1− yx)−1

). (7)


9.8. The relations B(x, y). Let G be a group over V and let (x, y) ∈ V be aquasi-invertible pair. We say G satisfies the relations B(x, y) if the formulas

b(x, y) · x+(z) · b(x, y)−1 = x+(B(x, y)z), (1)

b(x, y)−1 · x−(v) · b(x, y) = x−(B(y, x)v) (2)

hold for all (z, v) ∈ V + × V −. Since B(x, y)z = z − xyz+QxQyz, an equivalentformulation is (((((((

b(x, y), x+(z))))))))

= x+

(− xyz+QxQyz

), (3)(((((((

b(x, y)−1, x−(v))))))))

= x−(− yxv+QyQxv

), (4)

for all (z, v) ∈ V . From (9.7.3) and (9.2.1) we see that

G satisfies B(x, y) ⇐⇒ b(x, y) ∈ N. (5)

In particular, it follows from (7.7.2) and (7.7.3) that G satisfies B(x, y) for all quasi-invertible pairs (x, y). The same is true for elementary groups of special Jordanpairs:

E(M, V ) satisfies the relations B(x, y) for all quasi-invertible (x, y) ∈ V . (6)

Indeed, this follows by a simple matrix computation from (9.7.7).

Suppose G′ is a group over a Jordan pair V ′ as well and f : V → V ′ is a surjectivehomomorphism of Jordan pairs which lifts to a group homomorphism ϕf : G→ G′

as in 9.5. Then by (9.7.5) and (9.5.1),

if G satisfies B(x, y) then G′ satisfies B(f+(x), f−(y)). (7)

In particular, if G satisfies B(x, y) and a = (a+, a−) ∈ NormA(G) then G alsosatisfies B(a+(x), a−(y)).

9.9. Lemma. Let (x, y) and (u, v) be quasi-invertible and let (s, t) ∈ V withthe property that (s+x, y) and (x, y+ t) are quasi-invertible. If G satisfies B(x, y)then the following formulas hold:

b(s+ x, y) = b(s, yx) · b(x, y), (1)

b(x, y + t) = b(x, y) · b(xy, t), (2)

b(x, y)−1 = b(−x, yx) = b(xy, −y), (3)

b(x, y) = b(Qxy − x, −yx

)= b

(− xy, Qyx− y

), (4)

b(x, y) · b(u, v) · b(x, y)−1 = b(B(x, y)u, B(y, x)−1v

). (5)

Proof. By (6.11.4) applied to V op, (s, yx) is quasi-invertible and the formulas

(s+ x)y = xy +B(x, y)−1(s(yx)

), y(x+s) = (yx)s (6)

hold. From (9.7.1) we therefore obtain


x+(s+ x) · x−(y) = x−(ys+x) · b(s+ x, y) · x+((s+ x)y)

= x−(ys+x) · b(s+ x, y) · x+

(xy +B(x, y)−1s(yx)

).

On the other hand, (9.8.1) shows

x+(s+ x) · x−(y) = x+(s) · x−(yx) · b(x, y) · x+(xy)

= x−((yx)s) · b(s, yx) · x+(s(yx)) · b(x, y) · x+(xy)

= x−(yx+s) · b(s, yx) · b(x, y) · x+

(B((x, y)−1s(yx) + xy

)so that (1) follows by comparison. Formula (2) follows similarly from (9.8.2).

We obtain (3) by setting s = −x and t = −y in (1) and (2) and using (9.7.4).To prove (4) first note that (3) shows G satisfies the relations B(−x, yx) and

B(xy,−y) since b(x, y)−1 ∈ N . Hence by (3),

b(x, y) =(b(x, y)−1

)−1= b(−x, yx)−1

= b((−x)(yx),−yx

)= b(Qxy − x,−yx),

because (6) yields, for s = −x, the Hua-type relation

(−x)(yx) = −B(x, y)xy = −(x−Qxy). (7)

The second formula of (4) is proved similarly. Finally, replace (x, y) in (9.7.6) by(u, v) and put h = b(x, y) and a = β(x, y). Then (5) follows from (9.7.6) and(9.8.5).

9.10. Groups over V op. Let V op = (V −, V +) be the opposite of a Jordanpair V and let G be a group over V . Then G can and will be considered as a groupover V op by switching the roles of U+ and U−, i.e., by setting

xopσ (v) = x−σ(v) (v ∈ V −σ, σ ∈ +,−); (1)

more precisely,

(G, x+, x−, π)op = (G, xop+ , xop

− , πop) = (G, x−, x+, π)

in the notation of 9.1. This is in line with the identification of PE(V ) and PE(V op)in 7.5. The assignment (G, x+, x−, π) 7→ (G, x−, x+, π) from st(V ) to st(V op) isthen an isomorphism of categories.

Let (x, y) ∈ V be quasi-invertible. By the symmetry principle (6.11.1), this isequivalent to (y, x) being quasi-invertible in V op. Invert (9.7.1), use the fact that(−IdV + ,−IdV −) is an automorphism of V and replace (x, y) by (−x,−y). Theresult is

x−(y) · x+(x) = x+(xy) · b(−x,−y)−1 · x−(yx) (2)

which, when read in (G, x+, x−, π)op, and with the obvious definition of bop inanalogy to (9.7.1), says

bop(y, x) = b(−x,−y)−1. (3)

With the aim of achieving greater symmetry in formulas, we will often use thenotation


b+(x, y) = b(x, y), b−(y, x) = b(−x,−y)−1. (4)

Let (x, y) ∈ V σ × V −σ be quasi-invertible. Then (3) implies

bσ(x, y)−1 = b−σ(−y,−x), (5)

and (9.7.1) and (2) can be subsumed under the single formula

xσ(x) · x−σ(y) = x−σ(yx) · bσ(x, y) · xσ(xy), (6)

or the equivalent commutator formula(((((((x−σ(−y), xσ(x)

)))))))= x−σ(yx − y) · bσ(x, y) · xσ(xy − x). (7)

In analogy to 9.8 we say G satisfies the relations Bσ(x, y) if bσ(x, y) normalizesU+ and U−, thus

B+(x, y) ⇐⇒ B(x, y), and B−(x, y) ⇐⇒ B(−y,−x). (8)

Explicitly, G satisfies Bσ(x, y) if the formulas

bσ(x, y) · xσ(z) · bσ(x, y)−1 = xσ(B(x, y)z)

bσ(x, y)−1 · x−σ(v) · bσ(x, y) = x−σ(B(y, x)v)

(9)

or the commutator formulas(((((((bσ(x, y), xσ(z)

)))))))= xσ

(− xyz+QxQyz

)(((((((bσ(x, y)−1, x−σ(v)

)))))))= x−σ

(− yxv+QyQxv

) (10)

hold for all (z, v) ∈ V σ × V −σ.

9.11. The elements te and we. Let e = (e+, e−) ∈ V be an idempotent andconsider a group G over V . In analogy to 7.8, we introduce the notations

te =(x−(e−), x+(e+), x−(e−)

)∈ U− × U+ × U−, (1)

we = x−(e−) · x+(e+) · x−(e−) ∈ G. (2)

Again as in 7.8, we put

teop =(x+(e+), x−(e−), x+(e+)

)∈ U+ × U− × U+, (3)

weop = x+(e+) · x−(e−) · x+(e+) ∈ G. (4)

Since −e = (−e+,−e−) is an idempotent as well, we have the following formula forthe inverse:

w−1e = w−e. (5)

Clearly, π(te) = θe and π(we) = ωe are the elements defined in 7.8. However,even when V has invertible elements and hence θe is a Weyl triple for G consideredas a group with A1-commutator relations (see 7.11), this is in general no longer thecase for G. We now discuss this question in more detail.


9.12. Proposition. Let V be a Jordan pair and let G be a group over V ,considered as a group with A1-commutator relations and root subgroups U±1 = U±.

(a) For a triple

x =(x−(v), x+(u), x−(v′)

)∈ U− × U+ × U−

(where u ∈ V +, v, v′ ∈ V −) the following conditions are equivalent:

(i) x is a Weyl triple for the root α = 1 (5.1),

(ii) u is invertible in V +, so e = (e+, e−) = (u, u−1) is an idempotentof V . Furthermore, putting z := v− e− ∈ V −, z′ := v′− e− ∈ V −,n := x−(z) and n′ := x−(z′), and with we as in (9.11.2), we have

µ(x) = nwe n′, (1)

we · U− · w−1e = n−1 · U+ · n, w−1

e · U− · we = n′ · U+ · (n′)−1. (2)

If these conditions hold then z and z′ belong to the extreme radical and satisfy2z = 2z′ = 0.

(b) Conversely, if G = PE(V ) then any triple(exp−(z + u−1), exp+(u), exp+(u−1 + z′)

),

where u ∈ V + is invertible and z, z′ ∈ Extr(V −), is a Weyl triple for α = 1.

Proof. (a) (i) =⇒ (ii): Let w = µ(x) = x−(v)x+(u)x−(v′) be the Weyl elementdetermined by x. Since π is surjective on root groups,

ω := π(w) = exp−(v) exp+(u) exp−(v′) (3)

is a Weyl element for α = 1 in PE(V ). Hence

ω · exp−(y) · ω−1 = exp+

(f(y)

)(4)

for all y ∈ V −, where f : V − → V + is an isomorphism of additive groups. Equation(4) is equivalent to

exp+(u) · exp−(y) · exp+(u) · exp−(−v) = exp−(−v) · exp+(f(y)). (5)

By applying both sides of (5) to the element ζ ∈ L0(V ) of the Tits-Kantor-Koecheralgebra and comparing the terms in V − we obtain, by a lengthy but straightforwardcomputation using (JP4), the formula

f(y) = Qu(y −Qy(u−Quv)

). (6)

Since f is surjective, this shows that Qu: V − → V + is surjective. In particular,there exists b ∈ V − such that u = Qub. But Qu is injective as well: indeed, Quy = 0and (6) imply

f(y) = −QuQyu+QuQyQuv = QuQyQu(v − b) = Q(Quy)(v − b) = 0,


and therefore y = 0, because f is a group isomorphism. Thus Qu is invertible.It follows that u is an invertible element with inverse u−1 = Q(u−1)u ∈ V −. Bydefinition, v = z + e− and v′ = e− + z′, so that w = n ·we · n′. Now the conditionswU−w−1 = U+ and wU+w−1 = U− are easily seen to be equivalent with (2). Thisalso shows that (ii) implies (i).

It remains to prove the last statement. By 7.11, π(we) = ωe is a Weyl elementof PE(V ), so ωe · U− · ω−1

e = U+ = ω−1e · U− · ωe. Applying π to (2) shows that

n = exp−(z) and n′ = exp−(z′) normalize U+ = exp+(V +). Thus z, z′ ∈ Extr(V −)by 8.7(b). Finally, D(z, V +) = 0 implies 2z = ze+e− = 0 and in the same way2z′ = 0.

(b) This follows easily from the fact that ωe is a Weyl element in PE(V ) by7.11, and that exp+(Extr(V +)) normalizes U−, by Theorem 8.7(a).

9.13. Corollary. The following conditions on a Jordan pair V are equivalent.

(i) V is a Jordan division pair,

(ii) PE(V ) is a rank one group.

In this case, PE(V ) is a special rank one group: all elements of Uσ = Uσ 1satisfy the equivalent conditions of Corollary 5.23(b).

Proof. We write G = PE(V ) and U± instead of U± for simpler notation.

(i) =⇒ (ii): As observed in 8.6, a Jordan division pair has trivial extremeradical. Now Proposition 9.12 shows that the set T1 of Weyl triples for α = 1is precisely the set of all

(exp−(u−1), exp+(u), exp−(u−1)

)where 0 6= u ∈ V +.

Hence pr2: T1 → U+ = U+ 1 is bijective. By passing to V op, one sees thatpr2: T−1 → U− is bijective as well, so G is a rank one group by Proposition 5.22.Since the Weyl triples θe are balanced by 7.11, G is special by definition.

(ii) =⇒ (i): By Proposition 5.22, pr2: T1 → U+ is bijective. Hence, every1 6= x ∈ U+ is the second component of a Weyl triple. By Proposition 9.12 thisshows that every element 0 6= v ∈ V − has the form v = z+u−1 where z ∈ Extr(V −)and u ∈ V + is invertible. From the definition of the extreme radical in 8.6 we havez, V +, V − = 0, whence Q(z, V −) = 0. Hence Qv = Qz +Qz,u−1 +Qu−1 = Qu−1

is invertible. Also, U+ 6= 1 is part of the definition of a rank one group. HenceV ± 6= 0, so V is a Jordan division pair.

If G is the elementary group of a special Jordan pair, we have the followingmore precise description of the Weyl elements and Weyl triples generalizing 5.2.

9.14. Proposition. Let V be a special Jordan pair, embedded in a Morita con-text M = (R,M+,M−, S), and let G = E(M, V ) be the corresponding elementarygroup as in 9.3. We consider G as a group with A1-commutator relations and rootgroups U±1 = U±. Then G has Weyl elements for α = 1 if and only if V hasinvertible elements. In this case, the Weyl triples and Weyl elements are given by

t+(u) =(x−(u−1), x+(u), x−(u−1)

), (1)

wu = µ(t+(u)) =

(1R − uu−1 u−u−1 1S − u−1u

), (2)


where x± is defined in (9.3.1) and u ∈ V + is invertible in V with inverse u−1 ∈ V −.In particular, every Weyl triple is balanced and the multiplication map µ: T1 →W1

is bijective.

We emphasize that u−1 in (1) and (2) is the inverse in the Jordan pair V . It isin general not true that uu−1 = 1R or u−1u = 1S . For example, let R′ 6= 0 be anarbitrary ring. Then V is also a special Jordan pair embedded in the Morita contextM′ = (R ⊕ R′,M+,M−, S) or M′′ = (R,M+,M−, S ⊕ R′) in which R′ acts by 0on the original M. Even if uu−1 = 1R it will not be the case that uu−1 = 1R⊕R′ ,but see Corollary 9.15 for a situation where these equations hold.

Proof. Let u ∈ V + and v, v′ ∈ V − and consider the element

w = w(v, u, v′) = x−(v)x+(u)x−(v′) =

(1− uv′ u

vuv′ − v − v′ 1− vu

)(3)

of G. A matrix computation shows that

w

(1 0−y 1

)w−1 =

(1− uy(1− uv) uyu

−(1− vu)y(1− uv) 1 + (1− vu)yu

), (4)

for all y ∈ V −.First suppose that u is invertible in V with inverse u−1 = v = v′ and abbreviate

wu := w(u−1, u, u−1). Then u−1 = u−1uu−1 = vuv′ so (3) shows that wu has theform claimed in (2). Also, since Qu−1 : V + → V − is bijective, we have y = Qu−1x =u−1xu−1 for a unique x ∈ V +. Hence

(1− vu)y = y − u−1u(u−1xu−1) = y − (u−1uu−1)xu−1 = y − u−1xu−1 = 0, (5)

and similarlyy(1− uv) = 0 (6)

for all y ∈ V −, so (4) yields wu

(1 0−y 1

)w−1u =

(1 uyu0 1

)whence wuU

−w−1u = U+.

By a similar computation as before we have wu

(1 x0 1

)w−1u =

(1 0−vxv 1

). This

proves that wu is indeed a Weyl element and hence t+(u) is a Weyl triple forα = 1. It is balanced since one easily sees by direct computation that also wu =x+(u)x−(u−1)x+(u).

Conversely, suppose that w as in (3) is a Weyl element for α = 1. Then inparticular wU−w−1 = U+. Hence (4) shows that Qu: V − → V +, y 7→ uyu, isbijective (i.e., u is invertible in V ) and, for all y ∈ V −,

(1− vu)yu = 0 = uy(1− uv), (7)

(1− vu)y(1− uv) = 0. (8)

From (7) we obtain, for y = v, that uv = uvuv and hence Quv = uvu = uvuvu =QuQvu. By injectivity of Qu this implies v = Qvu. From (8) and the fact thatB(v, u)y = (1 − vy)y(1 − uv) (cf. (6.10.1)) we conclude with (6.13.2) that 0 =B(v, u) = Q(v − u−1)Qu and therefore Q(v − u−1) = 0. This implies 0 = Q(v −u−1)u = Qvu − vuu−1 + Q(u−1)u = v − 2v + u−1 whence v = u−1. It remains


to show that v′ = v. Let z = v′ − v and n =(

1 z0 1

)= w−1

u · w. Since both w and

wu are Weyl elements for α = 1, it follows from Proposition 5.4 that n ∈W−11 ·W1

normalizes U+. Now (3) implies

n ·(

1 u0 1

)· n−1 = w(z, u,−z) =

(1 + uz u−zuz 1− zu

),

and this belongs to U+ if and only if uz = zu = zuz = 0. Hence uzu = Quz = 0and therefore z = 0 because Qu is invertible.

Under suitable hypotheses on the Morita context the form of the Weyl elementwu in (2) simplifies as follows. For the notion of an associative pair see [52, §6].

9.15. Corollary. With the assumptions and notations of Proposition 9.14, letM = (R,M+,M−, S) so M = (M+,M−) is an associative pair. If M is generatedas an associative pair by V , and R and S act faithfully on M+ and M− in the sensethat r ∈ R, s ∈ S, and rM+ = M−r = 0 and M+s = sM− = 0 imply r = s = 0,then wu is given by

wu =

(0 u−u−1 0

). (1)

The conditions on faithfulness hold in particular if M is the standard embedding ofM in the sense of [58, 2.3].

Proof. Let u ∈ V + be invertible with inverse u−1 ∈ V −, and put r = 1R−uu−1

and s = 1S − u−1u. From (9.14.5) and (9.14.6) we see that sV − = V −r = 0, andsimilarly one shows rV + = V +s = 0. Since M is generated by V as an associativepair, this implies rM+ = M+s = sM− = M−r = 0, and thus r = s = 0, byour faithfulness assumption. In case of the standard embedding, faithfulness isimmediate from the definition in [58, 2.3].

9.16. Lemma. Let e ∈ V be an idempotent with Peirce spaces V σi = V σi (e).Let G be a group over V and define Uσi = xσ

(V σi (e)

), for i = 0, 1, 2 and σ ∈ +,−.

Consider the following conditions:

we xσ(z2) w−1e = x−σ

(Qe−σz2

)(z ∈ V σ2 , σ ∈ +,−), (1)

weop xσ(z2) (weop)−1 = x−σ(Qe−σz2

)(z ∈ V σ2 , σ ∈ +,−), (2)

we Uσ2 w−1

e = U−σ2 (σ ∈ +,−), (3)

weop Uσ2 (weop)−1 = U−σ2 (σ ∈ +,−), (4)

we = weop . (5)

Then (1) – (4) are all equivalent and imply (5).

Proof. In the presence of (5), it is clear that (1)⇐⇒ (2). Hence the equivalenceof (1) and (2) will follow once we have shown (1) =⇒ (5) ⇐= (2). By puttingz2 = eσ in (1) and using Qe−σeσ = e−σ we see

we xσ(eσ) w−1e = x−σ(e−σ).

This implies we weop w−1e = we and therefore (5), as required. A similar argument

shows (2) =⇒ (5).


Evidently, (1) implies (3). Conversely, if (3) holds and z2 ∈ V σ2 (e) thenwe xσ(z2) w−1

e = x−σ(v2) for some v2 ∈ V −σ2 (e). Applying π to this relation andcomparing with (7.10.1) and (7.10.2) yields v2 = Q(e−σ)z2, so we have (1). In thesame way, one proves the equivalence of (2) and (4).

9.17. The Weyl relations. Let G be a group over V and let e ∈ V be anidempotent. We say G satisfies the Weyl relations W(e) if the equivalent conditions(9.16.1) – (9.16.4) of Lemma 9.16 hold. Clearly, by that lemma and (9.11.5), therelations W(e), W(−e) and W(eop) are equivalent and imply we = weop .

By Proposition 7.10, the projective elementary group satisfies these relations.Also, the elementary group G = E(M, V ) of a special Jordan pair satisfies the Weylrelations. Indeed, since we = w(e−, e+, e−) as in (9.14.3) one obtains

we =

(1R − e+e− e+

−e− 1S − e−e+

).

Now let y2 ∈ V −2 (e), so y2 = Qe−Qe+y2 = e−e+y2e+e−. Since e+e− is an idempo-tent in A, this implies

y2(1− e+e−) = e−e+y2e+e−(1− e+e−) = 0.

We also have (1− e−e+)y2 = (1− e−e+)e−e+y2e+e− = 0. Now (9.14.4) shows that(9.16.1) holds for σ = −, and the case σ = + follows similarly.

On the other hand, the Weyl relations do not hold in all groups over V . Forexample, the free product Fr(V ) = V + ∗V − of the additive groups V + and V − is agroup over V , namely the Steinberg group of G = PE(V ), cf. 4.15(a). The relation(9.16.5) is not satisfied in Fr(V ).

For an automorphism h ∈ NormA(G) we have:

if G satisfies W(e) then G also satisfies W(h(e)). (1)

Indeed, let e′ = h(e). Then for u ∈ V σ2 (e′),

we′xσ(u)w−1e′ = ϕh(wexσ(h−1

σ (u))w−1e ) = ϕh

(x−σ(Q(e−σ)h−1

σ (u)))

= x−σ(Q(e′−σ)u

),

i.e., W(h(e)) holds in G.

9.18. The Steinberg groups Stn(A), n > 3. Let A be an associative unitalk-algebra, let p > 1, q > 1 and n = p + q > 3, and consider the Morita contextM = (Matpp(A),Matpq(A),Matqp(A),Matqq(A)) of matrices of size p × p, p × q,q × p and q × q over A as in 6.1. Then

V = (Matpq(A),Matqp(A)) = Mpq(A)

is the Jordan pair of p× q matrices over A as in 6.6(a). We have already observedin 6.2 that the group E(M, V ) is the usual elementary group En(A) in the sense of[32, 1.2C], see also Example 3.16(c).


Recall [74, 32] that the Steinberg group Stn(A) is the group presented bygenerators xij(a) (a ∈ A, i 6= j, i, j ∈ 1, . . . , n) and relations

xij(a)xij(b) = xij(a+ b), (1)(((((((xij(a), xjl(b)

)))))))= xil(ab) for i 6= l, (2)(((((((

xij(a), xkl(b))))))))

= 1 for j 6= k, i 6= l, (3)

where a, b ∈ A. To see that Stn(A) is indeed a group over V in the sense of 9.1,define xσ: V σ → Stn(A) by

x+(u) =∏

16i6p16j6q

xi,p+j(uij), x−(v) =∏

16i6p16j6q

xp+j,i(−vji), (4)

for u = (uij) ∈ V + und v = (vji) ∈ V −. From the defining relations of Stn(A) itfollows easily that the order of the factors in (4) is immaterial, and that xσ is infact a homomorphism of the additive group V σ into Stn(R).

It is well known [32, 1.4C] that there is a homomorphism ψ: Stn(A)→ En(A) =E(M, V ) satisfying ψ(xij(a)) = 1n+aEij . Combining this with the map π: E(M, V )→ PE(V ) of 9.3, we have a homomorphism π = π ψ: Stn(A)→ PE(V ) satisfyingπ xσ = expσ. Hence Stn(A) is a group over V . Moreover, Stn(A) is alreadygenerated by x+(V +)∪ x−(V −). Indeed, the generators of Stn(A) not contained inthis set are the xij(a) where i, j belong to 1, . . . , p or to p+ 1, . . . , p+ q. In thefirst case we have, by (2),

xij(a) =(((((((

xin(a), xnj(1))))))))

=(((((((

x+(aEiq), x−(−Eqj)))))))).

The missing generators of the second type are recovered similarly.

Notes

§6. Our basic reference for Jordan pairs is [52]. For the closely related Jordan triple systems

(6.6(g)) see [51, 73, 76], and for quadratic Jordan algebras [35, 37, 80]. For the theory ofJordan algebras over rings containing 1/2 (linear Jordan algebras) we refer to [34, 71, 112]. Theconnection with the theory of bounded symmetric domains is elaborated in [53], and with algebraic

groups and homogeneous algebraic varieties in [55, 54]. Structural transformations (6.12) wereintroduced in [56].

§7. As the name suggests, the construction of the Tits-Kantor-Koecher algebra L(V ) of a

Jordan pair V goes back to the almost simultaneous work of Tits, Kantor and Koecher [97, 98,40, 41, 42, 45], albeit in the setting of Jordan algebras. The extension to Jordan triple systemsand Jordan pairs is straightforward. Various versions of L(V ) have been used. Apart from the

minus sign in (7.1.2), they differ in the definition of L0(V ). An incomplete list of papers fordifferent TKK-constructions is [3, 10, 21, 28, 43, 73, 79]. The TKK-construction can also be

done for Jordan superpairs and superalgebras [31, 39, 67].

Our definition of the projective elementary group PE(V ) in 7.5 is taken from [58]. It differsslightly from Faulkner’s [28], see 8.8 and 8.9 for a comparison. Theorem 7.7 is based on [58,Theorem 1.4].

§8. Generalized Bergmann operators and quasi-inverses of quadruples were introduced in[58], and later generalized to arbitrary tuples by Faulkner [30] with a recursive definition. The

equivalence of Faulkner’s definition with the one of 8.4 follows from [30, Lemma 2]. The formulasin 8.4, such as (8.4.3), can be found in [58]. The fact that generalized Bergmann operators giverise to structural transformations (8.4.4) is proved in [58, Lemma 4.3] for quadruples and in [30,

Corollary 4(a)] in the general case. The formulas (8.4.5) can also easily be deduced from Faulkner’s


recursive definition of the generalized Bergmann operators. The equivalence (8.5.2) is shown in[58, Theorem 5.1] for quadruples and in [30, Theorem 1] in general.

§9. The elementary group of a special Jordan pair (9.3) is studied in more detail in [58, §2].

For example, the equality Ω = U−G0U+, established in 9.2(a) in general, is also shown in [58,Theorem 2.8] for G = E(M, V ). The elements b(x, y), the relations B(x, y) as well as the Weyl

relations W(e) appeared first in [59, Definition 1.4] and were used there to define a Steinberg

group St(V ) for an arbitrary Jordan pair V .

The theory of Steinberg groups Stn(A) over rings goes back to [94, 74], and is described in

several other sources, e.g. in [32].

CHAPTER III

STEINBERG GROUPS FOR PEIRCE GRADED JORDAN PAIRS

Summary. The chapter begins with an auxiliary section §10 on Peirce gradings of Jordan

pairs. A Peirce grading P of a Jordan pair V is a decomposition V = V2⊕V1⊕V0 into the directsum of subpairs satisfying the same composition rules as the Peirce spaces of an idempotent. These

gradings are the simplest non-trivial examples of the more general root gradings investigated in

depth in Chapter V, and are an indispensable tool for the study of the latter.Given a group G over V and a Peirce grading P of V , we define in §11 subgroups Uα of

G indexed by the roots α of the classical root system C2 and introduce the full subcategoryst(V,P) of st(V ) consisting of those groups which have C2-commutator relations with respect to

the Uα. In the fundamental Theorem 11.2 we give several characterizations of these groups. As a

consequence, the projective elementary group of a Peirce-graded Jordan pair has C2-commutatorrelations in a natural way. We also show in 11.11 that st(V,P) has an initial object, called the

Steinberg group of P. Its relation with the Steinberg groups defined in §4 will be elucidated in

§21.In the next §12 we consider Peirce gradings defined by an idempotent e of V , denoted P(e)

and called idempotent Peirce gradings. The idempotent e yields a natural candidate we for a

Weyl element in any group G ∈ st(V,P(e)). Accordingly, we define a full subcategory st(V, e) ofst(V,P(e)) consisting of those groups for which this is the case, and characterize its objects in

Theorem 12.5.

The last section §13 deals with Steinberg categories and groups for Jordan pairs endowed with

a set of idempotents instead of a single idempotent.

§10. Peirce gradings

10.1. Peirce gradings. In this section we let V = (V +, V −) be a Jordan pairover a unital associative and commutative ring k, see §6. A Z-grading of V [57] isa family (Vi)i∈Z of pairs Vi = (V +

i , V−i ) of k- submodules such that V σ =

⊕i∈Z V

σi

(σ ∈ +,−) (direct sum of k-submodules) and the following multiplication ruleshold:

V σi V −σj V σl ⊂ V σi−j+l, Q(V σi )V −σj ⊂ V σ2i−j . (1)

The convention for numbering the V −i differs from that of [57] by a sign. Ahomomorphism of Z-graded Jordan pairs is a Jordan pair homomorphism h: V →V ′ satisfying h(Vi) ⊂ V ′i . From (1) it follows immediately that the Vi = (V +

i , V−i )

are subpairs of V .A Peirce grading of V is a Z-grading with Vi = 0 for i /∈ 0, 1, 2 and the

additional orthogonality relations

D(V σ2 , V−σ0 ) = D(V σ0 , V

−σ2 ) = 0. (2)

We usually write P: V = V2 ⊕ V1 ⊕ V0 or simply V = V2 ⊕ V1 ⊕ V0 to specify aPeirce grading of V . If V = (Vi)i∈Z is a Z-grading with Vi = 0 for i /∈ 0, 1, 2,then formula (1) implies V σi V

−σj V σl = 0 if i− j + l /∈ 0, 1, 2. Therefore, (2) is

equivalent toV σ2 V −σ0 V σ0 = V σ0 V −σ2 V σ2 = 0.

133

134 STEINBERG GROUPS FOR PEIRCE GRADED JORDAN PAIRS [Ch. III

The following properties are immediate from the definition.

V σ0 and V σ2 are inner ideals, (3)

V ∗i = V2−i defines another Peirce grading, (4)

B(V σi , V−σj ) = Id for |i− j| = 2. (5)

We call the Peirce grading of (4) the reverse of P.

The automorphism group Aut(V ) acts on the set of Peirce gradings of V in theobvious way: if f ∈ Aut(V ) and P : V = V2⊕ V1⊕ V0 is a Peirce grading, then thePeirce grading f(P) : V = V2 ⊕ V1 ⊕ V0 is given by Vi = f(Vi).

10.2. Examples. (a) For any Jordan pair V and a fixed i ∈ 0, 1, 2 there isalways the trivial Peirce grading Vi = V and Vj = 0 for j 6= i. A Peirce gradingwith V1 = 0 is the same as a direct sum decomposition V = V0 ⊕ V2 of V intoideals. If PU : U = U2 ⊕U1 ⊕U0 and PW : W = W2 ⊕W1 ⊕W0 are Peirce gradedJordan pairs then the direct sum V = U ⊕W has a Peirce grading PU ⊕PW givenby Vi = Ui ⊕Wi.

(b) The main examples of Peirce gradings are the ones defined by an idempotente of V . Indeed, from 6.14 it is clear that the Peirce spaces Vi = Vi(e) define a Peircegrading of V , denoted by P(e). A Peirce grading is called idempotent if it is of theform P(e) for some idempotent e.

(c) However, there are important examples of Peirce gradings which are not in-duced by an idempotent, for instance the decomposition of the Jordan pair Mp,q(A)of p× q and q × p matrices over a ring A as in 6.6(a), given by the following blockdecomposition:

V + =

s︷︸︸︷ q−s︷︸︸︷r

V +

2 V +1

p−r

V +

1 V +0

, V − =

r︷︸︸︷ p−r︷︸︸︷s

V −2 V −1

q−s

V −1 V −0

. (1)

If r = s then this Peirce grading is idempotent, where e+ can be any unit ofMatr(A) with inverse e−. Conversely, assume it comes from an idempotent, saye = (e+, e−) ∈ Matr,s(A) ×Mats,r(A). Then e+ is invertible in Matr,s(A) in theJordan pair sense, so by the example in 6.13, e+: As → Ar is an isomorphism ofright A-modules. If A is a ring with invariant basis number, this implies r = s, butnot in general.

(d) If P: V = V2 ⊕ V1 ⊕ V0 is a Peirce grading and e ∈ V2 is an invertibleidempotent of V2 then P 6= P(e) in general. Indeed, let U be a Jordan pair withan idempotent e 6= 0 inducing the Peirce grading P(e) of U and let W be anarbitrary non-trivial Jordan pair with the trivial Peirce grading PW : W = W1.The Peirce grading PU ⊕PW of the Jordan pair V = U ⊕W as defined above hasV1 = U1(e)⊕W ' V1(e) = U1(e), yet e is invertible in V2 = V2(e) = U2(e).

§10] Peirce gradings 135

10.3. Lemma. Let V = V2 ⊕ V1 ⊕ V0 be a Peirce grading.

(a) If e ∈ V is an invertible idempotent with e ∈ V1 then

QeVj = V2−j for j ∈ 0, 1, 2. (1)

(b) Assume V0 = V2(e0) and V2 = V2(e2) for idempotents e0, e2 ∈ V , andV1 = V1(e0) ∩ V1(e2). Then

V1 = e0 V1 e2 = V0 V1 V2. (2)

Proof. (a) By invertibility of e, the operator Q(eσ)Q(e−σ) is the identity onV σ. Since Q(eσ)V −σj ⊂ V σ2−j by (10.1.1), the map Q(eσ) induces isomorphisms

V −σj∼= V σ2−j , whence (1).

(b) By 6.16 and our assumptions, e = e0 + e2 is an idempotent of V withV = V2(e). Thus e is an invertible idempotent. By (10.1.1) we have Q(eσ)V −σ0 =Q(eσ0 )V −σ = V σ0 , Q(eσ)V −σ2 = Q(eσ2 )V −σ2 = V σ2 and Q(eσ)V −σ1 = eσ0V −σ1 eσ2 ⊂V σ1 . Since Q(eσ): V −σ → V σ is invertible, this implies (2)

In 10.7 we will establish some multiplication rules for Jordan pairs with a Peircegrading. These formulas hold in fact in a more general setting, which we reviewnow.

10.4. Kernels and annihilators. We recall from [62] and [52, 10.3] thedefinition of the kernel KerS and annihilator AnnS of a subset S ⊂ V σ:

KerS = v ∈ V −σ : QSv = QSQvS = 0,AnnS = v ∈ V −σ : D(v, S) = D(S, v) = QvS = QSv

= QvQS = QSQv = 0,

where QSv = 0 means of course Qxv = 0 for all x ∈ S. Clearly AnnS ⊂ KerS. IfS = x consists of a single element, we simply write Kerx and Annx. Note thesymmetry in the definition of the annihilator:

v ∈ Annx ⇐⇒ x ∈ Ann v. (1)

For example, if P is a Peirce grading of V then it follows easily from the multipli-cation rules that

V −σ2−i ⊕ V−σ1 ⊂ KerV σi , V −σ2−i ⊂ AnnV σi (2)

for i ∈ 0, 2. If P = P(e) is idempotent then, by [52, 10.3], V −σ0 = AnnV σ2 =Ann eσ.

10.5. Proposition. (a) Let (x, v) ∈ V σ × V −σ and v ∈ Kerx, i.e., Qxv =QxQvx = 0. Then (x, v) is quasi-invertible with quasi-inverses xv = x, vx =v + Qvx. For all y ∈ V −σ we have yxv ∈ Kerx and v ∈ KerQxy, and thefollowing “shift formulas” hold:


D(Qxy, v) = D(x, yxv), D(v,Qxy) = D(yxv, x), (1)

Q(Qxy)Qv = QxQyxv, QvQ(Qxy) = QyxvQx, (2)

B(Qxy, v) = B(x, yxv), B(v,Qxy) = B(yxv, x). (3)

(b) Let (x, v) ∈ V σ × V −σ and v ∈ Annx. Then (x, v) is quasi-invertible withquasi-inverses xv = x and vx = v, and B(x, v) and B(v, x) is the identity. For all(z, y) ∈ V σ × V −σ we have yzv ∈ Kerx and xyz ∈ Ker v, and the followingshift formulas hold:

D(x, yzv) = D(xyz, v), D(yzv, x) = D(v, xyz), (4)

QxQyzv = QxQyQzQv = QxyzQv, (5)

QvQxyz = QvQzQyQx = QyzvQx, (6)

B(x, yzv) = B(xyz, v), B(yzv, x) = B(v, xyz), (7)

Qxyzv = QxQyQzv. (8)

Proof. (a) We have Bx,vx = x−2Qxv+QxQvx = x, so (x, v) is quasi-invertibleby (iv) of 6.11 with quasi-inverse xv = x. By the symmetry principle (6.11.1) andformula (6.11.2), we have (v, x) quasi-invertible, with vx = v +Qvx

v = v +Qvx.From the fundamental formula (JP3) it follows easily that v ∈ KerQxy. Before

showing yxv ∈ Kerx, we establish the shift formulas (1)–(3). By (JP8) we haveD(Qxy, v) = −D(Qxv, y) + D(x, yxv) = D(x, yxv), proving the first formulaof (1), and the second one follows similarly from (JP7). Furthermore, the identities(JP20), (JP3), (JP13) and (JP2) yield

QxQyxv = QxQyQxQv +QxQvQxQy +QxQy,vQxQy,v −QxQ(Qvx,Qyx)

= Q(Qxy)Qv +Q(Qxv)Qy +Q(Qxy,Qxv)Qy,v −QxQ(Qvx,Qyx)

= Q(Qxy)Qv −Qx[D(x,Qvx)D(x,Qyx)−D(QxQvx,Qyx)

]= Q(Qxy)Qv −QxD(Qxv, v)D(x,Qyx) = Q(Qxy)Qv.

This establishes the first formula of (2). The second formula is proved by a similarcomputation. Finally, (3) is immediate from (1) and (2) and the definition of theB-operators.

We can now show yxv ∈ Kerx. Indeed, Qxyxv = D(x, y)Qxv = 0 by (JP1),and QxQyxv = QxQy

(QxQvx

)= 0 by (2).

(b) From the definition of the annihilator it is clear that B(x, v) and B(v, x) arethe identity and that xv = x and vx = v. As before, we first prove the shift formulas.Since D(x, v) = 0, (JP15) yields 0 =

[Dz,y, Dx,v

]= D(zyx, v)−D(x, yzv), and

similarly one shows the second formula of (4). Next, by (JP20),

QxQyzv = QxQyQzQv +QxQvQzQy +QxQy,vQzQy,v −QxQ(Qyz,Qvz).

The second term on the right vanishes by definition of the annihilator. Furthermore,QxQy,v = D(x, v)D(x, y) − D(Qxv, y) = 0 by (JP13), and again by (JP13) and(JP7), for any t ∈ V −σ,

QxQ(t, Qvz) = D(x,Qvz)D(x, t)−D(QxQvz, t) = D(x,Qvz)D(x, t)

=(−D(z,Qvx) +D(xvz, v)

)D(x, t) = 0.


This establishes the first formula of (5). The second one is proved similarly, and(6) follows from annihilator symmetry (10.4.1). As before, (7) is a consequenceof the definition of the Bergmann operators. For (8) we obtain from (JP20) thatQxQyQzv = Qxyzv + Qxy, v,Qzy. But, by (JP8), D(Qxy, v) = −D(Qv, y) +D(x, vxy) = 0. Finally we show yzv ∈ Kerx which, again by annihilatorsymmetry, also establishes xyz ∈ Ker v by switching the roles of x, z and v, y.We have Qxyzv = −D(z, y)Qxv + xvzyx = 0 by (JP12), and QxQyzvx =QxQyQzQvx = 0 by (5).

10.6. Corollary. If (x, v) ∈ V + ×Kerx then for all y ∈ V −,

β(Qxy, v) = β(x, yxv). (1)

If (x, v) ∈ V + ×Annx then for all (z, y) ∈ V + × V −,

β(x, yzv) = β(xyz, v). (2)

Proof. (1) follows from the definition of an inner automorphism in 6.12 and(10.5.3), while (2) follows from (10.5.7).

10.7. Corollary. Let V = V2 ⊕ V1 ⊕ V0 be a Peirce grading of V and letsubscripts indicate membership in the corresponding Peirce space.

(a) If i 6= j then (xi, yj) is quasi-invertible with quasi-inverses

xyji = xi +Q(xi)yj , yxij = yj +Q(yj)xi, (1)

where either Q(xi)yj = 0 or Q(yj)xi = 0. In particular, if i, j = 0, 2 thenQ(xi)yj = 0 = Q(yj)xi and hence

xyji = xi and yxij = yj (|i− j| = 2), (2)

while for i = 1 6= j we obtain

xyj1 = xi +Qx1yj and yx1

j = yj (i = 1 6= j). (3)

Moreover,β(xi, yj)

−1 = β(xi,−yj). (4)

(b) For i ∈ 0, 2 we have the formulas

D(y, xi)D(v1, xi) = Q(y, v1)Q(xi), (5)

D(xi, v1)D(xi, z) = Q(xi)Q(v1, z), (6)

D(Qxiy, v1) = D(xi, yxiv1), D(v1, Qxiy) = D(v1xiy, xi), (7)

D(xi, yzv2−i) = D(xiyz, v2−i), (8)

D(v2−izy, xi) = D(v2−i, zyxi), (9)

Q(Qxiy)Qv1 = QxiQyxiv1, Qv1Q(Qxiy) = Qv1xiyQxi , (10)

QxiQyzv2−i = QxiQyQzQv2−i = QxiyzQv2−i , (11)

Qv2−iQzyxi = Qv2−iQzQyQxi = Qv2−iyzQxi , (12)


Qxiyzv2−i = QxiQyQzv2−i (13)

B(Qxiy, v1) = B(xi, yxiv1), B(v1, Qxiy) = B(v1xiy, xi), (14)

B(xi, yzv2−i) = B(xiyz, v2−i), (15)

B(v2−iyz, xi) = B(v2−i, zyxi), (16)

β(xi, v2−i) = Id, (17)

β(Q(xi)y, v1) = β(xi, yxiv1), (18)

β(xiyz, v2−i) = β(xi, yzv2−i). (19)

Proof. (a) For i ∈ 0, 2 we have yj ∈ Kerxi by (10.4.2), so that

xyji = xi and yxij = yj +Qyjxi (20)

follows from 10.5. In particular, this proves (2). For i = 1 we switch the roles of xiand yj : we have x1 ∈ Ker yj which implies (3). The formula (1) is a consequenceof (2) and (3). For the proof of (4) we distinguish the same cases: if i ∈ 0, 2then β(xi, yj)

−1 = β(xyji ,−yj) = β(xi,−yj) by (6.12.3) and (20). For i = 1 we

use the second formula of (6.12.3) and (3) to obtain β(x1, yj)−1 = β(−x1, y

x1j ) =

β(−x1, yj) = β(x1,−yj).(b) Since Q(xi)v1 = 0, (5) follows from (JP9). Similarly, (6) follows from

(JP13) while (17) is immediate from v2−i ∈ Annxi. The remaining formulas areall special cases of 10.5 since, by (10.4.2), V −2−i ⊕ V −1 ⊂ KerV +

i ⊂ Kerxi and

V −2−i ⊂ AnnV +i ⊂ Annxi.

10.8. Corollary. Let P = P(e) be the Peirce grading determined by an idem-potent e of V . Then in addition to the formulas of 10.7(b), we have the followingrelations, where j ∈ 0, 1 and again subscripts indicate membership in the corre-sponding Peirce space:

D(eσ, y2) = D(Qeσy2, e−σ), D(x2, e−σ) = D(eσ, Qe−σx2), (1)

D(xj+1, yj) = D(eσ, e−σxj+1yj), D(uj , vj+1) = D(ujvj+1eσ, e−σ), (2)

Qxj+1Qyj = QeσQe−σxj+1yj, QujQvj+1

= Qujvj+1eσQe−σ , (3)

B(xj+1, yj) = B(eσ, e−σxj+1yj), B(uj , vj+1) = B(ujvj+1eσ, e−σ), (4)

β(xj+1, yj) = β(e+, e−xj+1yj), β(uj , vj+1) = β(ujvj+1e+, e−). (5)

Proof. For (1) we use the identity (JP8) and get

D(Qeσy2, e−σ) = −D(Qeσe−σ, y2) +D(eσ, e−σeσy2)= −D(eσ, y2) + 2D(eσ, y2) = D(eσ, y2).

The second formula can be proved similarly using (JP7). The remaining formulasnow all follow from 10.7(b). We will prove the first formula in (2) and leave theproof of the rest, which follows a similar pattern, as an exercise. For j = 1 we have,using (10.7.7) and (1), D(x2, y1) = D(QeσQe−σx2, y1) = D(eσ, Qe−σx2, eσ, y1) =D(eσ, e−σ, x2, y1. For j = 0 we use (10.7.8) and obtain that D(x1, y0) =D(eσe−σx1, y0) = D(eσ, e−σx1y0).

In the following lemma we use the abbreviations

ViVjVk =(V +

i V−j V

+k , V

−i V

+j V

−k )

and QViVj =(Q(V +

i )V −j , Q(V −i )V +j

).


10.9. Lemma. Let P: V = V2 ⊕ V1 ⊕ V0 be a Peirce grading of V .

(a) The ideal of V generated by V2 ⊕ V1 is

I(V2 ⊕ V1) = V2 ⊕ V1 ⊕(V1V1V0+QV1V2 +QV0QV1V2

). (1)

(b) If P = P(e) is an idempotent Peirce grading then the ideals generated byV2, V1 ⊕ V0 and V1 are

I(V2) = V2 ⊕ V1 ⊕QV1V2, (2)

I(V1 ⊕ V0) =(V2 V1 V1+QV1

V0 +QV2QV1

V0

)⊕ V1 ⊕ V0 (3)

=(e V1 V1+QV1

V0 +QeQV1V0

)⊕ V1 ⊕ V0, (4)

I(V1) =(V2 V1 V1+QV1

V0 +QV2QV1

V0

)⊕ V1 ⊕ QV1

V2 (5)

=(e V1 V1+QV1

V0 +QeQV1V0

)⊕ V1 ⊕ QV1

V2. (6)

In particular, if 2 ∈ k× then

I(V1) =(V2 V1 V1+ V1 V0 V1

)⊕ V1 ⊕ V1 V2 V1. (7)

(c) Let again P = P(e) the Peirce grading defined by an idempotent e, so thatVi = Vi(e), and suppose that either V is simple and V1 6= 0, or there exists anidempotent which is collinear to e or which governs e. Then V2 = V2 V1 V1 +QV1V0 +QV2QV1V0.

Proof. (a) Let I denote the right hand side of (1). Since obviously I ⊂I(V2 ⊕ V1), it remains to show that I is an ideal, i.e.,

(i) V V I ⊂ I, (ii) Q(V )I ⊂ I, (iii) Q(I)V ⊂ I.

Let Ii := I ∩Vi be the Peirce components of I. Obviously only the 0-component in(i) – (iii) is of interest.

(i) It suffices to check that Vi Vj Il ⊂ I0 for i− j+ l = 0. Because of (10.1.2),this leads to the condition Vi Vi+1 V1 ⊂ I0, which holds by definition of I0, andto Vi Vi I0 ⊂ I0. Concerning this last condition, we have Vi Vi V1 V1 V0 ⊂V1 V1 V0 by (JP15) and Vi Vi Vj ⊂ Vj . Moreover, Vi, Vi, QV1

V2 ⊂ QV1V2 by

(JP12). A second application of (JP12) then shows

Vi, Vi, QV0QV1V2 ⊂ QV0Vi, Vi, QV1V2+ Vi Vi V0, QV1V2, V0⊂ QV0QV1V2.

(ii) It is immediate that Vi Ij Vl ⊂ I0 for i − j + l = 0, i 6= l. Hence it isenough to prove QViIj ⊂ I0 for 2i = j. The case QV1

V2 ⊂ I0 holds by definition,so only QV0

I0 ⊂ I0 has to be checked, and this follows from (10.7.5) and (10.7.13):

QV0V1 V1 V0 = QV0

Q(V1, V0)V1 ⊂ D(V0, V1)D(V0, V0)V1 ⊂ V0 V1 V1,QV0

QV0QV1

V2 = Q(V0 V0 V1)V2 ⊂ QV1V2.


(iii) In view of (i) it suffices to verify QIiVj ⊂ I0 for 2i = j, and since QV1V2 ⊂ I0by definition, only the case i = 0 = j is of interest. Again by (i) it is sufficient tocheck the following cases:

Q(QV1V2)V0 ⊂ QV1QV2QV1V0 ⊂ QV1V2,

Q(QV0QV1V2)V0 ⊂ QV0QV1QV2V ⊂ QV0QV1V2,

which follow from the fundamental formula (JP3), and

Q(V0 V1 V1)V0 ⊂ Q(QV0

QV1V1, V1

)V0 +QV1

QV1QV0

V0

+QV0QV1QV1V0 + V0, V1, QV1V1 V0 V0⊂ 0 +QV1V2 +QV0QV1V2 + V0 V1 V1

which follows from (JP21) and QV0V1 = 0.

(b) Now assume Vi = Vi(e). Then eσe−σx1 = x1 for all x1 ∈ V σ1 , whence theideals generated by V2 and V2 ⊕ V1 agree. Also,

V1 V1 V0 = V0 V1 V1 = V0 V1 e e V1 ⊂ QV1V2 (by (10.8.2)),

QV0QV1V2 = Q(V0 V1 e)QeV2 ⊂ QV1V2 (by (10.8.3))

which proves (2). For the proof of (3) we apply (1) to the reverse Peirce gradingV ∗i = V2−i(e) and obtain that(

V1V1V2+QV1V0 +QV2QV1V0

)⊕ V1 ⊕ V0

is an ideal, whence the ideal generated by V1⊕V0. However, V1 V1 V2 = e V1 V1by (10.8.2) and QV2

QV1V0 = QeQ(e V2 V1)V0 ⊂ QeQV1

V0 by (10.8.3). This proves(4). The ideal (5) is the intersection of the ideals (2) and (3). Formula (6) followsin the same way.

Finally, suppose 2 ∈ k×. Then QV1V0 = V1V0V1 and

QV2QV1V0 = QV2QV1,V1V0 ⊂ V2V1V2V1V0( by (10.7.6)) ⊂ V2V1V1

by the Peirce rules. This proves (7).

(c) LetI2 = V2 V1 V1+QV1

V0 +QV2QV1

V0.

We claim that I2 = V2 under the assumptions of (c). If V is simple and V1 6= 0 thenthis follows from (5). Now assume that f is an idempotent collinear to e. Recallfrom 6.15 that V has a simultaneous Peirce decomposition with respect to e and fgiven by

V =⊕

i,j∈0,1,2

V(ij)

where V(ij) = Vi(f) ∩ Vj(e). In particular,

V2(e) = V(22) ⊕ V(12) ⊕ V(02). (8)


Since Q(f) vanishes on V1(f)⊕ V0(f) the Peirce multiplication rules show

QfV2 = QfV(22) ⊂ V(20), QfV1 = QfV(21) ⊂ V(21), QfV0 = QfV(20) ⊂ V(22).

Because V(2i) ⊂ QfV for i ∈ 0, 1, 2, it follows that

V(22) = QfV(20) ⊂ QV1V0. (9)

An analogous argument for the action of Qe on V2(e) yields

V(02) = Qe · V(22) = QeQfV(20) ⊂ QV2QV1V0. (10)

Finally, every x ∈ V σ(12) satisfies

x = x f−σ fσ ∈ V σ2 V −σ1 V σ1 (11)

by (6.14.7). Now (8) – (11) show that V2 = QV1V0 +QV2QV1V0 + V2 V2 V1 = I2.

Finally, assume that g is an idempotent governing e. Then V2 = Q(e)V2 =Q(e)Q(g)Q(g)V2 ⊂ Q(V2)Q(V1)V0 shows again that I2 = V2.

10.10. Definition. Let P be a Peirce grading of V . The group of P-element-ary automorphisms of V is the subgroup

EA(V,P) =⟨β(V +

i , V−j ) : i 6= j

⟩of the inner automorphism group Inn(V ). If P = P(e) is an idempotent Peircegrading, we write EA(V, e) := EA(V,P(e)). It follows from (10.8.5) and (10.7.17)that

EA(V, e) =⟨β(e+, V

−1 ) ∪ β(V +

1 , e−)⟩, (1)

which shows that EA(V, e) is generated by two subgroups: because of (6.12.2),(6.12.3) and (10.7.3) the maps

x+: V +1 → EA(V, e) : u1 7→ β(−u1, e−),

x−: V −1 → EA(V, e) : v1 7→ β(e+, v1)

are group homomorphisms.

The next results give examples of P(e)-elementary automorphisms. Recall thedefinition of the element ωe ∈ PE(V ) in 7.8.

10.11. Lemma. Let e and f be idempotents with f ∈ V1(e), and put

ωf,e := ωe ωf ω−1e ∈ PE(V ). (1)

Then ωf,e ∈ EA(V, e) ∩ EA(V, f), in fact,

ωf,e = β(e+, f−)β(−f+, e−)β(e+, f−) = β(−f+, e−)β(e+, f−)β(−f+, e−), (2)

ω−1f,e = ω−f,e = ωf,−e. (3)

Proof. This follows easily from (7.10.3), (7.10.4) and (7.10.6).

The assumption of Lemma 10.11 naturally occurs for idempotents which arecollinear or where one governs the other, as defined in 6.15. In both cases, theidempotents e and f are compatible. Thus V admits a joint Peirce decomposi-tion ((6.15.3)): V =

⊕i,j∈0,1,2 V(ij) for V(ij) = Vi(f) ∩ Vj(e). In the next two

propositions we identify the action of ωf,e on the joint Peirce spaces V(ij).


10.12. Proposition. Let e and f be collinear idempotents in V , thus f ∈ V1(e)and e ∈ V1(f). Let ωf,e = ωe ωf ω

−1e ∈ EA(V, e) as in (10.11.1). Then

ωf,e = ωe,−f (1)

and the action of ωf,e on x(ij) ∈ V σ(ij) = V σi (f) ∩ V σj (e) is given as follows:

ωf,e · x(22) = x(22),

ωf,e · x(11) = x(11) − fσ e−σ eσ f−σ x(11) ∈ V σ(11),

ωf,e · x(00) = x(00),

ωf,e · x(12) = fσ e−σ x(12) ∈ V σ(21),

ωf,e · x(01) = fσ e−σ x(01) ∈ V σ(10),

ωf,e · x(21) = −eσ f−σ x(21) ∈ V σ(12),

ωf,e · x(10) = −eσ f−σ x(10) ∈ V σ(01),

ωf,e · x(02) = QfσQe−σx(02) ∈ V σ(20),

ωf,e · x(20) = QeσQf−σx(20) ∈ V σ(02).

Hence, ωf,e maps V σ(ii) bijectively onto itself and induces isomorphisms V σ(ij)∼= V σ(ji)

for i 6= j. Moreover,

ωf,e(e) = f and ωf,e(f) = −e, (2)

ω2f,e

∣∣V σ(ij) = (−1)i+j · Id, (3)

ω4f,e = Id, (4)

x(ij) = fσ e−σ eσ f−σ x(ij) for (ij) = (21) or (10). (5)

Proof. To prove (1) we use the first formula in (10.11.2) to calculate ωe,−f =β(−f+, e−)β(−e+,−f−)β(−f+,−e−). This equals ωf,e by the second formula in(10.11.2) and (6.9.1).

We prove the formulas describing the action of ωf,e by using (7.9.1) repeatedly.Also recall ω−1

e = ω−e by (7.8.1). If i = j = 0, it is clear from (7.9.1) thatωf,e ·x = x. If i = j = 2, thus x ∈ V σ2 (f)∩V σ2 (e), then ω−1

e ·x = ω−e ·x = Qe−σx ∈V −σ0 (f) ∩ V −σ2 (e). This implies ωfω

−1e · x = ω−1

e · x and therefore ωf,e · x = x.Before doing the remaining cases, we observe

ωe · fσ = [e−σ, fσ] = −[fσ, e−σ] = −ωf · e−σ, (6)

since f ∈ V1(e) and e ∈ V1(f). Now let x ∈ V σ1 (f) ∩ V σ1 (e). Then, since ωf is anautomorphism of L(V ),

ωfω−1e · x = ωf · [−e−σ, x] = [−ωf · e−σ, ωf · x] = [ωe · fσ, ωf · x] (by (6)),

and therefore, since ω2e = se is the Peirce reflection with respect to e by Lemma 7.9,

ωf,e · x = [ω2e · fσ, ωeωf · x] = −[fσ, ωeωf · x].

We compute the second term in the bracket by using the Jacobi identity and thedefinition of the Tits-Kantor-Koecher algebra, in particular, (7.1.3) and (7.1.4), aswell as the fact that e ∈ V1(f):


ωeωf · x = ωe[f−σ, x] = [ωe · f−σ, ωe · x] = [[eσ, f−σ], [e−σ, x]]

= [[[eσ, f−σ], e−σ], x] + [e−σ, [[eσ, f−σ], x]]

= [e−σeσf−σ, x]− [e−σ, eσf−σx] = [f−σ, x]− [e−σ, eσf−σx].

This implies finally

ωf,e · x = −[fσ, [f−σ, x]] + [fσ, [e−σ, eσf−σx]]= fσf−σx − fσe−σeσf−σx = x− fσe−σeσf−σx.

Next, let x ∈ V σ(12). Then ω−1e · x = Qe−σx ∈ V −σ(12), hence ωfω

−1e · x =

[fσ, Qe−σx] = [fσ, ω−1e · x] and therefore

ωf,e · x = [ωe · fσ, ωeω−1e · x] = [[e−σ, fσ], x] = fσe−σx.

For x ∈ V σ(01) we have ω−1e · x = −[e−σ, x] and ωfω

−1e · x = −[ωf · e−σ, ωf · x] =

[ωe · fσ, x] (by (6) and since x ∈ V σ0 (f)), which implies

ωf,e · x = [ω2e · fσ, ωe · x] = [−fσ, [e−σ, x]] = fσe−σx.

Now let x ∈ V σ(21). Then ω−1e · x = −[e−σ, x] whence

ωfω−1e · x = [−ωf · e−σ, ωf · x] = [ωe · fσ, Qf−σx] (by (6)).

This implies

ωf,e · x = [ω2e · fσ, ωe ·Qf−σx] = [−fσ, [eσ, Qf−σx]] = −eσ, Qf−σx, fσ

= −eσ, f−σ, fσ, f−σ, x+ eσ, Qf−σfσ, x (by (JP9))

= −eσ, f−σ, 2x+ eσ, f−σ, x = −eσ, f−σ, x.

The next case is x ∈ V σ(10). Then ω−1e · x = x, ωfω

−1e · x = [f−σ, x] and therefore

ωf,e · x = [ωe · f−σ, ωe · x] = [[eσ, f−σ], x] = −eσ, f−σ, x.

The remaining cases are x ∈ V σ(02) and x ∈ V σ(20). In the first of these, ω−1e · x =

ω−e · x = Qe−σx ∈ V −σ(22), hence ωfω−1e · x = QfσQe−σx ∈ V σ(20) which implies

ωf,e · x = QfσQe−σx. In the second one, we have ω−1e · x = x, ωfω

−1e · x = ωf · x =

Qf−σx ∈ V −σ(22) and therefore ωf,e · x = QeσQf−σx.

Since e ∈ V(12) we get ωf,e(eσ) = fσ e−σ eσ = fσ, and similarly ωf,e(f) = −e.We prove (3) for (ij) = (12) and leave the proof in the other cases, which followsthe same pattern, to the reader. Since y(21) := ωf,e · x(12) ∈ V(21) it follows from(10.11.3) and the formulas proved above that

ω2f,e · x(12) = −eσ f−σ y(21) = −ω−f,e · y(21) = −ω−1

f,eωf,e · x(12) = −x(12).

Obviously, (3) implies (4), and (5) follows from (3) and the formulas describing theaction of ωf,e.


10.13. Proposition. Let g and e be idempotents in V with g governing e, thusg ∈ V1(e) and e ∈ V2(g), and let ωg,e = ωe ωg ω

−1e ∈ EA(V, e) as in (10.10.1). Put

f = Qge. Then f is an idempotent orthogonal to e and governed by g, and e+ f isan idempotent associated with g. The action of ωg,e on x ∈ V σ(ij) := V σi (g) ∩ V σj (e)is given as follows:

ωg,e · x(21) = −eσ, Qg−σx(21), fσ ∈ V σ(21),

ωg,e · x(22) = QgσQe−σx(22) ∈ V σ(20),

ωg,e · x(20) = QeσQg−σx(20) ∈ V σ(22),

ωg,e · x(11) = fσ g−σ x(11) ∈ V σ(10),

ωg,e · x(10) = −eσ g−σ x(10) ∈ V σ(11),

ωg,e · x(00) = x(00).

Hence ωg,e maps V σ(21) bijectively onto itself, induces isomorphisms V σ(22)∼= V σ(20)

and V σ(11)∼= V σ(10) and fixes V σ(00). Moreover,

ωg,e · e = f, ωg,e · f = e, ωg,e · g = −g, (1)

ω2g,e

∣∣V σ(ij) = (−1)i · Id, (2)

ω4g,e = Id. (3)

Remark. The associated idempotents g and e+f have the same Peirce spacesby 6.17. Hence V2(e) ⊂ V2(e+ f) = V2(g) and V0(g) = V0(e+ f) ⊂ V0(e) by 6.16.Therefore the joint Peirce spaces V(ij) considered above are the only ones which arepossibly non-zero.

Proof. The proof follows the pattern of that of 10.12. Since (Qg+ , Qg−) is anisomorphism from V op

2 (g) to V2(g) and e ∈ V2(g), it is clear that f ∈ V2(g) is anidempotent, and f ∈ QV1(e)V2(e) ⊂ V0(e) by (6.14.8). We postpone the proof thatg governs f until we have shown (1).

To prove the formulas describing the action of ωg,e, we use (7.9.1) and ω−1e = ω−e

by (7.8.1). For x ∈ V σ(22) we have Qe−σx ∈ V −σ2 (e) ⊂ V −σ2 (g) and QgσQe−σx ∈V σ0 (e), hence

ωg,e · x = ωeωgω−e · x = ωeωg ·Qe−σx = ωe ·QgσQe−σx = QgσQe−σx.

Let x ∈ V σ(21). Then similarly, and since Qgσe−σ = fσ ∈ V σ0 (e) and Qg−σx ∈V −σ1 (e),

ωg,e · x = ωeωg · [−e−σ, x] = −ωe · [ωg · e−σ, ωg · x]

= −ωe · [Qgσe−σ, Qg−σx] = −ωe · [fσ, Qg−σx]

= −[ωe · fσ, ωe ·Qg−σx] = −[fσ, [eσ, Qg−σx]] = −eσ, Qg−σx, fσ.

Next, let x ∈ V σ(20). Then ω−1e · x = x and ωg · x = Qg−σx ∈ V −σ2 (e), whence

ωg,e · x = ωe ·Qg−σx = QeσQg−σx.

§11] Groups defined by Peirce gradings 145

For x ∈ V σ(11), we have

ωg,e · x = ωeωg[−e−σ, x] = −ωe · [Qgσe−σ, [g−σ, x]] = ωe · fσ g−σ x = fσ g−σ x,

since fσ g−σ x ∈ V σ0−1+1(e) = V σ0 (e). If x ∈ V σ(10) then

ωg,e · x = ωeωg · x = ωe · [g−σ, x] = [ωe · g−σ, ωe · x] = [[eσ, g−σ], x] = −eσ g−σ x.

The last case x ∈ V σ(00) is clear.

From these formulas we see that ωg,e exchanges e and f and maps gσ to

−eσ, Qg−σgσ, fσ = −eσ, g−σ, fσ = −eσ, g−σ, Qgσe−σ= Qgσg−σ, eσ, e−σ − eσ, g−σ, gσ , e−σ, gσ (by (JP12))

= Qgσg−σ − 2eσ, e−σ, gσ = gσ − 2gσ = −gσ,

since e ∈ V2(g) and g ∈ V1(e). This proves (1). We use this to show that g governsf . Since ωg,e is an automorphisms of V and g ` e, it follows that −g = ωg,e(e) `ωg,e(e) = f , but then also g ` f holds by (6.14.5). By 6.16, e+ f is an idempotent,obviously contained in V2(g). We also know g ∈ V1(e) ∩ V1(f) ⊂ V2(e+ f) by 6.16again. Thus, by definition in 6.17, g and e+ f are associated idempotents.

Now the remaining formulas (2) and (3) follow easily, using the method appliedin the proof of 10.12.

§11. Groups defined by Peirce gradings

11.1. Definition. Let P : V = V2 ⊕ V1 ⊕ V0 be a Peirce grading of a Jordanpair V as in Section 10. The spaces V σn may be indexed in a natural way by asubset of the root system R = C2 as follows. Recall from (2.16.4) that

R =± εi ± εj : i, j ∈ 0, 1

⊂ X = x0ε0 + x1ε1 ∈ Z2 : x0 + x1 ∈ 2Z.

Define f : X → Z by f(x0ε0 +x1ε1) = (1/2)(x0 +x1) and let Ri = α ∈ R : f(α) =i. Then

R = R1 ∪R0 ∪R−1,

where

R1 = 2ε0, ε0 + ε1, 2ε1, R0 = ε0 − ε1, 0, ε1 − ε0, R−1 = −R1.

In the following picture, the elements of R1, R0 and R−1 are represented by ⊕, and , respectively.


⊕ 2ε0

ε0 − ε1 ⊕ ε1 + ε0

−2ε1

OO

??

__???????//oo

??????? ⊕ 2ε1

−ε0 − ε1 ε1 − ε0 f = 1

− 2ε0 f = 0

f = −1

(This is a simple example of a 3-graded root system as defined in [63, §§17, 18], seealso §14.) Then R1 may be used to index the Peirce spaces V σn by putting

V σεi+εj := V σi+j , for i, j ∈ 0, 1, σ ∈ +,−. (1)

Let now G be a group over V as in 9.1. We use the notations introduced in §9freely. The formulas (1) suggest to define subgroups Uα of G for α ∈ R1 ∪R−1 by

Uσ(εi+εj) := xσ(V σi+j). (2)

From V σ = V σ0 ⊕ V σ1 ⊕ V σ2 we see that

Uσ = xσ(V σ) = U2σε0 · Uσ(ε0+ε1) · U2σε1 =∏

α∈Rσ1

Uα = URσ1 . (3)

We also define subgroups Uµ for µ ∈ R0 as follows. By Corollary 10.7(a), (xi, yj) ∈V +i × V −j is quasi-invertible for i 6= j, so the elements b(xi, yj) ∈ G of 9.7 are

well-defined. Put U0 = 1 and

Uε1−ε0 =⟨b(V +

1 , V −0 ) ∪ b(V +2 , V −1 )

⟩, (4)

Uε0−ε1 =⟨b(V +

0 , V −1 ) ∪ b(V +1 , V −2 )

⟩, (5)

UR0=⟨Uµ : µ ∈ R0

⟩. (6)

Let U = (Uα)α∈R be the family of subgroups of G defined in (2), (4) and (5). From(3) it is clear that G is generated by the subgroups Uα. Recall from 4.1 the categorygcR of groups with R-commutator relations. We define a full subcategory st(V,P)of st(V ) by

G ∈ st(V,P) ⇐⇒ (G,U) ∈ gcR, (7)


that is, G has C2-commutator relations with root groups (Uα)α∈C2 .Suppose G ∈ st(V,P). By 3.3(b), any homomorphic image of G has C2-

commutator relations with root groups the images of the Uα. In particular, this isso for G = PE(V ) = π(G), with root groups U = (Uα)α∈C2

given explicitly byUσ(εi+εj) = expσ(V σi+j)

Uε1−ε0 =⟨β(V +

1 , V −0 ) ∪ β(V +2 , V −1 )

⟩Uε0−ε1 =

⟨β(V +

0 , V −1 ) ∪ β(V +1 , V −2 )

⟩ , (8)

because π x± = exp± and π b = β, by (9.1.3) and (9.7.3). Thus to make surethat st(V,P) is not empty, we must show that PE(V ) satisfies the C2-commutatorrelations with these root group. This will be done in Corollary 11.8 of Theorem 11.2below. By Corollary 4.12, we have therefore a well-defined category st(G, U) whichturns out to be isomorphic to a full subcategory of st(V,P), see Proposition 21.23.

A morphism ϕ: G→ G′ of st(V,P) preserves the root groups Uα, more precisely,

ϕ: Uα → U ′α is

bijective for α ∈ R1 ∪R−1

surjective for α ∈ R0

. (9)

This follows immediately from the definitions and (9.1.4) and (9.7.5) (for f± = Id).Let a be an automorphism of V which normalizes G in the sense of 9.5. Then

G ∈ st(V,P) =⇒ G ∈ st(V, a(P)). (10)

Indeed, by definition, an automorphism a of V belongs to NormA(G) if and onlyif there exists an automorphism ϕa of G such that ϕa(xσ(u)) = xσ(aσ(u)) for allu ∈ V σ. As mentioned above, G has then C2-commutator relations with respectto the root subgroups U ′α = ϕa(Uα). It is easily seen that the root subgroups U ′αare given by (2) and, because of (9.7.5), by (4) and (5), taken with respect to thePeirce grading a(P) instead of P.

If P is trivial in the sense of 10.2, then U± are the only possibly non-trivialroot subgroups for any G ∈ st(V,P), so that st(V,P) = st(V ) in this case. Adescription of st(V,P) for P: V = V2 ⊕ V1 will be given in 11.9.

We now characterize the groups in st(V,P). Recall the definition of the elementsb(x, y) in 9.7 and of the relation B(x, y) in 9.8.

11.2. Theorem. Let P: V = V0⊕V1⊕V2 be a Peirce grading of a Jordan pairV and let G be a group over V . For α ∈ R = C2 define subgroups Uα of G by(11.1.2), (11.1.4) and (11.1.5). Then the following conditions are equivalent:

(i) G has C2-commutator relations, i.e., G ∈ st(V,P),

(ii) the commutator relations(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) hold for all α ∈ R0, β ∈

R1 ∪R−1, and for (α, β) = (±2ε0,∓2ε1),

(iii) UR0normalizes U+ and U−, and

(((((((U2ε0 , U−2ε1

)))))))=(((((((U−2ε0 , U2ε1

)))))))= 1,

(iv) for all i, j ∈ 0, 1, 2 and xi ∈ V +i , yj ∈ V −j , the relations

B(xi, yj) for |i− j| = 1, (StP1)

b(xi, yj) = 1 for |i− j| = 2 (StP2)

hold in G.


Proof. (i) =⇒ (ii): This is obvious.

(ii) =⇒ (iii): For α ∈ R0, β ∈ R1 we have(((((((α, β

)))))))⊂ R1 because an element

γ = pα + qβ ∈(((((((α, β

)))))))(where p, q ∈ N+) satisfies f(γ) = pf(α) + qf(β) = n, and

f has only the values 0,±1 on R. Now(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) implies Int(Uα) · Uβ ⊂

U(((((α,β))))) ·Uβ ⊂ U+ by (11.1.3). In the same way, one shows that Uα normalizes U−.The remaining conditions hold because ±(2ε0−2ε1) /∈ R, so the corresponding rootgroups commute.

(iii) =⇒ (iv): By the definitions (11.1.4) and (11.1.5), b(xi, yj) ∈ U±(ε1−ε0)

whenever |i − j| = 1. Hence b(xi, yj) normalizes U+ und U− which, by (9.8.5),is equivalent to the relations B(xi, yj). Now let |i − j| = 2 . Then either i = 2,j = 0 or i = 0, j = 2, so 2i − j /∈ 0, 1, 2 and x

yji = xi and yxij = yj by (10.7.2).

Therefore, by (9.7.1),

b(xi, yj) = x−(−yj) x+(xi) x−(yj) x+(−xi) =(((((((

x−(−yj), x+(xi))))))))

belongs to(((((((U−2ε0 , U2ε1

)))))))or(((((((U−2ε1 , U2ε0

)))))))and is therefore equal to 1.

The proof of (iv) =⇒ (i), which requires some preparation, will be given in 11.7below.

11.3. Equivalent versions of (StP1) and (StP2). It will be useful to havemore symmetrical versions of (StP1) and (StP2). Using the conventions of 9.10,the relations (StP1) are equivalent to any one of the relations

bσ(xi, yj) · xσ(z) · bσ(xi, yj)−1 = xσ

(B(xi, yj)z

), (StP1′)(((((((

bσ(xi, yj), xσ(z))))))))

= xσ(− xiyjz+QxiQyjz

), (StP1′′)

understood to hold for all σ ∈ +,−, |i − j| = 1, and all (xi, yj) ∈ V σi × V−σj ,

z ∈ V σ. Using (10.7.4) and (9.10.5), one finds that (StP1) is also equivalent to(((((((bσ(xi, yj), x−σ(v)

)))))))= x−σ

(yjxiv+QyjQxiv

), (StP1′′′)

where v ∈ V −σ. Similarly, the relations (StP2) are equivalent to x+(xi)x−(y2−i)= x−(y2−i)x+(xi), thus to (((((((

xσ(xi), x−σ(y2−i))))))))

= 1 (StP2′)

for all (xi, y2−i) ∈ V σi × V−σ2−i, i ∈ 0, 2 and σ ∈ +,−.

Finally, the following commutator formula is a consequence of (StP1) and(StP2): (((((((

b(xi, yj), b(u, v))))))))

= b(B(xi, yj)u ,B(yj ,−xi)v

)· b(u, v)−1, (1)

for i 6= j and all quasi-invertible (u, v) ∈ V . This is immediate from (9.9.5) and(10.7.4).

In the following three lemmas we assume that V is a Jordan pair with a Peircegrading P: V2 ⊕ V1 ⊕ V0. We will write xi to indicate that xi ∈ V σi .


11.4. Lemma. Let x = x2 +x1 +x0 and y = y2 + y1 + y0 be the decompositionof (x, y) ∈ V σ × V −σ relative to P, and let i ∈ 0, 2.

(a) Quasi-invertibility of (x, yi), (xi, yi) and (xi, y) are all equivalent. In thiscase xyi = xyii .

(b) The following relations hold in G for quasi-invertible pairs (x2, y) and(x, y2):

bσ(xi, y) = bσ(xi, y1) · bσ(xi, yi), bσ(x, yi) = bσ(xi, yi) · bσ(x1, yi). (1)

Proof. (a) Possibly after replacing V by V op and interchanging the roles of V2

and V0, i.e., after passing to the reverse grading (10.1.4), we may assume σ = +and i = 2. From the multiplication rules for a Peirce grading it follows that theinclusion f : V +

2 → V + and projection g: V − → V −2 form a structural transforma-tion (f, g): V2 V . Hence by (6.12.4), (x2, y) = (f(x2), y) is quasi-invertible in Vif and only if (x2, y2) = (x2, g(y)) is quasi-invertible in V2, and then

xy22 = f(xy22 ) = f(xg(y)2

)= f(x2)y = xy2.

In particular, (x2, yj) is always quasi-invertible with quasi-inverse xyj2 = x2, for

j ∈ 0, 1.

(b) Since the relations (StP1) and (StP2) hold in G, (9.9.2) yields

b(x2, y) = b(x2, y0 + (y1 + y2)) = b(x2, y0) · b(xy02 , y1 + y2)

= b(x2, y1 + y2) = b(x2, y1) · b(xy12 , y2) = b(x2, y1) · b(x2, y2).

The second formula is proved similarly from (9.9.1) or follows from (9.10.3) afterpassing to V op.

11.5. Lemma. Let |i − j| = 1. Then bσ(V σi , V−σj ) generates an abelian sub-

group of G, and the map bσ: V σi × V−σj → G is bi-multiplicative:

bσ(xi, yj+vj) = bσ(xi, yj)·bσ(xi, vj), bσ(xi+ui, yj) = bσ(xi, yj)·bσ(ui, yj). (1)

In particular,bσ(xi, yj)

−1 = bσ(xi,−yj) = bσ(−xi, yj). (2)

Proof. The possibilities for (i, j) are (2, 1), (1, 2), (1, 0), and (0, 1). Hence, pos-sibly after passing to V op and/or the reverse Peirce grading, it suffices to considerthe case σ = + and (i, j) = (2, 1). By the Peirce multiplication rules, B(x2, y1)u2

= u2 and y1x2v1 ∈ V −0 . Hence (11.3.1) and (11.4.1) yield(((((((b(x2, y1),b(u2, v1)

)))))))= b(B(x2, y1)u2, B(−y1, x2)v1) · b(u2, v1)−1

= b(u2, v1 + y1x2v1) · b(u2, v1)−1 = b(u2, v1) · b(u2, v1)−1 = 1.

Finally, the first formula of (1) follows from (9.9.2) and 11.4:

b(x2, y1 + v1) = b(x2, y1) · b(xy12 , v1) = b(x2, y1) · b(x2, v1),

and the second one follows analogously from (9.9.1).


11.6. Lemma. The following formulas hold for i ∈ 0, 2 and elements x1, yietc. in the corresponding Peirce spaces:(((((((

bσ(x1, yi), xσ(u2−i))))))))

= 1 =(((((((

bσ(zi, v1), xσ(ti)))))))), (1)(((((((

xσ(x1), x−σ(yi))))))))

= xσ(Q(x1)yi

)· bσ(x1, yi) = bσ(x1, yi) · xσ

(Q(x1)yi

). (2)

Proof. Formula (1) follows from the relation (StP1′) since B(x1, yi)u2−i = u2−iand B(zi, v1)ti = ti by the Peirce rules. For (2), we work out the commutator onthe left, using (9.10.6) on the two middle factors as well as 11.4, (10.7.1), (11.5.2)and (9.10.5):(((((((

xσ(x1), x−σ(yi))))))))

= xσ(x1) · x−σ(yi) · xσ(−x1) · x−σ(−yi)= xσ

(x1 + (−x1)yi

)· b−σ(yi,−x1) · x−σ(y−x1

i − yi)= xσ(x1 − x1 +Q(x1)yi) · b−σ(−yi,−x1)−1 · x−σ(yi − yi)= xσ(Q(x1)yi) · bσ(x1, yi).

11.7. Proof of (iv) =⇒ (i) of Theorem 11.2. For α ∈ R1 ∪ R−1 the rootgroups Uα are abelian because they are contained in the abelian groups U+ resp.U−, while for α ∈ R0 this follows from (11.3.1) and (11.4.1):(((((((

bσ(x2, y1),bσ(z1, v0))))))))

= bσ(B(x2, y1)z1, B(−y1, x2)v0

)· bσ(z1, v0)−1

= bσ(z1 − x2y1z1, v0) · bσ(z1, v0)−1

= bσ(−x2y1z1, v0) · bσ(z1, v0) · bσ(−z1, v0) = 1,

For the proof of (3.2.3) we may now assume α 6= β. Since(((((((g, h)))))))−1 =

(((((((h, g)))))))

itsuffices to consider the following cases.

(a) α, β ∈ R1 or α, β ∈ R−1: Then(((((((α, β

)))))))= ∅ but also

(((((((Uα, Uβ

)))))))= 1 since

Uα, Uβ ⊂ Uσ and Uσ is abelian.

(b) α ∈ R1, β ∈ R−1: If (α, β) = (2εi,−2εj) for i 6= j then(((((((α, β

)))))))= ∅

and(((((((Uα, Uβ

)))))))= 1 by (StP2′). If α = 2εi, β = −εi − εj with i 6= j then(((((((

α, β)))))))

= εi − εj ,−2εj while by (11.6.2),(((((((x+(xi), x−(y1)

)))))))=(((((((

x−(y1), x+(xi))))))))−1 = b−(y1, xi)

−1 · x−(Q(y1)xi)−1

= b(−xi, y1) · x−(−Q(y1)xi

)∈ Uεi−εj · U−2εj .

The remaining case of (b) is α = εi + εj , β = −2εi which follows in the same way.

(c) α ∈ R0, β ∈ R±1: It is straightforward to show, using (StP1′′) and (StP1′′′)and the Peirce multiplication rules, that (3.2.3) holds for the generators of Uα andall elements of Uβ . The extension to general elements of Uα then follows fromLemma 3.8.

11.8. Corollary. The projective elementary group of a Jordan pair V withPeirce grading P belongs to st(V,P); in particular, st(V,P) is not empty.

Proof. By Theorem 11.2, it suffices to verify the relations (StP1′) and (StP2)in PE(V ). These follow from (7.7.1) and (10.7.17).


11.9. Corollary. Let P: V = V2 ⊕ V1 be a Peirce grading of a Jordan pair Vwith V0 = 0, and let A2 =

εi−εj : i, j ∈ 0, 1, 2

as in (2.16.6). Let G be a group

over V . Then G ∈ st(V,P) if and only if G has A2-commutator relations with rootgroups (Uα)α∈A2

defined by

Uσ(εi−ε0) := xσ(V σi ) for i ∈ 1, 2, Uσ(ε2−ε1) :=⟨bσ(V σ2 , V

−σ1 )

⟩.

Proof. Let R′ = C2 ±2ε0. The linear map f : Z(ε0 + ε1) ⊕ Z(ε1 − ε0) →spanZ(A2) ⊂ Z3 satisfying f(ε0 + ε1) = ε1 − ε0 and f(ε1 − ε0) = ε2 − ε1 is anisomorphism in SF between R′ and A2. It follows that G has A2-commutatorrelations relative to the root groups (Uα)α∈A2

defined above if and only if G hasR′-commutator relations relative to the root groups defined in 11.1. Because V0 = 0this is equivalent to G having C2-commutator relations.

11.10. Corollary. Let (Uα)α∈C2be the root subgroups of G = PE(V ) defined

in (11.1.8), and let EA(V,P) be the group of P-elementary automorphisms of Vas in 10.10. Then

UR0 =⟨Uε1−ε0 ∪ Uε0−ε1

⟩= EA(V,P), (1)

and EA(V,P) ⊂ NormA(G) for any G ∈ st(V,P). Hence

st(V,P) = st(V, h(P)) (2)

for any h ∈ EA(V,P).

Proof. (1) follows immediately from the definitions. For the remaining state-ments, it is enough to prove that G ∈ st(V,P) implies G ∈ st

(V, β(xi, yj)(P)

)for (xi, yj) ∈ V +

i × V −j , i 6= j. By condition (iv) of Theorem 11.2 and (9.8.5),

b(xi, yj) ∈ G0 normalizes U+ and U−. Hence, by (9.5.2), π(b(xi, yj)) = β(xi, yj) ∈NormA(G) whence G ∈ st(V, β(xi, yj)(P)) by (11.1.10).

11.11. The group St(V,P). Let P be a Peirce grading V = V0 ⊕ V1 ⊕ V2 of

a Jordan pair V , and let G be the quotient of the free product Fr(V ) = V + ∗ V −of the additive groups V + and V − by the (normal subgroup generated by the)

relations (StP1) and (StP2) and let Uσ = can(V σ) ⊂ G. Since G satisfies these

relations by Corollary 11.8, there is a canonical homomorphism π: G→ G inducedfrom the canonical homomorphism ηG, see 9.4, making the diagram

V σxσ=can //

expσ !!BBBBBBBB Uσ

π||||||||

Uσ

commutative. Hence (G, U+, U−, π) is a group over V , and it follows from Theo-

rem 11.2(iv) that for every G ∈ st(V,P) there exists a unique morphism κ: G→ G,more precisely, a commutative diagram


Gκ //

π >>>>>>>> G

π

G

We define

St(V,P) = (G, U+, U−, π),

called the Steinberg group of (V,P). Then St(V,P) is an initial object in thecategory st(V,P), hence uniquely determined up to unique isomorphism.

The relation between St(V,P) and the Steinberg group of (G, U) in the senseof 4.10 will be discussed in §21 in the more general context of root graded Jordanpairs.

Let h: (V,P) → (V ′,P′) be a homomorphism of Peirce-graded Jordan pairs,

and let St(V ′,P′) = (G′, U ′±, π′). Then h induces a group homomorphism

ϕh: G→ G′

which is a lift of h as defined in (9.5.1). This can be seen as follows. Let

η: Fr(V ) → G and η′: Fr(V ′) → G′ be the canonical homomorphisms as in 9.4.The homomorphism h: V → V ′ extends uniquely to a group homomorphismFr(h): Fr(V ) → Fr(V ′), so that ϕ = η′ Fr(h): Fr(V ) → G′ is a lift of h. We

show that the normal subgroup K of Fr(V ) defining G is contained in the kernel ofϕ, that is, that the relations (StP1) and (StP2) are annihilated by ϕ. By (9.7.5)

we know ϕ(b(x, y)

)= b

(h+(x), h−(y)

)∈ G′ for any quasi-invertible pair. Hence

the relation B(xi, yj) for |i − j| = 1 is mapped under ϕ onto B(h+(xi), h−(yj))

which holds in G′ by its definition. From ϕ(xσ(v)

)= xσ(hσ(v)) it follows that

the relation (StP2) in Fr(V ) is mapped to 1G′

. Now it is easily checked that the

assignments (V,P) 7→ G and h 7→ ϕh define a functor from Peirce graded Jordanpairs to groups. We refer to Lemma 22.2 for a more precise statement in the generalsetting of root graded Jordan pairs.

If P is trivial in the sense that V0 = V then G = Fr(V ) is the free product, and

if V1 = 0 then G is the direct product of Fr(V2) and Fr(V0). In these cases, othertypes of Steinberg groups contain more information, see [59] or §12 below.

Let h ∈ NormA(St(V,P)). Then we know from (11.1.10) that St(V,P) ∈st(V, h(P)). In fact, it is easy to see that

St(V,P) = St(V, h(P)) for all h ∈ NormA(St(V,P)). (1)

In particular, by 11.10, we have EA(V,P) ⊂ NormA(St(V,P)).

11.12. Perfectness. Let G ∈ st(V,P), let(((((((G,G

)))))))be its commutator sub-

group, and put

Kσ = v ∈ V σ : xσ(v) ∈(((((((G,G

))))))).

Clearly Kσ is an additive subgroup of V σ. When studying perfectness of G, thefollowing formulas will be useful. The index i is assumed in 0, 2.

§12] Weyl elements for idempotent Peirce gradings 153

V σi V −σ1 V σ1 + V σ1 V −σ2−iVσ1 ⊂ Kσ ∩ V σi , (1)

Q(xi)Q(y1)z2−i + xiy1z2−i ∈ Kσ ∩(V σi ⊕ V σ1

), (2)

Q(x1)Q(yi)zi + x1yizi ∈ Kσ ∩(V σ2−i ⊕ V σ1

). (3)

By (StP1′′), we have for (xj , yk) ∈ V σj × V−σk , j 6= k, and arbitrary z ∈ V σ that

a = −xjykz+Q(xj)Q(yk)z ∈ Kσ.

Special choices now yield (1)–(3). Namely, for (j, k) = (i, 1) we see that −xiy1z1∈ Kσ and −xiy1z2−i+Q(xi)Q(y1)z2−i ∈ Kσ, while for (j, k) = (1, i) we obtain−x1yizi+Q(x1)Q(yi)zi ∈ Kσ and −x1yiz1 ∈ Kσ.

In general, G is not perfect. We leave it to the reader to formulate conditionswhich force perfectness of G. For a special type of Peirce grading we will use theresult above to prove perfectness of a related Steinberg group, see 27.8.

§12. Weyl elements for idempotent Peirce gradings

12.1. Definition. Let e be an idempotent of the Jordan pair V with associatedPeirce grading P(e) : V = V2(e) ⊕ V1(e) ⊕ V0(e). In 11.1 we have introduced thecategory st(V,P(e)) whose objects are groups with C2-commutator relations withrespect to the root subgroups (Uα)α∈C2

defined in (11.1.2), (11.1.4) and (11.1.5).Also recall the element

we = x−(e−) · x+(e+) · x−(e−) (1)

defined in (9.11.2) for any group over V . Contrary to what the notation we mightsuggest, we is in general not a Weyl element for the root 2ε1 in the sense of 5.1.Therefore, we define a full subcategory st(V, e) of st(V,P(e)) by

G ∈ st(V, e) ⇐⇒ G ∈ st(V,P(e)) and we is a Weyl element for 2ε1. (2)

If G ∈ st(V, e) and ϕ: G → G′ is a morphism of st(V,P(e)), then by 5.1 and(11.1.9), G′ belongs again to st(V, e). Thus st(V, e) is not empty if and only ifPE(V ) belongs to st(V, e). This will be shown in Corollary 12.8.

Since P(e) = P(−e), w−1e = w−e and generally w is a Weyl element for a root

α if and only if w−1 is (cf. 5.4(a)), we have

st(V,−e) = st(V, e). (3)

Recall that for any automorphism h of V , the element h(e) = (h+(e+), h−(e−))is again an idempotent with h(P(e)) = P(h(e)). We have

G ∈ st(V, e) =⇒ G ∈ st(V, h(e)) for h ∈ NormA(G). (4)

Indeed, by (11.1.10) we know G ∈ st(V,P(h(e))

). Let ϕh: G → G be the induced

group automorphism of G satisfying ϕh(xσ(u)) = xσ(hσ(u)) for u ∈ V σ. Since ϕhis bijective on root groups, it follows again from 5.1 that ϕh(we) = wh(e) is a Weylelement for 2ε1.


The main result of this section is Theorem 12.5. Our first task is to derive anumber of identities satisfied by we in a group G ∈ st(V,P(e)). We first introducethe following notation. As usual, subscripted symbols vi indicate that vi belongsto the Peirce space V σi = V σi (e), i ∈ 0, 1, 2. The index j takes values in 0, 1.We define

z(xj+1, yj) := b(e+, e−xj+1yj)−1 · b(xj+1, yj), (5)

z(xj , yj+1) := b(xjyj+1e+, e−)−1 · b(xj , yj+1), (6)

d+(x2) := x−(Qe−x2

)·(wex+(−x2)w−1

e

), (7)

d−(y2) :=(wex−(−y2)w−1

e

)· x+(Qe+y2). (8)

It should be kept in mind that this notation is incomplete; z and d depend not onlyon the x and y but also on the idempotent e. If necessary, we therefore write zeand d

(e)σ to indicate this dependence.

12.2. Lemma. Let G ∈ st(V,P(e)). We use the notations introduced above.Then z(xj+1, yj) and z(xj , yj+1) are central in G, and the following formulas hold:

we · xσ(v0) · w−1e = xσ(v0), (1)

we · x+(x1) · w−1e = b(x1,−e−) = b(−x1, e−), (2)

we · b(x1, e−) · w−1e = x+(x1), (3)

we · x−(y1) · w−1e = b(e+, y1), (4)

we · b(e+, y1) · w−1e = x−(−y1), (5)

we · b(xj+1, yj) · w−1e = x−(−e−xj+1yj) · z(xj+1, yj), (6)

we · b(xj , yj+1) · w−1e = x+(xjyj+1e+) · z(xj , yj+1), (7)

we · b(x1, y0) · w−1e = x−

(− e−x1y0

)· d+(Qx1

y0), (8)

we · b(x0, y1) · w−1e = d−(Qy1x0) · x+

(e+y1x0

), (9)

z(x1, y0) = d+(Qx1y0), (10)

z(x0, y1) = d−(Qy1x0), (11)

we · x+

(xj+1yjz1

)· w−1

e = x−(Qe−xj+1yjz1

), (12)

we · x−(yj+1xjv1

)· w−1

e = x+

(Qe+yj+1xjv1

). (13)

Proof. By (10.8.5), β(xj+1, yj) = β(e+, e−xj+1yj) in PE(V ), and by condition(iii) of Theorem 11.2, b(xj+1, yj) and b(e+, e−xj+1yj) normalize U+ and U−.Hence it follows from Lemma 9.2(c) that these two elements are congruent modulothe centre of G, so z(xj+1, yj) is central. The proof for z(xj , yj+1) is similar.

Formula (1) follows immediately from (StP2′) in 11.3. For (2), we use (−e−)x1 =

−e− and x−e−1 = x1−Q(x1)e− by (10.7.1), and B(x1, e−)e+ = e+−x1 +Q(x1)e−,

as well as B(x1,−e−)B(x1, e−) = Id by (10.7.4). Putting g = x−(e−) · x+(e+) weget

we · x+(x1) · w−1e = g · x−(e−) ·

(x+(x1) · x−(−e−)

)· g−1

= g · x−(e− + (−e−)x1) · b(x1,−e−) · x+(x−e−1 ) · g−1 (by (9.7.1))

= g · b(x1,−e−) · x+(x1 −Q(x1)e− − e+) · x−(−e−)


= g · b(x1,−e−) · x+(B(x1, e−)(−e+)) · x−(−e−)

= x−(e−) · x+(e+ −B(x1,−e−)B(x1, e−)e+) · b(x1,−e−) · x−(−e−)

(by (StP1′) of 11.3)

= x−(e−) · b(x1,−e−) · x−(−e−) = b(x1,−e−),

since(((((((

x−(e−),b(x1,−e−))))))))∈(((((((U−2ε1 , Uε1−ε1

)))))))= 1. The second equality follows

from (11.5.2). For (3) we replace e by −e in (2) and use w−1e = w−e (by (9.11.5))

and (11.5.2).

The proof of (4) is similar to that of (2):

we · x−(y1) · w−1e = x−(e−) ·

(x+(e+) · x−(y1)

)· x+(−e+) · x−(−e−)

= x−(e− + ye−1 ) · b(e+, y1) · x+(ey1+ − e+) · x−(−e−) (by (9.7.1))

= x−(e− + y1 +Q(y1)e+) · b(e+, y1) · x−(−e−)

= x−(B(−y1, e+)e− −B(y1, e+)−1e−) · b(e+, y1) (by (StP1′))

= b(e+, y1),

and (5) follows from (4) in the same way as (3) from (2).

Replace y1 in (5) by e−xj+1yj. Then, by definition of z and since z(xj+1, yj)is central,

x−(−e− xj+1 yj) = we · b(e+, e− xj+1 yj) · w−1e

= we ·(

b(xj+1, yj)z(xj+1, yj)−1)· w−1

e

=(

we · b(xj+1, yj) · w−1e

)· z(xj+1, yj)

−1

which yields (6). The proof of (7) follows similarly from (3).

We prove (8). By (11.6.2), we have(((((((x+(x1), x−(y0)

)))))))= b(x1, y0) · x+(Qx1

y0).

Conjugating this with we and using (1) and (2) yields(((((((b(x1,−e−), x−(y0)

)))))))=(

we · b(x1, y0) · w−1e

)·(

we · x+(Qx1y0) · w−1e

).

On the other hand,(((((((b(x1,−e−), x−(y0)

)))))))= x−(−e−x1y0) · x−(Qe−Qx1y0)

by (StP1′′′), from which (8) follows. The proof of (9) is again similar.

The formulas (10) and (11) follow by comparing (6) and (8) resp. (7) and (9),in case j = 0.

We prove (12). By the Peirce rules, Qxj+1Qyjz1 ∈ V +

2j+2−2j+1 = V +3 = 0, hence

centrality of z(xj+1, yj) and (StP1′′) yield(((((((b(e+, e−xj+1yj), x+(z1)

)))))))=(((((((

b(xj+1, yj), x+(z1))))))))

= x+(−xj+1yjz1).Conjugating this with we and using (5) as well as (2) yields(((((((

x−(−e−xj+1yj), b(z1,−e−))))))))

= we · x+(−xj+1yjz1) · w−1e .

By (StP1′′′), the left hand side equals

x−(− e−z1 e−xj+1yj −Qe−Qz1e−xj+1yj

).

Here the first term is Qe−xj+1yjz1 by (JP12), and the second term vanishes bythe Peirce rules, so we have (12). The proof of (13) is similar.


12.3. The shift relations. Let G ∈ st(V,P(e)). We say G satisfies the shiftrelation S10 if

S10 : b(x1, y0) = b(e+, e− x1y0) for all (x1, y0) ∈ V +1 × V

−0 .

for all (x1, y0) ∈ V +1 × V

−0 . This is equivalent to the conjugation relations

we · x+

(Qx1y0

)· w−1

e = x−(Qe−Qx1y0

),S′10 :

we · b(x1, y0) · w−1e = x−(−e− x1 y0),S′′10 :

for all (x1, y0) ∈ V +1 (e)× V −0 (e). Indeed, using the definitions of z and d in 12.1 as

well as (12.2.10),

S10 ⇐⇒ z(V +1 , V −0 ) = 1 ⇐⇒ d+(QV +

1V −0 ) = 1 ⇐⇒ S′10,

and S10 ⇐⇒ S′′10 follows from (12.2.6) (in case j = 0).Similarly, we say G satisfies the shift relation S21 if

S21 : b(x2, y1) = b(e+, e−, x2, y1) for all (x2, y1) ∈ V +2 × V

−1 .

This is equivalent to z(V +2 , V −1 ) = 1 (by definition of z) and also to the relations

b(Qx2y2, v1) = b(x2, y2x2v1),S′21 :

we · b(x2, y1) · w−1e = x−(−e− x2 y1),S′′21 :

for all (x2, y1) ∈ V +2 × V

−1 . Indeed, if S21 holds then

b(Qx2y2, v1) = b(e+, e−, Qx2

y2, v1) (by S21)

= b(e+, e−, x2, y2x2v1) (by (10.7.7))

= b(x2, y2x2v1), (by S21)

which is S′21. The implication S′21 =⇒ S21 follows from

b(x2, y1) = b(Qe+Qe−x2, y1) = b(e+, Qe−x2, e+, y1) (by S′21)

= b(e+, e−, x2, y1) (by (10.8.1)).

Finally, the equivalence of S21 and S′′21 follows from (12.2.6) (for j = 1).

We define dual shift relations S01 and S12 by requiring

b(x0, y1) = b(x0, y1, e+, e−),S01 :

b(x1, y2) = b(x1, y2, e+, e−),S12 :

for all elements in the appropriate Peirce spaces. By definition of z, Sj,j+1 isequivalent to z(V +

j , V−j+1) = 1 for j ∈ 0, 1. Moreover, the reader will easily

prove the following equivalent versions of these relations, for example by passing toV op:

we · x−(Qy1x0) · w−1e = x+

(Qe+Qy1x0

),S′01 :

we · b(x0, y1) · w−1e = x+(x0 y1 e+),S′′01 :

b(x1, Qy2z2) = b(x1, y2, z2, y2),S′12 :

we · b(x1, y2) · w−1e = x+(x1 y2 e+).S′′12 :

As before, S′01 is equivalent to d−(QV −1V +

0 ) = 1.

We say G satisfies all shift relations if all four relations Sij hold in G.


12.4. Lemma. We use the notations and conventions of 11.1 and 12.2, andlet G ∈ st(V,P(e)).

(a) The following conditions are equivalent:

(i) we · Uε1−ε0 · w−1e = U−ε1−ε0 ,

(ii) Uε1−ε0 = b(e+, V−1 ),

(iii) π: Uε1−ε0 → Uε1−ε0 is bijective,

(iv) G satisfies the shift relations S10 and S21.

If these conditions are satisfied, then the formulas

b(x2yj+1zj, v0) = b(x2, yj+1zjv0) (1)

hold in G for all elements in the respective Peirce spaces.

(b) The following conditions are equivalent:

(i) we · Uε0−ε1 · w−1e = Uε0+ε1 ,

(ii) Uε0−ε1 = b(V +1 , e−),

(iii) π: Uε0−ε1 → Uε0−ε1 is bijective,

(iv) G satisfies the shift relations S01 and S12.

If these conditions are satisfied, then the formulas

b(x0yjzj+1, v2) = b(x0, yjzj+1v2) (2)

hold in G for all elements in the respective Peirce spaces.

Proof. We prove (a) and leave the analogous proof of (b) to the reader.

(i) =⇒ (ii): Conjugating (i) with w−1e yields Uε1−ε0 = w−1

e ·U−ε1−ε0 ·we. NowU−ε1−ε0 = x−(V −1 ) = we · b(e+, V

−1 ) · w−1

e by (12.2.5), so Uε1−ε0 = b(e+, V−1 )

follows.

(ii) =⇒ (iii): The restriction of π to b(e+, V−1 ) is injective, because π(b(e+, v1))

= β(e+, v1) = (B(e+, v1), B(−v1, e+)) and the Peirce 1-component of B(−v1, e+)e−is v1e+e− = v1. Surjectivity is always true and follows from (11.1.9).

(iii) =⇒ (iv): Since both sides of the asserted equations lie in Uε1−ε0 and theimages under π are equal by (10.8.5), the assertion follows.

(iv) =⇒ (i): If (iv) holds, then the inclusion we · Uε1−ε0 · w−1e ⊂ U−ε1−ε0

follows from S′′10, S′′21 and the definition of Uε1−ε0 in (11.1.4) and the fact thate− xj+1yj ∈ V −1 . The other inclusion is clear from (12.2.5).

Finally, the formulas (1) follow from the corresponding formulas (10.7.15) forthe β’s instead of the b’s and the injectivity of π on Uε1−ε0 .

12.5. Theorem. Let e be an idempotent of a Jordan pair V , and let P(e) :V = V2⊕V1⊕V0 be the Peirce decomposition with respect to e. Let G ∈ st(V,P(e))and let we = x−(e−) x+(e+) x−(e−) ∈ G. Consider the following conditions:

(i) we is a Weyl element for the root 2ε1, i.e., G ∈ st(V, e),

(ii) the Weyl relations W(e) hold, i.e., weU±2ε1w−1e = U∓2ε1 ,

(iii) G satisfies all shift relations.


Then (i) ⇐⇒ (ii) =⇒ (iii). If V2 = V2V1V1+QV1V0 then also (iii) =⇒ (ii).

Proof. (i) =⇒ (ii): Trivial.

(ii) =⇒ (iii): The relations S10 and S01 are equivalent to S′10 and S′01 by 12.3,and therefore hold in G because of the Weyl relation W(e). Let us prove S21. From(9.7.2) and (10.7.1) we have

b(x2, y1) = x−(−Qy1x2

)·(((((((

x−(−y1), x+(x2)))))))).

Here Qy1x2 ∈ V −0 by the Peirce rules. We conjugate this with we, use W(e) as wellas (12.2.1), (12.2.4) and (StP1′′′) and obtain:

we · b(x2, y1) · w−1e = x−

(−Qy1x2

)·(((((((

b(e+,−y1), x−(Qe−x2

))))))))= x−

(−Qy1x2 − y1, e+, Qe−x2+Qy1Qe+Qe−x2

)= x−

(− e− x2y1

). (1)

In the last step, we used the formula y1, e+, Qe−x2 = e− x2y1 which followsfrom (10.8.1). Comparing this with (12.2.7) shows z(x2, y1) = 1. But as pointedout in 12.3, z(V +

2 , V −1 ) = 1 is an equivalent version of S21.The proof of S′12 is similar and left to the reader.

(ii) =⇒ (i): We must show we Uβ w−1e = Us2ε1 (β) for all β ∈ C2, where s2ε1 acts

bys2ε1(εj) = (−1)jεj (j ∈ 0, 1). (2)

If β = ±2ε0 this is clear from (12.2.1), and if β = ±2ε1, this is W(e). Theremaining cases β = ±ε0± ε1 are an easy consequence of (12.2.2), (12.2.4), and theshift relations which hold by (ii) =⇒ (iii). The details are left to the reader.

Finally, suppose the shift relations hold and V2 = V2V1V1+QV1V0. Then S′10,S′01 and (12.2.12) and (12.2.13) (in case j = 1) show that W(e) holds.

Remark. Rather than requiring that G satisfy all shift relations and V2 =V2V1V1+QV1

V0, one can require the condition

V2 = V2V1V1+ V1V0V1 (3)

alone, to establish the Weyl relation W(e). Indeed, assuming (3), W(e) follows byapplying (12.2.12) and (12.2.13) for j = 0, 1. Condition (3) holds in the strongerform V2 = V2V1V1 whenever there exists an idempotent f which is rigid collinearwith e in the sense of [70], i.e., V2(e) ⊂ V1(f). If 2 is invertible in the base ringk then (3) becomes V2 = V2V1V1 + QV1

V0 so that one does not need the shiftrelations to prove W(e) in this case.

12.6. Example: st(V, e) is a proper subcategory of st(V,P(e)). Indeed, letV = V2(e)⊕ V1(e)⊕ V0(e) where V1(e) = V0(e) = 0. Then st(V,P(e)) = st(V ), aswe have already noted at the end of 11.1 (this is also clear from Theorem 11.2 sincethe conditions (iii) or (iv) hold automatically). On the other hand, the conditionW(e) of Theorem 12.5 is not automatically true for an arbitrary G ∈ st(V ).

For a minimal example, let V + = V − = F2, the field with two elements, ande = (1, 1). Then the free product F of V + and V −, i.e., the free product of twocopies of Z/2Z, belongs to st(V ) but does not satisfy W(e). Indeed, F is generatedby two elements a, b satisfying a2 = b2 = 1 and no other relations. The relationW(e) amounts to bab ·a · bab = b, equivalently, (ab)3 = 1, which does not hold in F .

§13] Groups defined by sets of idempotents 159

12.7. Corollary. If G ∈ st(V, e) then π: G → PE(V ) is bijective on all rootgroups Uα, α ∈ R = C2.

Proof. For α ∈ R±1 this holds for any group G over V , and for µ ∈ R0 it followsfrom Theorem 12.5(iii) and Lemma 12.4.

12.8. Corollary. The projective elementary group of a Jordan pair V with anidempotent e belongs to st(V, e). Hence the category st(V, e) is not empty.

Proof. This is an immediate consequence of Theorem 12.5, the fact thatPE(V ) ∈ st(V,P(e)) by Corollary 11.8, and that it satisfies all Weyl relationsW(e) by Proposition 7.10.

12.9. Corollary. Let V be a special Jordan pair embedded in a Morita contextM, and let e ∈ V be an idempotent. Then the elementary group E(M, V ) belongsto st(V, e).

Proof. We know from 9.3 that G = E(M, V ) ∈ st(V ). We prove next thatG ∈ st(V,P(e)). Let us abbreviate Vi = Vi(e) for i = 0, 1, 2. By (9.8.6), therelations B(x, y) hold in G for any quasi-invertible pair (x, y) ∈ V . Since (x, y) ∈V +i ×V

−j for |i− j| = 1 is quasi-invertible, this proves the condition (StP1) of 11.2.

Hence, by 11.2, G ∈ st(V,P(e)) as soon as we have shown condition (StP2) of 11.2:b(x, y) = 1 for (x, y) ∈ V +

i × V−j and |i− j| = 2.

To prove this, let M = (R,M+,M−, S) as in 6.1. We have noted in 6.14 thatc+ = e+e− ∈ R and c− = e−e+ ∈ S are idempotents of the associative algebras Rand S respectively. Assume first i = 2, thus x ∈ V +

2 (e) and y ∈ V −0 (e). Then

0 = Q(e−)Q(e+)y = e+e−ye−e+ = c−yc+,

0 = e+ e− y = e+e−y + ye−e+ = c−y + yc+.

Hence, multiplying the last equation from the right with c+ = c2+ gives 0 =c−yc+ + yc2+ = yc+, and then also 0 = c−y. For x ∈ V +

2 (e) we know x =Q(e+)Q(e−)x = c+xc−. But then xy = c+xc−y = 0 and yx = yc+xc− = 0. Nowb(x, y) = 1 follows from the formula (9.7.7) for b(x, y) ∈ G. The case i = 0 canbe dealt with in an analogous way. We have now established that G ∈ st(V,P(e)).Then G ∈ st(V, e) is immediate from Theorem 12.5 since we have shown in 9.17that the Weyl relations W(e) hold in G.

§13. Groups defined by sets of idempotents

13.1. Definition. Let S be a set of idempotents of the Jordan pair V . Gen-eralizing (12.1.2), we define a full subcategory of st(V ) by

G ∈ st(V,S ) ⇐⇒ G ∈ st(V, e) for all e ∈ S .

Thus in particular, st(V, e) = st(V, e). If S is empty or consists of the trivialidempotent 0 = (0, 0) then clearly st(V,S ) = st(V ); in fact, we have st(V,S ) =st(V,S 0). Also, st(V, e) = st(V, e,−e) by (12.1.3). Finally, always PE(V ) ∈st(V,S ) by Corollary 12.8.

Similarly to 11.11, we define the Steinberg group


St(V,S ) = (G, U±, π)

to be the quotient of the free product Fr(V ) = V + ∗ V − by the normal subgroupgenerated by the relations W(e) and the relations (StP1) and (StP2) for all e ∈ S ,

where Uσ are the canonical images of V σ ⊂ Fr(V ). Then St(V,S ) is an initialobject in the category st(V,S ): for every G ∈ st(V,S ) there is a unique morphismSt(V,S )→ G. This follows immediately from Theorem 11.2 and Theorem 12.5. IfS = e is a singleton we simply write St(V, e) instead of St(V, e).

The group St(V,S ) depends functorially on (V,S ) in the following sense: sup-pose f : V → V ′ is a homomorphism of Jordan pairs, S and S ′ are sets of idem-potents in V and V ′, respectively, and f(S ) ⊂ S ′. Then there is a unique grouphomomorphism St(f): St(V,S ) → St(V ′,S ′) sending xσ(v) to x′σ(fσ(v)). Theproof is analogous to the proof of functoriality of Steinberg groups for Peirce gradedJordan pairs in 11.11.

In the following proposition we use the notation of 11.1 and let U = (Uα)α∈C2

be the family of subgroups of a group G ∈ st(V,P) defined in (11.1.2), (11.1.4) and(11.1.5). Then by definition in 12.1, G belongs to st(V, e) if and only if the element

we = x−(e−) x+(e+) x−(e−)

is a Weyl element for the root 2ε1 and the root groups U.

13.2. Proposition. Let e and f be associated idempotents in V , see 6.17, andlet G ∈ st(V, e). Then the following conditions are equivalent:

(i) wf is a Weyl element for the root 2ε1, i.e., G ∈ st(V, f),

(ii) the relations W(f) hold in G,

(iii) Int(wf ) · U±2ε1 = U∓2ε1 ,

(iv) the relation B(e+ + f+, f−) holds in G,

(v) the relation B(e+, e− + f−) holds in G.

If these conditions are satisfied then

we wf = b(e+ + f+, f−) = b(e+, e− + f−), (1)

and this element normalizes all root subgroups (Uα)α∈C2.

Proof. Since e and f are associated we have P(e) = P(f), so the equivalenceof (i) – (iii) is evident from Theorem 12.5. In preparation for the proof of (i) =⇒(iv) =⇒ (ii), we compute wewfop ∈ G where wfop = x+(f+) x−(f−) x+(f+) as in(9.11.4), without assuming (i) or (iv). Since G ∈ st(V, e), the relation W(e) holdsby Theorem 12.5 so we = weop by (9.16.5). We also use the formulas (9.7.1) and(6.18.1) with e and f interchanged:

we wfop = x+(e+) x−(e−)(

x+(e+ + f+) x−(f−))

x+(f+)

= x+(e+) x−(e− − e−)b(e+ + f+, f−) x+(−f+ −Qf+e− + f+)

= x+(e+) b(e+ + f+, f−) x+

(−Qf+e−

). (2)


Since U+ normalizes itself, (2) shows that b(e+ +f+, f−) normalizes U+ if and onlyif wewfop does so, and in this case, the formulas (6.18.2) and (9.8.1) show that

we wfop = x+

(e+ +B(e+ + f+, f−)(−Qf+e−)

)b(e+ + f+, f−)

= x+

(e+ −Qe+Qf−Qf+e−

)b(e+ + f+, f−) = b(e+ + f+, f−). (3)

Now suppose (i) holds. Then we and wf are both Weyl elements for the root 2ε1,so wf = wfop as before, and (5.4.3) shows that wewf normalizes all root subgroupsand therefore also U± =

⊕α∈R±1

Uα. Thus (3) holds, i.e., the first formula in (1),

and b(e+ +f+, f−) normalizes U±. This proves (iv), in view of (9.8.5). Conversely,if (iv) holds, then b(e+ + f+, f−) normalizes U+ and U−, so by (3),

wfop = w−1e b(e+ + f+, f−). (4)

It follows immediately from (9.8.1) and (9.8.2) that, for any quasi-invertible pair(x2, y2) ∈ V2(e),

b(x2, y2) normalizes U±2ε1 , (5)

so by (4) and since we is a Weyl element for 2ε1, we see that condition (iii) issatisfied for wfop in place of wf . But by Lemma 9.16 the latter is one of theequivalent versions of the Weyl relations W(f), so that (ii) holds.

For the proof of (i) =⇒ (v) =⇒ (iii), we calculate wewf ∈ G, using the formulas(9.7.1) and (6.18.1) interpreted in V op:

we wf = x−(e−)(

x+(e+) x−(e− + f−))

x+(f+) x−(f−)

= x−(e− + (e− + f−)e+) b(e+, e− + f−) ·· x+

((e+)e−+f− + f+

)x−(f−)

= x−(−Q(e−)f+) b(e+, e− + f−) x−(f−). (6)

As in the proof above it follows that wewf normalizes U− if and only if b(e+, e−+f−) does so, and in this case

we wf = x−(−Q(e−)f+ +B(e− + f−, e+)−1f−) b(e+, e− + f−)

= b(e+, e− + f−) (7)

since, by (6.18.2) for V op,

B(e− + f−, e+)−1f− =(Q(f−)Q(e+)

)−1f− = Q(e−)Q(f+)f− = Q(e−)f+.

Now assume (i). As above, wewf then normalizes all root subgroups, whencealso U±. By (7) we thus have the second formula in (1) and also know thatb(e+, e− + f−) normalizes U±. Hence (v) follows again from (9.8.5). Conversely, if(v) is satisfied, then b(e+, e− + f−) normalizes U−, hence so does wewf , and (7)holds. Thus wf = w−1

e b(e+, e−+ f−). Since we is a Weyl element for the root 2ε1,(5) implies (iii).


13.3. Corollary. In any group G ∈ st(V, e), we have

w2e = b(2e+, e−) = b(e+, 2e−), (1)

and this element normalizes all root subgroups. The action of Int(w2e) on the root

groups is

Int(w2e) · xσ(ui) = xσ

((−1)iui

)for ui ∈ V σi (e), i = 0, 1, 2, (2)

Int(w2e) · b = b−1 for b ∈ Uµ, µ ∈ R×0 . (3)

Proof. The special case e = f of Proposition 13.2 shows that w2e normalizes

all root groups of G. To determine the action of Int(w2e) on the root groups, let

π: G → PE(V ) be the canonical projection. We have shown in Lemma 7.9 and(6.14.6) that π(w2

e) = ω2e = se is the Peirce reflection with respect to e, so that (2)

follows from (9.2.1). By definition in (11.1.5), Uε0−ε1 =⟨b(V +

0 , V −1 )∪b(V +1 , V −2 )

⟩.

Hence by (9.7.6) and (11.5.2),

Int(w2e) · b(uj , vj+1) = b

(se(uj), se(vj+1)

)=

b(u0,−v1) if j = 0b(−u1, v2) if j = 1

= b(uj , vj+1)−1.

Since Uε0−ε1 is abelian by 11.5, this implies (3) for µ = ε0−ε1. The case µ = ε1−ε0

follows in the same way.

13.4. Corollary. For G ∈ st(V, e) the following conditions are equivalent:

(i) W(f) holds for all idempotents f associated with e,

(ii) B(x, y) holds in G for all quasi-invertible pairs (x, y) ∈ V2(e) with x or yinvertible in V2(e).

Moreover, if e is a division idempotent, i.e., V2(e) is a division pair, then (i) and(ii) are equivalent to

(iii) B(x, y) holds in G for all quasi-invertible pairs (x, y) ∈ V2(e).

Proof. Let S be the set of idempotents associated with e.

(i) =⇒ (ii): By Proposition 13.2 we know that G ∈ st(V,S ). Let (x, y) ∈V2(e) be quasi-invertible, and suppose first that y is invertible in V2(e). Then(y−1, y) ∈ S , and there is no harm in assuming y = e−. By 6.18, (x, y) = (x, e−)is quasi-invertible if and only if there exists f ∈ S with x = e+ + f+. But thenB(x, y) = B(e+ + f+, e−) holds in G by 13.2(iv). The case where x is invertible inV2(e) follows analogously (or by passing to the opposite Jordan pair).

(ii) =⇒ (i): Let f ∈ S . Then (e+ + f+, f−) is quasi-invertible by 6.18. Sincef− is invertible in V2(e), B(e+ + f+, f−) holds by assumption, but then also W(f)by Proposition 13.2.

Finally, let e be a division idempotent, and let (x, y) ∈ V2(e) be quasi-invertible.If x is invertible, B(x, y) holds by (ii). If x is not invertible then x = 0 andb(x, y) = 1 by (9.7.4) so B(x, y) holds trivially.

The following example shows that for associated idempotents e and f , the Weylrelation W(e) in general does not imply W(f).


13.5. Example. Let k = F2 be the field with two elements, let A = k(ε) bethe ring of dual numbers over k, and let V = (A,A). Then V has precisely twonontrivial idempotents, namely e = (1, 1) and f = (1 + ε, 1 + ε), and they areassociated. Let G ∈ st(V, e) be the Steinberg group St(V, e) as in 13.1. SinceV = V2(e) the relations (StP1) and (StP2) of Theorem 11.2 are vacuous, so thatst(V ) = st(V,P(e)). Hence G is the quotient of the free product Fr(V ) = V + ∗V −by the Weyl relations W(e). Thus G has the following generators and relations.The abelian group A being generated by 1 and 1 + ε, the generators of G are

a = x+(1), b = x−(1), c = x+(1 + ε), d = x−(1 + ε).

The relations defining Fr(V ) are:

a2 = b2 = c2 = d2 = 1, (1)(((((((a, c)))))))

=(((((((b, d)))))))

= 1, (2)

and the additional relations W(e) are as follows. First, define

w := bab.

Then (1) shows already w2 = bab · bab = bab2ab = ba2b = b2 = 1. Hence therelations W(e) are, by Lemma 9.16,

Int(w) · a = x−(Q1 · 1) = x−(1) = b, (3)

Int(w) · c = x−(Q1 · (1 + ε)) = x−(1 + ε) = d, (4)

Int(w) · b = x+(Q1 · 1) = x+(1) = a. (5)

Int(w) · d = x+(Q1 · (1 + ε)) = x+(1 + ε) = c. (6)

Because of w2 = 1, the relations (5) and (6) follow from (3) and (4), and aretherefore superfluous. Moreover, we have

bab = aba, (7)

equivalently, (ab)3 = 1 or (ba)3 = 1. Indeed, by (3),

waw = b ⇐⇒ bab · a · bab = b ⇐⇒ (ba)3 = 1 ⇐⇒ (ab)3 = 1.

From (4) we see that G is already generated by the three elements a, b, c, andthen the relations (1) – (4) are equivalent to

a2 = b2 = c2 = (ab)3 = (ac)2 = 1. (8)

Indeed, define d := Int(w) · c by (4). Then(((((((a, c)))))))

= aca−1c−1 = acac = (ac)2

by (1) and (2), and d2 = 1 follows from c2 = 1 while the relation(((((((b, d)))))))

= 1 is aconsequence of(((((((

b, d)))))))

=(((((((

Int(w) · a, Int(w) · c)))))))

= Int(w) ·(((((((a, c)))))))

= 1.


From (8) we see that G is a Coxeter group with generators a, b, c and Coxeterdiagram a b ∞ c [18, IV, §1.3].

We now discuss the question whether the Weyl relation W(f) holds in G. Let

v = wf = x−(1 + ε) x+(1 + ε) x−(1 + ε) = d c d.

Then v2 = dcd2cd = dc2d = d2 = 1, and the Weyl relations W(f) are

Int(v) · a = Int(v) · x+(1) = x−(Q1+ε1) = x−(1) = b,

Int(v) · c = Int(v) · x+(1 + ε) = x−(Q1+ε(1 + ε)) = x−(1 + ε) = d,

because, as before, these imply the relations Int(v) ·x−(y) = x+(Q1+εy) for y ∈ V −.By Proposition 13.2, the relations W(f) are equivalent to the relations B(e+, e−+f−). Now e− + f− = 1 + (1 + ε) = ε, so we have to work out the relations B(1, ε).By (9.8.3) and (9.8.4), and since all triple products vanish and Q(ε) = 0, theserelations just say that b(1, ε) is central in G. By (9.7.1) we have

b(1, ε) = x−(ε1) x+(1) x−(ε) x+(1ε)

where ε1 and 1ε denote the quasi-inverses of (ε, 1) and (1, ε). By (6.10.2), thequasi-inverses are given by

ε1 =ε

1− ε= ε(1 + ε) = ε, 1ε =

1

1− ε= 1 + ε.

Therefore,b(1, ε) = x−(ε) x+(1) x−(ε) x+(1 + ε) = db · a · db · c,

and we get:G ∈ st(V, f) ⇐⇒ z := dbadbc ∈ Z (G).

We claim that:z ∈ Z (G) ⇐⇒ (bc)3 ∈ Z (G). (9)

Indeed, by (7) and (8),

wb = bab · b = ba, aw = a · aba = ba,

wbaw = (ba)2 = ab.

Hence, using the above formulas as well as(((((((a, c)))))))

= 1,

z = dbadbc = wcw · ba · wcw · bc = wc · wbaw · cwbc= wc · ab · cwb · c = abac · abc · wbc = abac · abc · bac= abac · abc · bca = a · bc · a2 · bc · bc · a = Int(a) · (bc)3,

which proves (9).A necessary condition for v to be a Weyl element is the symmetry condition

(9.16.5)v = d c d = x+(1 + ε) x−(1 + ε) x+(1 + ε) = c d c,

equivalently, (dc)3 = 1. This condition is equivalent to


(bc)6 = 1. (10)

Indeed, dc = wcw · c = (wc)2, and by (8),

(wc)2 = abac · abac = abca · abca= a · bca2 · bc · a = a · (bc)2 · a = Int(a) · (bc)2,

and therefore (dc)3 = (wc)6 = Int(a) · (bc)6. By a standard fact on Coxeter groups,see [18, V, §4.3, Proposition 4], the element bc has infinite order in G, so (10) doesnot hold, and therefore the relation W(f) is not a consequence of W(e).

For the convenience of the reader we give an elementary argument that bc hasindeed infinite order in G. To do so, we define a homomorphism ϕ from G into asuitable group H for which ϕ(bc) has infinite order.

Let H be the subgroup of GL4(Z) of all matrices of the form(

1 0x D

)where x is

a column vector of size 3 and D ∈ GL3(Z). Let Pij denote the 3 × 3-permutationmatrix corresponding to the transposition of i and j in S3, and let e1, e2, e3 be thestandard column basis vectors of Z3. Consider the following elements of H:

A =

(1 00 P12

), B =

(1 00 P23

), C =

(1 0e3 −13

).

By simple matrix computation, one shows that

A2 = B2 = (AB)3 = 1.

Moreover, P12e3 =

(0 1 01 0 00 0 1

)(001

)= e3, and therefore

C2 =

(1 0e3 −13

)·(

1 0e3 −13

)=

(1 0

e3 − e3 13

)= 14,

AC =

(1 00 P12

)·(

1 0e3 −13

)=

(1 0

P12e3 −P12

)=

(1 0e3 −P12

),

(AC)2 =

(1 0e3 −P12

)·(

1 0e3 −P12

)=

(1 0

e3 − P12e3 (−P12)2

)= 14.

Since G is presented by generators a, b, c and relations (8), there is a unique homo-morphism ϕ: G→ H mapping a 7→ A, b 7→ B and c 7→ C.

It remains to show that BC has infinite order. We compute

BC =

(1 00 P23

)·(

1 0e3 −13

)=

(1 0e2 −P23

),

(BC)2 =

(1 0e2 −P23

)·(

1 0e2 −P23

)=

(1 0

e2 − e3 13

),

and conclude from the second formula

(BC)2n =

(1 0

n(e2 − e3) 13

)for all n ∈ N, so BC has infinite order in H.


13.6. The linear Steinberg group St2(A). Let A be a unital associative k-algebra and denote by A× the invertible elements of A. Recall from 5.14 that thelinear Steinberg group St2(A) is the group presented by generators xσ(a), a ∈ A,and relations

xσ(a+ b) = xσ(a)xσ(b), (1)

wσ(u)x−σ(a)wσ(u)−1 = xσ(uau), (2)

where σ = ±, a, b ∈ A, u ∈ A× and wσ(u) := x−σ(u)xσ(u−1)x−σ(u).We show how this group fits into our setup. Let V = (A,A) be the Jordan pair

with quadratic operators Qxy = xyx. Then e = (1, 1) is an invertible idempotentof V , thus V = V2(e). By 6.17 the set S of idempotents associated with e consistsexactly of the pairs f = (u, u−1), u ∈ A×. As V2(f) = V for all f ∈ S , therelations B(xi, yj) (i 6= j) are vacuous, while the relations W(f) are precisely therelations (2). Thus we see that

St2(A) = St(V,S ). (3)

This together with 13.4 implies that the relations B(x, y) hold in St2(A) for allquasi-invertible pairs (x, y) ∈ (A,A) with x or y ∈ A×. For example, let V = (Z,Z)and e = (1, 1)). Then

St2(Z) = St(V, e). (4)

Indeed, St2(Z) = St((V, e,−e) by (3) so that the claim follows from St(V, e) =St(V, e,−e) by (12.1.3).

13.7. Proposition. Let e and f be associated idempotents and put Vi =Vi(e) = Vi(f). We use the abbreviation

QST = (QS+T−, QS−T+) and QgS = (Qg+S

−, Qg−S+) (1)

for subpairs S and T of V and g = (g+, g−) ∈ V . Also let

I2 = V2 V1 V1+QV1V0 +QV2

QV1V0 (2)

be the Peirce 2-component of the ideal generated by V1, cf. (10.9.3). Consider agroup G ∈ st(V, e). Then the relations

wf · xσ(v2) · w−1f = x−σ(Q(f−σ)v2) (3)

hold for all v2 ∈ Iσ2 , σ ∈ +,−. Hence, if I2 = V2, then the relation W(f) holdsautomatically in G, and therefore G ∈ st(V, f). This is in particular the case if

(i) V is simple and V1 6= 0, or

(ii) there exists an idempotent which is collinear to e or which governs e.

Proof. Since e and f are associated, we have P(e) = P(f). Hence (12.2.12)and (12.2.13) show that (3) holds for v2 ∈ V σ2 V −σ1 V σ1 . We claim that G satisfiesthe shift relations S10 and S01 for f . Indeed, since G ∈ st(V, e) it follows fromTheorem 12.5(iii) that all shift relations for e hold in G. In particular, G satisfies


(12.4.1). Putting (x2, y2) = (f+, f−) in this formula and using Vi(e) = Vi(f) fori = 0, 1, 2, proves the shift relation S10 for the idempotent f . Analogously, theshift relation S01 for f follows from (12.4.2) for e. Hence the assertion followsfor elements in Q(V σ1 )V −σ0 from S′10 and S′01 of 12.3. Finally, since −f is alsoassociated with e and w−f = w−1

f while Q(−fσ) = Q(fσ), we have, by what wejust proved,

w−f · x−σ(Q(y1)z0) · wf = xσ(Q(fσ)Q(y1)z0)

and hence

wf · xσ(Q(fσ)Q(y1)z0) · w−1f = x−σ(Q(y1)z0) = x−σ(Q(f−σ)Q(fσ)Q(y1)z0),

because Q(f−σ)Q(fσ) is the identity on V −σ2 . By (10.8.3), Q(V σ2 )Q(V −σ1 )V σ0 =Q(fσ)Q(V −σ1 )V σ0 , so the proof is complete.

It remains to show I2 = V2 if either (i) or (ii) hold. This is proved in 10.9(c).

We now turn to st(V,S ) where S = e, f consists of two compatible butnot associated idempotents. The following are consequences of Corollary 11.10 and(12.1.4):

EA(V, e) ⊂ NormA(G) for all G ∈ st(V, e), (4)

st(V, e) = st(V, h(e)) for all h ∈ EA(V, e), (5)

where EA(V, e) is the group of e-elementary automorphisms of V as defined in10.10. We will describe the relation between st(V, e) and st(V, f) for collinear andgoverning idempotents. Before dealing with the first case, we establish the followingpreliminary results.

13.8. Lemma. Let (x, y) ∈ V with Qxy = 2x. Then (x, y) is quasi-invertibleand the inner automorphism β(x, y) =

(B(x, y), B(y, x)

)is involutorial. If in

addition QxQyx = 4x then

xy = −x. (1)

Proof. By (JP25) we have B(x, y)2 = B(2x − Qxy, y) = B(0, y) = Id andB(y, x)2 = B(y, 2x − Qxy) = Id, which implies the first claim. Now let alsoQxQyx = 4x. Then the quasi-inverse is xy = B(x, y)−1(x − Qxy) = −B(x, y)x =−x+ xyx −QxQyx = 3x−QxQyx, completing the proof.

13.9. Lemma. Let a and b be elements of a group G satisfying the braid rela-tion

aba = bab (1).

Then the following relations in G are equivalent:(aba−1

)−1 = bab−1, (2)

ba2b = a2, (3)

ab2a = b2, (4)

(ab)3 = a4 = (aba−1)4 = (bab−1)4 = b4 = (ba)3. (5)


Proof. (2) ⇐⇒ (3): We have (2) ⇐⇒ ab−1a−1 = bab−1. On the otherhand, (1) is equivalent to bab−1 = a−1ba. Hence (2) ⇐⇒ ab−1a−1 = a−1ba ⇐⇒a2b−1 = ba2 ⇐⇒ (3).

(2) ⇐⇒ (4) follows from what we just proved and the observation that (1) and(2) are symmetric in a and b.

(3) =⇒ (5): (ab)3 = aba · bab = aba2ba (by (1)) = aa2a (by (3)) = a4.Furthermore, (ba)3 = bab · aba = bab2ab (by (1)) = bb2b (by (4)) = b4, and(ab)3 = aba · bab = (bab) · (bab) (by (1)) = bab2ab = (ba)3. Finally, (aba−1)4 =ab4a−1 = aa4a−1 = a4, and the rest follows by symmetry.

(5) =⇒ (3): a4 = (ab)3 = aba · bab = aba · aba (by (1)) = aba2ba impliesa2 = ba2b.

13.10. Lemma. Let e, f ∈ V be two idempotents with f ∈ V1(e), and letG ∈ st(V,P(e)). Recall the elements we, wf and wfop = x+(f+) x−(f−) x+(f+)defined in (9.11.2) for any idempotent, and put

wf,e := Int(we) · wf , wopf,e := Int(we) · wfop . (1)

Then

wf,e = b(e+, f−) b(−f+, e−) b(e+, f−), (2)

wopf,e = b(−f+, e−) b(e+, f−) b(−f+, e−), (3)

and these elements belong to N ∩ G0 (notations of Lemma 9.2). For z ∈ V σ wehave

Int(wf,e) · xσ(z) = Int(wopf,e) · xσ(z) = xσ

(ωf,e · z

)(4)

with ωf,e ∈ EA(V, e) as in (10.11.1).

Proof. Formula (2) is immediate from (12.2.2) and (12.2.4):

Int(we) · wf = Int(we) ·(

x−(f−) x+(f+) x−(f−))

= b(e+, f−) b(−f+, e−) b(e+, f−),

and the proof of (3) follows similarly since wfop = x+(f+) x−(f−) x+(f+). Theo-rem 11.2(iii) shows that these elements normalize U+ and U−. From (2) and (9.7.3)we have π(wf,e) = π(wop

f,e) = ωf,e ∈ EA(V, e) ⊂ Inn(V ). Now Inn(V ) ⊂ H by 7.7

and H = PE0(V ) by 10.1, so the definition G0 = π−1(PE0(V )

)in 9.2 shows that

wf,e and wopf,e belong to N ∩G0. Hence (4) follows from 9.8.

13.11. Proposition. Let e and f be collinear idempotents in V .

(a) Then

st(V, e) = st(V, f) = st(V, e, f). (1)

(b) Let wf,e = we wf w−1e be as in (13.10.1). Then the following relations hold

in any group G ∈ st(V, e):


we wf we = b(e+ + f+, e− + f−) = wf we wf , (2)

wf,e = b(e+ + f+, e− + f−) b(2e+, e−)−1 (3)

= b(2f+, f−)−1 b(e+ + f+, e− + f−), (4)

w−1f,e = we,f , (5)

w2e = wfw2

ewf , (6)

w2f = wew

2fwe, (7)

(wf we)3 = w4

f = w4e,f = w4

f,e = w4e = (we wf )3, (8)

and (wf we)3 is central in G.

(c) Let ωf,e = π(wf,e

)∈ EA(V, e) as in (10.11.1), and let se and sf be the

Peirce reflections with respect to the idempotents e and f respectively, cf. (6.14.6).Then

τe,f = τf,e := β(e+ + f+, e− + f−) = ωf,ese = sfωf,e (9)

is an involutorial automorphism satisfying τe,f (e) = f and τe,f (f) = e.

Proof. (a) For collinear idempotents e and f the automorphism h = ωf,e of10.12 lies in EA(V, e) and maps e to f . Hence (a) follows from (13.7.5).

(b) Let us put x = e+ + f+ and y = e− + f− for short. Then a simplecomputation using the Peirce rules and the fact that e and f are collinear showsthat Qxy = 2x and QxQyx = 4x. Hence xy = −x and yx = −y by (13.8.1) andsymmetry. Now we use the definition of b(x, y) in formula (9.7.1) and compute

b(x, y) = x−(− yx) x+(x) x−(y) x+(−xy) = x−(y) x+(x) x−(y) x+(x)

= x−(f−)

x−(e−)x+(e+)

x+(f+)x−(e−)

x−(f−)x+(f+)

x+(e+)

= x−(f−) we x−(−e−) x+(f+) x−(e−) x+(−f+) wf x+(e+)

(since wf = x+(f+)x−(f−)x+(f+) by (9.16.5) and G ∈ st(V, f))

= we ·

w−1e x−(f−) we

·(((((((

x−(−e−), x+(f+))))))))· wf x+(e+)

= we · b(−e+, f−) · b(f+, e−) · wf x+(e+)

(by (9.11.5), (12.2.4), (9.7.2) and (10.7.3))

= we ·(wf x+(e+) w−1

f

)·(wf x−(e−) w−1

f

)·(wf x+(e+)

)(by (12.2.2) and (12.2.4))

= we · wf ·(x+(e+) x−(e−) x+(e+)

)= we wf weop = we wf we,

(by (9.16.5) and G ∈ st(V, e)).

The second formula of (2) follows by symmetry. Since w2e = b(2e+, e−) by (13.3.1),

(3) follows from (2) and wf,e = wewfwew−2e . Using the symmetry in (2) we obtain

(4):

wewfw−1e = w−1

f wewf = w−2f wfwewf = b(2f+, f−)−1b(f+ + e+, f− + e−).

By (2) and Lemma 13.9, (5) – (8) are all equivalent. We prove (6) in the form

w2e w−1

f w−2e = wf . (10)


Recall that w2e = b(e+, 2e−) normalizes all root groups by Corollary 13.3, whence

B(e+, 2e−) holds in G by (9.8.5). Since π(b(e+, 2e−)

)= β(e+, 2e−) = se is the

Peirce reflection with respect to e, an involutorial automorphism of V , we getw2exσ(z1)w−2

e = xσ(se(z1)

)= xσ(−z1) for all z1 ∈ V σ1 (e). Keeping in mind that

w−1f = w−f by (9.11.5), (10) now follows easily.

(c) The equations in (9) follow from (3) and (4). They imply τf,e(e) = f andτf,e(f) = e in view of (10.12.2). That τ2

f,e = Id follows from Lemma 13.8.

One of the goals of the remainder of this section is to describe st(V, e, g) forgoverning idempotents e a g (Proposition 13.14). We will use the following tworesults in the proof of Proposition 13.14.

13.12. Lemma. Let (e1, . . . en) be an orthogonal system of idempotents in V .Then

st(V, e1, . . . , en) ⊂ st(V, e1 + · · ·+ en).

Proof. By induction it suffices to prove this for n = 2. Thus, by 6.16, we havea Peirce decomposition V =

⊕i,j∈0,1,2 Vij of V such that the Peirce spaces of

c = e1 + e2 are V2(c) = V11 ⊕ V12 ⊕ V22, V1(c) = V01 ⊕ V02 and V0(c) = V00. Wemust verify that any group G ∈ st(V, e1, e2) satisfies the Peirce relations (StP1)and (StP2) for P = P(c) and also the Weyl relation W(c).

To check (StP1), i.e., B(V +i , V

−j ) for |i − j| = 1, we have to consider the

cases (ij) ∈ (21), (10), (12), (01). By (9.8.5) and (9.10.3) it suffices to show thatbσ(xi, yj) normalizes U± for (ij) = (21) and (ij) = (10), and this will follow if wecan write bσ(xi, yj) as a product with factors of type bσ(u, v) for u, v in differentPeirce spaces with respect to some ei. We consider only the case σ = +, the caseσ = − then follows by passing to V op. For x = x11 + x12 + x22 ∈ V +

2 (c) withxij ∈ Vij and y = y01 + y02 ∈ V −1 (c) we get, using (9.9.1),

b(x, y) = b(x12 + x22, (y01 + y02)x11

)· b(x11, y01 + y02).

Here b(x11, y01 + y02) = b(x11, y10) ∈ N = NormG(U+) ∩ NormG(U−) by (StP1)and (StP2) for P = P(e1), and

(y01 + y02)x11 = y01 + y02 + y00 for y00 = Q(y01)x11 ∈ V −00

by (10.7.1). Thus it remains to consider

b(x12 + x22, (y01 + y02)x11

)= b(x12 + x22, y01 + y02 + y00)

= b(x22, (y01 + y02 + y00)x12) · b(x12, y01 + y02 + y00). (1)

By (10.7.2) for the Peirce grading P(c), we have yx1200 = y00 and xy0012 = x12. Hence,

by (6.11.4), (10.7.17) and (10.7.3),

(y01 + y02 + y00)x12 = yx1200 +B(y00, x12)−1 · (y01 + y02)(x

y0012 )

= y00 + (y01 + y02)x12

= y00 + y01 + y02 +Q(y01 + y02)x12

= y00 + y01 + y02 + y01x12y02.


Therefore, by y01x12y02 ∈ V00 and (11.4.1) for P = P(e2), the first factor on theright hand side of (1) becomes b(x22, y01+y02+y00+y01x12y20) = b(x22, y02) ∈ N .We evaluate the second factor, using (9.9.2) and (11.5.1) for P = P(e1):

b(x12, y01 + y02 + y00) = b(x12, y01) · b((x12)y01 , y02 + y00)

= b(x12, y01) · b(x12, y02 + y00)

= b(x12, y01) · b(x12, y02) · b(x12, y00).

Since all three factors lie in N , we have established the relations B(V +2 (c), V −1 (c)).

We leave it to the reader to show that also B(V +1 (c), V −0 (c)) holds.

Having shown (StP1) we now turn to (StP2). By symmetry it suffices to provebσ(V σ2 (c), V −σ0 (c)) = 1. Again we will do the case σ = + and leave σ = − to thereader:

b(x11 + x12 + x22, y00) = b(x12 + x22, y00x11) · b(x11, y00)

= b(x12 + x22, y00) = b(x12, y00x22) · b(x22, y00) = b(x12, y00).

We now use the shift relation S10 which holds in G to conclude that G satisfies(StP2) for P = P(c): b(x12, y00) = b(e+

1 , e−1 x12 y00) = 1.

It remains to verify the relation W(c), that is, wc xσ(u2) w−1c = x−σ

(Q(c−σ)u2

)for u2 ∈ V σ2 (c). By (StP2) for the idempotent e1 we have wc = w2 w1 wherewi = wei . By linearity in u2, it suffices to check the Weyl relation W(c) foru ∈ V σij ⊂ V σ2 (c). We have, using the Weyl relation (9.16.1) and (12.2.1),

Int(wc) · xσ(u11) = Int(w2)(

Int(w1) · w1xσ(u11))

= Int(w2) · x−σ(Q(e−σ1 )u11)

= x−σ(Q(e−σ1 )u11) = x−σ(Q(c−)u11).

The case u ∈ V σ22 is similar. Finally,

Int(wc) · x+(u12) = Int(w2)(

Int(w1) · x+(u12))

= Int(w2) · b(−u12, e−1 ) (by (12.2.2))

= x−(e−2 u12 e−1 ) (by (12.2.6) and z(u12, e

−1 ) = 1)

= x−(Q(c−)u12).

13.13. Lemma. Let e and g be idempotents in V with e governed by g. Thenst(V, e) ∩ st(V,P(g)) ⊂ st(V, g).

Proof. By Theorem 12.5 we must prove that W(g) holds in every group G ∈st(V, e) ∩ st(V,P(g)), that is,

Int(wg) · xσ(z) = x−σ(Qg−σz), (1)

for all z ∈ V σ2 (g). We do this for σ = + and leave the case σ = − to the reader.

Recall that g governs e if and only if g ∈ V1(e) and e ∈ V2(g). In particular,we can form the element wg,e = Int(we) ·wg of 13.10. After conjugating both sideswith we, (1) for σ = + is equivalent to


Int(wg,e) ·(

Int(we) · x+(z))

= Int(we) · x−(Qg−z). (2)

Since e and g are compatible in the sense of 6.15, an element z ∈ V +2 (g) decomposes

z = z2 +z1 +z0 with zi ∈ Vi(e)∩V2(g), and it suffices to prove (2) for the zi. Hencewe distinguish the following cases.

Case 0: z = z0 ∈ V +0 (e) ∩ V +

2 (g). Then Int(we) · x+(z0) = x+(z0) by (12.2.1).By Lemma 13.10, wg,e normalizes U+ and U−, and the action of π(wg,e) = ωg,e ∈Aut(V ) on V is described in Proposition 10.13. Hence the left hand side of (2) is

Int(wg,e) · x+(z0) = x+(ωg,e · z0) = x+(Qe+Qg−z0).

On the right of (2) we have, since g ∈ V1(e) and hence Qg−z0 ∈ V −2 (e) by the Peircerules, that

Int(we) · x−(Qg−z0) = x+(Qe+Qg−z0)

by W(e), as desired.

Case 1: z = z1 ∈ V +1 (e)∩V +

2 (g). Then Int(we)·x+(z1) = b(−z1, e−) by (12.2.2).Hence the left hand side becomes, by (9.7.6) and 10.13,

Int(wg,e) · b(−z1, e−) = b(− ωg,e · z1, ωg,e · e−

)= b

(e+, Qg−z1, Qg+e−, Qg−Qe+e−

)=: (∗).

Here Qe+e− = e+, and y1 := Qg−z1 ∈ V −1 (e) while u0 := Qg+e− ∈ V +0 (e) and

v0 := Qg−e+ ∈ V −0 (e). Hence by (12.4.1),

(∗) = b(e+, y1, u0, v0

)= b

(e+, y1, u0, v0

).

Now (Qg+ , Qg−) is an anti-automorphism of V2(g). Hence

y1, u0, v0 = Qg−z1, Qg+e−, Qg−e+ = Qg−z1, e−, e+ = Qg−z1.

On the right of (2) we have Qg−z1 ∈ V −1 (e), so by (12.2.4),

Int(we) · x−(Qg−z1) = b(e+, Qg−z1),

and the assertion follows.

Case 2: z = z2 ∈ V +2 (e) ∩ V +

2 (g). Then we have on the left of (2), since W(e)holds for G, that we · x+(z2) · w−1

e = x−(Qe−z2) and therefore, by 10.13,

Int(wg,e) · x−(Qe−z2) = x−(ωg,e ·Qe−z2) = x−(Qg−Qe+Qe−z2) = x−(Qg−z2),

since z2 ∈ V +2 (e). On the right of (2), we have, since Qg−z2 ∈ V −0 (e), by (12.2.1),

Int(we) · x−(Qg−z2) = x−(Qg−z2),

as desired.


13.14. Proposition. Let V be a Jordan pair, and let e and g be idempotentsin V with g governing e. Then

st(V, e) = st(V, e, g). (1)

Moreover, if f ∈ V is an idempotent which is orthogonal to e and governed by g,then f is associated with the idempotent Qge and

st(V, e) = st(V, f) = st(V, e, g, f). (2)

Proof. Recall from 10.13 that f ′ = Qge is an idempotent governed by g andorthogonal to e. In particular, e+ f ′ is an idempotent, too. We claim

e+ f ′ ≈ g. (3)

Indeed, e+f ′ ∈ V2(g) since e a g ` f ′ and, for the same reason, g ∈ V1(e)∩V1(f ′) ⊂V2(e+ f ′) by (6.16.7), which shows e+ f ′ ≈ g.

The automorphism ωg,e of 10.13 lies in EA(V, e) and maps e to f ′. Hencest(V, e) = st(V, f ′) by (13.7.5), so that st(V, e) = st(V, f ′) = st(V, e, f ′). FromLemma 13.12 we get st(V, e, f ′) ⊂ st(V, e+ f ′), while st(V, e+ f ′) ⊂ st(V,P(e+f ′)) by definition of st(V, e+f ′). Since P(e+f ′) = P(g) by (3), we have st(V, e) ⊂st(V, e)∩ st(V,P(g)), and the latter category is a subcategory of st(V, g) by 13.13.Hence, st(V, e) = st(V, e, f ′) ⊂ st(V, e, g), which proves (1).

For the proof of the second part we first observe that f is associated with f ′.Indeed, f ′ = Qge ∈ V2−0(f) = V2(f) by the Peirce multiplication rules, whilef ∈ V2(g) ∩ V0(e) = V2(e + f ′) ∩ V0(e) = V2(f ′) by (6.16.7). This shows f ≈ f ′.But then st(V, f ′) = st(V, f) because of 13.7(ii). Now (2) follows from (1).

Notes

§10. Peirce gradings were introduced in [75] and further investigated in [62]. The kernel

KerS (10.4) is used in [62] to define a complement of an inner ideal leading to a characterizationof simple Artinian Jordan pairs in terms of complementation of all inner ideals.

For Jordan triple systems, (10.9.2) and (10.9.6) are due to K. McCrimmon [69, 2.13] with a

different proof. He also shows in [69, 3.8] that the Peirce-2 and Peirce-0 space of a tripotent in asimple Jordan triple system are simple or zero.

In view of the correspondence between Jordan pairs and polarized Jordan triple systems, ananalogous result holds for idempotents in simple Jordan pairs. This is still true for arbitrary

Peirce gradings by [4, Theorem 1.4(iii)]. On the other hand, V1(e) need not be simple if V is, see(10.2.1) where V1 is the direct product Mr,q−s(A) ×Mp−r,s(A). For other properties of V thatare inherited by V2 and V0 see [62, 4.1] and [4, 5].

The group EA(V,P) was introduced in [59, 3.1] when P is defined by an idempotent. Because

of (10.10.1), our definition of EA(V, e) agrees with the one there.§11 and §12. Lemma 11.6 and the conjugation formulas of Lemma 12.2 generalize [59,

Proposition 1.9]. For the case of an idempotent Peirce grading, Corollary 11.8 was proved in[28, Theorem 7] and Corollary 11.9 in [28, Corollary 8].§13. The shift of scalars β(λx, y) = β(x, λy), valid for the inner automorphisms β(x, y) and a

scalar λ in the base ring does not hold for the more general b(x, y) ∈ G, except in special cases as

in (13.3.1) above or in (11.5.2). For example, (13.6.4) shows that for V = (Z,Z) and e = (1, 1),we have St2(Z) = St(V, e). Here b(−2e+,−e−) = w2

−e = w−2e by (13.3.1) and (9.11.5). By [74,

p. 82], we has infinite order in St2(Z), so b(−2e+,−e−) 6= b(2e+, e−).

In the setting of Jordan triple systems, the existence of the exchange automorphism τf,e in

(13.11.9) was first established by McCrimmon and Meyberg [72, 1.1].

CHAPTER IV

JORDAN GRAPHS

Summary. This chapter, comprising §§14 – 19, contains the combinatorial foundation of the

book. We consider locally finite root systems R in the sense of [63], endowed with a 3-grading,that is, a decomposition R = R−1 ∪R0 ∪R1 satisfying conditions similar to the short Z-gradings of

Lie algebras, such as the Tits-Kantor-Koecher algebra of a Jordan pair. 3-graded root systems areequivalent to a class of partially directed graphs, called Jordan graphs. The graphical description

is helpful in visualizing 3-graded root systems.

In §14 we associate with every 3-graded root system a graph whose vertex set is R1 and whoseedges reflect the relations between the roots. If the root system is simply laced, the graphs are

well known; they are the rectangular, triangular and octahedral graphs, possibly infinite, together

with the Clebsch and the Schlafli graph on 16 and 27 vertices, respectively. In the doubly-lacedcases, the graphs contain arrows as well as simple edges. For example, the graph associated with

the (essentially unique) 3-grading of the root system C3 is a hexagram:

?????

?? //

?????

__?????oo

In §15 abstract Jordan graphs are introduced as a class of partially directed graphs in termsof the behaviour of the following three basic figures,

// oo

?????

?????

called collision, square and kite, respectively. The defining conditions are of two types: local-global

conditions and closure conditions. To explain the first, given a partially directed graph with vertexset Γ , define a Γ × Γ -matrix

(p(α, β)

)by p(α, β) = 0 if α and β are not connected, p(α, β) = 1 if

α β or α← β, and p(α, β) = 2 if α = β or α→ β. This is a variant of the adjacency matrix.

Then a collision α→ β ← γ must satisfy the condition p(α, δ)−2p(β, δ)+p(γ, δ) = 0 for all δ ∈ Γ .

To illustrate the closure conditions, any arrow α→ β in Γ must generate a collision α→ β ← γ.Similar conditions are imposed on squares and kites. The fundamental Theorem 15.11 establishesa categorical equivalence between 3-graded root systems and Jordan graphs.

In §16 we study subgraphs on three and four vertices (Proposition 16.1) and establish a numberof graph-theoretical properties of Jordan graphs, for example, the fact that they are claw-free and

have diameter at most two. We also prove an analog of the Peirce decomposition with respect toa system of pairwise non-connected vertices (Proposition 16.9).

The study of Jordan graphs continues in §17. We show that a connected Jordan graph is ofone of four types called linear, isolated, hermitian and orthogonal. Independently of the (known)classification of 3-graded root systems, we classify Jordan graphs of hermitian and orthogonal typein Propositions 17.10 and 17.12, and characterize the octahedral graphs in Proposition 17.11.

The usual bases of finite root systems are not the proper tool for 3-graded root systems—for

one thing, they need not even exist in the infinite case. Therefore, we introduce in §18 in their

stead grid bases, following [77], and establish their basic properties. The final §19 deals with

triangles in Jordan graphs and their embeddability in graphs of higher rank.

174

§14] 3-graded root systems 175

§14. 3-graded root systems

14.1. Definition. Let R be a root system in the free abelian group X as in2.8. The reader is reminded that the term “root system” is an abbreviation for“locally finite root system”. A 3-grading of R is a partition

R = R1 ∪R0 ∪R−1

satisfying the following conditions:

(i) if α ∈ Ri, β ∈ Rj and α + β ∈ R then α + β ∈ Ri+j ; in particular,α+ β 6∈ R whenever i+ j 6∈ 1, 0,−1,

(ii) R−1 = −R1,

(iii) every element of R0 is the difference of two elements of R1.

A 3-graded root system is a root system R together with a specified 3-grading. Sincethe abelian group X is generated by R and since by (ii) and (iii) a 3-grading of R isuniquely determined by the subset R1, it is appropriate to denote a 3-graded rootsystem by (R,X,R1) or just (R,R1).

The abelian group X is generated by R. Hence it follows from (ii) and (iii) thatfor a 3-graded root system, X is already generated by R1.

Let (R,R1) and (S, S1) be 3-gradings of root systems (R,X) and (S, Y ), re-spectively. By a morphism from (R,R1) to (S, S1) we mean a homomorphism ofabelian groups f : X → Y satisfying f(Ri) ⊂ Si, for i = 0,±1. Then in particularf(R) ⊂ S, so f is a morphism of root systems in the sense of 2.8. We denoteby RS3 the category whose objects are 3-graded root systems, with morphisms asdefined above. There is an obvious forgetful functor from RS3 to RS.

Isomorphisms are defined as usual. It is easy to see that an isomorphism inRS3 is the same as an isomorphism f : X → Y of abelian groups with the propertyf(Ri) = Si for i = −1, 0, 1.

14.2. 3-graded root systems over the reals. A 3-grading of a root systemover R and morphisms between such root systems are defined in the same way asabove, replacing Z by R, cf. [63, 17.6, 17.1]. We denote the category of 3-gradedroot systems over R thus obtained by RS3,R. The mechanism of extending andrestricting scalars from Z to R and vice versa expounded in Proposition 2.9 worksjust as well for 3-graded root systems and results in an equivalence between thecategories RS3 and RS3,R. The details are left to the reader. This equivalenceallows us to transfer results of [63] for 3-graded real root systems to 3-graded rootsystems over Z. A first example is the following criterion for morphisms. We usethe term embedding of 3-graded root systems to mean a morphism of 3-graded rootsystems which is an embedding of root systems in the sense of 2.15.

14.3. Lemma. Let (R,X,R1) and (S, Y, S1) be 3-graded root systems, and letf : X → Y be homomorphism of abelian groups satisfying f(R1) ⊂ S1. Then thefollowing conditions are equivalent:

(i) f is a morphism of 3-graded root systems,

(ii) for all α, β ∈ R1,

176 JORDAN GRAPHS [Ch. IV

〈α, β∨〉 6= 0 =⇒ 〈f(α), f(β)∨〉 6= 0, (1)

(iii) for all α, β ∈ R1,

〈α, β∨〉6 〈f(α), f(β)∨〉. (2)

Moreover, f is an embedding of 3-graded root systems if and only if equality holdsin (2).

Proof. The equivalences (i) – (iii) are shown in [63, Corollary 18.10]. The lastpart is a special case of [63, Proposition 11.7], because P = R0 ∪R1 is an effectiveparablic subset of R with unipotent part Pu = R1 ([63, 17.7]).

14.4. Examples of 3-graded root systems. (a) Let (R,X) be a root systemand let R = R1 ∪R0 ∪R−1 be a partition of R satisfying the conditions (i) and (ii)of the Definition 14.1. Then this partition is a 3-grading if and only if C ∩R1 6= ∅for every connected component C of R [63, 17.2]. In particular, if R is irreduciblethen 3-gradings are partitions satisfying (i) and (ii) of 14.1.

(b) Let (R,X,R1) be a 3-graded root system over Z and let (S, Y ) be asubsystem of the root system (R,X). Then S = S1 ∪ S0 ∪ S−1 is a partitionsatisfying (i) and (ii) of 14.1. In general, (iii) will not be satisfied so this is nota 3-grading. However, if this is the case, we call it the induced 3-grading . Forinstance, this happens if S = R ∩ spanZ(Σ) for some subset Σ ⊂ R1 ∪ R−1 [63,17.3], or, by (a), if S is irreducible.

(c) Let R =⊕

j∈J R(j) be a direct sum of root systems R(j), j ∈ J , cf. 2.4. If

all R(j), j ∈ J , are 3-graded then R is 3-graded by

R1 =⋃R

(j)1 . (1)

Conversely, if R is 3-graded then each R(j) inherits a 3-grading from R. This isimmediate from the definitions. It is also easily seen that (R,R1) together with the

natural inclusion morphisms (R(j), R(j)1 )→ (R,R1) is a a coproduct in RS3.

(d) The root system (R,X) underlying a 3-graded root system (R,X,R1) isirreducible if and only if the set R1 is connected in the sense of 2.11 [63, 11.9].

14.5. 3-gradings of the classical root systems. We now give the list (upto isomorphism) of 3-gradings of the classical root systems, using the notationsintroduced in 2.16. By (ii) of 14.1 it suffices to specify R1 and R0.

(a) Let R = AK ⊂ X = L0(K) as in (2.16.1), let ∅ 6= I & K, and put J = K I.Then AK has a 3-grading, denoted AI

K or AII∪J and called a rectangular grading ,

given by

R1 = εi − εj : i ∈ I, j ∈ J, R0 = εi − εj : i, j ∈ I or i, j ∈ J ∼= AI × AJ .

Clearly, AIK depends up to isomorphism only on the cardinalities of K and I. For

finite |K| = n and |I| = p we put accordingly, and in analogy to (2.16.6),


AIK = Ap

n = Apn−1. (1)

For historic reasons, a 3-grading AIK with |I| = 1 is called a collinear grading of

AK and denoted by AcollK or A1

K . (The terminology is explained by the fact thatany two roots in (Acoll

K )1 are collinear, as defined in (14.8.7)). In particular, letK = 1, 2 and I = 1. Then R1 = ε1 − ε2 and R0 = 0. After identifyingR with −1, 0, 1 ⊂ Z, this means Ri = i for i ∈ −1, 0, 1. We denote this3-graded root system by Acoll

1 .

(b) R = BI ⊂ X = L (I), |I| > 2: assume that I contains a distinguishedelement, denoted 0. Then

R1 = ε0 ∪ ε0 ± εi : 0 6= i ∈ I, R0 = BI 0

defines a 3-grading, denoted BqfI and called the odd quadratic form grading.

(c) For |I|> 2 the root system R = CI ⊂ X = L2(I) has a 3-grading given by

R1 = εi + εj : i, j ∈ I, R0 = εi − εj : i, j ∈ I ∼= AI ,

called the hermitian grading and denoted CherI . The case I = 0, 1, denoted Cher

2 ,was already used in 11.1. We will often abbreviate εi + εj = εij .

(d) R = DI ⊂ X = L2(I), |I|>3: there are two types of 3-gradings in this case,both of them arising as induced gradings in the sense of 14.4(b) on the subsystemDI of CI and BI , respectively.

(d1) First, we have a 3-grading, denoted DaltI and called the alternating grading.

It is defined by

R1 = εi + εj : i, j ∈ I, i 6= j, R0 = εi − εj : i, j ∈ I ∼= AI ,

and is of course nothing but the 3-grading induced by the hermitian grading CherI

on the subsystem DI of CI .

(d2) Second, let us assume again 0 ∈ I. Then

R1 = ε0 ± εi : 0 6= i ∈ I, R0 = DI 0

defines a 3-grading, called the even quadratic form grading, and denoted DqfI . This

is the 3-grading induced by BqfI on the subsystem DI of BI .

The well-known isomorphism between low rank root systems yield the followingisomorphisms of 3-graded root systems

Bqf2∼= Cher

2 , A14 = Acoll

3∼= Dalt

3 , A24 = A2

3∼= Dqf

3 , Dalt4∼= Dqf

4 . (2)


14.6. 3-gradings and minuscule coweights. Let (R,X) be a root systemand let q: X → Z be a linear form which does not vanish on any connectedcomponent of R and satisfies q(R) ⊂ 0,±1, the restriction of a minuscule coweightof (R,XR) in the terminology of [63, 7.14] to X. (Since C 6= 0 for a connectedcomponent C of R, we have q(R) = 0 if R = 0, while q(C) = 0,±1.) Forsuch a coweight the decomposition

R = R1 ∪R0 ∪R−1 with Ri = q−1(i) (1)

is a 3-grading of R. Conversely, any 3-grading of R arises in this way from a uniqueminuscule coweight of R [63, 17.6].

Examples. (a) We exhibit the minuscule coweights determining the 3-gradingsof the classical root systems introduced in 14.5.

(i) AIK : q is the restriction to L0(K) of the linear form q: Z(K) → Z defined

by q(εi) = 1 if i ∈ I and q(εi) = 0 if i /∈ I.

(ii) In the cases BqfI and Dqf

I , the coweight is given by q(ε0) = 1, q(εi) = 0 fori 6= 0.

(iii) For the cases CherI and Dalt

I , the coweight is the restriction to L2(I) ofthe linear form q: Z(I) → Q given by q(εi) = 1

2 .

(b) Let R be a finite irreducible root system, let B be a root basis of R in thesense of 2.10, and let β ∈ B such that

the β-coefficient of the highest root with respect to B is 1. (2)

Then X =⊕

α∈B Zα and the linear form qβ : X → Z defined by

qβ(α) =

1 α = β,0 α 6= β,

is a minuscule coweight, and q−1β (1) consists of all roots containing β with coeffi-

cient 1 when written as a linear combination of B. Every minuscule coweight arisesin this way from a pair (B, β) satisfying (2) [63, 17.9].

One can now use the well-known description of minuscule weights [63, VIII,§7.3] or, equivalently, the description of highest roots in [18, Planches] to obtain aclassification of the 3-gradings of the finite irreducible root systems (see for example[63, 17.9]). We leave it to the reader to work out the classification of the 3-gradingsof the finite classical root systems and only present the exceptional cases, using theenumeration of root bases from [18, Planches].

(i) R = E6, β = α1 or β = α6; both 3-gradings are isomorphic and aredenoted Ebi

6 ; the 1-part of Ebi6 has 16 roots.

(ii) R = E7, β = α7: this 3-grading is denoted Ealb7 , its 1-part has 27 roots.

The root systems BCn, E8, F4 and G2 do not have 3-gradings. The superscriptsin the notation Ebi

6 and Ealb7 refer to the connection of these 3-graded root systems

with the exceptional Jordan pairs M12(C) (the “bi-Cayley pair”) and H3(C) (the“Albert pair”), see Example (e) and (h) in 6.6: if C is a split octonion algebra thenthese Jordan pairs have root gradings (20.1) of type Ebi

6 and Ealb7 , respectively.


14.7. Classification of 3-graded root systems. The classification of 3-graded real root systems is given in [63, 17.8, 17.9]. In view of the equivalencebetween the categories RS3 and RS3,R (14.2) this also yields a classification of3-graded root systems. With the notation introduced in 14.4 and 14.6(d), anirreducible 3-graded root system is isomorphic to precisely one of the following:

(i) AII∪J , 16 |I|6 |J |,

(ii) BqfI , |I|> 3,

(iii) CherI , |I|> 2,

(iv) DqfI , |I|> 4,

(v) DaltI , |I|> 5,

(vi) Ebi6 or Ealb

7 .

See (14.5.2) for isomorphisms in low ranks.

14.8. Some basic properties of 3-graded root systems. Let (R,R1) be a3-graded root system. We refer to [63, Lemma 18.5] for the proofs of the followingproperties.

For α, β ∈ R1 we have

〈α, β∨〉 ∈ 0, 1, 2, (1)

α− β ∈ R0 ⇐⇒ 〈α, β∨〉 > 0, (2)

α− β /∈ R ⇐⇒ α ⊥ β. (3)

For µ = α− β ∈ R0 the coroot µ∨ is given by

(α− β)∨ = 〈α, β∨〉α∨ − 〈β, α∨〉β∨, (4)

and hence the coroot lattice X∨ is spanned by R∨1 :

X∨ = spanα∨ : α ∈ R1. (5)

Also, in obvious notation,

|〈R0, R∨1 〉|6 1 and |〈R,R∨〉|6 2. (6)

It is immediate from 2.11, (2.10.2) and (1) that for α, β ∈ R1 the possible valuesof 〈α, β∨〉 and 〈β, α∨〉 are as listed in the following table. We also give the symbolicnotation used in the sequel. For easy comparison, the notation and terminologyused in [63] is listed as well.

〈α, β∨〉〈β, α∨〉 Symbol Symbol of [63] Terminology of [63]

0 0 α ⊥ β α ⊥ β α orthogonal to β

1 1 α β α > β α collinear to β

2 1 α→ β α a β α is governed by β

1 2 α← β α ` β α governs β

2 2 α = β α = β α equals β

(7)


In this book, the symbols >, ` and a as well as the terminology “collinear” and“governs” are reserved for relations between idempotents in Jordan pairs (see 6.15)while the graph theoretical symbols α β and α → β are used for elements ofR1.

Let again α, β ∈ R1. It follows easily from the definition of a 3-grading that theopen root intervals (cf. (1.6.3)) satisfy(((((((

α, β)))))))

= ∅ and(((((((α,−β

)))))))⊂ α− 2β, α− β, 2α− β. (8)

More precisely,(((((((α,−β

)))))))has the following structure:

(((((((α,−β

)))))))=

∅ if α ⊥ βα− β if α β

α− β, α− 2β if α→ β

α− β, 2α− β if α← β

α, 0, −α if α = β

. (9)

Indeed, using (1) we get 2α−β ∈ R ⇐⇒ 2α−β ∈ R1 which implies 〈2α−β, α∨〉 =4−〈β, α∨〉62 and hence 〈β, α∨〉 = 2, and similarly α−2β ∈ R implies 〈α, β∨〉 = 2.Thus, by (2.10.2), 2α − β ∈ R and α − 2β ∈ R implies α = β, and in this caseobviously

(((((((α,−β

)))))))= α, 0,−α. If α → β then sα(β) = β − α, sβ(α) = α − 2β

together with the previous case shows(((((((α,−β

)))))))= α−β, α− 2β. The case α← β

follows by symmetry. These last two cases easily imply that(((((((α,−β

)))))))= α− β if

α β. Finally,(((((((α,−β

)))))))= ∅ if α ⊥ β by (3).

14.9. Examples of morphisms. (i) The minuscule coweight q defining a3-grading of R defines a morphism q: (R,R1)→ Acoll

1 .

(ii) Let Cher2 as in 14.5(c) with I = 0, 1. Let (R,R1) be any 3-graded

root system and fix α ∈ R1. Then there exists a unique morphism fα: (R,R1)→ (S, S1) = Cher

2 with the property

R ∩ f−1α (εi + εj) = β ∈ R1 : 〈β, α∨〉 = i+ j (i, j ∈ 0, 1). (1)

Indeed, let q be the minuscule coweight defining the 3-grading of R. Then

fα(x) = q(x) · 2ε0 + 〈x, α∨〉(ε1 − ε0) (2)

satisfies f(Ri) ⊂ Si in view of (14.8.1) and (14.8.6).

(iii) Let I be an index set, partitioned as I = J ∪J ′ and let j 7→ j′ be a bijectionJ → J ′. Consider the 3-grading AJ

I of R = AI as in 14.5(a). Also let S = CJ withthe 3-grading Cher

J of 14.5(c). Then there is a morphism f : AJI → Cher

J given byf(εj) = εj , f(εj′) = −εj for all j ∈ J .

(iv) Let I =⋃j∈J Ij be a partition of I indexed by J . Then there is a morphism

f : CherI → Cher

J with f−1(εj) = εi : i ∈ Ij.


14.10. Mixed graphs. Recall that a mixed graph, also called a partially di-rected graph [33, p. 6, D23], [7, 1.8], is a graph that can have directed and undirectededges. A graph without directed edges will also be called simply laced.

By abuse of notation, we often write Γ both for a graph and its set of vertices.The possible relations between two vertices α, β ∈ Γ are

α β (no edge), α β, α→ β, α← β, α = β.

We will write α ⊥ β in the first case, and α ∼ β otherwise. With Γ we associatethe integral Γ × Γ -matrix

P =(p(α, β)

)(α,β)∈Γ×Γ

defined by

p(α, β) =

0 if α ⊥ β1 if α β or α← β2 if α = β or α→ β

. (1)

Then the graph Γ is uniquely determined by the set Γ of vertices and the matrixP because

α ⊥ β ⇐⇒ p(α, β) = p(β, α) = 0, (2)

α β ⇐⇒ p(α, β) = p(β, α) = 1, (3)

α→ β ⇐⇒ p(α, β) = 2, p(β, α) = 1, (4)

α = β ⇐⇒ p(α, β) = p(β, α) = 2. (5)

Conversely, given a matrix P : Γ × Γ → 0, 1, 2 satisfying (5) and

p(α, β) = 0 ⇐⇒ p(β, α) = 0, (6)

it is clear that there exists a unique mixed graph structure on Γ satisfying (1).Replacing P by its transpose corresponds to reversing all the arrows in Γ . In

particular, P is symmetric if and only if Γ is simply laced. In this case, the matrixP is related to the adjacency matrix A [19, 3.2] by P = 2IΓ +A where IΓ denotesthe identity matrix on the set Γ .

The disjoint union of a finite or infinite family Γj of graphs (with functions pjas above) is the disjoint set-theoretical union Γ =

∐Γj of their vertex sets, and

relations given by

p(α, β) =pj(α, β) if α, β ∈ Γj for some j0 otherwise

.

We will frequently use the concept of an induced subgraph on a subset Γ ′ of thevertex set of a mixed graph Γ , which is the graph with vertex set Γ ′ and all edgesof Γ , directed or not, beginning and ending in Γ ′. If not necessarily all edges of Γare also edges of Γ ′ we will speak of a subgraph.

We give two very simple examples and refer to 14.17 – 14.20 for more.

(a) The graph without edges on a set Γ has P = 2IΓ , thus p(α, β) = 2δα,β(Kronecker’s delta).

(b) The complete graph Γ = KI on a set I has vertex set I and α β for allα 6= β in Γ , with matrix

p(α, β) = 1 + δα,β , (7)

for all α, β ∈ Γ .


14.11. Lemma. Let Γ be a mixed graph with associated matrix P . For a vertexα ∈ Γ we define maps Pα: Γ → Z and Pα: Γ → Z by

Pα(β) = p(α, β), Pα(β) = p(β, α), (1)

for all β ∈ Γ . Also put

X•(Γ ) = spanZPα : α ∈ Γ ⊂ ZΓ , (2)

X •(Γ ) = spanZPα : α ∈ Γ ⊂ ZΓ . (3)

(a) Then X•(Γ ) and X •(Γ ) are free abelian groups.

(b) The maps α 7→ Pα and α 7→ Pα from Γ to X•(Γ ) and X •(Γ ), respectively,are injective.

(c) There exists a unique bi-additive pairing 〈 , 〉: X•(Γ )×X •(Γ )→ Z suchthat

〈Pα, P β〉 = p(α, β) (4)

for all α, β ∈ Γ . This pairing is non-degenerate: if 〈x,X •(Γ )〉 = 0 for somex ∈X•(Γ ) then x = 0, and similarly on the other side.

Proof. (a) By (14.10.1), X•(Γ ) and X •(Γ ) are subgroups of the group B ⊂ZΓ of bounded maps from Γ to Z. By a theorem of Specker and Nobeling [11,Corollary 1.2], B is free abelian, hence so are X•(Γ ) and X •(Γ ).

(b) Suppose Pα = Pβ . Then 2 = Pα(α) = Pβ(α) and 2 = Pβ(β) = Pα(β), soα = β by (14.10.5). The proof for the Pα is analogous.

(c) Since the Pα and Pα generate X•(Γ ) and X •(Γ ) respectively, uniquenessis clear. To prove existence, let x ∈ X•(Γ ) and y ∈ X •(Γ ), so y =

∑β nβP

β

where (nβ)β∈Γ is a family of integers of finite support. We show that

〈x, y〉 :=∑β

x(β)nβ (5)

is well-defined (recall that x ∈ ZΓ so x(β) makes sense). It suffices to show thaty = 0 implies the right hand side of (5) vanishes. Now

y = 0 ⇐⇒ 0 = y(α) =∑β

nβ Pβ(α) =

∑β

p(α, β)nβ (6)

for all α ∈ Γ . Let x =∑αmα Pα where (mα)α∈Γ is a family of integers of finite

support. Then by (1) and (6),∑β

nβ x(β) =∑β

nβ

(∑α

mα Pα

)(β) =

∑β

nβ

(∑α

mα p(α, β))

=∑α

mα

(∑β

p(α, β)nβ

)= 0,

as desired. Specializing x = Pα and y = P β in (5) shows that (4) holds.If 〈x,X •(Γ )〉 = 0 then in particular 〈x, P β〉 = 0 for all β. By (5) this says

x(β) = 0 for all β, and therefore x = 0. The proof on the other side is analogous.


14.12. Rank and morphisms of mixed graphs. By Lemma 14.11(b), themaps α 7→ Pα and α 7→ Pα are injective. To simplify notation, we will identifyα ∈ Γ with Pα ∈X•(Γ ) and write

α∨ := Pα ∈X •(Γ ).

The rank of a mixed graph Γ is defined as the rank of the free abelian groupX•(Γ ):

rankΓ = rank X•(Γ ).

Let Γ and ∆ be mixed graphs. A map f : Γ → ∆ of the vertex sets is called amorphism if it satisfies the following two conditions:

(i) 〈α, β∨〉6 〈f(α), f(β)∨〉 for all α, β ∈ Γ ,

(ii) f extends to a homomorphism X•(f): X•(Γ )→X•(∆) of the associatedabelian groups satisfying Pα 7→ Pf(α).

Condition (i) implies in particular that connected vertices cannot be “torn apart”by a morphism:

α ∼ β =⇒ f(α) ∼ f(β). (1)

But also an arrow α→ β cannot be mapped to an edge f(α) f(β). Rather,

α→ β =⇒ f(a)→ f(β) or f(α) = f(β). (2)

The necessary conditions (1) and (2) are in fact sufficient for (i). Indeed, (i)obviously holds if 〈α, β∨〉 = 0 and follows from (1) if 〈α, β∨〉 = 1. If 〈α, β∨〉 = 2,i.e., α = β or α→ β, then (2) implies (i). On other hand, an edge can be mappedto an arrow and f need not be injective, as Example (a) of 14.16 shows. SinceX•(Γ ) is generated by the Pα, α ∈ Γ , the homomorphism X•(f) in (ii) is unique.

It is clear that the composition of two morphisms is a morphism. We denoteby mgraph the category of mixed graphs, with morphisms defined as above. Thenthe assignments Γ 7→ X•(Γ ) and f 7→ X•(f) define a functor X• from mgraphto abelian groups.

As usual, an isomorphism f : Γ → ∆ is a morphism for which there exists amorphism g: ∆→ Γ such that f g and g f are the identity. Then it is clear from(i) that an isomorphism satisfies

〈α, β∨〉 = 〈f(α), f(β)∨〉 (3)

for all α, β ∈ Γ . Conversely, suppose f : Γ → ∆ is a bijection of the vertex sets oftwo graphs which satisfies (3). Then it follows easily from the definition of X• in(14.11.2) that (ii) holds, so f is an isomorphism. The automorphism group of Γ isdenoted Aut(Γ ).

Examples. (a) Let E = Γ ∪∆ be the disjoint union of two graphs, thus γ ⊥ δfor all γ ∈ Γ and δ ∈ ∆. Then the inclusion map s: Γ → E is a morphism. Indeed,condition (i) is trivially satisfied. To show (ii), let ni ∈ Z and αi ∈ Γ , for i in a finiteindex set I, and assume that

∑i niPαi = 0 in X•(Γ ). Then also

∑niPs(αi) = 0 in

X•(E), i.e.,∑niPs(αi)(ε) = 0 for all ε ∈ E. Indeed, if ε ∈ Γ this is clear, and if

ε ∈ ∆ it follows from the fact that Γ ⊥ ∆.


(b) Let Γ be the graph without edges. Then X•(Γ ) = 2Z(Γ ) is free with basisPα, α ∈ Γ (because Pα(β) = 2δαβ by Example (a) of 14.10). Hence rankΓ =CardΓ , and for an arbitrary map f : Γ → ∆ conditions (i) and (ii) are triviallysatisfied.

(c) The situation is similar for a complete graph Γ as in Example (b) of 14.10.Here, too, X•(Γ ) is free with basis Pα, α ∈ Γ . Indeed, let α1, . . . , αn ∈ Γ bedistinct and assume

∑ni=1miPαi = 0 for mi ∈ Z. Then

n∑i=1

mi p(αi, αj) = 0 (4)

for all j = 1, . . . , n, and the matrix A =(p(αi, αj)

)equals, by (14.10.7), I + E

where I is the n×n unit matrix and E is the n×n matrix with all entries equal to1. It is an exercise in linear algebra to show that detA = n+ 1. Hence (4) impliesmi = 0 for all i = 1, . . . , n. Let ∆ be an arbitrary graph and let f : Γ → ∆ be amap such that α β in Γ implies f(α) ∼ f(β) for all α, β ∈ Γ . Then (i) and (ii)are satisfied, so f is a morphism from Γ to ∆.

(d) Condition (ii) is fairly strong. For example, consider the induced subgraphon a subset Γ of a mixed graph ∆. Then the inclusion map Γ ⊂ ∆ is in generalnot a morphism. While (i) is trivially satisfied, this need not be so for (ii). For aconcrete example, let

Γ = α1// α2 α3oo ⊂ ∆ = α1

// α2 α3oo

α4

OO

Then the matrices P and Q associated with Γ and ∆ are given by

P =

2 2 01 2 10 2 2

, Q =

2 2 0 01 2 1 10 2 2 00 2 0 2

.

In X•(Γ ) we have the relation Pα1+ Pα3

= 2Pα2for the row vectors of P , but the

corresponding relation for the first three row vectors of Q fails. Hence there is nohomomorphism X•(Γ )→X•(∆) satisfying (ii).

(e) A morphism f : Γ → ∆ does in general not induce a group homomorphismh: X •(Γ )→X •(∆) satisfying h(Pα) = P f(α) for all α ∈ Γ . For example, let Γ beas in (c) and let ∆ = δ be the graph with one vertex. Then there is a morphismf : Γ → ∆ given by f(αi) = δ for all i = 1, 2, 3. Indeed, condition (i) is triviallysatisfied. To see that condition (ii) holds observe that X•(Γ ) is free abelian withbasis Pα1

, Pα2, and the relation Pα3

= 2Pα2− Pα1

, while X•(∆) is free abelianwith basis Pδ. Hence there is a homomorphism X•(Γ ) → X•(∆) mapping Pαi toPδ. But a glance at the column vectors of P shows that X •(Γ ) is free abelian withbasis Pα1 , Pα2 and the relation Pα3 = Pα2 −Pα1 . Since X •(∆) is free with basisP δ, there is no homomorphism h: X •(Γ ) → X •(∆) mapping Pαi → P δ for alli = 1, 2, 3.


14.13. Lemma. Let Γ =∐i∈I Γi be the disjoint union of a family (Γi)i∈I of

mixed graphs, see 14.10. Then Γ together with the inclusion morphisms si: Γi → Γof Example 14.12(a) is a coproduct in mgraph. The functor X• of 14.12 commuteswith coproducts:

X•(∐

Γi

)∼=⊕

X•(Γi). (1)

Proof. It is easily seen that the map(⊕

i X•(si))

:⊕

i X•(Γi) → X•(Γ ) isan isomorphism of abelian groups. A straightforward verification then proves that∐i Γi is a coproduct in mgraph.

Remark. A proof analogous to the one in Example (a) of 14.12 shows that theinclusion map si: Γi → Γ extends to a homomorphism X •(si): X •(Γi)→X •(Γ )of abelian groups such that

⊕i X

•(si) :⊕

i X•(Γi)→X •(Γ ) is an isomorphism.

Hence, in addition to (1) we also have

X •(∐

Γi

)∼=⊕

X •(Γi).

14.14. Definition. Let (R,R1) be a 3-graded root system. Comparing (14.8.7)with (14.10.2)–(14.10.5) shows that we obtain a mixed graph Γ with vertex set R1

by definingp(α, β) := 〈α, β∨〉, (1)

for all α, β ∈ R1. We denote the graph thus obtained from (R,R1) by

Γ = G (R,R1).

The class of graphs obtained from 3-graded root systems in this way will be char-acterized in the next section.

14.15. Lemma. Let (R,R1) be a 3-graded root system in X and let Γ be theassociated graph as in 14.14. Then there are unique isomorphisms ϕ: X → X•(Γ )and ψ: X∨ → X •(Γ ) of abelian groups mapping α ∈ R1 to Pα ∈ X•(Γ ) andα∨ ∈ X∨ to Pα ∈X •(Γ ), compatible with the natural pairing X×X∨ → Z of 2.14and the pairing X•(Γ )×X •(Γ )→ Z of (14.11.4). In particular,

rank G (R,R1) = rankR.

Proof. To prove that ϕ is well defined, suppose x =∑niαi = 0 where ni ∈ Z

and αi ∈ R1 for i in a finite index set I. Then by (14.14.1), for all β ∈ Γ ,(∑niPαi

)(β) =

∑ni〈αi, β∨〉 =

⟨∑niαi, β

∨⟩

= 〈x, β∨〉 = 0,

so∑niPαi = 0. This shows that ϕ is well-defined.

By 14.1, X is spanned by R1, so ϕ is unique and surjective. It remains to showthat ϕ is injective. Suppose ϕ(x) =

∑niPαi = 0. Then for all β ∈ Γ = R1,

0 =∑

niPαi(β) =∑

ni〈αi, β∨〉 = 〈x, β∨〉 =2(x |β)

(β |β)


where ( | ) is the normalized inner product of 2.12. Hence also (x |β) = 0 for allβ ∈ R1. Since the normalized inner product is non-degenerate and X is spannedby R1, it follows that x = 0.

Similarly, we have∑i∈I

ni Pαi = 0 (in X •(Γ )) ⇐⇒

∑i∈I

niα∨i = 0 (in X∨).

Indeed,∑ni P

αi = 0 if and only if∑ni P

αi(β) =∑ni p(β, αi) =

∑ni〈β, α∨i 〉 = 0

for all β ∈ Γ . Since X is spanned by R1 = Γ , this is equivalent to∑ni α

∨i = 0 in

X∨. By (14.8.5), X∨ is spanned by R∨1 . As before, we therefore have a well-definedisomorphism ψ: X∨ →X •(Γ ) mapping α∨ to Pα. Finally, it follows from (14.14.1)that these isomorphisms are compatible with the pairing 〈 , 〉 on X•(Γ )×X •(Γ )and the natural pairing on X ×X∨, respectively.

14.16. Proposition. There is a functor G : RS3 →mgraph given by (R,R1)7→ Γ as in (14.14.1) on objects, and by f 7→ f

∣∣R1 on morphisms f : (R,R1) →(S, S1). This functor commutes with coproducts.

Proof. By 14.14, G (R,R1) is a mixed graph. Let f : (R,R1) → (S, S1) be amorphism of 3-graded root systems, and let Γ = G (R,R1) and ∆ = G (S, S1) be theassociated graphs. We show that f

∣∣R1 : Γ → ∆ is a morphism of mixed graphs. Itfollows from (14.3.2) and (14.14.1) that condition (i) of 14.12 is satisfied. To showcondition (ii), let ϕX : X → X•(Γ ) and ϕY : Y → X•(∆) be the isomorphismsof Lemma 14.15, and define a group homomorphism h: X•(Γ ) → X•(∆) by thecommutative diagram

Xf //

ϕX ∼=

Y

∼= ϕY

X•(Γ )

h// X•(∆)

Then h(Pα) = h(ϕX(α)) = ϕY (f(α)) = Pf(α), so h is a group homomorphismextending the map Pα 7→ Pf(α).

It is easy to verify the functor properties of G . That G commutes with coprod-ucts follows from (14.4.1) and the fact that R(j) ⊥ R(k) for j 6= k, see 2.4 and2.11.

Examples. (a) If J has two elements, the morphism of 14.9(iii) induces onthe associated graphs the following map:

ε1 − ε2′

oooooooOOOOOOO

ε1 − ε1′

OOOOOOO

ε2 − ε2′

ooooooo

ε2 − ε1′

2ε1// ε1 + ε2 2ε2

oo


(b) Let α ∈ R1 and let fα: (R,R1) → (S, S1) = Cher2 be the morphism of

14.9(ii). Then gα = G (fα): R1 → S1 = 2ε0, ε0 + ε1, 2ε1 is given by

gα(β) = 2ε0 + 〈β, α∨〉(ε1 − ε0) (1)

for all β ∈ R1. This follows from (14.9.2).

In the following subsections we describe the graphs obtained from the 3-gradedroot systems listed in 14.7 in more detail.

14.17. The complete graphs and the rectangular graphs. The completegraph KI on a set I was introduced in 14.10(b). If I is finite of cardinality n wesimply write KI = Kn.

Now let J be a second set. The rectangular graph KI KJ on the set I × J isthe undirected graph with vertex set I×J , and (i, j) (i′, j′) if and only if eitheri = i′ and j 6= j′, or i 6= i′ and j = j′. When arranging the elements of I × J asa rectangular matrix, two vertices are connected if and only if they lie in the samerow or the same column. If I or J is a singleton then it is clear that KI KJ isisomorphic to KJ resp. KI .

In particular, K2 is an edge, K3 is a triangle, K4 is (the graph of vertices andedges of) a tetrahedron, and so on:

K2 = 1 2 K3 =3

????

1 2

K4 =

4

HHHHHHHHH

1FFFF 3

2

mmmmmmmm

The first proper rectangular graphs are the square and the prism:

K2 K2 =

(1, 1) (1, 2)

(2, 1) (2, 2)

K2 K3 =

(1, 1)MMMM

(1, 3)

(1, 2)

kkkkkkkk

(2, 1)MMMM

(2, 3)

(2, 2)

kkkkkkkkk

(1)

The rectangular graphs are associated with the 3-graded root systems (R,R1) =AII∪J as in 14.5(a):

G (AII∪J) ∼= KI KJ . (2)

Indeed, εi − εj and εi′ − εj′ in R1 are connected by an edge if and only if eitheri = i′ and j 6= j′, or i 6= i′ but j = j′. In the special case where I = i hascardinality one we have AI

I∪J = AcollJ and therefore

G (AcollJ ) = G (A

iI∪J) ∼= K1 KJ ∼= KJ .

In the finite case we get, using the abbreviation (14.5.1),

G (App+q)

∼= Kp Kq and G (Acolln ) ∼= Kn. (3)


14.18. The triangular and extended triangular graphs. Denote by P(I)the power set of a set I and by Pc(I) the set of all elements of P(I) of cardinalityc. The triangular graph TI on the set I is the undirected graph with vertex setP2(I) and matrix

p(α, β) = Card(α ∩ β),

equivalently,

i, j k, l ⇐⇒ Card(i, j ∩ k, l) = 1,

see [106, Example 21.1]. Thus T1 is empty, T2 is a singleton, T3∼= K3 is a triangle,

and

T4 =

14

ooooooooOOOOOOOO

444444444444444

12

444444444444444 13

24 34

23

OOOOOOOOoooooooo

(1)

is the edge graph of an octahedron in 3-space. Here i, j has been replaced by ijto improve readability. The name “triangular graph” stems from the visualizationof the vertices of Tn as the entries of a strict upper triangular matrix (not all edgesbetween vertices are drawn):

T5 =

12

BBBBBB 13 14 15

23

BBBBBB 24 25

34

BBBBBB 35

45

The triangular graphs are the ones obtained from the 3-graded root systems DaltI :

TI ∼= G (DaltI ). (2)

Indeed, it follows easily from 14.5(d1) that an isomorphism is obtained by mappingi, j ∈ TI to εi + εj ∈ R1.

Let TI = TI ∪ P1(I) be the mixed graph with vertex set P1(I) ∪ P2(I) andrelations

p(α, β) =

0 if α ∩ β = ∅2 if α = β or α & β1 otherwise

,

called the extended triangular graph because it contains TI as the induced subgraphon the subset P2(I). For small values of n = |I| we have


T2 = 11 // 12 22oo , T3 =

11

????

12 13

22

??// 23

???? 33

__????oo

(3)

called a collision and a hexagram, respectively, and

T4 =

11

444444444444444

14

ooooooooOOOOOOOO

444444444444444

12

444444444444444 13

44

OO

wwoooooooo

''OOOOOOOO

24 34

22 //

77oooooooo

DD23

OOOOOOOOoooooooo

33oo

ggOOOOOOOO

ZZ444444444444444

(4)

Again, the symbol ij stands for the subset i, j of I.

The extended triangular graphs are obtained from the 3-graded root systemsCherI of 14.5(c),

TI ∼= G (CherI ), (5)

an isomorphism being given by mapping i, j ∈ TI to εi + εj .

14.19. The octahedral and extended octahedral graphs. Let I be a non-empty set. The octahedral graph OI , also known as the cocktail party graph [33,p. 719], is the simply-laced graph with vertices I × +,− and edges

(i, σ) (j, τ) ⇐⇒ i 6= j,

equivalently,

(i, σ) ⊥ (j, τ) ⇐⇒ i = j and σ 6= τ.

Let us define a map α 7→ α′ of the vertex set by (i, σ)′ = (i,−σ). Then α′′ = α and

α β ⇐⇒ α 6= β′ ⇐⇒ α′ β′,

for α 6= β in OI ; equivalently, α ⊥ β if and only if β = α′ if and only if α′ ⊥ β′. Inparticular, α 7→ α′ is an automorphism of OI .

For I finite of cardinality n we write simply OI = On. Then O0 is empty, O1

consists of two not connected vertices, O2∼= K2 K2 is a square, and O3 is the

graph of vertices and edges of a regular octahedron in 3-space, whence the name:


O2 =

(1,+) (2,+)

(2,−) (1,−)

O3 =

(1,+)

oooo OOOO

44444444444444

(2,+)

44444444444444(3,+)

(3,−) (2,−)

(1,−)

OOOO oooo

(1)

A comparison with (14.18.1) shows that T4∼= O3. Let 0 be an element not contained

in I and put formally I + 1 = I ∪ 0. Then these graphs are obtained from the

3-graded root systems DqfI+1:

OI ∼= G (DqfI+1). (2)

Indeed, by 14.5(d2), R1 = ε0 ± εi : i ∈ I, with relations

α ⊥ β ⇐⇒ (α |β) = 0, α β ⇐⇒ (α |β) = 1,

so an isomorphism is obtained by

(i, σ) 7→ ε0 + σεi. (3)

Let ω be an element not in I×+,−. We define the extended octahedral graph

by OI = OI ∪ ω with the additional relations α → ω for all α ∈ OI . As before

we write OI = On if I is finite of cardinality n. Then O0 = ω consists of a

single vertex, O1∼= T2 is a collision, and for n = 2 we have, putting α = (1,+) and

β = (2,+),

O2 =

α

???? β

ω

β′

??α′

__????(4)

called a pyramid. For n = 3 and with γ := (3,+) the picture is

O3 =

α

4444444

OOOOOOOOOOOOOOO β

ooooooooooooooo

4444444

γ′ //

555555

OOOOOOOOOOOOOOO ω γoo

β′

EE

oooooooooooooooα′

ZZ4444444

(5)

These are the graphs obtained from the 3-graded root systems BqfI+1 of 14.5(b):

OI ∼= G (BqfI+1), (6)

an isomorphism being given by (3) and ω 7→ ε0.


14.20. The Clebsch graph and the Schlafli graph. Recall that the sym-metric difference of two sets α, β is

α4 β = (α ∪ β) (α ∩ β) = (α β) ∪ (β α).

Let I be a set with five elements and let Γ be the undirected graph whose verticesare the subsets of even cardinality of I,

Γ = Pev(I) = ∅ ∪P2(I) ∪P4(I),

and with edgesα β ⇐⇒ Card(α4 β) = 2. (1)

Thus Γ has(

50

)+(

52

)+(

54

)= 1 + 10 + 5 = 16 vertices. It is called the Clebsch graph

and denoted by Cl. This is the graph obtained from the 3-graded root system Ebi6 :

Cl ∼= G (Ebi6 ). (2)

For the proof of (2), we let I = 1, . . . , 5. By definition in 14.6 and 14.14, thevertex set of G (Ebi

6 ) is R1 = α ∈ R : q(α) = 1 where q is a minuscule coweight ofR = E6. Using the description of E6 in [18, VI, §4.12] and taking $1 of loc. cit. asminuscule coweight, we find that R1 consists of the roots

α =1

2

(ε8 − ε7 − ε6 +

∑i∈I

(−1)ν(i)εi

),∑i∈I

ν(i) even, (3)

where ν is a map from I to 0, 1, that is, the characteristic function of the subsetν−1(1) of I. Since ∑

i∈Iν(i) = Card ν−1(1),

(3) yields a bijection α 7→ ν−1(1) between R1 and Pev(I). We use (14.12.3) to showthat this bijection is an isomorphism of graphs and calculate 〈α, β∨〉 for α as in (3)and β given analogously with µ. Using (2.12.1) and (β |β) = 2, and identifying αand β with the subsets ν−1(1) and µ−1(1) of I, respectively, we get

〈α, β∨〉 = 2(α |β)

(β |β)= (α |β) =

1

4

(3 +

∑i∈I

(−1)ν(i)+µ(i))

=1

4

(3 + |I (α ∪ β)| − |α β| − |β α|+ |α ∩ β|

)=

1

4

(3 + (5− |α ∪ β|)− |α4 β|+ |α ∩ β|

)=

1

4

(8− 2|α4 β|),

whence 〈α, β∨〉 = 1 ⇐⇒ |α4 β| = 2.There is some confusion about the name “Clebsch graph”. Our terminology

follows [86] and [19], whereas in [106, Example 21.4] the name Clebsch graph isused for the complement of Γ .


Let J be a set of six elements and let Sch be the simply laced graph with vertexset P1(J) ∪P2(J) ∪P5(J) and edges

α β ⇐⇒ Card(α4 β) ∈ 2, 3, 6,

known as the Schlafli graph. It has(

61

)+(

62

)+(

65

)= 6 + 15 + 6 = 27 vertices, and

is the graph associated with Ealb7 :

Sch ∼= G (Ealb7 ). (4)

To prove (4), we let J = 1, . . . , 6 and consider the root system E8 as in [18,§4.10], with non-zero roots

± εi ± εj , i < j, (5)

1

2

( 8∑i=1

(−1)ν(i) εi

),

8∑i=1

ν(i) ≡ 0 ( mod 2), (6)

where ν: 1, . . . , 8 → 0, 1.For any α ∈ E×8 , we have E7

∼= β ∈ E8 : 〈β, α∨〉 = 0. Indeed, by [18], this istrue for α = ε7 + ε8 and hence, by transitivity of the Weyl group on E×8 , it holdsfor all α ∈ E×8 . For our purposes, it is most convenient to choose

α =1

2(ε1 + · · ·+ ε6 − ε7 − ε8),

so thatR := β ∈ E8 : 〈β, α∨〉 = 0 ∼= E7.

Let ε∗i be the linear form given by 〈εi, ε∗j 〉 = δij . We claim that

q = ε∗1 + · · ·+ ε∗6

is a coweight of E8 which induces a minuscule coweight on E7. Indeed, it is clearthat q takes integral values on roots of type (5). For a root of type (6) we have

q(1

2

8∑1

(−1)ν(i)εi

)=

1

2

6∑1

(−1)ν(i) =1

2(6− 2|K|),

where K = i ∈ J : ν(i) = 1, so q takes integral values on E8.To prove the second statement, we must show that q takes values in 0,±1 on

E7. Letf = 2α∨ = ε∗1 + · · ·+ ε∗6 − ε∗7 − ε∗8 = q − ε∗7 − ε∗8. (7)

Then β ∈ E7 if and only if f(β) = 0.The roots in E7 of type (5) are

± (εi − εj) (16 i < j 6 6), ±(ε7 − ε8), (8)

± (εi + ε7), ±(εi + ε8) (i = 1, . . . , 6). (9)


Since q = f + ε∗7 + ε∗8 by (7), q vanishes on the roots of type (8), and has value ±1on those of type (9).

For a root β as in (6) we have

2q(β) =

6∑i=1

(−1)ν(i), (10)

2f(β) = 2q(β)− (−1)ν(7) − (−1)ν(8). (11)

Because β ∈ E7 if and only if f(β) = 0 we get from (10) and (11) that 2q(β) =(−1)ν(7) + (−1)ν(8). Since q takes integer values on E8, it follows that 2q(β) ∈0,±2, so q is a minuscule coweight of E7.

We determine R1 = β ∈ E7 : q(β) = 1. For roots of type (8) we have q(β) = 0,and the roots of type (9) with q(β) = 1 are εi + ε7 and εi + ε8, i = 1, . . . , 6.For roots of type (6) in R1 we have 2q(β) = 2 and 2f(β) = 0, whence by (11),

(−1)ν(7) + (−1)ν(8) = 2 so that ν(7) = ν(8) = 0. By (10),∑6

1(−1)ν(i) = 2.

Also, (6) and ν(7) = ν(8) = 0 show∑6

1 ν(i) ≡ 0 ( mod 2). This easily impliesCardi ∈ 1, . . . , 6 : ν(i) = 1 = 2, so that ν corresponds to a two-element subsetof 1, . . . , 6 of which there are

(62

)= 15. Altogether, R1 consists of the 27 roots

εi + ε7, εi + ε8 (i = 1, . . . , 6), βi,j = γ − εi − εj (16 i < j 6 6),

where γ := 12 (ε1 + · · · + ε8). Hence we obtain a bijection Sch → R1 by mapping

subsets of J = 1, . . . , 6 to R1 as follows:

i 7→ εi + ε7, J i 7→ εi + ε8, i, j 7→ βi,j .

It remains to show that this mapping is an isomorphism of graphs. This is astraightforward verification (use (14.12.3) and the fact that β γ in R1 if andonly if (β | γ) = 1 since R is simply laced).

We collect the 3-graded root systems and graphs discussed so far in the followingtable:

Γ Name of Γ CardΓ rankΓ (R,R1)

KI complete |I| |I| AcollI

KI KJ rectangular |I × J | |I|+ |J | − 1 AII∪J

TI triangular(|I|

2

)|I| Dalt

I

TI extended triangular(|I|+1

2

)|I| Cher

I

OI octahedral 2|I| |I|+ 1 DqfI+1

OI extended octahedral 2|I|+ 1 |I|+ 1 BqfI+1

Cl Clebsch 16 6 Ebi6

Sch Schlafli 27 7 Ealb7

(12)


§15. Jordan graphs and 3-graded root systems

15.1. Notation. Let Γ be a mixed graph as in 14.14, and let X•(Γ ) andX •(Γ ) as in Lemma 14.11. We use the conventions introduced in 14.12, so weidentify α ∈ Γ with Pα ∈X•(Γ ) and write α∨ := Pα ∈X •(Γ ). Then by (14.11.4)and (14.10.1),

〈α, β∨〉 = 〈Pα, P β〉 = p(α, β) =

0 if α ⊥ β1 if α β or α← β2 if α = β or α→ β

. (1)

Thus Γ now appears as embedded in the abelian group X•(Γ ), and α 7→ α∨ is awell-defined injective map ∨: Γ →X •(Γ ). The non-degenerate pairing

〈 , 〉: X•(Γ )×X •(Γ )→ Z

of Lemma 14.11(c) induces injective homomorphisms X •(Γ ) → X•(Γ )∗ andX•(Γ ) → X •(Γ )∗ into the respective duals. In particular, we view β∨ for β ∈ Γas a linear form on X•(Γ ) satisfying (1). Since Γ and Γ∨ span X•(Γ ) and X •(Γ ),respectively, Γ∨ and Γ are total subsets, i.e.,

〈x, Γ∨〉 = 0 implies x = 0 for all x ∈X•(Γ ), (2)

and similarly on the other side.If f : Γ → ∆ is a morphism of mixed graphs, the induced homomorphism

X•(f): X•(Γ ) → X•(∆) of 14.12(ii) is then simply given by α 7→ f(α) for allα ∈ Γ ⊂X•(Γ ).

Recall that the induced subgraph on a subset Γ ′ of the vertex set of Γ is thegraph with vertex set Γ ′ and all edges, directed or not, beginning and ending inΓ ′. If P is the matrix associated with Γ as in (14.10.1) then P ′ = P

∣∣Γ ′ ×Γ ′ is thematrix associated with Γ ′.

15.2. Basic configurations. In the sequel, the following mixed graphs onthree and four vertices will play a special role. A triple (α, β, γ) of elements of Γis called a collision if α→ β ← γ and α ⊥ γ, equivalently, if the induced subgraphΓ ′ on α, β, γ and the corresponding matrix P ′ are

Γ ′ = α // β γoo , P ′ =

α β γ

α 2 2 0β 1 2 1γ 0 2 2

. (1)

If (α, β, γ) is a collision then so is (γ, β, α). Collisions occurred already in (14.18.3)as the graphs associated with the 3-graded root system Cher

2 .A quadruple (α, β, γ, δ) of elements in Γ is called a square if α β γ

δ α and α ⊥ γ and β ⊥ δ, equivalently, if the induced subgraph Γ ′ and thematrix P ′ are

Γ ′ =

α β

δ γ

, P ′ =

α β γ δ

α 2 1 0 1β 1 2 1 0γ 0 1 2 1δ 1 0 1 2

. (2)

§15] Jordan graphs and 3-graded root systems 195

Evidently, if (α, . . . , δ) is a square then so is (t(α), . . . , t(δ)) for any permutation tof the dihedral group of order eight on the four letters α, . . . , δ. By (14.17.1) and(14.17.2), squares are the graphs associated with the 3-graded root systems A2

4.Finally, a quadruple (δ, α, β, γ) in Γ is called a kite if δ ⊥ β, α ← δ → γ and

α β γ α, equivalently, the induced subgraph Γ ′ and its matrix P ′ are

Γ ′ =

δ

????

α???? γ

β

, P ′ =

δ α β γ

δ 2 2 0 2α 1 2 1 1β 0 1 2 1γ 1 1 1 2

. (3)

It is clear that δ plays a distinguished role among the vertices making up a kite.This is the reason for listing it first. If (δ, α, β, γ) is a kite then so is (δ, γ, β, α). Inparticular, α, β, γ form a triangle as in 14.17. Since the order of the vertices in atriangle, unlike the case of a square or a kite, is unimportant, we usually write atriangle as α, β, γ.

In [63], a different terminology for the basic configurations introduced here wasemployed, as indicated in the following table.

graph graphical terminology terminology of [63]

// oo collision triangle

????

triangle collinear family of 3 elements

square quadrangle

????

????

kite diamond

(4)

15.3. Jordan graphs. We keep the notations introduced in 15.2. A mixedgraph Γ , embedded in X•(Γ ) as in 15.1, is said to be a Jordan graph if it satisfiesthe following conditions:

(a) Local-global conditions: for a collision (α, β, γ),

α− 2β + γ = 0, (1)

and for a square (α, β, γ, δ) or a kite (δ, α, β, γ),


α− β + γ − δ = 0. (2)

(b) Closure conditions:

(C1) Given an arrow α → β in Γ , there exists γ ∈ Γ such that (α, β, γ)is a collision.

(C2) Given an induced subgraph α β γ (thus α ⊥ γ), there existsδ ∈ Γ such that (α, β, γ, δ) is a square.

(C3) Given an induced subgraph δ → α β (thus δ ⊥ β), there existsγ such that (δ, α, β, γ) is a kite.

The conditions (a) are indeed local-global conditions in the following sense. Aglance at the matrix P ′ of (15.2.1) shows that its row vectors satisfy the relationP ′α − 2P ′β + P ′γ = 0, equivalently,

〈α− 2β + γ, ξ∨〉 = 0 for all ξ ∈ α, β, γ,

while (1) is equivalent to

〈α− 2β + γ, ξ∨〉 = 0 for all ξ ∈ Γ .

Similar remarks apply to (2) and the matrices in (15.2.2) and (15.2.3).The axioms in (b) say that certain configurations can be completed as indicated,

so they have the character of existence statements. It is immediate from (1) and(2) that the vertices whose existence is asserted in (b) are unique.

We denote by jgraph the full subcategory of mgraph whose objects are Jordangraphs.

15.4. Examples. (a) A graph without edges or a complete graph (see theexamples in 14.10) is a Jordan graph by default because it contains no collisions,squares or kites.

(b) A disjoint union Γ =∐Γj of graphs is Jordan if and only if each Γj is so.

(c) Evidently, a collision is a Jordan graph and so is a square.

(d) Given a subgraph

α

???? β

ε

(1)

of a Jordan graph, we can complete the arrows α → ε and β → ε to collisions,obtaining a pyramid as in (14.19.4):

α

???? β

ε

δ

??γ

__????

(2)

This will be shown in 15.7. It is easy to see (and also follows from Proposition 15.5in combination with (14.19.6)) that a pyramid is itself a Jordan graph.


(e) A kite (δ, α, β, γ) in a Jordan graph Γ is itself not a Jordan graph becausethe closure condition (C1) is violated. Since Γ satisfies (C1), the arrows δ → α andδ → γ can be completed to collisions. It will be shown in Lemma 16.13 that thisyields the hexagram

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

(3)

as in (14.18.3). As the reader may easily check, a hexagram is a Jordan graph.This also follows from Proposition 15.5 and (14.18.5).

(f) A prism as in (14.17.1) is a Jordan graph, again by Proposition 15.5, becauseit is the graph associated with the 3-graded root system A2

5.

We now show that the mixed graph associated with a 3-graded root system in14.14 is a Jordan graph. The main result of this section (Theorem 15.11) will bethat, conversely, all Jordan graphs arise in this way.

15.5. Proposition. Let (R,X) be a root system and let R = R−1 ∪R0 ∪R1 bea 3-grading as in 14.1. Then the mixed graph Γ with vertex set R1 defined in 14.14is a Jordan graph. Hence the functor G : RS3 → mgraph of Proposition 14.16takes values in jgraph.

Proof. This is proved in [63, Lemma 18.4] but we include a proof for theconvenience of the reader.

We verify the axioms in 15.3(a) and begin with the following observation. LetI be a finite index set and let αi ∈ R1 and ni ∈ Z for i ∈ I. Let P ′ = (p′ij) =(〈αi, α∨j 〉) ∈ Matn(Z), and let P ′i ∈ Zn be the ith row vector of P ′. Put x =

∑niαi.

Then ∑niP

′i = 0 =⇒ x = 0. (1)

Indeed, let ( | ) be the normalized inner product as in 2.12. Then the jth coordinateof∑niP

′i is ∑

i

ni〈αi, α∨j 〉 = 〈x, α∨j 〉 =2(x |αj)(αj |αj)

= 0.

This implies (x |αj) = 0 for all j = 1, . . . , n and therefore also (x |x) = 0. Sincethe inner product is positive definite, the assertion follows.

Let (α, β, γ) = (α1, α2, α3) be a collision, and let P ′ be the corresponding 3×3-matrix as in (15.2.1), with row vectors P ′1, P

′2, P

′3. Evidently, P ′1 − 2P ′2 + P ′3 = 0 so

(1) yields (15.3.1). If (α1, . . . , α4) is a square or a kite then by (15.2.2) and (15.2.3),the alternating sum of the row vectors of P ′ vanishes, whence we have (15.3.2).

It remains to verify the closure conditions 15.3(b). We start with (C1). Givenα → β, put γ := −sβ(α) = −α + 〈α, β∨〉β = 2β − α ∈ R. Observe that sα(β) =β − α ∈ R, and since R is 3-graded, β − α ∈ R0. Hence γ = β + (β − α) ∈(R1 +R0)∩R ⊂ R1, again since R is 3-graded. To show that (α, β, γ) is a collision,it remains to verify α ⊥ γ and γ → β. Now 〈γ, α∨〉 = 〈2β − α, α∨〉 = 2 · 1− 2 = 0.Also, 〈γ, β∨〉 = 〈2β−α, β∨〉 = 4−2 = 2, and 〈β, γ∨〉 = 〈−sβ(β), α∨〉 = 〈β, α∨〉 = 1.


For (C2), let α β γ ⊥ α and put δ = sαsβ(γ) ∈ R. Then sβ(γ) =γ − β ∈ R0 since R is 3-graded, and hence

δ := sα(γ − β) = γ − (β − α) = (γ − β) + α ∈ R ∩ (R0 +R1) ⊂ R1,

again since R is 3-graded. We claim that (α, β, γ, δ) is a square. Indeed, 〈δ, β∨〉 =〈γ − β + α, β∨〉 = 1 − 2 + 1 = 0, 〈δ, α∨〉 = 0 − 1 + 2 = 1 and 〈δ, γ∨〉 = 2 −1 + 0 = 1. Moreover, 〈α, δ∨〉 = 〈sβsα(α), γ∨〉 = −〈α − β, γ∨〉 = 1, and 〈γ, δ∨〉 =〈sβsα(γ), γ∨〉 = 〈γ − β, γ∨〉 = 2− 1 = 1 which proves our claim.

For (C3), let δ → α β ⊥ δ and put γ = sδsα(β) ∈ R. Then sα(β) = β−α ∈R0 since R is 3-graded, and as before,

γ = sδ(β − α) = β − (α− δ) = (β − α) + δ ∈ R ∩ (R0 +R1) ⊂ R1,

again since R is 3-graded. We claim that (δ, α, β, γ) is a kite. Indeed, 〈γ, β∨〉 =〈β−α+ δ, β∨〉 = 2− 1 + 0 = 1, 〈γ, α∨〉 = 1− 2 + 2 = 1, and 〈γ, δ∨〉 = 0− 1 + 2 = 1.Furthermore,

〈β, γ∨〉 = 〈sαsδ(β), β∨〉 = 〈sα(β), β∨〉 = 〈β − α, β∨〉 = 2− 1 = 1,

〈α, γ∨〉 = 〈sαsδ(α), β∨〉 = 〈sα(α− δ), β∨〉= 〈−α− δ + 2α, β∨〉 = 1− 0 = 1,

〈δ, γ∨〉 = 〈sαsδ(δ), β∨〉 = −〈sα(δ), β∨〉 = 〈2α− δ, β∨〉 = 2− 0 = 2.

This shows that (δ, α, β, γ) is a kite.

We return to a Jordan graph Γ embedded in X•(Γ ) as before and prove a seriesof lemmas.

15.6. Lemma. A connected induced subgraph on three vertices of a Jordangraph is one of the following:

// // oo (1)

????

????

??

__???? (2)

Proof. It is straightforward to see that the list of all possible connected inducedsubgraphs on three vertices in a mixed graph consists of the six cases listed in (1)and (2), and the following seven additional cases:

// //

????

??

????

?? //

????

??oo

(3)

// oo //

????

(4)

We exclude the subgraphs listed in (3) by showing that


a subgraph α→ β → γ is impossible in a Jordan graph. (5)

Indeed, suppose α → β → γ, complete α → β to a collision α → β ← δ, andcomplete β → γ to a collision β → γ ← ε:

α // β // γ

δ

OO

ε

OO

Then α + δ = 2β and 2γ = β + ε by (15.3.1), and β ⊥ ε. Hence 0 = 2〈β, ε∨〉 =〈α, ε∨〉+ 〈δ, ε∨〉, so ε ⊥ δ. It follows that 2〈γ, δ∨〉 = 〈β, δ∨〉+ 〈ε, δ∨〉 = 1 + 0, whichis impossible since 〈γ, δ∨〉 ∈ N by (15.1.1).

We eliminate the first two cases in (4) by showing that

an induced subgraph α ∼ β → γ is impossible in a Jordan graph. (6)

Indeed, complete β → γ to a collision β → γ ← δ. Then

0 < 〈β, α∨〉 = 2〈γ, α∨〉 − 〈δ, α∨〉 = 0− 〈δ, α∨〉6 0,

contradiction. In the last case of (4), where now α β → γ α, completeagain β → γ to a collision β → γ ← δ. Then

2 = 2〈γ, α∨〉 = 〈β, α∨〉+ 〈δ, α∨〉 = 1 + 〈δ, α∨〉,

so 〈δ, α∨〉 = 1. We also have 〈α, δ∨〉 = 1, else 〈α, δ∨〉 = 2 and we would haveα → δ → γ, which was already shown to be impossible. Thus β α δ ⊥ β,so there exists ε completing (β, α, δ) to a square:

α

>>>>>>

β //

?????? γ δoo

ε

Then by (15.3.2), 〈α, γ∨〉 + 〈ε, γ∨〉 = 1 + 〈ε, γ∨〉 = 〈β, γ∨〉 + 〈δ, γ∨〉 = 2 + 2 = 4,contradiction.

15.7. Corollary. A subgraph as in 15.4(d) generates a pyramid (15.4.2) bycompleting the arrows α→ ε and β → ε to collisions.

Proof. By construction all relations among the five vertices pictured in (15.4.2)are known, except α δ, β γ and γ δ. We show that α δ, theremaining relations then follow by symmetry. From (15.3.1) we have 2ε = α+ γ =β+δ, whence α−β+γ−δ = 0. This implies 〈δ, α∨〉 = 〈α−β+γ, α∨〉 = 2−1+0 = 1,so either α δ or α→ δ. But the second alternative leads to α→ δ → ε, whichis impossible by (15.6.5).


15.8. Lemma. Let Γ be a mixed graph satisfying (15.3.1) and (C1). Then(C3) of 15.3 is equivalent to the condition

(C4) Given an induced subgraphδ

????

α γthere exists β such that (δ, α, β, γ)

is a kite.

Proof. (C3) =⇒ (C4): Complete δ → α to a collision δ → α ← δ′. Then by(15.3.1),

〈δ′, γ∨〉 = 2〈α, γ∨〉 − 〈δ, γ∨〉 = 2 · 1− 2 = 0,

so we have an induced subgraph δ′ → α γ, to which we can apply (C3). Thisyields β in the following diagram:

δ

????

α???? γ

δ′

??// β

????

Then β ⊥ δ because by (15.3.1), 〈δ, β∨〉 = −〈δ′, β∨〉+ 2〈α, β∨〉 = −2 + 2 · 1 = 0.

(C4) =⇒ (C3): Given an induced subgraph δ → α β, complete δ → α to acollision δ → α← δ′:

δ

α????

δ′

??β

Then we have an arrow δ′ → β. Indeed, by (15.3.1),

〈δ′, β∨〉 = 2〈α, β∨〉 − 〈δ, β∨〉 = 2− 0

(since δ ⊥ β), and we cannot have δ′ = β, else 2 = 〈δ′, α∨〉 = 〈β, α∨〉 = 1. Now wehave the diagram

δ

α????

δ′ //

??β

and we apply (C4) to the induced subgraph δ′ → α β ← δ′. So there exists γsuch that (δ′, α, γ, β) is a kite:

δ

α???? γ

δ′ //

??β

It remains to show that δ → γ. By (15.3.1), 〈δ, γ∨〉 = 2〈α, γ∨〉 − 〈δ′, γ∨〉 = 2 − 0,and γ = δ is impossible because 〈δ, α∨〉 = 2 while 〈γ, α∨〉 = 1.


15.9. Definition. Let Γ be a Jordan graph, embedded in X•(Γ ) as in 15.1.Define a subset R0 of X•(Γ ) by

R0 := β − α : α, β ∈ Γ, β ∼ α. (1)

Recall here from 14.10 that β ∼ α means β and α are not orthogonal. Since α = βis allowed in the definition of R0 it is clear that 0 ∈ R0. Also, µ ∈ R0 implies−µ ∈ R0. Every 0 6= µ ∈ R0 can be written in the form

µ = β − α, 〈β, α∨〉 = 1 (2)

for 〈β, α∨〉 defined in (15.1.1). Indeed, if µ = β − α with 〈β, α∨〉 = 2 then β → α.Completing this to a collision β → α ← γ, we have 2α = β + γ by (15.3.1), andtherefore µ = α− γ where 〈α, γ∨〉 = 1.

Next, putR1 := Γ and R−1 := −Γ, (3)

as subsets of X•(Γ ). We claim that

R(Γ ) := R := R−1 ∪R0 ∪R1 (4)

is a disjoint union. Indeed, assume α = −β ∈ Γ ∩ (−Γ ). Then 0 = 〈α + β, α∨〉 =2+〈β, α∨〉>2, contradiction. Next, assume that µ = γ ∈ R0∩Γ . Since 〈γ, γ∨〉 = 2,µ 6= 0, so we may assume it has the form (2). Then 1 = 〈β, α∨〉 = 〈α, α∨〉+〈γ, α∨〉 =2 + 〈γ, α∨〉> 2, contradiction. This also shows R0 ∩ (−Γ ) = ∅ because R0 = −R0.

15.10. Lemma. Let γ ∈ Γ = R1. Then the formula

sγ(x) := x− 〈x, γ∨〉γ

defines a reflection sγ of X•(Γ ) satisfying sγ(γ) = −γ and sγ(R) ⊂ R. Moreover,for all µ ∈ R0,

i := −〈µ, γ∨〉 ∈ 0,±1 and sγ(µ) = µ+ iγ ∈ Ri. (1)

Proof. Since 〈γ, γ∨〉 = 2, it is clear from 2.1 that sγ is a reflection mapping γto −γ. We discuss the possibilities for sγ(δ), δ ∈ R. First let δ ∈ R1.

(i) If δ = γ ∈ R1 then sγ(δ) = −γ ∈ R−1.

(ii) If γ ⊥ δ and δ ∈ R1 then sγ(δ) = δ ∈ R1.

(iii) If δ → γ then 〈δ, γ∨〉 = 2. There is a collision δ → γ ← ε by (C1). Hencesγ(δ) = δ − 2γ = −ε ∈ R−1 by (15.3.1).

(iv) If γ → δ or γ δ then 〈δ, γ∨〉 = 1, and therefore sγ(δ) = δ − γ ∈ R0.

The cases treated so far prove that sγ(R1) ⊂ R and, since sγ is linear, alsosγ(R−1) ⊂ R. It remains to discuss sγ(µ) for µ ∈ R0 and to prove (1).

We may assume 0 6= µ = β − α ∈ R0 as in (15.9.2). From 〈β, α∨〉 = 1 wehave 〈α, β∨〉 ∈ 1, 2. If γ = α then 〈µ, γ∨〉 = 〈β − α, α∨〉 = 1 − 2 = −1 andsγ(µ) = sα(β − α) = (β − α) + α = β ∈ R1, while if γ = β then


i = −〈µ, γ∨〉 = 〈α− β, β∨〉 = 〈α, β∨〉 − 2 =

−1 if 〈α, β∨〉 = 1,0 if 〈α, β∨〉 = 2

,

sγ(µ) = µ+ iβ = (1 + i)β − α =

−α if i = −1,µ if i = 0

∈ Ri.

Also, if 〈µ, γ∨〉 = 0 then sγ(µ) = µ.Now we may assume α, β, γ pairwise distinct and 〈µ, γ∨〉 6= 0, i.e., 〈β, γ∨〉 6=

〈α, γ∨〉. Since 〈α, β∨〉 ∈ 1, 2, either α β or α→ β. Accordingly, we distinguishthe cases (v) α β and (vi) α→ β. In each case, (1) will follow from the proof.

(v) α β. By Lemma 15.6 the induced subgraph on the set α, β, γ is oneof the following:

(1)

α β

δ γ

(2)

α β

γ δ

(3)

γ

β δ

α

????

(4)

γ

α δ

β

====

In case (1), 〈µ, γ∨〉 = 1. By (C2), there exists δ ∈ Γ such that (α, β, γ, δ) isa square. Hence sγ(µ) = µ − γ = β − α − γ = −δ ∈ R−1 by (15.3.2). Incase (2), we complete γ, α, β to a square (α, β, δ, γ) and then have 〈µ, γ∨〉 = −1and sγ(µ) = β − α + γ = δ ∈ R1. In cases (3) and (4), we use (C3) andcomplete to a kite. Then in the first case, 〈µ, γ∨〉 = 1 − 0 = 1, and thereforesγ(µ) = β − α− γ = −δ ∈ R−1, by (15.3.2). In the second case, 〈µ, γ∨〉 = −1 andsγ(µ) = δ ∈ R1.

(vi) α→ β. Here the possible induced subgraphs on α, β, γ are the following:

(5)

α

β δ

γ

====(6)

α

????

γ β

δ

We use (C3) and (C4) and complete to a kite. Then in (5) we get 〈µ, γ∨〉 = 1−0 = 1and sγ(µ) = β − α − γ = −δ ∈ R−1, while in (6) we obtain 〈µ, γ∨〉 = 1 − 2 = −1and sγ(µ) = β − α+ γ = δ ∈ R1, both times using (15.3.2) for a kite.

15.11. Theorem. (a) Let Γ be a Jordan graph and let X•(Γ ) and R(Γ ) be asin 14.11 and (15.9.4). Then (R(Γ ),X•(Γ )) is a root system, and the decomposition(15.9.4) is a 3-grading of R(Γ ), denoted by

R(Γ ) = (R(Γ ),X•(Γ ), R1(Γ ) = Γ ).

(b) Let Γ and ∆ be Jordan graphs and let f : Γ → ∆ be a morphism as in14.12. Then X•(f): X•(Γ ) → X•(∆) is a morphism of 3-graded root systems.The assignments Γ 7→ R(Γ ) and f 7→ R(f) define a functor R: jgraph→ RS3.

(c) The functors R of (b) and G of 14.16 are up to natural isomorphismsinverses of each other, so RS3 and jgraph are equivalent categories.


Proof. (a) By Lemma 14.11(a) and the conventions of 15.1, X•(Γ ) is a freeabelian group generated by R = R(Γ ). We show that R is locally finite. LetV = X•(Γ ) ⊗ Q and embed X•(Γ ) into V in the natural way. Let U ⊂ V be asubspace of dimension n. By Lemma 2.6, it suffices to show that R ∩ U is finite.By 15.1, Γ∨ ⊂ X •(Γ ) is a total subset, and the same holds for the Q-linearextensions of the γ∨ to linear forms on V , and then also for the restrictions γ∨|U .By elementary linear algebra (see, e.g., [15, Chapter II, §7, No. 5, Corollary 1of Theorem 7]), there exist γ1, . . . , γn ∈ Γ such that the linear forms γ∨i

∣∣U are abasis of U∗. Let y1, . . . , yn ∈ U be the dual basis. Then for every % ∈ U ∩ Rwe have % =

∑ni=1 yici with coordinates ci = 〈%, γ∨i 〉. If % ∈ R1 ∪ R−1 then

ci ∈ 0,±1,±2 by (15.1.1) and (15.9.3). If % ∈ R0 then by (15.10.1), ci ∈ 0,±1.Hence Card(R ∩ U)6 5n <∞.

We show next that for every % ∈ R× there exists a reflection s satisfyings(%) = −% and s(R) ⊂ R. If % ∈ Γ = R1 this was shown in Lemma 15.10.For % = −γ ∈ R−1 = −Γ simply put s−γ = sγ . Finally, let % = µ = β − α ∈ R0

as in (15.9.2) and observe sα(β) = β − 〈β, α∨〉α = β − α = µ. Put s = sαsβsα.Since sα = s−1

α , s is conjugate to the reflection sβ and therefore itself a reflection.By Lemma 15.10, sα and sβ leave R invariant, hence so does s. Finally, s(µ) =sαsβsα(sα(β)) = sα(sβ(β)) = sα(−β) = −sα(β) = −µ. This completes the proofthat (R,X•(Γ )) is a root system.

To show that (15.9.3) and (15.9.4) define a 3-grading of R, it remains to verifythat (Ri + Rj) ∩ R ⊂ Ri+j , where Rk = ∅ if k /∈ 0,±1. A priori, this means sixcases:

(i, j) = (1, 1), (1,−1), (1, 0), (0, 0), (0,−1), (−1,−1).

However, since R−i = −Ri, it suffices to deal with the first four cases.

Case (1, 1): Let α, β ∈ R1 = Γ . We show that α + β /∈ R. Indeed, assumeα + β = γ ∈ R. Then 0 = 〈α + β − γ, α∨〉 = 2 + 〈β, α∨〉 − 〈γ, α∨〉 implies〈γ, α∨〉 = 2 + 〈β, α∨〉 > 2. Therefore γ 6∈ R−1 because 〈R−1, R

∨1 〉 6 0 and γ 6∈ R0

because of (15.10.1). But for γ ∈ R1 we have γ ∼ α since 〈γ, α∨〉 = 2 by theprevious inequality and so β = γ −α ∈ R1 ∩R0 by (15.9.1), contradicting (15.9.4).

Case (1,−1): For α, β ∈ R1 with α ∼ β we have α − β ∈ R0 by definition ofR0. We show α − β 6∈ R if α ⊥ β. Assume to the contrary that α − β = γ ∈ R.Then 0 = 〈α − β + γ, α∨〉 = 2 − 〈γ, α∨〉, whence 〈γ, α∨〉 = 2. As in the case (1, 1)this implies γ ∈ R1 and then the contradiction β = α− γ ∈ R1 ∩R0.

Case (1, 0): Let µ = β − α ∈ R0 as in (15.9.2) and γ ∈ R1. We show thatγ + µ /∈ R−1 ∪ R0. Indeed, if γ + β − α = −δ ∈ R−1 then α = β + γ + δ and2 = 〈α, α∨〉 = 〈β + γ + δ, α∨〉 = 1 + 〈γ, α∨〉 + 〈δ, α∨〉. Hence γ ⊥ α or δ ⊥ α, andtherefore 0 = 〈α, γ∨〉 = 〈β + γ + δ, γ∨〉 > 2 or 0 = 〈α, δ∨〉 = 〈β + γ + δ, δ∨〉 > 2,contradiction.

Now assume γ+µ = ν ∈ R0. Then 2 = 〈γ, γ∨〉 = 〈ν−µ, γ∨〉 = 〈ν, γ∨〉−〈µ, γ∨〉.Since 〈µ, γ∨〉 and 〈ν, γ∨〉 are in 0,±1 by (15.10.1), it follows that 〈µ, γ∨〉 = −1.Hence by (15.10.1), γ + µ = sγ(µ) ∈ R1, contradiction.

Case (0, 0): Let λ, µ ∈ R0 and assume λ + µ = γ ∈ R1. Since R0 = −R0, thisimplies γ+ (−µ) = λ ∈ R0 which we just excluded in Case (1, 0). In the same way,λ+ µ = −γ ∈ R−1 leads to the contradiction γ + µ = −λ ∈ R0.


(b) By the definition of morphisms of mixed graphs in 14.12, f induces ahomomorphism X•(f): X•(Γ ) → X•(∆) of abelian groups such that X•(f)(α) =f(α) for all α ∈ Γ . Hence X•(f) maps Γ = R1(Γ ) to ∆ = R1(∆). The condition (i)of 14.12 coincides with the condition (iii) of Lemma 14.3, (or use Corollary 15.12and condition (ii) of Lemma 14.3). So by that lemma, X•(f) is a morphism of3-graded root systems.

(c) This follows easily from the definitions and Lemma 14.15. The details areleft to the reader.

In the remainder of this section we draw several consequences of the equivalencebetween jgraph and RS3. We often identify a Jordan graph with the 1-part of a3-graded root system and conversely.

15.12. Corollary. Let f : Γ → ∆ be a map on the vertex sets of Jordan graphsΓ and ∆. Then f is a morphism of Jordan graphs if and only if

(i) α ∼ β implies f(α) ∼ f(β), for all α, β ∈ Γ , and

(ii) f extends to a homomorphism X•(f): X•(Γ )→X•(∆) of the associatedabelian groups.

Proof. Let (R,R1) and (S, S1) be the 3-graded root systems associated with Γand ∆ respectively. Then X•(f): X•(Γ )→ X•(∆) is a homomorphism of abeliangroups sending Γ to ∆. Now the assertion follows from Lemma 14.3 and thedefinition of morphisms of mixed graphs in 14.12.

15.13. Corollary. Let Γ be a Jordan graph, embedded in X•(Γ ) as in 15.1.Then there exists a positive definite quadratic form Φ with associated inner product( | ) on X•(Γ ) such that

p(α, β) = 〈α, β∨〉 =(α |β)

Φ(β)=

2(α |β)

(β |β)

for all α, β ∈ Γ .

Proof. The corresponding statement holds for root systems, see 2.12, so theassertion follows from Theorem 15.11.

15.14. Corollary. A 3-graded root system (R,R1) is irreducible if and only ifthe associated Jordan graph Γ is connected.

Proof. This is obvious from 14.4(d) and the definition of the graph associatedwith R1. Alternatively, one can use Proposition 14.16, saying that G commuteswith coproducts. Then the assertion follows from Theorem 15.11.

15.15. Corollary. A 3-graded root system is reduced.

Proof. Suppose µ and 2µ are in R×. We cannot have µ ∈ R1 ∪ R−1 since R is3-graded and therefore R2 = R−2 = ∅. Thus 2µ ∈ R0 so by (15.9.2) 2µ = β−α with〈β, α∨〉 = 1. Hence 2〈µ, α∨〉 = 〈2µ, α∨〉 = 〈β − α, α∨〉 = 1 − 2 = −1, contradicting〈µ, α∨〉 ∈ Z.


15.16. Jordan subgraphs. Let Γ be a Jordan graph. A subgraph Σ of Γ iscalled a Jordan subgraph if it is an induced subgraph of Γ and, with its inducedgraph structure, is itself a Jordan graph.

There may well be subgraphs of Γ which are Jordan graphs in their own right butare not induced; these do not qualify as Jordan subgraphs. For example, considera collision α → β ← γ which by 15.4(c) is a Jordan graph Γ . This contains thesubgraph ∆ consisting of the vertices α, β, γ and no edges, which by 15.4(a) isa Jordan graph. But ∆ is not an induced subgraph and therefore not a Jordansubgraph.

Let Γ ′ be an induced subgraph of a Jordan graph Γ . Then Γ ′ is itself a Jordangraph (and hence a Jordan subgraph) if and only if the closure conditions (C1) –(C3) hold in Γ ′. In more detail: whenever α → β is an arrow in Γ ′ and γ is thecompletion to a collision in Γ , then also γ ∈ Γ ′, and similarly for the other twocases in 15.3(b). This is immediate from the fact that the conditions in 15.3(a) holdin Γ ′ automatically. Indeed, the matrix associated with Γ ′ in (14.10.1) is obtainedfrom the matrix associated with Γ by restriction.

It is clear that the intersection of an arbitrary family of Jordan subgraphs of Γis itself a Jordan subgraph. Hence, given an induced subgraph Σ of Γ , it makessense to speak of the Jordan subgraph generated by Σ. If S is merely a subset of theset of vertices of Γ , we define the Jordan subgraph generated by S to be the Jordansubgraph generated by the induced subgraph on the set S. Finally, we say that asubset S generates Γ if Γ equals the Jordan subgraph generated by S.

For example, by 15.4(d),(e), the Jordan subgraph generated by (15.4.1) is thepyramid (15.4.2), and the Jordan subgraph generated by a kite as in (15.2.3) is thehexagram (15.4.3).

15.17. Automorphism groups of Jordan graphs. Let Γ be a Jordangraph, embedded in the 3-graded root system (R,R1 = Γ ) as in Theorem 15.11.The equivalence of the categories jgraph and RS3 described in loc. cit. impliesthat there is a natural isomorphism

Aut(Γ ) ∼= Aut(R,R1) (1)

for the respective automorphism groups, defined in 14.12 and 14.1. It is given bysending f ∈ Aut(R,R1) to f

∣∣Γ ∈ Aut(Γ ), and by extending an automorphism t ofΓ to R by defining t(−γ) = −t(γ) for γ ∈ Γ , and t(µ) = t(β) − t(α) for µ ∈ R0,written as the difference of two roots in Γ . In the sequel, we will identify these twogroups by means of (1).

Let µ ∈ R0. It is clear from (2.2.1) and the grading property (i) of 14.1that the reflection sµ belongs to Aut(R,R1) = Aut(Γ ). The subgroup of Aut(Γ )generated by these sµ is called the inner automorphism group of Γ , denoted Inn(Γ ).Restriction to R0 yields an isomorphism of this group with the Weyl group of thesubsystem R0 of R [63, Corollary 5.8]:

Inn(Γ ) ∼= W(R0). (2)

As for Aut(Γ ) and Aut(R,R1), we will in the future often identify Inn(Γ ) andW(R0) by means of (2).

By (15.9.2) every µ ∈ R×0 can be written in the form µ = β−α with 〈β, α∨〉 = 1.Hence sα(β) = β − 〈β, α∨〉α = β − α = µ and therefore


tα,β := sµ = ssα(β) = sαsβsα ∈W(R0). (3)

The Weyl group W(R) is generated by all sα, α ∈ R = R−1∪R0∪R1, and s−α = sαby (2.2.7). Hence (3) implies

W(R) =⟨sγ : γ ∈ Γ

⟩, (4)

W(R0) = Inn(Γ ) =⟨tα,β : α, β ∈ Γ, 〈β, α∨〉 = 1

⟩. (5)

We also haveAut(Γ ) ∩W(R) = W(R0). (6)

Indeed, by [63, 17.4] the subset P = R0∪R1 of R is parabolic as defined in [63, 10.5].Its symmetric part in the sense of [63, 10.6] is R0. Any w ∈ Aut(Γ ) = Aut(R,R1)stabilizes P , so that [63, Proposition 15.8] shows Aut(Γ ) ∩W(R) ⊂ W(R0). Theother inclusion is clear.

The following lemma describes the tα,β in more detail.

15.18. Lemma. Let Γ be a Jordan graph.

(a) Assume α, β ∈ Γ satisfy 〈β, α∨〉 = 1, so that µ = β−α ∈ R0 and sµ = tα,βas in (15.17.3). Then tα,β acts on γ ∈ Γ as follows. Let i = 〈γ, β∨〉, j = 〈γ, α∨〉and k = 〈α, β∨〉. Then

tα,β(γ) = γ − (i− jk)(β − α). (1)

In particular,

α β =⇒ tα,β(α) = β, (2)

α→ β =⇒ tα,β(α) = 2β − α, (3)

β γ ⊥ α =⇒ tα,β(γ) = α− β + γ, (4)

If (δ, α, β, γ) is a kite, embedded in a hexagram (15.4.3) then

tα,β(γ) = γ, tα,β(δ′) = δ′, tα,β(δ) = δ′′, (5)

(b) Let ∆ be an induced subgraph of Γ and let

T (∆) =⟨tα,β : α, β ∈ ∆, 〈β, α∨〉 = 1

⟩. (6)

Then T (∆) ⊂ Inn(Γ ) and ∆ is a Jordan subgraph if and only if it is invariantunder T (∆).

Proof. (a) We have sα(β) = β − 〈β, α∨〉α = β − α = µ ∈ R0. Hencesµ = ssα(β) = sαsβsα, and

sµ(γ) = sαsβsα(γ) = sαsβ(γ − jα) = sα(γ − iβ − j(α− kβ)

)= γ − jα− i(β − α)− j

(− α− k(β − α)

)= γ − (i− jk)(β − α).


For (2), we have j = 2 and i = k = 1, for (3), i = j = k = 2, while for (4), j = 0and i = 1. Now the first three formulas are immediate from (1). The proof of thefirst two formulas of (5) is similar. Finally, for the last case of (5), i = 〈δ, β∨〉 = 0,j = 〈δ, α∨〉 = 2 and k = 〈α, β∨〉 = 1, hence tα,β(δ) = δ + 2(β − α) = 2γ − δ = δ′′,since β − α = γ − δ by (15.3.2).

(b) By (15.17.5) we have T (∆) ⊂ Inn(Γ ). It is clear that a Jordan subgraph∆ of Γ is stable under T (∆). Conversely, suppose that ∆ is an induced subgraphinvariant under T (∆). By 15.16, ∆ is a Jordan subgraph if and only if the closureproperties (C1) – (C3) of 15.3 hold in ∆. For (C1) this means that an arrow α→ βin ∆ can be completed to a collision (α, β, γ) in ∆. By (15.3.1), γ = 2β − α, sotα,β(α) = 2β − α = γ ∈ ∆ by (3). The closure properties (C2) and (C3) follow inthe same way from (15.3.2) and (4).

15.19. Example. Let S(I) be the full symmetric group and SI the group offinitary permutations of a set I. Now consider a rectangular graph Γ = KI KJ asin 14.17, with associated 3-graded root system (R,R1) = AI

I∪J . Then S(I)×S(J)acts naturally on Γ by graph automorphisms. If I and J have equal cardinalityand ϕ: I → J is a bijection, then the map τ , defined by τ(i, j) =

(ϕ−1(j), ϕ(i)

), is

also an automorphism of Γ . Then one shows easily that

Aut(KI KJ) =

S(I)×S(J) if |I| 6= |J |(S(I)×S(J)

)o Id, τ if |I| = |J |

.

There are two types of edges in Γ , either (i, j) (i′, j) with i 6= i′ or (i, j)(i, j′) with j 6= j′. In the first case, t(i,j), (i′,j) = τi,i′ × IdJ where τi,i′ is thetransposition of i and i′, and in the second case τ(i,j), (i,j′) = IdI × τj,j′ . Therefore,

W(R0) = Inn(KI KJ) = SI ×SJ .

The following result shows that Jordan subgraphs correspond to subsystems of3-graded root systems.

15.20. Proposition. Let Γ be a Jordan graph with associated 3-graded rootsystem R(Γ ) = (R(Γ ),X•(Γ ), Γ ) = (R,X,R1) as in 15.11(a).

(a) Let Γ ′ ⊂ Γ be an induced subgraph. Then the inclusion map inc: Γ ′ → Γ is amorphism of mixed graphs. The induced group homomorphism X•(inc): X•(Γ ′) =X ′ → X is injective.

(b) Let ∆ ⊂ Γ be a Jordan subgraph and let R(∆) = (S, Y, S1 = ∆) be itsassociated 3-graded root system. Then the map X•(inc): Y → X of (a) is anembedding of 3-graded root systems.

Conversely, if f : (S, Y, S1) → R(Γ ) is an embedding of 3-graded root systems,then f(S1) is a Jordan subgraph of Γ .

(c) Let ∆ be a Jordan subgraph of Γ , and identify R(∆) with a subsystem of(R,X,R1) by means of (b). Let WR(S) and WR(S0) be the subgroups of W(R)generated by sβ : β ∈ S and sµ : µ ∈ S0, respectively, and recall the definitionof T (∆) in (15.18.6). Then these subgroups stabilize Y , and restriction to Y yieldsisomorphisms


WR(S) ∼= W(S), T (∆) = WR(S0) ∼= W(S0) = Inn(∆). (1)

Treating these isomorphisms as identifications, we have

Aut(Γ ) ∩W(S) = W(S0). (2)

Proof. (a) Condition (i) of 14.12 is trivially satisfied. As to condition (ii), itsuffices to show that any relation

∑i niP

′αi = 0 in X ′ implies

∑i niPαi = 0 in X.

By Corollary 15.13,

0 =∑i

niP′αi(αj) =

∑i

nip(αi, αj) =∑i

ni(αi |αj)Φ(αj)

for all j. Multiplying by njΦ(αj) and summing over j yields 0 = ‖∑i niαi‖2 and

hence∑i niαi = 0 in X, equivalently, by 15.1,

∑i niPαi = 0.

It remains to show that X•(inc) is injective. Let∑niP

′αi ∈ X

′ and suppose∑niPαi = 0. Then

∑niPαi(β) = 0 for all β ∈ Γ , hence a fortiori for all β ∈ Γ ′.

But this just says that∑niP

′αi = 0 in X ′.

(b) The first part is immediate from (a) and the definitions. The conversefollows from Proposition 15.5.

(c) The first isomorphism of (1) is a consequence of [63, Corollary 5.8], and thesecond follows from S0 ⊂ R0 and (15.17.3) for the equality T (∆) = WR(S0).

The inclusion from right to left in (2) follows from

W(S0) ⊂W(S) ∩W(R0) ⊂W(S) ∩Aut(Γ ).

For the inclusion from left to right, let w ∈ Aut(Γ ) ∩W(S). Then W(S) ⊂W(R)implies

w ∈ Aut(Γ ) ∩W(R) = W(R0)

by (15.17.6). Hence also w ∈ W(R0) ∩W(S). Thus w is an automorphisms ofS stabilizing R1, hence also S1 = S ∩ R1 = ∆, so w ∈ Aut(∆). It follows thatw ∈ Aut(∆) ∩W(S) = W(S0), again by (15.17.6).

15.21. Corollary. With the notation and identifications of Proposition 15.20we assume that ∆ is a Jordan subgraph of Γ generated by a subset D ⊂ ∆, as in15.16. Then

spanZ(D) = spanZ(∆), (1)

W(S0) = w ∈W(S) : w(D) ⊂ Γ. (2)

Proof. For the proof of (1) let us put Y ′ = spanZ(D) and ∆′ = Γ ∩ Y ′. Weclaim that ∆′ is a Jordan subgraph containing D and spanning Y ′. Indeed, it isclear that D ⊂ ∆′ ⊂ Y ′. Hence Y ′ = spanZ(D) ⊂ spanZ(∆′) ⊂ Y ′, which proves

Y ′ = spanZ(∆′).

The fact that ∆′ is a Jordan subgraph follows easily from the closure properties (C1)– (C3) of Γ , the relations (15.3.1) and (15.3.2), and the fact that Y ′ is a subgroup

§16] Local structure 209

of X. Hence ∆ ⊂ ∆′, and therefore Y ′ = spanZ(D) ⊂ spanZ(∆) ⊂ spanZ(∆′) = Y ′,which proves (1).

For the inclusion from left to right in (2), let w ∈W(S0). Then w is in particularan automorphism of ∆ by (15.17.2) applied to S, hence w(D) ⊂ w(∆) = ∆ ⊂ Γ ,while w ∈ W(S) holds by (15.17.6) applied to S. For the other inclusion, let q bethe minuscule coweight defining the 3-grading of R, see 14.6. By (1), every δ ∈ ∆ ⊂Γ = R1 can be written in the form δ =

∑ni=1miαi with mi ∈ Z and αi ∈ D. Hence

1 = q(δ) =∑imi and thus q

(w(δ)

)=∑imiq

(w(αi)

)=∑imi = 1, which proves

w(δ) ∈ R1. By linearity, w(Si) ⊂ Ri for i ∈ 1, 0,−1, and since w is in particularan automorphism of R, this implies w ∈ Aut(R,R1) ∼= Aut(Γ ) by (15.17.1). Hencew ∈ Aut(Γ ) ∩W(S) = W(S0) by (15.20.2).

§16. Local structure

16.1. Proposition. Let Γ be a Jordan graph, and let (α1, . . . , α4) be a quadru-ple of elements of Γ with alternating sum zero:

4∑i=1

(−1)i αi = 0. (1)

The indices 1, . . . , 4 are taken modulo 4.

(a) If αk = αk+1 for some k then also αk+2 = αk+3, so Cardα1, . . . , α46 2.

(b) If αk 6= αk+1 6= αk+2 for some k then αi 6= αi+1 and αi ∼ αi+1 for all i,and one of the following holds:

(b1) if not all αi are distinct then either α1 = α3 and (α2, α3, α4) is acollision, or α2 = α4 and (α1, α2, α3) is a collision,

(b2) if the αi are distinct then either (α1, . . . , α4) is a square, or a cyclicpermutation of the αi is a kite.

Proof. (a) This follows immediately from (1).

(b) Since (1) is equivalent to αk−1−αk+αk+1−αk+2 = 0, assuming αk−1 = αkforces αk+1 = αk+2, while αk+2 = αk+3 implies αk = αk+1, both of which areimpossible. Hence αi 6= αi+1 for all i.

Let pij = 〈αi, α∨j 〉. Evaluating α∨j on the equation (1) yields the four equations∑4i=1(−1)i+1pij = 0, explicitly,

2− p21 + p31 − p41 = 0, (2)

p12 − 2 + p32 − p42 = 0, (3)

p13 − p23 + 2− p43 = 0, (4)

p14 − p24 + p34 − 2 = 0. (5)

By symmetry, it suffices to show that α1 ∼ α2. Assume to the contrary thatα1 ⊥ α2, so p21 = p12 = 0. Then (2) and (3) imply

p41 = 2 + p31, p32 = 2 + p42


from which we conclude (since the pij can only have the values 0, 1, 2) that

p41 = p32 = 2, p31 = p42 = 0,

hence also p24 = p13 = 0 by (14.10.6). In particular, α2 ⊥ α4. Substituting this in(4) and (5) yields

p23 + p43 = 2, p14 + p34 = 2.

Since p32 = p41 = 2 it follows from α2 6= α3 and α4 6= α1 that p23 = p14 = 1, andtherefore p43 = p34 = 1, thus α3 α4. We also have α3 → α2 and α4 ⊥ α2,leading to the forbidden configuration α4 α3 → α2 ⊥ α4 (Lemma 15.6),contradiction.

If not all αi are distinct then, because of our assumption (b1), either α1 = α3 orα2 = α4. It follows from (1) that not both of these alternatives are possible. In thefirst case, (4) becomes p23 + p43 = 4 whence α2 → α3 and α4 → α3, and thereforep32 = 1 = p12. Equation (3) now says p42 = 0, so α2 ⊥ α4. Thus (α2, α3, α4) is acollision. The second case is analogous.

Let the αi be distinct. If (α1, . . . , α4) is not a square then it follows from (a)and the definition of a square in (15.2.2) that

(i) either αi αi+1 does not hold for some i,

(ii) or αi ∼ αi+2 holds for some i.

Case (i): Our assumptions are invariant under cyclic permutations, so we mayassume i = 4. Since α4 ∼ α1, either α4 → α1 or α4 ← α1. We claim that α4 → α1

implies (α4, α1, α2, α3) is a kite.Indeed, since p14 = 〈α1, α

∨4 〉 = 1, (5) implies 1 + p24 = p34 6 2, whence p24 = 1

or p24 = 0. In the first case p34 = 2 follows, so α3 → α4 because α3 6= α4. Butthen α3 → α4 → α1 contradicts (15.6.5). Hence α2 ⊥ α4.

Again by (15.6.5), α1 → α2 is impossible, so α1 ← α2 or α1 α2 follows fromα1 ∼ α2 6= α1. In the first case we have the collision α4 → α1 ← α2, and thereforeα4 + α2 = 2α1 = α1 + α3 (by (1)), contradicting the fact that the αi are distinct.Thus α4 → α1 α2 ⊥ α4. Now it follows from (C3) that there exists γ ∈ Γ suchthat (α4, α1, α2, γ) is a kite, and then γ = α3 follows from (1) and (15.3.2).

Next, assume α4 ← α1. The quadruple (α1, α4, α3, α2) satisfies (1) as well, soby the preceding argument, (α1, α4, α3, α2) is a kite. But then so is (α1, α2, α3, α4),as remarked in 15.2.

Case (ii): Again by symmetry, we may assume α1 ∼ α3. By what we provedin Case (i), we already know: if one of the relations α1 ∼ α2 and α2 ∼ α3 is anarrow then a cyclic permutation of (α1, . . . , α4) is a kite. Hence we may assumethat α1 α2 α3, and then Lemma 15.6 shows that also α1 α3. Thisimplies p21 = p31 = 1. Substituting this in (2) shows p41 = 2. Hence α4 → α1, andby Case (i) we are done.

16.2. Corollary. Let Γ be a Jordan graph and let α, β, γ be elements of Γsatisfying α 6= β 6= γ. Then the following assertions (a), (b) and (c) are equivalent:

(a) α− β + γ ∈ Γ ,


(b) the induced subgraph on α, β, γ is one of the following:

(b.i) α ∼ β ∼ γ ⊥ α,

(b.ii)β

????

α γ,

(b.iii)β

????

α γwith α− β + γ ∈ R1,

(b.iv) α = γ ← β,

(c) there exists δ ∈ Γ such that exactly one of the following holds:

(c.i) α = γ and (β, γ, δ) is a collision,

(c.ii) β = δ and (α, β, γ) is a collision,

(c.iii) (α, β, γ, δ) is a square,

(c.iv) a cyclic permutation of (α, β, γ, δ) is a kite.

In all cases of (c) the vertex δ is unique, namely δ = α− β + γ. Moreover,

α− β + γ ∈ R1, α 6= β 6= γ =⇒ α ∼ β ∼ γ. (1)

Proof. If (a) holds then we are in the situation of Proposition 16.1(b), and thestatements in (c) are (up to order) precisely those of 16.1(b). Hence we have (a) =⇒(c). The implication (c) =⇒ (a) is clear from the properties of collisions, squaresand kites ((15.3.1) and (15.3.2)).

(a) =⇒ (b): put δ = α − β + γ. Then (α, . . . , δ) satisfies the assumptions ofProposition 16.1(b). Discussing the cases (b1) and (b2) listed there, we have:

(b1): Cardα, . . . , δ = 3. Then either α = γ and β → γ ← δ, which is (b.iv),or β = δ and α→ β ← γ which is one of the four cases of (b.i).

(b2): Cardα, . . . , δ = 4. Then either (α, . . . , δ) is a square or a cyclic permu-tation is a kite. In case of a square, we have α β γ, one of the cases of (b.i).If (α, . . . , δ) is a kite then α → β γ, another of the cases (b.i). If (β, γ, δ, α) isa kite we have case (b.ii). If (γ, δ, α, β) is a kite we have one of the cases (b.i). If(δ, α, β, γ) is a kite we are in case (b.iii).

(b) =⇒ (a): here one uses the closure conditions (C1) – (C4), see Lemma 15.8,and the formulas (15.3.1) and (15.3.2) for a Jordan graph. The details are omitted.

Remarks. (a) This corollary is a corrected version of [63, Proposition 18.9],formulated for Jordan graphs instead of 3-graded root systems. In loc. cit. we haveerroneously combined (b.ii) and (b.iii) to “α γ and 〈α, β∨〉 = 1 = 〈γ, β∨〉”, thatis, α γ and either α← β → γ or α β γ. Since there may exist (α, β, γ)

with α β γ α but α − β + γ 6∈ Γ , for example in TI = G (CherI ) with

|I|> 4, the condition that α− β + γ ∈ Γ must be added in (b.iii).

(b) By (15.6.1) the case (b.i) α ∼ β ∼ γ ⊥ α breaks down into the followingfour subcases: α β γ, α→ β γ, α β ← γ, and α→ β ← γ.


In case (b.iii) not only α− β + γ ∈ Γ but also −α+ β + γ and α+ β − γ are inΓ , which follows by applying sα and sγ to α − β + γ ∈ Γ . We therefore have thefollowing alternative for a triangle α1, α2, α3 in Γ : either all αi − αj + αk ∈ Γ ,i, j, k = 1, 2, 3 belong to Γ or none of them, see Lemma 19.9(b).

16.3. Corollary. Let Γ be a Jordan graph, A an abelian group, and f : Γ → Aa map. Then the following conditions are equivalent:

(i) f extends to a group homomorphism f : X•(Γ )→ A,

(ii) for all quadruples (α1, . . . , α4) in Γ ,

4∑i=1

(−1)iαi = 0 =⇒4∑i=1

(−1)if(αi) = 0.

(iii) f(α1) − 2f(α2) + f(α3) = 0 for all collisions (α1, α2, α3), as well asf(α1)− f(α2) + f(α3) + f(α4) = 0 for all squares and kites (α1, . . . , α4)in Γ .

Proof. The implication (i) =⇒ (ii) is clear, and (ii) =⇒ (iii) follows from thedescription of quadruples (α1, α2, α3, α4) with vanishing alternating sum in Propo-sition 16.1. The implication (iii) =⇒ (i) is a consequence of Theorem 15.11 andthe fact that the root lattice of a 3-graded root systems is presented by gener-ators xα, α ∈ R1, and relations xα1

− 2xα2+ xα3

= 0 for all collisions, andxα1− xα2

+ xα3− xα4

= 0 for all squares and kites by [63, Corollary 18.11].

Example. The simplest case of this is A = Z and q(α) = 1 for all α ∈ Γ . Thenq: X•(Γ )→ Z is the coweight defining the 3-grading of the associated root systemas in 14.6.

16.4. Lemma. A Jordan graph is claw-free: there are no 3-claws, that is,induced subgraphs on four vertices of the form

α

=====

OOO

!a!a!a!a!a

ω1 ω2 ω3

Here the symbol α /o/o/o ωi stands for α ωi or α→ ωi or α← ωi.

Proof. Let (R,X,R1) be the 3-graded root system associated with Γ (The-orem 15.11) and assume we have an induced subgraph as indicated. Since theωi are pairwise orthogonal, the reflections sωi commute by Lemma 2.11, and

s := sω1sω2

sω3satisfies s(x) = x −

∑3i=1〈x, ω∨i 〉ωi (see also [63, Lemma 5.3]).

Put ni := 〈α, ω∨i 〉. Then ni > 1 and s(α) = α −∑3i=1 niωi ∈ R. Let q be

the minuscule coweight defining the 3-grading (Example in 16.3 or 14.6). Then

q(s(α)

)= 1 −

∑3i=1 ni 6 −2, which contradicts the fact that q can take only the

values 0 and ±1.

Recall that the distance between two vertices in a mixed graph is the numberof edges or arrows in a shortest path connecting them; the diameter of a graph isthe greatest distance between any pair of vertices.


16.5. Lemma. A connected Jordan graph has diameter at most two.

Proof. It suffices to show that any chain α ∼ β ∼ γ ∼ ω of length threeconnecting α and ω can be shortened. We may assume α ⊥ ω, α ⊥ γ and β ⊥ ω,else we would already have a connecting chain of length 62.

By (15.6.1), there are only the following possibilities for the induced subgraphon α, β, γ:

α β γ, α→ β γ, α β ← γ, α→ β ← γ. (1)

In the first three cases, it follows from 15.3(b) that there exists δ ∈ Γ such that(α, β, γ, δ) is a square or a cyclic permutation of these elements is a kite. In anycase, we have δ = α− β + γ by (15.3.2), and hence

〈δ, ω∨〉 = 〈α− β + γ, ω∨〉 = 0− 0 + 〈γ, ω∨〉> 1,

so that α ∼ δ ∼ ω is a shorter chain connecting α and ω. The fourth case isalso impossible because 2β = α + γ by (15.3.1), and therefore 0 6= 〈γ, ω∨〉 =2〈β − α, ω∨〉 = 0, contradiction.

16.6. Lemma. Let Γ be a Jordan graph embedded in X•(Γ ) as in 15.1, and letf : X•(Γ )→ A be a homomorphism into an abelian group A. Then for every a ∈ Athe induced subgraph Γ ∩ f−1(a) is a Jordan subgraph. In particular, if f : Γ → ∆is a morphism of Jordan graphs and δ ∈ ∆ then the induced subgraph f−1(δ) ⊂ Γis a Jordan subgraph.

Proof. Let Γ ′ = f−1(a) ∩ Γ . As mentioned in Definition 15.16, the conditions(15.3.1) and (15.3.2) of Γ are inherited by Γ ′. Let α1 → α2 in Γ ′ and let α3 ∈ Γbe the completion to a collision α1 → α2 ← α3. Then α3 = 2α2 − α1, whencef(α3) = 2a− a = a, so γ ∈ Γ ′. The closure conditions (C2) and (C3) are proved inthe same way.

16.7. Definition. Given a mixed graph Γ and ω ∈ Γ , we define, for i ∈0, 1, 2 , induced subgraphs Γi(ω) by

Γi(ω) = γ : 〈γ, ω∨〉 = i. (1)

Thus γ ∈ Γ0(ω) if and only if γ ⊥ ω, and γ ∈ Γ1(ω) if and only if either γ ω orω → γ. Finally,

Γ2(ω) = ω ∪ γ ∈ Γ : γ → ω (2)

consists of ω and all starting points of arrows ending in ω. Clearly

Γ = Γ2(ω) ∪ Γ1(ω) ∪ Γ0(ω). (3)

16.8. Lemma. Let Γ be a Jordan graph and let ω ∈ Γ . The subgraphs Γi(ω)defined above are Jordan subgraphs and have the following properties.

(a) Γ2(ω) is connected and Γ2(ω) ⊥ Γ0(ω).

(b) Γ1(ω) has at most two connected components.


(c) If Γ is connected then so is Γ0(ω).

(d) If Γ is connected then Γ2(ω) ∪ Γ1(ω) generates Γ .

Proof. By Lemma 14.11, the map ω∨: X•(Γ ) → Z, ω∨(x) = 〈x, ω∨〉, is awell-defined homomorphism of abelian groups. Hence Lemma 16.6 shows that theΓi(ω) = ω∨−1(i) are Jordan subgraphs of Γ .

(a) The description of Γ2(ω) given above shows that it is connected. Letα ∈ Γ0(ω) and β ∈ Γ2(ω), and assume α ∼ β. Since ω ⊥ Γ0(ω), we have β 6= ωand therefore β → ω. Hence we obtain an induced subgraph α ∼ β → ω, which isimpossible by Lemma 15.6.

(b) This follows immediately from Lemma 16.4.

(c) Let α, γ ∈ Γ0(ω) be orthogonal. By Lemma 16.5, there exists β ∈ Γsuch that α ∼ β ∼ γ. We claim that β ∈ Γ0(ω). Indeed, by (a) we cannot haveβ ∈ Γ2(ω). Assume β ∈ Γ1(ω). Since α and γ are orthogonal to ω this would resultin the 3-claw

β

?????

OOO

^^^^^

α ω γ

contradicting Lemma 16.4.

(d) In view of (16.7.3) it remains to show that every α ∈ Γ0(ω) lies in theJordan subgraph ∆ generated by Γ2(ω) ∪ Γ1(ω). By (16.5.1) there exists β ∈ Γsuch that the induced subgraph on α, β, ω is one of the following:

α β ω, α→ β ω, α β ← ω, α→ β ← ω.

Clearly 〈β, ω∨〉 = 1 and therefore β ∈ Γ1(ω) ⊂ ∆ in all four cases. We now discussthese in turn.

Case α β ω. By the closure property (C2) for Γ there exists γ ∈ Γ suchthat (α, β, ω, γ) is a square. Since ω γ we have γ ∈ Γ1(ω) ⊂ ∆ and hence aninduced subgraph β ω γ in ∆. Again by (C2), but this time for ∆, yieldsthe existence of α′ ∈ ∆ such that (β, ω, γ, α′) is square in ∆. From (15.3.2) for Γand ∆ we then obtain α = α′ and thus α ∈ ∆.

Case α→ β ω. Using (C1) for Γ , we complete the arrow α→ β to a collision(α, β, γ) in Γ . By (15.3.1), γ = 2β − α, whence 〈γ, ω∨〉 = 2〈β, ω∨〉 − 〈α, ω∨〉 =2 · 1− 0 = 2 proving γ ∈ Γ2(ω) ⊂ ∆. We now complete the arrow γ → β in ∆ to acollision (γ, β, α′) in ∆. By (15.3.1) again, α = α′ ∈ ∆.

The remaining two cases can be dealt with in the same way, using (C3) and(C4) in the third and (C1) in the fourth case.

16.9. Proposition. Let Γ be a Jordan graph and let Ω = ωi : i ∈ I ⊂ Γbe an orthogonal system: ωi ⊥ ωj for all i 6= j in I. We assume 0 /∈ I and

put I ′ = I ∪ 0. Let ∆ = TI′ be the extended triangular graph on the set I ′ as

in 14.18. Using the isomorphism (14.18.5), we identify the vertex i, j ∈ TI′ =P1(I ′) ∪P2(I ′) with the root εij := εi + εj of CI′ .


(a) For all α ∈ Γ , we have 〈α, ω∨i 〉 6= 0 for at most two i ∈ I.

(b) The formula

f(α) = ε00 +∑i∈I〈α, ω∨i 〉(ε0i − ε00) (1)

defines a morphism of Jordan graphs f : Γ → ∆. Hence Γ decomposes

Γ =⋃

p,q∈∆

Γpq (2)

where Γpq := f−1(εpq) = f−1(p, q) is a Jordan subgraph.

(c) Explicitly, the Γpq are given as follows: for i, j ∈ I, i 6= j,

Γii = Γ2(ωi), (3)

Γij = Γ1(ωi) ∩ Γ1(ωj), (4)

Γi0 = Γ1(ωi) ∩⋂j 6=i

Γ0(ωj), (5)

Γ00 =⋂i∈I

Γ0(ωi). (6)

These subsets satisfy the orthogonality relations

Γpq ⊥ Γrs for all p, q, r, q ∈ I ′ with p, q ∩ r, s = ∅. (7)

Proof. By Lemma 16.4, α ∼ ωi for at most two indices i, which proves (a) andthe finiteness of the sum in (1). We define Γ(ii), Γ(ij), Γ(0i) = Γ(i0) and Γ(00) to bethe right hand sides of (3) – (6), and show

Γ =⋃

p,q∈I′Γ(pq) (8)

Let α ∈ Γ . If 〈α, ω∨i 〉 = 2 for some i ∈ I then α ∈ Γ2(ωi). Otherwise, 06〈α, ω∨i 〉61for all i ∈ I and by (a), we even have 〈α, ω∨i 〉 = 1 for at most two i ∈ I and〈α, ω∨j 〉 = 0 for the remaining j ∈ I. This easily implies (8). Since

f(α) =(

2−∑i∈I〈α, ω∨i 〉

)ε0 +

∑i∈I〈α, ω∨i 〉εi (9)

it is immediate that f(Γ(pq)) ⊂ εpq for p, q ∈ I ′. Hence f(Γ ) ⊂ ∆, Γ(pq) =f−1(εpq) = Γpq for p, q ∈ I ′, and (2) holds.

Now we show that f : Γ → ∆ is a graph morphism in the sense of 14.12. Fromthe formula (1) and the fact that any x ∈ X•(Γ ) is a finite linear combination ofelements of Γ it follows easily that f extends to a homomorphism X•(Γ )→ L2(I ′)of abelian groups given by

f(x) = q(x)ε00 +∑i∈I〈x, ω∨i 〉(ε0i − ε00) (10)


where q is the coweight defining the 3-grading of the associated root system (see14.6 and the example in 16.3). It remains to verify condition (i) of Corollary 15.12,equivalently, that f(α) ⊥ f(β) implies α ⊥ β, for all α, β ∈ Γ .

In ∆ we have εpq ⊥ εrs if and only if p, q ∩ r, s = ∅, for p, q, r, s ∈ I ′.Therefore, we have to verify (7). This leads to the following seven cases, where theindices i, j, k, l are different and run over I:

(a) Γii ⊥ Γjj , (b) Γii ⊥ Γ00, (c) Γii ⊥ Γj0, (d) Γii ⊥ Γjk,(e) Γij ⊥ Γkl, (f) Γij ⊥ Γk0, (g) Γij ⊥ Γ00.

Case (a): We apply (9) for α ∈ Γii and obtain

f(α) =(−∑i 6=j∈I

〈α, ω∨j 〉)ε0 + 2εi +

∑i 6=j∈I

〈α, ω∨j 〉εj

Since f(α) ∈∆ necessarily 〈α, ω∨j 〉 = 0 for all j ∈ I, j 6= i. Thus α ∈ Γ0(ωj). Now(3) and Lemma 16.8(a) shows α ⊥ Γ2(ωj) = Γjj .

Cases (b) and (c): By definition in (6) and (5) we know Γ00 ∪ Γj0 ⊂ Γ0(ωi).Orthogonality therefore follows from Γii = Γ2(ωi) and Lemma 16.8(a).

Case (d): Let α ∈ Γ2(ωi) and β ∈ Γjk = Γ1(ωj) ∩ Γ1(ωk), by (3) and (4), andassume α ∼ β. The case α = ωi leads to a 3-claw

β

~>~>~>~>~>

OOO

` ` ` ` `

ωj ωi ωk

which is impossible by Lemma 16.4. If α→ ωi then β ∼ α→ ωi implies β ∼ ωi byLemma 15.6, so we obtain a 3-claw and therefore a contradiction as before.

Cases (e), (f) and (g): A β ∈ Γkl ∪ Γk0 ∪ Γ00 satisfies ωi ⊥ β ⊥ ωj by cases (d),(c) and (b) respectively. Assuming β ∼ α for α ∈ Γij leads to the 3-claw

α

?????

OOO

` ` ` ` `

ωi β ωj

which again contradicts Lemma 16.4. Thus α ⊥ β, so Corollary 15.12 is applicable,and f is a morphism of Jordan graphs. Finally, it follows from Lemma 16.6 thatthe Γpq are Jordan subgraphs.

16.10. Examples. (a) If Ω = ∅ then f is the coweight q defining the 3-gradingof the root system associated with Γ , see (16.9.10).

(b) If Ω = ω is a singleton then f is the morphism of 14.9(ii).

(c) Let Γ = (α1, . . . , α4) ∼= K2K2 be a square; i.e., the 1-part of the 3-gradedroot system A2

4, see (14.17.3). Here Ω = α1, α3 is an orthogonal system. The

morphism f : Γ → ∆ = T3 satisfies


f(α1) = ε11 = 2ε1, f(α2) = f(α4) = ε12 = ε1 + ε2, f(α3) = ε22 = 2ε2.

Hence f collapses the square Γ to the collision (2ε1, ε1 + ε2, 2ε2), see also Ex-ample (a) of 14.16. However, it is not true that every morphism A2

4 → Cher3 of

3-graded root systems is induced from an orthogonal system as in Proposition 16.9.For example, it is easily checked that the assignments

g(αi+1) = ε0i = ε0 + εi for i = 0, 1, 2, g(α4) = ε12 = ε1 + ε2

define a morphism g: A24 → Cher

3 which is not of the type of Proposition 16.9.Remarkably, g is an isomorphism of the associated abelian groups and maps thesquare bijectively onto the kite (ε00, ε01, ε12, ε20).

16.11. Lemma. The following conditions on three distinct vertices α, β, γ ofa Jordan graph Γ are equivalent:

(i) α→ γ ← β is a collision,

(ii) α+ β = 2γ,

(iii) α∨ + β∨ = γ∨.

Proof. (i) =⇒ (ii) follows from (15.3.1).

(ii) =⇒ (i): This follows from Proposition 16.1(b1) for α1 = α3 = γ, α2 = α,α4 = β.

(i) =⇒ (iii): Let sγ be the reflection in the root γ in the 3-graded root systemassociated with Γ by Theorem 15.11. We have sγ(α) = α−〈α, γ∨〉γ = α−2γ = −β(by (15.3.1)). Since sγ is the orthogonal reflection with respect to the invariantinner product ( | ) (see 2.12), α and β have the same length. Moreover, α ⊥ β,so (α |β) = 0 and therefore 4(γ | γ) = (α + β |α + β) = 2(α |α). Now (iii) followseasily from (2.12.1).

(iii) =⇒ (i): By assumption 〈α, γ∨〉 = 〈α, α∨〉 + 〈α, β∨〉 = 2 + 〈α, β∨〉, whichimplies 〈α, β∨〉 = 0 and 〈α, γ∨〉 = 2. Thus α ⊥ β and α→ γ. By symmetry β → γ,so we have a collision.

16.12. Lemma. Let (α1, α2, α3, α4) be a quadruple of four distinct vertices ina Jordan graph Γ . The indices 1, . . . , 4 are taken modulo 4. Then the followingconditions are equivalent:

(i) (α1, . . . , α4) is a square,

(ii)∑4i=1(−1)iαi = 0 and αi ⊥ αi+2 for all i,

(iii)∑4i=1(−1)iα∨i = 0, and αi ∼ αi+1 for all i.

Proof. (i) =⇒ (ii): The first relation is (15.3.2) and the orthogonality relationsare part of the definition of a square in 15.2.

(ii) =⇒ (i): By Proposition 16.1(b2), either (α1, . . . , α4) is a square or a cyclicpermutation of these vertices is a kite. But the latter possibility contradicts αi ⊥αi+2.


(i) =⇒ (iii): Since αi αi+1 by definition of a square, it remains to show∑4i=1(−1)iα∨i = 0. From αi αi+1 we have 1 = 〈αi, α∨i+1〉 = 〈αi+1, α

∨i 〉. Let

( | ) be the invariant inner product of 2.12. Then by (2.12.1)

1 =(αi |αi+1)

Φ(αi+1)=

(αi+1 |αi)Φ(αi)

,

which shows that all αi have the same length. Now∑4i=1(−1)iα∨i = 0 follows from

(15.3.2) and (2.12.1).

(iii) =⇒ (i): We evaluate the equation α∨1 + α∨3 = α∨2 + α∨4 (of linear forms onX•(Γ )) on αi. Putting pij = 〈αi, α∨j 〉 = 〈αi, α∨j 〉, we obtain pi1 + pi3 = pi2 + pi4,explicitly,

2 + p13 = p12 + p14, (1)

p21 + p23 = 2 + p24, (2)

p31 + 2 = p32 + p34, (3)

p41 + p43 = p42 + 2. (4)

Let us first show that all relations αi ∼ αi+1 must be αi αi+1. For reasons ofsymmetry, it suffices to show that α1 → α2 is impossible.

Assume to the contrary that α1 → α2, thus p12 = 2 and p21 = 1. Then (1)shows p13 = p14 6= 0, since α1 ∼ α4. Now α1 ∼ α3 and α2 ∼ α3, so Lemma 15.6shows that the induced subgraph on α1, α2, α3 must be one of the following:

(a)

α1

||yyyy""EEEE

α2 α3

(b)

α2

α1

<<yyyyα3

bbEEEE

In case (a), we read off the diagram that p31 = p32 = 1. From (3) we get p34 = 2,hence an arrow α3 → α4, leading to α1 → α3 → α4 and contradicting Lemma 15.6.In case (b), we have p21 = p23 = 1, hence p24 = 0 by (2), thus α2 ⊥ α4 and thereforean induced subgraph α4 ∼ α1 ← α2, which contradicts (15.6.6).

Now we have pi,i+1 = 1 = pi+1,i for all i (indices mod 4). Hence (1) showsp13 = 0, and (2) yields p24 = 0, so α1 ⊥ α3 and α2 ⊥ α4, so that (α1, . . . , α4) isindeed a square.

16.13. Lemma. Let (α4, α1, α2, α3) be a kite in a Jordan graph Γ as in (15.2.3)and write β2 := α4. Complete the arrows β2 → α1 and β2 → α3 to collisionsβ2 → α1 ← β3 and β2 → α3 ← β1. Then the six vertices αi, βi are distinct and theinduced subgraph on them is the hexagram

β2

????

α1 α3

β3

??// α2

???? β1

__???oo

(1)

Proof. By (15.3.1) we have β2 + β3 = 2α1 and β2 + β1 = 2α3. Hence


β3 + β1 = 2α1 − β2 + 2α3 − β2 = 2(α1 + α3 − β2).

By (15.3.2), α2 = α1 + α3 − β2. It follows that β3 + β1 = 2α2, so (β1, α2, β3) is acollision by Lemma 16.11. Also, β3 ⊥ α3 and β1 ⊥ α1. Indeed, by (15.3.1),

〈β3, α∨3 〉 = 2〈α1, α

∨3 〉 − 〈β2, α

∨3 〉 = 2 · 1− 2 = 0,

and the second assertion is proved similarly. By construction, β1 6= α3, β2. More-over, β1 ⊥ α1 and β1 → α2 shows β2 6= α1 and β2 6= α2. In the same way,β3 /∈ β2, α1, α2, α3. Finally, β1 6= β3 is clear from the collision β1 → α2 ← β3.Hence the six vertices are distinct, and (1) is indeed the subgraph induced on theαi, βi, that is, it has all edges inherited from Γ .

16.14. Lemma. Let α1, . . . , α4 be four distinct vertices in a Jordan graph Γ .Then the following conditions are equivalent:

(i) (α4, α1, α2, α3) is a kite,

(ii)∑4i=1(−1)iαi = 0 and α2 ⊥ α4, but α1 ∼ α3,

(iii) 2α∨4 = α∨1 − α∨2 + α∨3 .

Proof. (i) =⇒ (ii): This is immediate from (15.3.2) and the definition of a kite.

(ii) =⇒ (i): By Proposition 16.1(b2), either (α1, . . . , α4) is a square or a cyclicpermutation of these vertices is a kite. The first possibility contradicts α1 ∼ α3.In the second case the conditions α2 ⊥ α4, α1 ∼ α3 are only fulfilled for a kite(α4, α1, α2, α3).

(i) =⇒ (iii): We apply Lemma 16.13, including its notation. If (i, j, k) is a cyclicpermutation of (1, 2, 3) then (βi, αj , βk) is a collision. By Lemma 16.11,

α∨j = β∨i + β∨k .

Henceα∨1 − α∨2 + α∨3 = β∨2 + β∨3 − (β∨3 + β∨1 ) + β∨1 + β∨2 = 2β∨2 .

(iii) =⇒ (i): Let pij = 〈αi, α∨j 〉. Evaluating (iii) on α4 yields

2p44 = 4 = p41 − p42 + p43.

Since the pij take only the values 0, 1, 2, this implies p41 = p43 = 2 and p42 = 0.Hence α4 ⊥ α2, and p14 = p34 = 1, so we have arrows α4 → α1 and α4 → α3. FromLemma 15.6 we now get α1 α3. Evaluating (iii) on αi for i = 1, 2, 3 yields

2 = 2p14 = p11 − p12 + p13 = 2− p12 + p13,

0 = 2p24 = p21 − p22 + p23 = p21 − 2 + p23,

2 = 2p34 = p31 − p32 + p33 = p31 − p32 + 2.

Hencep12 = p13 = 1, p21 + p23 = 2, p31 = p32 = 1. (1)

The first and third equation imply p21 > 0 and p32 > 0, whence p21 = 1 = p23 fromthe second equation, proving α1 α2 α3. Thus (α4, α1, α2, α3) is a kite.


16.15. Corollary. Let (α1, . . . , α4) be a quadruple of vertices of Γ whose al-ternating sum is zero. Then

〈α1, α∨2 〉α∨1 − 〈α2, α

∨1 〉α∨2 + 〈α3, α

∨4 〉α∨3 − 〈α4, α

∨3 〉α∨4 = 0. (1)

Proof. If αi = αi+1 for some i then this is easily seen. Otherwise, we are in thesituation of Proposition 16.1(b), and (1) follows by a straightforward verificationfrom Lemma 16.11(iii), Lemma 16.12(iii) and Lemma 16.14(iii).

§17. Classification of arrows and vertices

17.1. Lemma. A Jordan graph does not contain embedded arrows: a (notnecessarily induced) subgraph α β → γ δ is impossible.

Proof. Assume we have a subgraph as indicated. Then first of all, the fourvertices α, . . . , δ are distinct. Indeed, it is clear that α, β, γ are distinct and so areβ, γ, δ. If α = δ we would have a subgraph

α

????

β // γ

which is impossible by Lemma 15.6.The relations between α and γ, α and δ and β and δ remain to be determined.

We will repeatedly use the following consequence of Lemma 15.6: given a subgraphλ µ→ ν, then necessarily λ→ ν:

λ µ→ ν =⇒λ

???? µ

ν

. (1)

Applying this to α β → γ, we have α→ γ, so we obtain

α

???? β

γ

δ

symmetric in α and β. We show next that β is connected to δ. Assume to thecontrary that β ⊥ δ. Then β → γ δ generates a kite by (C3) of 15.3:

β

???? α

γ

???? ε

δ

(2)

We show that a configuration as in (2) is impossible. Indeed, by (1), α β → εimplies α → ε. By Lemma 16.14 (iii), 2β∨ = γ∨ − δ∨ + ε∨. Evaluating this at αyields

§17] Classification of arrows and vertices 221

2 = 2〈α, β∨〉 = 〈α, γ∨〉 − 〈α, δ∨〉+ 〈α, ε∨〉 = 2− 〈α, δ∨〉+ 2,

so we have 〈α, δ∨〉 = 2, which implies α → δ. Again by (1), β α → δ impliesβ → δ, contradicting β ⊥ δ.

Now we have β ∼ δ and by symmetry in α and β, also α ∼ δ. In fact, byLemma 15.6, the induced subgraphs on α, γ, δ and β, γ, δ are

α

????

γ δand

β

????

γ δ

By (C4) of Lemma 15.8, the second diagram generates another kite (β, γ, ζ, δ)

β

???? α

γ???? δ

ζ

(3)

In particular, we have β ⊥ ζ. This is the same configuration as in (2) (replace εand δ by δ and ζ), contradiction.

17.2. Definition. Let Γ be a Jordan graph. An arrow δ → α is of hermitiantype or simply hermitian if there exists γ such that the induced subgraph on δ, α, γis

δ

????

α γ(1)

In fact, it suffices that there exist a second arrow δ → γ with γ 6= α, since then, by15.6, necessarily α γ.

Clearly, δ → γ is then of hermitian type as well. Moreover, by (C4) andLemma 16.13, the configuration (1) can be completed to a kite (δ, α, β, γ) andfurther to a hexagram

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

(2)

An arrow α → ε is of orthogonal type if there exists β such that the inducedsubgraph on α, β, ε is

α

???? β

ε

(3)

Again, it is clear that β → ε is of orthogonal type, too. By Corollary 15.7 thisgenerates a pyramid:


α

???? β

ε

β′

??α′

__????(4)

Finally, an arrow α → β is called of isolated type or simply isolated if thecollision (α, β, α′ = 2β − α) which it generates is a connected component of Γ .

As we have seen above, a hermitian arrow generates a subgraph (2) which is theJordan graph of the hermitian 3-grading Cher

3 of the root system C3, see 14.14(c),thus explaining the name “hermitian type”. Similarly, an arrow of orthogonal typegenerates a subgraph (4) which is the Jordan graph of the 3-grading Bqf

3 of the rootsystem B3, belonging to the orthogonal group SO7, see 14.14(b).

17.3. Lemma. Let ξ → η be an arrow in a Jordan graph and let ξ → η ← ξ′ =2η − ξ be the collision generated by ξ → η.

(a) If ξ → η is of hermitian (orthogonal, isolated) type then so is ξ′ → η.

(b) The three types of arrows are mutually exclusive and complete: every arrowis of exactly one of the three types defined in 17.2.

Proof. (a) If δ → α is of hermitian type then it generates a subgraph (17.2.2),and the lower left hand triangle shows that δ′ → α is of hermitian type. An arrowα → ε of orthogonal type generates a subgraph (17.2.4), in which the right handtriangle shows that α′ → β is of orthogonal type, too. Finally, it is evident fromthe definition that if ξ → η is of isolated type then so is ξ′ → η.

(b) We first show that the three types are mutually exclusive. From (17.2.1)and (17.2.3) it is clear that an arrow of isolated type cannot be of hermitian ororthogonal type. Assume → is both of hermitian and orthogonal type. Thenthere exist vertices • such that

???? •

•

leading to an embedded arrow • → • which is impossible byLemma 17.1.

Next, we show that every arrow is of one of the three types. For this, it sufficesto show that an arrow which is not isolated is either of hermitian or orthogonaltype. Let α→ β not be of isolated type and put C = α, β, α′ = 2β − α. Since Cis not a connected component of Γ , there exists δ ∈ Γ C which is connected toone of the vertices of C.

If δ ∼ α then δ ∼ α → β and (15.6.6) shows δ ∼ β. It then follows fromLemma 15.6 that we have one of the following two cases:

α

????

β δor

β

α

??δ

__????

In the first case α → is of hermitian type, and in the second of orthogonal type.We can now assume δ ⊥ α, so that δ ∼ β or δ ∼ α′. But since α∨ + α′∨ = β∨ by


16.11, we have 〈δ, α′∨〉 = 〈δ, β∨〉, and therefore always α′ ∼ δ. By the argumentsjust used, this implies that α′ → β is of hermitian resp. orthogonal type, and by(a) so is α→ β.

17.4. Definition. A vertex of a Jordan graph Γ is said to be of hermitian typeif it is the initial or end point of a hermitian arrow, and it is of orthogonal type ifit is the initial or end point of an orthogonal arrow. It is called isolated if it is notconnected to any other vertex, and of isolated type if it is either isolated or it isthe initial or end point of an arrow of isolated type. A vertex which is not of oneof these types is said to be of linear type. We denote by Γher, Γorth, Γiso and Γlin

the set of vertices of hermitian, orthogonal, isolated and linear type, respectively.Finally, we say Γ is of hermitian (orthogonal, linear, isolated) type if Γ = Γher

(Γ = Γorth, Γ = Γlin, Γ = Γiso), respectively.

17.5. Proposition. Let Γ be a Jordan graph. The sets Γher, Γorth, Γiso andΓlin are pairwise orthogonal, in particular, they are disjoint. Hence they are Jordansubgraphs and Γ is their coproduct as in Lemma 14.13:

Γ = Γher q Γorth q Γiso q Γlin. (1)

Proof. Let ∆ = Γher ∪Γorth ∪Γiso. By definition, Γlin = Γ ∆ so Γlin is disjointfrom the other three subsets. Also, it follows directly from the definition of Γiso

and that of an arrow of isolated type that Γiso ∩ (Γher ∪Γorth) = ∅. Hence, to showdisjointness, it remains to prove that Γher and Γorth are disjoint.

Assume ξ ∈ Γher ∩Γorth. Then ξ is the initial or end point of a hermitian arrowand also the initial or end point of an orthogonal arrow. But since → → isimpossible by (15.6.5), there are only two cases to consider:

(i) δ → ξ ← β where δ → ξ and β → ξ are, respectively, of hermitian andorthogonal type. Then β 6= δ by Lemma 17.3(a), whence δ β by Lemma 15.6,so δ → ξ is also of orthogonal type, contradicting Lemma 17.3(a).

(ii) α← ξ → ε where ξ → α is of hermitian type and ξ → ε is of orthogonal type.Then Lemma 15.6 implies α ε, hence ξ → ε is of hermitian type, contradiction.Thus the vertex sets Γher, Γorth, Γiso and Γlin are pairwise disjoint.

(a) We show next that Γher is absorbing in the following sense: if ξ ∼ Γher forsome ξ ∈ Γ then ξ ∈ Γher. Indeed, the elements of Γher are the initial or end pointsof hermitian arrows. Thus if ξ is connected to Γher then there exists a hermitianarrow δ → α as in (17.2.1), i.e., α γ ← δ for some γ ∈ Γ , such that ξ ∼ δ orξ ∼ α.

Case 1: ξ ∼ δ. If ξ = α or ξ = γ then ξ ∈ Γher (since δ → γ is also a hermitianarrow). Thus we may assume ξ 6= α, γ. By (15.6.5) ξ → δ is impossible, and byLemma 17.1 we cannot have ξ δ which would imply ξ δ → α γ. Henceξ ← δ. But then it follows from Lemma 15.6 that ξ α, so δ → ξ is a hermitianarrow which proves ξ ∈ Γher.

Case 2: ξ ∼ α. By Case 1 we can assume ξ ⊥ δ. We complete the hermitianarrow δ → α to a hexagram as in (17.2.2),

δ

????

α

δ′

??//

????

__????oo


and apply Lemma 16.11: α∨ = δ∨ + δ′∨. It follows that 0 6= 〈ξ, α∨〉 = 〈ξ, δ′〉.Since δ′ → α is also a hermitian arrow, Case 1 can be applied to ξ ∼ δ′ and showsξ ∈ Γher.

(b) Now we show that Γorth is absorbing as well. Thus let again ξ ∼ Γorth, soξ is connected to the initial or end point of an arrow α → ε of orthogonal type asin (17.2.3). We may obviously assume ξ 6= α, β, ε because these three vertices arein Γorth.

Case 3: ξ ∼ α. By (15.6.5) ξ → α is impossible, so either ξ ← α or ξ α.Case 3.1: ξ ← α. Then Lemma 15.6 shows ξ ε which implies α → ε is a

hermitian arrow, contradicting Lemma 17.3(b).Case 3.2: ξ α. Then Lemma 15.6 (applied to ξ, α, ε) shows ξ → ε, so we

have ξ α→ ε← ξ, and therefore ξ → ε is an arrow of orthogonal type, whenceξ ∈ Γorth.

Case 4: ξ ∼ ε. By Lemma 17.3, ε← α′ = 2ε− α is also an arrow of orthogonaltype and (α, ε, α′) is a collision. Hence, 0 < 〈2ε, ξ∨〉 = 〈α, ξ∨〉 + 〈α′, ξ∨〉, so thatξ ∼ α or ξ ∼ α′. As we have just seen in Case 3, this implies ξ ∈ Γorth.

(c) Now it is easy to show that the four subsets are pairwise orthogonal. Indeed,assume ξ ∈ Γher is connected to Γorth. Then by (b), ξ ∈ Γher ∩ Γorth = ∅,contradiction. In the same way, ξ ∈ Γlin and ξ ∼ Γher implies ξ ∈ Γher by(a), contradiction, and the same argument shows that Γlin is orthogonal to Γorth.Finally, it follows directly from the definition of Γiso that it is orthogonal to theother three subsets.

From 15.4(b) we conclude that the four subsets are Jordan graphs.

17.6. Definition. Let Γ be a mixed graph. A vertex α of Γ is called internalif there exists β ∈ Γ such that α β, and external otherwise. We introduce thenotations Γ 0 and ∂Γ for the set of internal and external vertices of Γ , respectively,so that

Γ = Γ 0 ∪ ∂Γ. (1)

As we shall see below, there is some similarity with the corresponding notions intopology.

If ∆ ⊂ Γ is a (not necessarily induced) subgraph then it is immediate from thedefinition that ∆0 ⊂ Γ 0, and this implies for a coproduct Γ =

∐j Γj that

(∐j

Γj

)0

=⋃j

Γ 0j and ∂

(∐j

Γj

)=⋃j

∂(Γj). (2)

It is also easily seen that

(Γ 0)0 = Γ 0.

Examples. (a) If Γ is a hexagram (15.4.3) then Γ 0 consists of the innertriangle α, β, γ and ∂Γ consists of the outer three corners δ, δ′, δ′′. This givessome justification for the terminology “internal” and “external”. If Γ is a pyramid(15.4.2), Γ 0 is the outer square and ∂Γ is the “tip” of the pyramid (seen fromabove).


(b) If Γ is simply laced (contains no arrows) and has no connected componentconsisting of an isolated vertex, then Γ = Γ 0.

(c) If Γ consists of a single vertex or is a collision then obviously Γ = ∂Γ .

We describe Γ 0 and ∂Γ for the Jordan graphs Γher, Γorth, Γlin and Γiso in thefollowing proposition.

17.7. Proposition. Let Γ be a Jordan graph. We use the notations introducedin 17.2 and 17.4.

(a) If δ → α is a hermitian arrow then δ ∈ ∂Γ and α ∈ Γ 0. Hence ∂(Γher) isthe set of all initial points and (Γher)

0 the set of all endpoints of hermitian arrows.

(b) If α → ε is an arrow of orthogonal type, then α ∈ Γ 0 and ε ∈ ∂Γ . Hence∂(Γorth) is the set of all end points and (Γorth)0 the set of all initial points of arrowsof orthogonal type.

(c) Γlin is a simply laced Jordan graph consisting entirely of internal vertices:(Γlin)0 = Γlin, and Γiso consists entirely of external vertices: ∂(Γiso) = Γiso.

(d) In terms of (17.5.1), we have

Γ 0 = (Γher)0 ∪ (Γorth)0 ∪ Γlin, ∂Γ = ∂(Γher) ∪ ∂(Γorth) ∪ Γiso. (1)

In particular, Γ 0 = ∅ if and only if Γ = Γiso, while ∂Γ = ∅ if and only if Γ = Γlin.

Proof. (a) and (b): If δ → α is of hermitian type or α→ ε is of orthogonal type,it is clear from (17.2.1) and (17.2.3) that α ∈ Γ 0. If δ resp. ε were also internal,there would exist edges ξ δ resp. ε ξ in Γ and hence subgraphs

ξ δ

????

α γresp.

α

???? γ

ε ξ

leading to embedded arrows ξ δ → α γ resp. γ α → ε ξ,contradicting 17.1.

(c) We first observe that Γlin is simply laced (contains no arrows), becauseby Lemma 17.3(b) any arrow is of hermitian, orthogonal, or isolated type, andtherefore contained in Γher ∪ Γorth ∪ Γiso. Since (by definition) all isolated verticesare also contained in Γiso, it follows that Γlin consists of internal vertices. Thestatement concerning Γiso is clear from the definitions.

(d) This follows from (17.5.1), (17.6.2) and (a)–(c).

17.8. Corollary. Let Γ be a Jordan graph.

(a) If Γ contains no isolated arrows then ∂Γ is an orthogonal system.

(b) Let Γiso = ∅. Then α∨ : α ∈ Γ 0 is a total subset; i.e., if x ∈ X = X•(Γ )and 〈x, α∨〉 = 0 for all α ∈ Γ 0 then x = 0.

Proof. (a) By (17.7.1) it suffices to show that the external points of the threesubsets listed there form an orthogonal system. By assumption, Γiso is an orthog-onal system. So it remains to consider Γher and Γorth. Let δ → α and ε → β be


hermitian arrows and assume δ ∼ ε. Since δ and ε are both external by 17.7(a),δ ε is impossible. But δ → ε or δ ← ε leads immediately to the situation → → , contradicting (15.6.5). The same argument shows that two externalpoints of Γorth are orthogonal.

(b) By 15.1, Γ∨ is total. Since Γ = Γ 0 ∪∂Γ , it suffices to show that 〈x, (Γ 0)∨〉 =0 implies 〈x, (∂Γ )∨〉 = 0. By (17.7.1), our assumption Γiso = ∅ and 17.7(a) and17.7(b), an external vertex of Γ is either the initial point of a hermitian arrow orthe end point of an arrow of orthogonal type.

A hermitian arrow δ → α creates a hexagram (17.2.2), in particular a kite(δ, α, β, γ). Then α, β, γ ∈ Γ 0, 2δ∨ = α∨ − β∨ + γ∨ by Lemma 16.14, and hence2〈x, δ∨〉 = 〈x, α∨ − β∨ + γ∨〉 = 0.

Now let α→ ε be an arrow of orthogonal type. By Lemma 17.3(a), α′ = 2ε−αis of orthogonal type as well, so we have α, α′ ∈ Γ 0 by Proposition 17.7(b). Henceby Lemma 16.11, 〈x, ε∨〉 = 〈x, α∨ + α′∨〉 = 0.

17.9. Proposition. Let Γ be a Jordan graph.

(a) The set of internal points Γ 0 of Γ is itself a simply laced Jordan graph. Inparticular, Γ 0 is a Jordan subgraph of Γ , as defined in 15.16.

(b) If Γ is connected then so is Γ 0.

(c) Let (R,R1) be the 3-graded root system associated with Γ as in Theo-rem 15.11. Then every 0 6= µ ∈ R0 can be written in the form µ = β − α where〈β, α∨〉 = 1, β ∈ Γ and α ∈ Γ 0 ∪ Γiso.

Proof. (a) First, Γ 0 contains no arrows. Indeed, assume β, γ ∈ Γ 0 with β → γ.Then there exist α, δ ∈ Γ such that α β → γ δ, which is impossible byLemma 17.1. To show that Γ 0 is a Jordan graph (with the induced structure),we use 15.16. Since Γ 0 contains no arrows, we need only show that Γ 0 is closedunder completion of squares. Let α β γ ⊥ α in Γ 0 and let (α, . . . , δ) be thecompletion to a square in Γ . Then also δ ∈ Γ 0 since δ α.

(b) Let Γ be connected, and let α, α′ ∈ Γ 0. By Lemma 16.5, Γ has diameter atmost two. If α ∼ α′ then necessarily α α′ because Γ 0 is simply laced. If α ⊥ α′,there exists ε such that α ∼ ε ∼ α′. By Lemma 15.6, we have α ε α′ orα → ε α′ or α ε ← α′ or α → ε ← α′. In the first case, α and α′ areconnected in Γ 0. In the second case, it would follow that ε ∈ Γ 0 so Γ 0 wouldcontain an arrow, contradicting (a). Likewise, the third case is impossible. Sinceα ∈ Γ 0, the arrow α→ ε can, by Proposition 17.7(a), not be of hermitian type, andsince α ∈ Γ 0, it is not of isolated type. Hence by Lemma 17.3(b), it is of orthogonaltype. It follows that there exists β such that (17.2.3) holds. Completing this to apyramid as in (17.2.4), we have α β α′, so α and α′ are connected in Γ 0.

(c) By (15.9.2), we can write µ = β − α where 〈β, α∨〉 = 1. If 〈α, β∨〉 = 1 thenα β and hence α ∈ Γ 0. Otherwise 〈α, β∨〉 = 2, so that we have an arrow α→ βwhich by Lemma 17.3 is of hermitian, orthogonal or isolated type. In the secondalternative, α ∈ Γ 0 by Proposition 17.7(b), and in the third, α ∈ Γiso by definition.It remains to consider the case where α → β is of hermitian type. By 17.2 thereexist γ, δ ∈ Γ such that (α, β, δ, γ) is a kite. Then µ = β − α = δ − γ by (15.3.2),and γ δ shows γ ∈ Γ 0 and 〈δ, γ∨〉 = 1.


In the remainder of this section we give a first application of the theory developedso far, and present a quick classification of the connected non-simply laced Jordangraphs as well as a characterization of the Jordan graphs of octahedral type, see14.19. We emphasize that 17.10–17.12 are independent of 15.11 and 14.7.

17.10. Proposition. Let Γ = Γher be a connected non-empty Jordan graph ofhermitian type. Then I = ∂Γ has at least three elements, Γ is isomorphic to theextended triangular graph TI ∼= G (Cher

I ) of 14.18 where I is an index set of thesame cardinality as ∂Γ , and Γ 0 is isomorphic to TI .

Proposition 17.10 is indeed a classification of non-empty connected Jordangraphs of hermitian type: they are precisely those isomorphic to TI , |I|>3, since it

is immediate from the definition of TI in (14.18.3) that such a graph is a connectedJordan graph of hermitian type.

Proof. Since Γ is not empty, ∂Γ contains at least one hermitian arrow δ → α.By 17.2, this embeds in a hexagram (17.2.2), so by Proposition 17.7(a), ∂Γ containsat least three elements δ, δ′, δ′′ in the notation of (17.2.2).

Next we show: for every α ∈ Γ 0 there exist exactly two vertices δ, δ′ ∈ ∂Γ suchthat δ → α ← δ′ ⊥ δ, and the map α 7→ δ, δ′ is a bijection Φ between Γ 0 andP2(∂Γ ), the set of two-element subsets of ∂Γ .

Indeed, by Proposition 17.7(a), α is the endpoint of a hermitian arrow δ → α,and putting δ′ = 2α − δ we have δ ⊥ δ′ and δ′ → α hermitian by Lemma 17.3(a).Suppose we also have ε, ε′ ∈ ∂Γ such that ε→ α← ε′ ⊥ ε, and assume, aiming for acontradiction, that δ, δ′ 6= ε, ε′, say, ε 6= δ, δ′. Since ∂Γ is an orthogonal systemby Corollary 17.8(a), each of the three orthogonal vertices δ, δ′, ε is connected toα, contradicting Lemma 16.4. Hence we have a well-defined map Φ: Γ 0 → P2(∂Γ )sending α to δ, δ′.

This map is injective: if Φ(α) = δ, δ′ = Φ(β) then 2α = δ + δ′ = 2β andtherefore α = β, because Γ is embedded in the free abelian group X•(Γ ). It is alsosurjective: let δ 6= δ′ be two vertices in ∂Γ . Then δ ⊥ δ′ because ∂Γ is an orthogonalsystem. Since Γ is connected, there exists by Lemma 16.5 a vertex α ∈ Γ such thatδ ∼ α ∼ δ′, and again since ∂Γ is an orthogonal system, α ∈ Γ ∂Γ = Γ 0. Becauseδ and δ′ are external, δ α and δ′ α are impossible. Hence, by (15.6.1),δ → α← δ′, so we have δ, δ′ = Φ(α).

For better readability, let us write ∂Γ = δi : i ∈ I where i 7→ δi is a bijectionbetween the index set I and ∂Γ , but of course this bijection could be the identity.Then it easy to see that the map i 7→ δi and i, j 7→ Φ−1(δi, δj) (for i 6= j) is

an isomorphism between TI and Γ . This also easily implies the statement aboutΓ 0.

17.11. Proposition. For a non-empty connected Jordan graph Γ the followingconditions are equivalent:

(i) for every α ∈ Γ there exists a unique α′ ∈ Γ such that α ⊥ α′,(ii) Γ is isomorphic to the octahedral graph OI ∼= G (Dqf

I+1) of 14.19 where Iis a set of cardinality at least two.

Proof. (i) =⇒ (ii): The map α 7→ α′ is of period two and fixed point free onthe set Γ . We show first that Γ is simply laced. Assume to the contrary that Γ


contains an arrow. It cannot be of hermitian type since any such arrow generatesa hexagram as in (17.2.2) containing three orthogonal vertices, which contradicts(i). Nor can it be of isolated type since then Γ , being connected, is a collision forwhich (i) does not hold either. Hence, by Lemma 17.3(b), the arrow, say α→ ε, isof orthogonal type, and therefore generates a pyramid as in (17.2.4) whose notationwe will use. By assumption there exists ε′ ∈ Γ with ε′ ⊥ ε. Since (α, ε, α′) is acollision, we have ε∨ = α∨ + α′∨, whence 0 = 〈ε′, ε∨〉 = 〈ε′, α∨〉 + 〈ε′, α′∨〉 andconsequently 〈ε′, α∨〉 = 0 = 〈ε′, α′∨〉, i.e., (α, α′, ε) is an orthogonal system, whichagain contradicts (i).

Let P ⊂ P2(Γ ) be the set of unordered pairs α, α′, α ∈ Γ . Then the mapp: Γ → P sending α to α, α′ is surjective and has fibres of cardinality two. Lets: P → Γ be a section of p. Then Γ = s(P ) ∪ s(P )′. Now let I be an indexset and let I 7→ s(P ) be a bijection, written i 7→ αi. (The introduction of I justserves to make notation easier, I = s(P ) and the identity map would be perfectlyacceptable). Then Γ = αi, α′i : i ∈ I, and it is clear that the map OI → Γ sending(i,+) to αi and (i,−) to α′i is an isomorphism of graphs. Since Γ is connected andO1 consists, by 14.19, of two not connected vertices, it is clear that |I|> 2.

(ii) =⇒ (i): This follows immediately from the description of OI in 14.19.

17.12. Proposition. Let Γ = Γorth be a non-empty connected Jordan graphof orthogonal type. Then Γ is isomorphic to the extended octahedral graph OI ∼=G (Bqf

I+1) as in 14.19 where I is a set of cardinality at least two, ∂Γ is a singleton,and Γ 0 is isomorphic to OI .

The proposition above provides a classification of non-empty connected Jordangraphs of orthogonal type: they are precisely those isomorphic to OI , |I|>2. Indeed,

it follows from the definition, see for example (14.19.4), that any vertex of OI with

|I|> 2 is of orthogonal type and hence the same is true for OI .

Proof. We show first that ∂Γ consists of a single vertex. Assume to the contrarythat ε 6= ε′ are in ∂Γ . By Proposition 17.7(d), ε ⊥ ε′. Since Γ is connected, thereexists, by Lemma 16.5, ξ ∈ Γ such that ε ∼ ξ ∼ ε′. Discussing the possible relationsbetween these three vertices and keeping in mind that ε and ε′ are external, one seesthat either ε→ ξ ← ε′ or ε← ξ → ε′. In the first case, since ε is the end point ofan orthogonal arrow by Proposition 17.7(b), there exists α→ ε→ ξ, contradicting(15.6.5). In the second case, it follows from Lemma 15.6 that ε ε′, contradictingε ⊥ ε′. Thus we have ∂Γ = ε.

We show next: if α, α′ ∈ Γ 0 are orthogonal then α → ε← α′ is a collision andhence α′ = 2ε− α is uniquely determined by α. Indeed, by Proposition 17.7(b), αis the initial point of an orthogonal arrow whose end point is in ∂Γ = ε, so wehave α→ ε. By the same argument, α′ → ε, and since α ⊥ α′, we have the collisionas indicated. Hence, for every α ∈ Γ 0 there exists a unique α′ ∈ Γ 0 orthogonal toα. It follows from (17.2.4) that for α, α′ there exists β such that α β α′, soΓ 0 is connected. By Proposition 17.11, there is an isomorphism f : OI → Γ 0 andone checks that f extends to an isomorphism f : OI → Γ by defining f(ω) = ε.

17.13. Lemma. Let Γ be a Jordan graph, and let Γ 0 be the set of internalvertices as defined in 17.6, which by Proposition 17.9 is a Jordan subgraph of Γ .As in (15.17.2) and (15.20.1) we view Inn(Γ 0) as a subgroup of Inn(Γ ).


(a) Inn(Γ 0) stabilizes each connected component Σ of Γ and acts transitivelyon Σ0.

(b) If Σ is a component which is not of isolated type then Inn(Γ 0) induces thegroup of all finitary permutations on ∂Σ; in particular, it acts transitively on ∂Σ.

(c) Inn(Γ 0) acts faithfully on Γ 0.

Proof. (a) By Lemma 15.18(b) applied to ∆ = Γ 0, the group Inn(Γ 0) is gen-erated by all tα,β = sαsβsα, where α, β ∈ Γ 0 and 〈β, α∨〉 = 1. The latter conditionimplies α β or α→ β, but since Γ 0 is simply laced by Proposition 17.9(b), thesecond case is impossible.

An edge α β lies in some component Σ and therefore satisfies α ⊥ Θ ⊥ βfor all components Θ 6= Σ. Hence tα,β acts like the identity on all Θ 6= Σ andstabilizes Σ. To prove transitivity of Inn(Γ 0) on all components, we may thereforeassume Σ = Γ connected.

If Γ 0 is empty there is nothing to prove. Otherwise, Γ 0 contains, by its def-inition, at least two vertices. Let α 6= β in Γ 0. By induction on the length of achain connecting α and β in Γ 0 it suffices to show that for α β there existst ∈ Inn(Γ 0) such that t(α) = β. This follows from (15.18.2).

(b) We may assume Σ = Γ connected. Clearly, any automorphism of Γ sta-bilizes Γ 0 and ∂Γ . By (17.5.1), Γ can be of linear, orthogonal or hermitian type.In the first case, ∂Γ = ∅, and in the second case, ∂Γ is a singleton by Proposi-tion 17.12, so we are done. If Γ is of hermitian type then by Proposition 17.10 and14.18, Γ ∼= TI ∼= G (Cher

I ) for |I|> 3, and ∂Γ = εii : i ∈ I. Let δ = εii 6= δ′′ = εjjin ∂Γ , and choose k 6= i, j. Let α = εik, β = εjk and γ = εij . Then (δ, α, β, γ) is akite. Hence tα,β(δ) = 2γ − δ = δ′′ = εjj by (15.18.5). Moreover, it is easily verifiedthat tα,β fixes εll for all l 6= i, j, and therefore induces the transposition of εii andεjj in the set ∂Γ . Hence Inn(Γ 0) induces the full group of finitary permutations of∂Γ , that is, the permutations which act like the identity outside of a finite set.

(c) Assume t ∈ Inn(Γ 0) acts trivially on Γ 0, and let Σ be a connected compo-nent of Γ . By (a), t stabilizes Σ. Also, Σ0 ⊂ Γ 0, so t acts trivially on Σ0. Nowit suffices to show that t acts like the identity on Σ. If ∂Σ is empty or consists ofa single vertex, this is clear. By Proposition 17.7, the following possibilities for Σremain.

Case 1. Σ = ∂Σ, so Σ0 = ∅. Since t is a product of tα,β where α β, all ofthese are in components different from Σ and therefore are orthogonal to Σ. Hencet is the identity on Σ.

Case 2. Σ = Σher, so Σ ∼= TI for |I|>3 by Proposition 17.10. By (b), t inducesa permutation τ on ∂Σ ∼= I, and Σ0 ∼= P2(I), the set of two-element subsets ofI. The action of t on Σ0 is equivalent to the induced action of τ on P2(I). Since|I| > 3, a permutation which acts trivially on P2(I) is the identity, so t

∣∣Σ0 = Id

implies t∣∣Σ = Id.

Remark. Let (R,R1) be the 3-graded root system associated with Γ . Ingeneral, Inn(Γ 0) is a proper subgroup of the Weyl group W(R0) which, by 15.17,acts by automorphisms of Γ . For example, for a collision Γ = α→ β ← γ we haveΓ 0 = ∅ so Inn(Γ 0) = 1, while R0 = α − β, β − α consists of two orthogonalroots. Hence W(R0) has order 2, the non-trivial element exchanging α and γ.


17.14. Definition. Let Γ be a connected Jordan graph. We define a numericalinvariant c(Γ ) as follows. If Γ 0 is empty we put c(Γ ) = 0. Otherwise, let α ∈ Γ 0

and putc(Γ ) = Cardβ ∈ Γ 0 : β ⊥ α. (1)

By Lemma 17.13, c(Γ ) is well-defined, independent of the choice of α ∈ Γ 0. Sincewe saw in 17.6 that (Γ 0)0 = Γ 0, it is clear that

c(Γ ) = c(Γ 0). (2)

Let us stress that c(Γ ) is not defined for disconnected Jordan graphs.We now determine c(Γ ) for all connected Jordan graphs resp. irreducible 3-

graded root systems, using the list 14.7 and the description of the associated graphsof 14.17 – 14.20.

(i) Γ = KI KJ ∼= G (AII∪J), the Cartesian product of two complete graphs,

with vertex set I × J and (i, j) (i′, j′) if and only if i = i′ or j = j′, see 14.17.Let i0 ∈ I and j0 ∈ J , and put α = (i0, j0). Then, using the definition 16.7,

Γ0(α) = (i, j) : i ∈ I i0, j ∈ J j0 ∼= KI i0 KJ j0,

so we havec(AI

I∪J) = c(KI KJ) = (|I| − 1)(|J | − 1). (3)

If |J | = 1 then KI KJ ∼= KI is the complete graph on the set I, and thereforec(KI) = 0. This is still true in case |I| = 1 where Γ is the graph with one vertex,so Γ 0 = ∅, and therefore c(Γ ) = 0.

(ii) Γ = TI = P2(I) and Γ = TI = P1(I)∪P2(I), the triangular and extended

triangular graph as in 14.18. For |I| = 1 we have Γ = ∅ and Γ is a single vertex,

so c(Γ ) = c(Γ ) = 0. For |I| = 2, Γ is a single vertex and Γ is a collision, so again

c(Γ ) = c(Γ ) = 0. Now assume |I|> 3 and let α ∈ Γ . Then Γ0(α) = P2(I α) and

|I α| = |I| − 2. By Proposition 17.10, (Γ )0 = TI = Γ 0. Hence

c(TI) = c(TI) =

(|I| − 2

2

). (4)

(If a is a cardinal,(a2

)denotes the cardinality of the set of two-element subsets of a

set of cardinality a. Hence(a2

)= 0 for a6 1 and

(a2

)= a if a is an infinite cardinal.

By abuse of notation, we put(a2

)= 0 for negative a.)

(iii) Γ = OI and Γ = OI , the octahedral and extended octahedral graph as in

14.19. If |I| = 1 then O1 consists of two not connected vertices, and O1∼= T2 is a

collision. Hence c(O1) is not defined, and c(O1) = 0. Otherwise,

c(OI) = c(OI) = 1 for |I|> 2 (5)

by Proposition 17.11 and 17.12 and (2).

(iv) For the exceptional cases Cl and Sch we use the descriptions given in 14.20.Let I = 1, . . . , 5 and Γ = Cl = ∅ ∪P2(I)∪P4(I) and choose α = ∅ ∈ Γ . Thenit follows easily from (14.20.1) that Γ1(α) = P2(I) and Γ0(α) = P4(I).


Let J = I ∪ 6 and ∆ = Sch = P1(J) ∪ P2(J) ∪ P5(J), and choose α = 6.It is straightforward to show that

∆1(α) = I ∪P2(I) ∪P1(I), Card(∆1(α)) = 1 + 10 + 5 = 16.

Hence Card(∆0(α)) = Card(∆ (α ∪∆1(α))

)= 27− (1 + 16) = 10. Thus

c(Cl) = 5, c(Sch) = 10. (6)

We collect these results in the following table:

Γ KI KI KJ TI , TI OI , OI Cl Sch

c(Γ ) 0 (|I| − 1)(|J | − 1)(|I|−2

2

)1 5 10

(7)

The above results together with the isomorphisms (14.5.2) yield:

c(Γ ) = 0 ⇐⇒ Γ ∼= Tn (n6 3) or Γ ∼= KI (|I|> 0), (8)

c(Γ ) = 1 ⇐⇒ Γ ∼= T4 or Γ ∼= OI or Γ ∼= OI (|I|> 2) (9)

⇐⇒ Γ 0 ∼= OI (|I|> 2). (10)

17.15. Lemma. (a) Let Γ be a Jordan graph with an induced subgraph

α

????

δ

γ

ε

(1)

Complete α, δ, γ and ε, δ, γ to squares (α, δ, γ, β) and (ε, δ, γ, ζ). Then (α, β, ζ, ε) isa square and the Jordan subgraph of Γ generated by (1) is a prism:

α

???? β

δ

γ

????

ε ζ

(2)

(b) Let α, β, γ, δ in Γ satisfy α β γ δ and γ, δ ∈ Γ0(α). Thenα, β, γ, δ are contained in a prism.

Proof. (a) The relations α β and ζ ε hold because (α, δ, γ, β) and(ε, δ, γ, ζ) are squares, and ε α holds by assumption. Hence it suffices to showβ ⊥ ε, α ⊥ ζ and β ζ. By (15.3.2), β = α− δ + γ which implies

〈β, ε∨〉 = 〈α− δ + γ, ε∨〉 = 1− 1 + 0 = 0.


Similarly, 〈ζ, α∨〉 = 〈γ − δ + ε, α∨〉 = 0 − 1 + 1 = 0. Finally, by Lemma 16.12,β∨ = α∨ − δ∨ + γ∨. Hence

〈ζ, β∨〉 = 〈ζ, α∨〉 − 〈ζ, δ∨〉+ 〈ζ, γ∨〉 = 0− 0 + 1 = 1.

In the same way, 〈β, ζ∨〉 = 〈β, ε∨〉− 〈β, δ∨〉+ 〈β, γ∨〉 = 0− 0 + 1 = 1, which finishesthe proof of β ζ.

(b) If β δ then (a) applies to

γ????

β

α

δ

If β ⊥ δ then β, γ, δ generate a square (β, γ, δ, ε). By (15.3.2),

〈ε, α∨〉 = 〈β − γ + δ, α∨〉 = 1− 0 + 0.

Since 〈α, ε∨〉 = 〈α, β∨〉 − 〈α, γ∨〉+ 〈α, δ∨〉 = 1− 0 + 0 = 1, we have α ε and ofcourse γ ⊥ ε, and the configuration

α????

β

γ

ε

which again by (a) embeds in a prism:

α????

β

γ

????

ε ζ

Since both (β, γ, δ, ε) and (β, γ, ζ, ε) are squares, δ = ζ also lies in the prism.

17.16. Lemma. Let Γ be a connected simply laced Jordan graph. Then thefollowing conditions are equivalent:

(i) c(Γ )> 2,

(ii) CardΓ > 2 and every two-element subset of Γ embeds in a prism.

Proof. (i) =⇒ (ii): CardΓ = 1 would imply c(Γ ) = 0 which is impossible. LetT be a two-element subset. There are two cases.

Case 1: T = α, γ where γ ⊥ α. Since Γ is connected, there exists byLemma 16.5 a vertex β such that α β γ. By Lemma 16.8(c), Γ0(α) is

§18] Bases 233

connected, and γ ∈ Γ0(α). Because |Γ0(α)| > 2, there exists δ ∈ Γ0(α) such thatγ δ. Now the assertion follows from 17.15(b).

Case 2: T = α, β where α β. Choose any γ ∈ Γ0(α). If β γ, chooseδ ∈ Γ0(α) with γ δ. This is possible since Γ0(α) contains at least two elementsand is connected. Then we are again in the situation of Lemma 17.15(b).

If β ⊥ γ, choose ε such that β ε γ and complete to a square (β, ε, γ, ζ).Then

0 = 〈β − ε+ γ − ζ, α∨〉 = 1− 〈ε, α∨〉+ 0− 〈ζ, α∨〉,

hence either 〈ε, α∨〉 = 0 or 〈ζ, α∨〉 = 0. By symmetry in ε and ζ, we may assumeα ⊥ ε. Then α, β, ε, γ satisfy the assumptions of Lemma 17.15(b).

(ii) =⇒ (i): A prism Σ is isomorphic to K2K3 and therefore has c(Σ) = 2 by(17.14.3). Hence c(Γ )> 2.

§18. Bases

18.1. Bases in free abelian groups. Let X be a free abelian group and letB be a basis of X. Then every b ∈ B yields a linear form b∗ ∈ X∗ by defining

〈a, b∗〉 = δab,

for a, b ∈ B. The representation of an element x ∈ X in terms of B is given by

x =∑b∈B

〈x, b∗〉 b

with only finitely many coefficients 〈x, b∗〉 6= 0. We define

P (x) = b ∈ B : 〈x, b∗〉 > 0, N(x) = b ∈ B : 〈x, b∗〉 < 0,

called the positive and negative support of x. The support of x is supp(x) =P (x) ∪ N(x). An element x ∈ X is called positive (negative) with respect to Bif N(x) = ∅ resp. P (x) = ∅. Define the positive and negative part of an elementx ∈ X by

x+ =∑

b∈P (x)

〈x, b∗〉 b, x− =∑

b∈N(x)

−〈x, b∗〉 b.

Then x is the difference of two positive elements,

x = x+ − x−, (1)

and this decomposition is uniquely determined by the requirement supp(x+) ∩supp(x−) = ∅. The element 0 is both positive and negative with 0+ = 0 = 0−.

There is a well-defined linear form t on X given by t(b) = 1 for all b ∈ B;equivalently,

t(x) =∑b∈B

〈x, b∗〉, (2)

called the trace form with respect to B. This is consistent with the terminology in2.16. Finally, the height of x is defined by


h(x) =∑b∈B

|〈x, b∗〉|.

Clearly, h defines a norm on X; in fact, it is just the L1-norm on X ∼= Z(B). Fromthe definitions, it is clear that

h(x) = t(x+ + x−),

in particular, h(x) = t(x) if x is positive. The elements of height 1 are preciselythe elements of B ∪ −B.

Conversely, suppose x = y − z is the difference of two positive elements y andz. Then it is clear that h(x)6 h(y) + h(z), and

h(x) = h(y) + h(z) =⇒ y = x+ and z = x−. (3)

Indeed, we must show supp(y) ∩ supp(z) = ∅. Assume to the contrary that b ∈supp(y) ∩ supp(z). Then y′ = y − b and z′ = z − b are positive, and x = y′ − z′.Hence h(x)6 t(y′) + t(z′) = t(y) + t(z)− 2, contradicting h(x) = t(y) + t(z).

18.2. Lemma. Let Γ be a Jordan graph, embedded in X = X•(Γ ) and let Bbe a basis of X with B ⊂ Γ .

(a) The set β∨ : β ∈ B ⊂ X∗ is total; that is, 〈x, β∨〉 = 0 for all β ∈ Bimplies x = 0.

(b) Let (R,R1) be the 3-graded root system associated with Γ as in Theo-rem 15.11. Then the minuscule coweight defining the 3-grading of R as in 14.6is the trace form t of (18.1.2). The height of an element of Γ is an odd positiveinteger, and the elements of height 1 in Γ are precisely the elements of B.

(c) For all α ∈ Γ B there exist β1, β2 ∈ B and γ ∈ Γ such that

α = β1 − β2 + γ or α = β1 − γ + β2, (1)

h(γ) = h(α)− 2. (2)

Proof. (a) Assume 〈x,B∨〉 = 0 and let P (x) ∪ N(x) = β1, . . . , βn be thesupport of x. Writing ni := 〈x, β∗i 〉, we have x =

∑i niβi, and hence∑

i

ni 〈βi, β∨j 〉 = 0 (3)

for all j = 1, . . . , n. By 2.12, 〈βi, β∨j 〉 = 2(βi |βj)/(βj |βj). Since ( | ) is positive

definite, the matrix(〈βi, β∨j 〉

)is non-singular. Hence (3) implies ni = 0 and

therefore x = 0.

(b) The minuscule coweight q defining the 3-grading is characterized by q(γ) =1, for all γ ∈ Γ . In particular, q(β) = 1 for all β ∈ B, so we have q = t. Letγ = γ+ − γ− be the decomposition of γ ∈ Γ as in (18.1.1). Then 1 = t(γ) =t(γ+)− t(γ−) and h(γ) = t(γ+) + t(γ−) implies h(γ) = 1 + 2t(γ−).

(c) Let α = α+ − α− be the decomposition of α ∈ Γ B into positive andnegative part. Then

§18] Bases 235

2 = 〈α, α∨〉 = 〈α+, α∨〉 − 〈α−, α∨〉. (4)

Since α± is a positive linear combination of elements of B and 〈β, α∨〉 ∈ 0, 1, 2,we have 〈α±, α∨〉 > 0. Hence (4) implies 〈α+, α∨〉 > 0, so there exists ε ∈ P (α)such that 〈ε, α∨〉 > 0 and therefore δ := α− ε ∈ R by [18, VI, §1.3, Theorem 1] or[63, A.3]. Also, δ 6= 0 because α /∈ B. Since ε ∈ P (α), the decomposition of δ inits positive and negative part is

δ+ = α+ − ε, δ− = α−.

Hence2 = 〈δ, δ∨〉 = 〈α+ − ε, δ∨〉 − 〈α−, δ∨〉,

so there are two cases, 〈α−, δ∨〉 < 0 and 〈α+ − ε, δ∨〉 > 0.

Case 1: 〈α−, δ∨〉 < 0. As α− = −∑β∈N(α)〈α, β∗〉β where 〈α, β∗〉 < 0, we have

〈α−, δ∨〉 = −∑

β∈N(α)

〈α, β∗〉〈β, δ∨〉,

so there exists β ∈ N(α) such that 〈β, δ∨〉 < 0. As before, this implies γ := β+ δ ∈R, and even γ ∈ Γ because t(γ) = t(β) + t(δ) = 1 + t(α)− t(ε) = 1 + 1− 1 = 1.

Now γ = α− ε+ β = (α+ − ε)− (α− − β). Since ε ∈ P (α) and β ∈ N(α), theelements α+−ε and α−−β are positive with disjoint supports. Hence γ+ = α+−εand γ− = α− − β are the positive and negative parts of γ. This implies

h(γ) = t(γ+ + γ−) = t(α+ + α− − ε− β) = h(α)− 2,

and we have α = ε− β + γ, so β1 = ε and β2 = β yields our assertion.

Case 2: 〈α+−ε, δ∨〉 > 0. Since δ+ = α+−ε is positive, there exists β ∈ P (α+−ε)such that 〈β, δ∨〉 > 0. As before, this implies γ := β − δ ∈ Γ , and therefore

γ = β − α+ ε = α− − (α+ − ε− β).

We have ε ∈ P (α) and β ∈ P (α − ε). This shows that supp(α+ − ε − β) ⊂supp(α+ − ε) ⊂ supp(α+) = P (α) and supp(α−) = N(α) are disjoint. Hence thepositive and negative part of γ is γ+ = α− and γ− = α+ − ε− β. It follows that

h(γ) = t(γ+ + γ−) = t(α− + α+ − ε− β) = h(α)− 2.

Since α = β − γ + ε, we obtain the desired decomposition by putting β1 = β andβ2 = ε.

18.3. Definition. Let Γ ⊂ X be a Jordan graph as in Lemma 18.2, and letB ⊂ Γ be a Z-basis of X. Define subsets B± of B by

B+ =⋃

γ∈Γ B

P (γ), B− =⋃α∈Γ

N(α). (1)

The condition γ /∈ B in the definition of B+ is important, because for β ∈ B weobviously have P (β) = β and N(β) = ∅. For the same reason it suffices in thedefinition of B− to take the union over all α ∈ Γ B. We say B is a grid basis ofΓ if

B+ ∩B− = ∅, (2)

which is obviously equivalent with

N(α) ∩ P (γ) = ∅ for all α, γ ∈ Γ B. (3)


18.4. Existence of grid bases. Let Γ =∐Γ (i) be the decomposition of Γ

in connected components. It is easily seen that B is a grid basis of Γ if and only ifB ∩ Γ (i) is a grid basis of Γ (i). We now show, using the classification of 3-gradedroot systems and the correspondence with Jordan graphs given by the functors Gand R of Proposition 14.16 and Theorem 15.11, that every Jordan graph admitsa grid basis. In the following cases (a) – (d), we use the notation of 14.5 withoutfurther reference.

(a) Let Γ = G (AII∪J) = εi − εj : i ∈ I, j ∈ J ⊂ X = L0(I ∪ J). Pick

elements i0 ∈ I and j0 ∈ J . We identify (i, j) ∈ I × J with εi − εj by means of theisomorphism Γ ∼= KI KJ of (14.17.2). Then

B = (i0, j0) ∪ (i, j0) : i ∈ I i0 ∪ (i0, j) : j ∈ J j0

is a grid basis of Γ . Indeed, one checks that B is a basis of the free abelian groupX. Let α ∈ Γ B. Then α = (i, j) where i ∈ I i0 and j ∈ J j0, and

(i, j) = (i, j0)− (i0, j0) + (i0, j).

Hence B− = (i0, j0) and B+ = (i, j0) : i ∈ I i0 ∪ (i0, j) : j ∈ J j0, soB is a grid basis.

(b) For Γ = G (BqfI ) = ε0 ∪ ε0 ± εi : 0 6= i ∈ I ⊂ X = L (I), |I|> 2, a grid

basis isB = ε0 ∪ ε0 + εi : 0 6= i ∈ I.

Indeed, ε0 − εi = 2ε0 − (ε0 + εi) shows that B is a Z-basis of X with B+ = ε0and B− = ε0 + εi : 0 6= i ∈ I.

(c) Let Γ = G (CherI ) = εij : i, j ∈ I ⊂ X = L2(I), |I| > 2. We choose an

element 0 ∈ I, and claim that

B = ε0i : i ∈ I

is a grid basis. Indeed, let γ ∈ Γ B, so γ = εij with 0 /∈ i, j. Then

γ = εij = εi0 − ε00 + εj0.

Since B is a Z-basis of X, this shows that B is a grid basis with B+ = εi0 : 0 6=i ∈ I and B− = ε00.

(d) The root system R = DI admits two 3-gradings. First, let Γ = G (DaltI ) =

εij : i 6= j, i, j ∈ I ⊂ X = L2(I). We have |I| > 3 and fix three distinctelements of I, say 1, 2, 3 ∈ I. Then

B = ε23 ∪ ε1i : i ∈ I 1 (1)

is a grid basis of Γ . Indeed, one easily checks that B is a basis of X, and obviouslyB ⊂ Γ . Moreover,

ε2l = (ε23 + ε1l)− ε13,

ε3l = (ε23 + ε1l)− ε12,

εlm = (ε23 + ε1l + ε1m)− (ε12 + ε13)

§18] Bases 237

for l 6= m in I 1, 2, 3 shows that B− = ε12, ε13 and B+ = ε23 ∪ εl : l ∈I 1, 2, 3, so B is a grid basis.

Now let Γ = G (DqfI ) = ε0 ± εi : 0 6= i ∈ I ⊂ L2(I), |I| > 4, and choose a

second element 1 6= 0 in I. Then

ε0 − εi = (ε0 + ε1)− (ε0 + εi) + (ε0 − ε1)

for i 6= 0, 1 shows that B = B+ ∪ B− with B+ = ε0 + ε1, ε0 − ε1 and B− =ε0 + εi : i ∈ I 0, 1 is a grid basis of Γ .

This finishes the classical root systems. The remaining two exceptional caseswill be done by a method applicable to any finite connected Jordan graph Γ withassociated 3-graded root system (R,R1). By Corollary 15.14, R is irreducible. ByExample (b) of 14.6 there exists a root basis (see 2.10(b)) of R which, by abuse oflanguage, we identify with its Dynkin diagram D, and an element β ∈ D havingcoefficient 1 in the highest root (with respect to D) such that γ ∈ Γ = R1 if andonly if the coefficient of β in the expression of γ as a linear combination of D is 1.In particular, β ∈ Γ .

For α ∈ D let Y (α) be the smallest connected subset of D containing both αand β. Then

S(α) :=∑

γ∈Y (α)

γ

belongs to R by [18, VI, §1.6, Corollary 3], and in fact to R1 by the description ofR1 given above. In particular, S(β) = β. We claim that

B =S(α) : α ∈ D

is a grid basis of Γ . Indeed, since D is a Z-basis of X•(Γ ) so is B. The verificationof (18.3.2) is straightforward (but tedious). The reader is invited to do this for theclassical finite root systems. As a result, up to isomorphisms, all examples of gridbases above for finite Γ are obtained in this way.

We now exhibit grid bases for the cases E6 and E7. First, let R = E6 andD = α1, . . . , α6:

D = α1 α3 α4 α5 α6

α2

We choose β = α6 and define βi = S(αi). Explicitly,

β1 = α1 + α3 + α4 + α5 + α6, β2 = α2 + α4 + α5 + α6,

β3 = α3 + α4 + α5 + α6, β4 = α4 + α5 + α6

β5 = α5 + α6, β6 = α6.

Any γ =∑6i=1 niαi ∈ Γ can be written in terms of B = β1, . . . , β6 ⊂ Γ as

γ = n1β1 + n2β2 + (n3 − n1)β3 + (n4 − n3 − n2)β4 + (n5 − n4)β5 + (1− n5)β6.


Since 06n16n3, 06n2, n2+n36n4 and n4>n5>n6 = 1 for all γ ∈ Γ by inspectionof [18, Planche V], B is a grid basis with B+ = β1, β2, β3, β4 and B− = β5, β6.

Now let R = E7 and D = α1, . . . , α7:

α1 α3 α4 α5 α6 α7

α2

Choose β = α7 and define βi = S(αi) and B = β1, . . . , β7 as before. Then for

γ =∑7i=1 niαi ∈ Γ ,

γ = n1β1 + n2β2 + (n3 − n1)β3 + (n4 − n3 − n2)β4

+ (n5 − n4)β5 + (n6 − n5)β6 + (1− n6)β7,

and B is a grid basis with B+ = β1, β2, β3, β4 and B− = β5, β6, β7.

18.5. Lemma. Let Γ be a Jordan graph and let B be a grid basis of Γ . Thenevery α ∈ Γ B has the form

α = β1 − β2 + γ where β1 ∈ P (α), β2 ∈ N(α) and h(γ) = h(α)− 2. (1)

The induced graph on the set β1, β2, γ, α is one of the following:

(i) β2 → β1 = γ ← α, (ii)

β1 β2

α γ

, (iii)

β2

????

β1

???? γ

α

, (2)

and B generates Γ .

Proof. By Lemma 18.2(c), either α = β1 − β2 + γ is already as asserted, orα = β1 − γ + β2. In this case, decompose γ = γ+ − γ− in its positive and negativepart. Then

α = (β1 + γ− + β2)− γ+

where y := β1 + γ− + β2 and z := γ+ are positive. Since

t(y) + t(z) = 1 + t(γ− + γ+) + 1 = 2 + h(γ) = h(α),

we have α+ = β1 + γ− + β2 and α− = γ+ by (18.1.3). Moreover, α ∈ Γ implies

1 = t(α) = t(α+)− t(α−) = t(β1 + γ− + β2)− t(γ+).

Here t(β1+γ−+β2) = 1+t(γ−)+1>2. We claim that γ ∈ B. Assume to the contraryγ ∈ Γ B. Then γ+ 6= 0 6= γ−, and from α− = γ+ we have N(α) = P (γ) 6= ∅,contradicting (18.3.3). Thus (1) holds.

§18] Bases 239

We are ready to prove (2). The four roots (β2, β1, α, γ) have alternating sumzero and satisfy β1 6= α because α /∈ B and α 6= γ because h(α) > h((γ). HenceProposition 16.1(b) shows that there are the following possibilities:

(b1) either β2 = α and β1 → α ← γ, or β1 = γ and β2 → β1 ← α. The firstcase is impossible because α /∈ B, so (i) of (2) holds.

(b2) either (β2, β1, α, γ) is a square, or a cyclic permutation of these is a kite.The first case is (ii) of (2). In the second case, we have to exclude the possibilities

(1)

β1

???

α

???? β2

γ

, (2)

α

????

γ???? β1

β2

, (3)

γ

????

β2

??? α

β1

.

We first remark:

if β ∈ B and δ → β for some δ ∈ Γ then β ∈ B+. (3)

Indeed, by the closure property (C1), the arrow δ → β generates a collision δ →β ← ε = 2β − δ ∈ Γ . To proceed, we distinguish the cases β ∈ P (δ) and β 6∈ P (δ).If β ∈ P (δ) then δ /∈ B, else P (δ) = P (β) = β and therefore δ = β, whichcontradicts δ → β. Hence β ∈ B+ by (18.3.1). If β /∈ P (δ), then decomposeδ = δ+ − δ− in its positive and negative part. Then

supp(2β + δ−) ∩ supp(δ+) =(β ∪N(δ)

)∩ P (δ) = ∅,

hence ε = (2β + δ−) − δ+ is the decomposition of ε in positive and negative part.It follows that β ∈ P (ε), and as before ε→ β implies ε /∈ B, so that β ∈ B+.

Now (1) or (3) are impossible because they would imply β2 ∈ B+, contradictingβ2 ∈ B−. In case (2), complete the kite to a hexagram

α

????

γ β1

δ

?? // β2

???? ε

__????oo

Then again by (3), β2 ∈ B+, contradiction.To prove the last statement, let Σ be the Jordan subgraph of Γ generated by

B. We use the notation of (1). Then (2) and the closure properties of Σ show thatγ ∈ Σ implies α ∈ Σ. Hence Σ = Γ follows by induction on the height.

18.6. Proposition. Let Γ and ∆ be Jordan graphs, embedded in X = X•(Γ )and Y = X•(∆), respectively. Let B be a grid basis of Γ and let f : B → ∆ be amap satisfying

〈f(β1), f(β2)∨〉 = 〈β1, β∨2 〉 (1)

for all βi ∈ B. Then f extends uniquely to an injective homomorphism f : X → Yof abelian groups satisfying f(Γ ) ⊂ ∆ and


〈f(x), f(α)∨〉 = 〈x, α∨〉 (2)

for all x ∈ X and α ∈ Γ . Hence f induces an isomorphism between Γ and theJordan subgraph of ∆ generated by f(B), as well as an embedding of the 3-gradedroot system R(Γ ) into R(∆).

Proof. Since B is a basis of X, f extends uniquely to a group homomorphismf : X → Y , and since 〈x, β∨〉 is linear in x, (1) implies

〈x, β∨〉 = 〈f(x), f(β)∨〉 (3)

for all x ∈ X and β ∈ B. Thus if f(x) = 0 then 〈x, B∨〉 = 0 and therefore x = 0since B∨ is total by Lemma 18.2(a). Hence f is injective.

To prove (2) in general, we use induction on the height of α ∈ Γ , the caseh(α) = 1, i.e., α ∈ B, being (3). Now let h(α) > 1 and write α = β1 − β2 + γ asin Lemma 18.5. Since h(βi) = 1 and h(γ) < h(α), we have 〈f(ξ), f(η)∨〉 = 〈ξ, η∨〉for all ξ, η ∈ β1, β2, γ by induction, so f restricts to an isomorphism between theinduced subgraphs on the sets β1, β2, γ and f(β1), f(β2), f(γ). We now discussthe three cases of (18.5.2).

Case (i): From β2 → β1 = γ follows f(β2) → f(β1) = f(γ). Since theclosure condition (C1) holds in the Jordan graph ∆, this generates a collisionf(β2)→ f(γ)← δ where, by (15.3.1),

δ = 2f(γ)− f(β2) = f(2γ − β2) = f(β1 − β2 + γ) = f(α),

hence f(α)→ f(γ)← f(β2) in ∆.By Lemma 16.11, α∨ = γ∨ − β∨2 and f(α)∨ = f(γ)∨ − f(β2)∨. Hence for all

x ∈ X,〈f(x), f(α)∨〉 = 〈f(x), f(γ)∨〉 − 〈f(x), f(β2)∨〉

= 〈x, γ∨〉 − 〈x, β∨2 〉 (by induction)

= 〈x, γ∨ − β∨2 〉 = 〈x, α∨〉.

Case (ii): From β1 β2 γ ⊥ β1 we conclude f(β1) f(β2) f(γ) ⊥f(β1) in ∆. By the closure condition (C2) for the Jordan graph ∆, the triple(f(β1), f(β2), f(γ)) generates a square in ∆ with fourth vertex

f(β1)− f(β2) + f(γ) = f(β1 − β2 + γ) = f(α)

by (15.3.2). From Lemma 16.12 we have α∨ = β∨1 − β∨2 + γ∨ and f(α)∨ = f(β1)∨−f(β2)∨ + f(γ)∨. Hence for all x ∈ X,

〈f(x), f(α)∨〉 = 〈f(x), f(β1)∨〉 − 〈f(x), f(β2)∨〉+ 〈x, f(γ)∨〉= 〈x, β∨1 〉 − 〈x, β∨2 〉+ 〈x, γ∨〉 (by induction)

= 〈x, β∨1 − β∨2 + γ∨〉 = 〈x, α∨〉.

Finally, Case (iii) of (18.5.2) follows in the same way, using the closure condition(C3) and Lemma 16.14.

So far, we have seen that f∣∣Γ is an isomorphism between Γ and the induced

subgraph f(Γ ) ⊂ ∆. Hence f(Γ ) is a Jordan subgraph isomorphic to Γ . NowLemma 18.5 shows that f(Γ ) is the Jordan subgraph generated by f(B). The laststatement follows from Theorem 15.11 and Lemma 14.3.

§19] Triangles 241

18.7. Example. In 18.6, the assumption that B be a grid basis cannot berelaxed, as the following example shows.

In Γ = G (Cher3 ) = εij : 1 6 i 6 j 6 3, the set B = ε12, ε13, ε23 is a Z-basis

of X•(Γ ) but not a grid basis since ε11 = ε12 − ε23 + ε13, ε22 = ε23 − ε13 + ε12 andthus ε23 ∈ N(ε11) ∩ P (ε22).

The induced subgraph on B is a triangle, whence the induced subgraph on B isa Jordan subgraph of Γ in the sense of 15.16. It is isomorphic to the Jordan graph∆ = K3. Any bijection f : B → ∆ satisfies (18.6.1) and extends to a homomorphismf : X•(Γ ) → X•(∆) satisfying f(Γ ) = ∆, but is not injective nor an isomorphismof Jordan graphs.

§19. Triangles

19.1. Lemma. Let Γ be a Jordan graph and let T and K be disjoint subgraphsof Γ with the following properties:

(i) T = α1, α2, α3 is a triangle,

(ii) the induced subgraph on α2, α3 ∪K is complete,

(iii) α1 ⊥ K:

T ∪K =

α2

KKKKKK

α1

ssssss

KKKKKK K

α3

ssssss

Then the Jordan subgraph of Γ generated by T ∪K is isomorphic to the triangulargraph TI ∼= G (Dalt

I ) where I is a set of cardinality |I| = 3 + |K|.

Proof. Let L be an index set disjoint from 1, 2, 3 and let l 7→ βl be a bijectionL→ K. Let B be the grid basis of TI as in (18.4.1), and define f : B → Γ by

f(εjk) = αi for i, j, k = 1, 2, 3 and f(ε1l) = βl for l ∈ L. (1)

Then (i) – (iii) show that f satisfies (18.6.1), so the lemma follows from Proposi-tion 18.6.

19.2. Corollary. An induced subgraph

α2

KKKKKK

α1

ssssss

KKKKKK β1

α3

ssssss

(1)

of a Jordan graph generates a subgraph T4∼= O3, the graph of an octahedron:


α2

NNNNNNNNNNNNNN β3

pppppppppppppp

666666

α1

7777777

OOOOOOOOOOOOOO β1

α3

ppppppppppppppβ2

(2)

Proof. This is the special case K = β1, a singleton, of Lemma 19.1.

19.3. Proposition. Let T be a triangle in a Jordan graph Γ , and let

Γn = β ∈ Γ T : Cardα ∈ T : α β = n, (1)

the set of vertices of Γ T which are connected by an edge with n vertices of T .Then

Γ = T ∪ Γ0 ∪ Γ1 ∪ Γ2 ∪ Γ3, (2)

∂Γ ⊂ Γ0, T ⊥ (Γ0 ∂Γ ), (3)

and T has the following embedding properties:

T embeds in

a prisman octahedrona tetrahedron

⇐⇒

Γ1

Γ2

Γ3

6= ∅. (4)

If Γ is connected and of rank >4 then Γ1∪Γ2∪Γ3 6= ∅, so at least one of the casesof (4) holds.

Proof. Since T has three elements, Γn is empty for n> 4, so we have (2). From(1) it is clear that T ∪ Γ1 ∪ Γ2 ∪ Γ3 ⊂ Γ 0, whence ∂Γ ⊂ Γ0 by (2). If δ ∈ Γ0 is notorthogonal to T then, by definition of Γ0, there cannot be an edge and thereforemust be an arrow between δ and an element of T , which shows δ ∈ ∂Γ . Thisfinishes the proof of (3).

The implications from left to right in (4) follow immediately from (14.17.1) and(19.2.2). To prove the implications from right to left, we put T = α1, α2, α3 anddistinguish the following cases.

Case 1: Γ1 6= ∅. Then a vertex β ∈ Γ1 yields, for a suitable permutation ijk of123, the induced subgraph

αi????

αk

β

αj

so T embeds in a prism by Lemma 17.15.

Case 2: Γ2 6= ∅. Let β ∈ Γ2. After renumbering the elements of T wemay assume that β = β1 satisfies (19.2.1), so T embeds in an octahedron byCorollary 19.2.

§19] Triangles 243

Case 3: Γ3 6= ∅. Then for β ∈ Γ3 we have T ⊂ T ∪ β, which is a tetrahedron.

Finally, assume Γ connected of rank >4 but Γ1 = Γ2 = Γ3 = ∅. Then Γ = T ∪Γ0

and Γ 0 = T ∪(Γ0 ∂Γ ) = T ∪(Γ0∩Γ 0) is an orthogonal decomposition by (3). SinceΓ 0 is connected along with Γ by Proposition 17.9(b), it follows that Γ0 ∩ Γ 0 = ∅,so Γ0 = ∂Γ . If ∂Γ = ∅ we have Γ = T of rank 3, contradiction. Otherwise,Γ = T ∪ ∂Γ with ∂Γ 6= ∅, so by Propositions 17.10 and 17.12, Γ ∼= T3, again ofrank 3, contradiction.

19.4. Definition. Let T be a triangle in a Jordan graph Γ . Let Γn = Γn(T )be defined as in (19.3.1), and let i, j, k be a permutation of 1, 2, 3. We say T is oftype i if Γi(T ) 6= ∅ and Γj(T ) = Γk(T ) = ∅. For example, by Proposition 19.3,T is of type 2 if and only if T embeds in an octahedron but not in a prism or atetrahedron, and similarly for the other types. From the definition it is clear that atriangle can be of at most one of the three types. As the following examples show,it is possible that a triangle is not of one of the three types.

19.5. Examples. (a) It is immediate from the definition that a triangle con-tained in a complete graph KI with |I|> 4 has type 3.

The prism Γ = K2 K3 as in (14.17.1) contains the triangle T = K1 K3 =(1, 1), (1, 2), (1, 3) with Γ1(T ) = Γ T and Γ2(T ) = Γ3(T ) = ∅, so T is of type1. Now embed Γ in ∆ = K2 K4. Then T is also contained in the tetrahedron1K4, so T has no definite type when considered in ∆.

(b) Let Γ be the be the triangular graph TI or the extended triangular graph

TI as in 14.18. Since triangles and the sets Γn are contained Γ 0 = TI we mayassume Γ = TI . Then Γ has vertices P2(I) and edges

i, j m,n ⇐⇒ |i, j ∩ m,n| = 1.

Let T = α, β, γ be a triangle in TI . Then necessarily |I| > 3, and for |I| = 3we have T = Γ , so all Γi(T ) are empty. Now assume |I| > 4. Since α β, wehave α = i, j and β = i, k for distinct i, j, k ∈ I. Writing γ = m,n withm 6= n, the criterion above applied to α γ yields |i, j ∩ m,n| = 1, whenceeither i ∈ m,n or j ∈ m,n, but not both. In the first case we have γ = i,m,possibly after exchanging m and n. Then β γ forces m /∈ i, j, k, while in thesecond case γ = j,m and therefore m = k. To summarize, a triangle T in TI isof exactly one of the following types:

(i) T = i, j, i, k, i, l involves four distinct indices i, j, k, l ∈ I.

(ii) T = i, j, j, k, k, i involves only three distinct indices i, j, k.

Since a bijection of I onto another set I ′ induces an isomorphism of TI ∼= TI′ , wemay assume that I ′ contains the elements 1, 2, 3, 4 and that a triangle T in TI hasone of the standard forms:

Tc = 1, 2, 1, 3, 1, 4, (1)

Th = 1, 2, 2, 3, 3, 1. (2)

(The choice of subscripts c and h will be explained in 19.8 and 19.10). We claimthat Th is of type 2. Indeed, assume β = l,m ∈ Γ1(Th) ∪ Γ2(Th) ∪ Γ3(Th). If,


say, β 1, 2 then either l = 1 or m = 1 or l = 2 or m = 2. By symmetry,we may assume that one of the first two alternatives holds. Then β 3, 1 butβ ⊥ 2, 3. This shows that β belongs to Γ2(Th) and that Γ1(Th) = Γ3(Th) = ∅ by(19.3.2).

Now let |I|> 5. Then Tc embeds in the tetrahedron Tc ∪ 1, 5, in the prism

2, 5NNNN

4, 5

3, 5

kkkkkkkk

1, 2NNNN

1, 4

1, 3

kkkkkkkkkk

(3)

and in the octahedron O3∼= T4 and hence (because of (19.3.4)) does not have a

definite type. However, for |I| = 4 the triangle Tc is of type 2. Indeed, in this caseΓ2(Tc) = 2, 3, 3, 4, 2, 4 and therefore Γ1(Tc) = Γ3(Tc) = ∅, because T4 hasonly six vertices.

(c) Let |I| > 3 and assume 1, 2, 3 ∈ I. In OI or OI as in 14.19, it suffices bysymmetry to consider the triangle T = (1,+), (2,+), (3,+). This is contained inthe octahedron O3, see (14.19.1). Since O3

∼= T4, T is of type 2 in case |I| = 3, aswe have seen in (b). On the other hand, if |I| > 4 and 1, 2, 3, 4 ⊂ I, then T isalso contained in the tetrahedron T ∪ (4,+), but not in any prism, because OIcontains no prisms. Indeed, in OI every vertex has precisely one vertex orthogonalto it, whereas in a prism, every vertex has two such vertices, as is evident from (3).Thus again T does not have a definite type.

The following Proposition 19.7 is in the spirit of the Propositions 17.10, 17.11and 17.12 which describe the isomorphism classes of 3-graded root systems in termsof properties of the associated Jordan graph. We only consider simply laced Jordangraphs in 19.7 since, by (19.3.3), T ∪ Γ1 ∪ Γ2 ∪ Γ3 ⊂ Γ 0 for a triangle T and hencethe type of T in Γ and in Γ 0 is the same. The proof requires the following lemma.

19.6. Lemma. Let Γ be a Jordan graph and let L and M be subsets of Γ withthe following properties:

(i) the induced subgraphs on L and M are complete,

(ii) L ∩M = α is a singleton,

(iii) (L α) ⊥ (M α).

Then the Jordan subgraph of Γ generated by L∪M is isomorphic to the rectangulargraph K|L| K|M |.

Proof. Let I and J be index sets of cardinality |L| and |M |, and let i 7→ βiand j 7→ γj be bijections I → L and J → M , respectively. Then α = βi0 = γj0for unique elements i0 ∈ I, j0 ∈ J . Let B be the grid basis of KI KJ defined in18.4(a) and define f : B → Γ by

f((i0, j0)) = α, f((i, j0)) = βi for i ∈ I i0, f((i0, j)) = γj for j ∈ J j0.

Then (i) – (iii) show that f satisfies (18.6.1), so the lemma follows from Proposi-tion 18.6.

§19] Triangles 245

19.7. Proposition. Let Γ be a connected simply laced Jordan graph. Then

Γ contains a triangle T of type

123

⇐⇒

Γ ∼= K3 KI , |I|> 2Γ ∼= TI , |I|> 4Γ ∼= KI , |I|> 4

.

Proof. The implications from right to left are easy consequences of the examplesin 19.5, with details left to the reader. We prove the implications from left to right.Let T = α1, α2, α3.

Case 1. By definition, Γ = T ∪ Γ1 ∪ Γ0 and Γ1 6= ∅. For i ∈ 1, 2, 3 let

Ki = β ∈ Γ1 : β αi = Γ1(αi) T. (1)

The second equality follows from the fact that β αi if and only if 〈β, α∨i 〉 = 1because Γ is simply laced, and the definition of Γ1(αi) in (16.7.1). Then Γ1 = K1 ∪K2 ∪K3. We claim that the induced subgraph on K1 is complete. Indeed, assumeto the contrary that there exist β, γ ∈ K1 with β ⊥ γ. Then γ α1 β ⊥ γ,so by (C1) there exists δ such that (α1, β, δ, γ) is a square:

α1 β

γ δ

(2)

This implies by (15.3.2) that 〈δ, α∨2 〉 = 〈γ − α1 + β, α∨2 〉 = 0 − 1 + 0 = −1,contradiction. Hence L = T and M = α1 ∪ K1 satisfies the assumptions ofLemma 19.6, so the Jordan subgraph ∆ of Γ generated by T ∪K1 is isomorphic toK3 KI where |I|> 2.

Since T ∪K1 = α1 ∪ Γ1(α1) = Γ2(α1) ∪ Γ1(α1) by (1), Lemma 16.8(d) shows

∆ = Γ .

Case 2. Here we have Γ = T ∪Γ2∪Γ0 and Γ2 6= ∅. For a permutation ijk of 123let Ki = β ∈ Γ2 : αj β αk. Then Ki ⊥ αi and Γ2 decomposes further in

Γ2 = K1 ∪K2 ∪K3.

We claim that the induced subgraph onK1 is complete. Assume to the contrary thatβ, γ ∈ K1 and β ⊥ γ. Then γ α2 β ⊥ γ, which by (C2) generates a square(α2, β, δ, γ) and implies β−α2 +γ ∈ Γ by (15.3.2). But 〈δ, α∨1 〉 = 〈γ−α2 +β, α∨1 〉 =0 − 1 + 0 = −1 which contradicts (15.1.1). Thus K1 is a complete graph, and wesee that T and K = K1 satisfy the assumptions of Lemma 19.1. Hence the Jordansubgraph ∆ of Γ generated by T ∪K1 is isomorphic to TI where |I| = 3 + |K1|. Itremains to prove that ∆ = Γ .

Again by Lemma 16.8(d), it suffices to show Γ2(α2) ∪ Γ1(α2) ⊂∆. We knowΓ2(α2) = α2 ⊂∆ and

Γ1(α2) =(Γ1(α2) ∩ Γ2(α3)

)∪(Γ1(α2) ∩ Γ1(α3)

)∪(Γ1(α2) ∩ Γ0(α3)

)= α3 ∪

(α1 ∪K1

)∪(Γ1(α2) ∩ Γ0(α3)

).


Since α1 ∪K1 ⊂∆, it is enough to prove Γ1(α2) ∩ Γ0(α3) ⊂∆.Let β3 ∈ Γ1(α2) ∩ Γ0(α3). Because β3 ∈ Γ (T ∪ Γ0) = K1 ∪K2 ∪K3, we get

β3 ∈ K3. Then the induced subgraph on T ∪ β3 is

α2

KKKKKK

α3

ssssss

KKKKKK β3

α1

ssssss

which by Corollary 19.2 generates an octahedron

α2

NNNNNNNNNNNNNN β1

pppppppppppppp

666666

α3

7777777

OOOOOOOOOOOOOO β3

α1

ppppppppppppppβ2

This shows β1 ∈ K1 ⊂ ∆. Moreover, (α1, α3, β1, β3) is a square and α1, α3, β1 ∈T ∪K1 ⊂ ∆. Since ∆ is the Jordan subgraph generated by T ∪K1, we have β3 ∈ ∆.

Case 3. By assumption, Γ = T ∪Γ3 ∪Γ0 and Γ3 6= ∅. We show that the inducedsubgraph on Γ3 is complete. Assume to the contrary that β, γ ∈ Γ3 and β ⊥ γ.This yields a square (α1, β, δ, γ) as in (2). Hence

〈δ, α∨2 〉 = 〈γ − α1 + β, α∨2 〉 = 1− 1 + 1 = 1,

so δ /∈ Γ0, and δ /∈ T because δ ⊥ α1. Hence δ ∈ Γ3 and therefore δ α1,contradiction. It follows that T ∪ Γ3 =∆ is a complete graph, hence a Jordansubgraph of Γ . From Γ = T ∪ Γ3 ∪ Γ0 we infer ∆ = Γ2(α1) ∪ Γ1(α1), so thatLemma 16.8(d) implies ∆ = Γ .

19.8. Lemma and Definition. Let T = α1, α2, α3 be a triangle in a Jordangraph Γ . Let Y = spanZ(T ) ⊂ X = X•(Γ ) and put βj := αi − αj + αk, for ijk a

permutation of 123. Then either Γ ∩ Y = T , or Γ ∩ Y = T ∪ β1, β2, β3 ∼= T3 is ahexagram:

β2

????

α3 α1

β1

??// α2

???? β3

__???oo

(1)

In the first case, T will be called closed. This is similar to the terminology used in1.5. In the second case, we say T is of hermitian type or simply hermitian.

Proof. Suppose β =∑3i=1 niαi ∈ (Γ ∩ Y ) T . Let f : X → Z be the homomor-

phism satisfying f(γ) = 1 for all γ ∈ Γ , see 14.6. Then

§19] Triangles 247

1 = f(β) = n1 + n2 + n3, (2)

and by (15.1.1), 〈β, α∨j 〉 ∈ 0, 1, 2, so

〈β, α∨j 〉 − 1 = ni + 2nj + nk − 1 = nj ∈ −1, 0, 1, (3)

for j ∈ 1, 2, 3.We claim that all ni 6= 0. Assume to the contrary that, say, n3 = 0, so that

β = n1α1 +n2α2. Then by (2) and (3), n1 +n2 = 1 implies either n1 = 0 or n2 = 0,and therefore β ∈ T , contradiction. Furthermore, again by (2), precisely one ofthe ni must be −1. Hence β is one of the βi, and by symmetry, we may assumeβ = β2 = α1 − α2 + α3. Then it follows from (3) that 〈β, α∨1 〉 = 〈β, α∨3 〉 = 2 and〈β, α∨2 〉 = 0, so (β2, α1, α2, α3) is a kite. By 16.13, this generates a hexagram (1)where β1 = α3 −α1 +α2 and β3 = α2 −α3 +α1 are in Γ ∩ Y . Hence we are in thesecond case.

19.9. Lemma. Let Γ be a Jordan graph with associated 3-graded root system(R,R1) as in Theorem 15.11, and let α1, α2, α3 ∈ Γ .

(a) If α1 − α2 + α3 /∈ Γ then also n(α1 − α2) + α3 /∈ R for all n> 1.

(b) Assume that the induced subgraph T on α1, α2, α3 is a triangle. For apermutation π = ijk ∈ S3 let απ = αijk = αi − αj + αk ∈X•(Γ ). Then

T is closed ⇐⇒ α1 − α2 + α3 /∈ Γ (1)

⇐⇒ απ /∈ Γ for all π ∈ S3. (2)

T is hermitian ⇐⇒ απ ∈ Γ for all π ∈ S3 (3)

⇐⇒ (απ, αi, αj , αk) is a kite for all π ∈ S3. (4)

If T is hermitian then the connected component of Γ containing T is isomorphic toTI , |I|> 3.

(c) A sufficient condition for a triangle with vertices α1, α2, α3 to be closed isthat it embed in a tetrahedron K4 or a prism K2 K3, see the examples in 14.17:

α4

KKKKKKKKKK

α1KKKK α3

α2

kkkkkkkkk

δ1HHHHH δ3

δ2

mmmmmmmmm

α1KKKK α3

α2

kkkkkkkkk

(5)

In the case of a tetrahedron,

αi − αj + α4 /∈ R, α1 − α2 + α3 − α4 /∈ R (6)

for all i 6= j in 1, 2, 3. In case of a prism,

αi − αj + δk /∈ R, αi − αj + αk − δl /∈ R (7)


for all permutations ijk of 123 and any l ∈ 1, 2, 3.

Proof. (a) Let f be the minuscule coweight defining the 3-grading of R. SinceR1 = R ∩ f−1(1) and f(n(α1 − α2) − α3) = 1 we have n(α1 − α2) + α3 /∈ Γ ifand only if n(α1−α2) +α3 /∈ R. Therefore, we may assume n> 2. If α1 ∼ α2 thenµ = α1 − α2 ∈ R×0 and α3 + nµ /∈ R for all n ∈ N+ by the root string propertyof R. If α1 ⊥ α2 then 〈n(α1 − α2) + α3, α

∨1 〉 = 2n + 〈α3, α

∨1 〉 > 2n > 4 and hence

n(α1 − α2) + α3 /∈ R because R is reduced by Corollary 15.15.

(b) (2) – (4) are immediate from (a) and Lemma 19.8. The remaining statementfollows from Proposition 17.5 and 17.10.

(c) Suppose we have (5) but T is hermitian. Then 2α∨π = α∨i − α∨j + α∨k byLemma 16.14. Hence, writing δ = α4 in the first and δ = δk in the second case,2〈δ, α∨π〉 = 〈δ, α∨i 〉 − 〈δ, α∨j 〉+ 1 = 1, which contradicts 〈δ, α∨π〉 ∈ Z.

If T embeds in a tetrahedron then so does the triangle (αi, αj , α4), which istherefore closed by the first part of the proof. This implies αi − αj + α4 /∈ R by(2). Assume β = α1 − α2 + α3 − α4 ∈ R. Then 〈β, α∨4 〉 = 1 − 1 + 1 − 2 = −1,and therefore, since R is a root system, sα4

(β) = β + α4 = α1 − α2 + α3 ∈ R,contradicting the fact that T is closed.

If T embeds in a prism then αj ⊥ δk and Proposition 16.1(b) shows αi−αj+δk /∈Γ , whence also αi − αj + δk /∈ R by (a). Assume that β = αi − αj + αk − δl ∈ R.If l = i then 〈β, δ∨i 〉 = 1− 0 + 0− 2 = −1, so sδi(β) = β + δi = αi − αj + αk ∈ R,which contradicts the fact that T is closed. By symmetry in i and k, we also haveαi−αj +αk− δk /∈ R. Finally, assume l = j. Then 〈β, α∨j 〉 = 1−2 + 1−1 = −1, sosαj (β) = β + αj = αi − δj + αk ∈ R, which again contradicts Proposition 16.1(b)since αi ⊥ δj .

19.10. Example. Let Γ be connected. By Lemma 19.9(b), non-closed trian-

gles can only occur if Γ = Γher is of hermitian type, see 17.4. In this case, Γ ∼= TIby Proposition 17.10, with Γ 0 ∼= TI . The triangles of TI have been described in19.5(b), so they have the standard forms (19.5.1) and (19.5.2). We claim that Tc isclosed while Th is hermitian. This follows from (19.9.1). Indeed, using the isomor-

phism TI ∼= G (CherI ) of (14.18.5), we have ε12 − ε13 + ε14 = ε2 − ε3 + ε1 + ε4 /∈ Γ .

For Th we get ε12 − ε23 + ε31 = 2ε1 = ε11 ∈ ∂Γ , so Th is hermitian.

19.11. Lemma. Let Γ be a connected Jordan graph of rank >4, and let α βbe an edge in Γ .

(a) Recall the invariant c(Γ ) defined in 17.14. If c(Γ ) 6 1 then α β iscontained in a closed triangle. In general, it is either contained in a closed triangle,or it is contained in a prism as follows:

α

???? β

ε

ζ

????

δ γ

(1)

where α, ε, δ and β, ζ, γ are closed triangles.

§19] Triangles 249

(b) Let Q = (α, β, γ, δ) be a square in Γ . If c(Γ ) = 1 then Q is contained inan octahedron

α

LLLLLLLLLLLLL β

qqqqqqqqqqqqq

3333333

ε′

4444444

MMMMMMMMMMMMM ε

δ

rrrrrrrrrrrrrγ

(2)

If c(Γ )> 2 then Q is contained in a prism as in (1) or as follows:

α

???? δ

ε

ζ

????

β γ

(3)

where α, β, ε and β, ζ, γ are closed triangles. In any case, either α, β or α, δ iscontained in a closed triangle.

Proof. By Proposition 17.9(b), Γ 0 is connected since Γ is so, and edges andsquares are contained in Γ 0. Recall that c(Γ ) = c(Γ 0) by (17.14.2). We prove (a)and (b) at the same time and distinguish cases according to the value of c(Γ ).

Case c(Γ ) = 0. Then by (17.14.8) and our assumption rankΓ>4, Γ is a completegraph of cardinality >4. Hence α, β are contained in a tetrahedron and thereforeby Lemma 19.9(c) in a closed triangle. This proves (a), and (b) is automatic sincea complete graph contains no squares.

Case c(Γ ) = 1. By (17.14.9) we have Γ ∼= OI or Γ ∼= OI or Γ ∼= T4, with |I|>3in the first two cases, and Γ 0 ∼= OI with |I|> 3 by (17.14.10) in all cases (observe

here that OI and OI have rank |I|+1 while T4 has rank 4). Since Γ 0 is simply lacedand has at least 6 vertices, every edge α β embeds in a triangle α, β, ε which

in the first two cases is closed by Lemma 19.9. If Γ ∼= T4 = i, j : 16 i, j 6 4 asin 19.10, then we may assume α = 1, 2 and β = 1, 3 which is contained in theclosed triangle α, β, 1, 4. This proves (a), and we come to the proof of (b).

Because of c(Γ ) = c(Γ 0) = 1, for every ε ∈ Γ 0 there exists a unique ε′ ∈ Γ 0

which is orthogonal to ε. In particular, for the vertices α, β in Q we have γ = α′

and δ = β′. As in the proof of (a) there exists ε ∈ Γ 0 satisfying α ε β, hencealso γ ε δ. Then Corollary 19.2 shows that Q ∪ ε, ε′ is an octahedron asin (2).

Case c(Γ ) > 2. Then Lemma 17.16 shows that α, β is contained in a prism Π.If α β is contained in a triangle of Π then we have the configuration (3) andhere α, β, ε is a closed triangle by Lemma 19.9(c). If α β is not contained ina triangle of Π, we have the configuration (1). This completes the proof of (a).

It remains to prove (b) in case c(Γ ) > 2. We may replace Γ by Γ 0 and thusassume Γ simply laced. By Lemma 16.8, Γ0(γ) is connected, contains α and hasmore than one element since c(Γ )>2. Hence there exists an edge α ε contained


in Γ0(γ). By (15.3.2), 〈α + γ, ε∨〉 = 1 = 〈β + δ, ε∨〉, so either δ ε ⊥ β orβ ε ⊥ δ. In the first case, Q is contained in a prism (1). Indeed, sinceγ δ ε ⊥ γ, there exists by (C2) ζ ∈ Γ such that (γ, δ, ε, ζ) is a square. Itremains to show that β ζ and α ⊥ ζ. This follows from (15.3.2):

〈ζ, β∨〉 = 〈ε− δ + γ, β∨〉 = 0− 0 + 1 = 1,

〈ζ, α∨〉 = 〈ε− δ + γ, α∨〉 = 1− 1 + 0 = 0.

Then α, δ, ε is a closed triangle, by Lemma 19.9(c) again. In the second case,an analogous proof shows that Q embeds in a prism (3) where α, β, ε is a closedtriangle.

Notes

The first part of §14 is based on [63, §17, 18]. The graphs associated with simply laced3-graded root systems are well known in graph theory, at least in the finite case. This does not

seem to be so for the extended triangular and extended octahedral graphs.The graph-theoretic language of §15 simplifies considerably the combinatorial description of

the 1-part of a 3-graded root system given in [78, §2], which was based on the notion of a closed

abstract cog. Theorem 15.11(a) corresponds to [78, 2.22]. The fact that a 3-graded root systemis reduced (Corollary 15.15), is also shown in [63, Lemma 18.5.b], with a different proof.

§16. Taking into account the equivalence between Jordan graphs and 3-graded root systems

(Theorem 15.11), Proposition 16.1 describes certain configurations of four roots in the 1-part ofa 3-graded root system whose sum is zero. In arbitrary root systems, such configurations are

classified in [100, §6].

The decomposition (16.7.3) is the analogue of the Peirce decomposition of a Jordan pairwith respect to a single idempotent (6.14), while Proposition 16.9 is the analogue of the Peirce

decomposition with respect to a family of orthogonal idempotents (6.16). From this perspective,

Lemma 16.8 is not surprising if one lets “connected Jordan graph” correspond to “simple Jordanpair”: it is known that in a simple Jordan pair V , the Peirce spaces V2(e) and V0(e) are simple

or zero, and under additional assumptions V1(e) is a sum of at most two simple ideals [75]. Thecorrespondence between “connected” and “simple” is a theorem for special classes of Jordan pairs,

see for example [76, IV, §1] where this is shown for Jordan triple systems covered by a grid.

The implications (i) =⇒ (ii) and (i) =⇒ (iii) in 16.11, 16.12 and 16.14 were proved in [63,Lemma 18.4]. Lemma 16.13 is stated in [63, 18.3]. In the setting of grids in Jordan triple systemsthis is [76, I, Theorem 2.11] in view of [78].

The purpose of §17 is to define a number of graph-theoretical notions for Jordan graphs andestablish their basic properties. These will be used extensively in the further development of the

theory. As a byproduct, one obtains an easy classification of non-simply laced connected Jordan

graphs in Proposition 17.10 and 17.12.In view of the categorical equivalence between Jordan graphs and 3-graded root systems and

the known classification of the latter (14.7), the classification results could also be derived, lesselegantly, by using 3-graded root systems. A second approach is based on the equivalence between3-graded root systems and grids in Jordan triple systems [78]. For example, the classification of

simply laced connected Jordan graphs can be easily achieved by translating the classification ofortho-collinear grids [76, II, §2, §3] into the language of Jordan graphs.

§18. The term “grid basis” goes back to [77]. The examples of grid bases in 18.4 are taken

from [77, 5]. Grid bases are conjugate under the action of Aut(Γ ) [77, 7, 8].

CHAPTER V

STEINBERG GROUPS FOR ROOT GRADED JORDAN PAIRS

Summary. In this chapter we take up the subject of Chapter III but in a more generalcontext. A root grading of a Jordan pair V by a Jordan graph Γ is a family R = (Vγ)γ∈Γ of

subpairs whose direct sum is V and which satisfy composition rules determined by Γ . Peirce

gradings are the special case where Γ is a collision, i.e., the graph belonging to the root systemC2. With obvious definitions, Γ -graded Jordan pairs form a category gradjpΓ .

An important property of root gradings is their behaviour under morphisms of the gradinggraph (Lemma 20.4): a morphism f : Γ → ∆ of Jordan graphs induces a ∆-grading f•(R) =

(Wδ)δ∈∆ of V , by

Wδ =∑

f(γ)=δ

Vγ .

This leads to the result that the categories gradjpΓ form the fibres of a category gradjp whichis a split opfibration over the category jgraph of Jordan graphs (Proposition 20.6).

Let (R,R1) be the 3-graded root system belonging to Γ as described in §15, and let R be aΓ -grading of V . Generalizing the procedure of §11, we use the root grading of V to introduce,

for any group G over V , a family U = (Uα)α∈R of subgroups, and define st(V,R) to be the full

subcategory of st(V ) consisting of those groups which have R-commutator relations with respectto U. The main result of §21 is Theorem 21.7 which characterizes these groups by a simpler set of

relations. In particular, it easily implies that the projective elementary group G = PE(V ) is anobject of st(V,R) (Corollary 21.12). We then compare the groups in st(V,R) with the coverings

of G in the sense of §4: up to a natural isomorphism, they are those groups (G,U) ∈ st(V,R) for

which the canonical morphism π: G→ G is bijective on all root groups (21.20 and 21.23).

In §22 we show first that st(V,R) has an initial object, called the Steinberg group of (V,R) and

denoted St(V,R). As long as Γ is fixed, St(V,R) depends functorially on (V,R) (Lemma 22.2),but difficulties arise when one tries to change the grading graph as well. Pursuing this further, we

introduce a split fibred category gradjp∗ over jgraphop whose fibres are the categories gradjpΓintroduced above, and show in Proposition 22.11 that the assignment (V,R) 7→ St(V,R) yields afunctor from gradjp∗ to a suitably defined fibred category of groups with commutator relations.

Following the procedure of Chapter III, we study next Weyl elements for the groups in st(V,R).Instead of considering a single idempotent related to a Peirce grading as in §12, we now deal withsystems of such idempotents, called cogs. A cog compatible with a Γ -grading R = (Vγ)γ∈Γ is a

family E = (eδ)δ∈∆ of non-zero idempotents of V , indexed by a subset ∆ of Γ , satisfying eδ ∈ Vδand conditions relating the Peirce spaces of the eδ with the root spaces Vγ . The idempotents of a

cog play the role of the unit element of the coordinate ring in the classical groups of elementarymatrices over a ring and their unitary analogues. This becomes evident in the discussion of the

main examples of cogs for rectangular and hermitian matrices in 23.23 and 23.24.

Given a Γ -grading R and a cog E = (eδ)δ∈∆ compatible with R, we define in §24 the categoryst(V,R, E ) of those groups G ∈ st(V,R) for which weδ is a Weyl element, for all δ ∈ ∆. The

main result is Theorem 24.2, the counterpart of Theorem 12.5, which characterizes these groupsby simpler conditions.

An analog of the monomial group of matrices is the subject of §25. Given a group G inst(V,R) with a sufficient supply of idempotents, we consider the subgroup M of G generated by

all Weyl elements obtained from all cogs compatible with R. The main result is Theorem 25.4

which states that M is an extension of the Weyl group of a subsystem S of R that can be describedin terms of R.

In the final §26 we consider the case where all connected components of Γ have infinite rank

and prove the important result that the Steinberg group St(V,R) is a central extension of PE(V )

(Corollary 26.6).

251

252 STEINBERG GROUPS FOR ROOT GRADED JORDAN PAIRS [Ch. V

§20. Root gradings

20.1. Definition. Let V = (V +, V −) be a Jordan pair and let A be an abeliangroup. An A-grading of V is a family (Vα)α∈A of pairs of submodules Vα =(V +α , V

−α ) of V such that V σ =

⊕α∈A V

σα (direct sum of submodules) and the

following multiplication rules hold for all α, β, γ ∈ A:

V σα , V −σβ , V σγ ⊂ V σα−β+γ , Q(V σα )V −σβ ⊂ V σ2α−β . (1)

From these relations it is clear that the Vα are subpairs of V .Let V and V ′ be Jordan pairs with A-gradings (Vα)α∈A and (V ′α)α∈A respec-

tively. A homomorphism h: V → V ′ of Jordan pairs is said to be graded ifh(Vα) ⊂ V ′α for all α ∈ A.

Now let Γ be a Jordan graph and let X = X•(Γ ) be the free abelian group asin Lemma 14.11. Also let (R,X,R1) be the associated 3-graded root system as inTheorem 15.11. In view of that theorem, we often identify Γ with R1 and X•(Γ )with X. We will use the notation and terminology introduced in §§14, 15 and 16freely. A grading of V by Γ , or a Γ -grading of V is an X-grading (Vα)α∈X asdefined above satisfying in addition

Vα = 0 if α /∈ Γ , (2)

V σα , V −σβ , V σ = 0 if α ⊥ β. (3)

Because of the equivalence between Jordan graphs and 3-graded root systems a Γ -grading is also called a root grading, and the Vα are called root spaces. As a matterof notation, and in view of (2), we often write R = (Vα)α∈Γ for a root grading. Asubscript α on an element x = xα of V σ indicates that xα ∈ V σα .

Assuming (1) and (2), condition (3) follows from the weaker condition

α ⊥ β =⇒ V σα V −σβ V σβ = 0. (4)

Indeed, we must show Vα Vβ Vγ = 0 whenever γ 6= β ∈ Γ . By (1), the lefthand side is in Vδ for δ = α − β + γ. But δ /∈ Γ , else it would follow fromProposition 16.1(b) (or (16.2.1)) that α ∼ β. Now the assertion follows from (2).

20.2. Examples. We now present a number of examples of Γ -graded Jordanpairs, where Γ is one of the graphs discussed in 14.17 – 14.19.

(a) Let Γ = KI KJ be the rectangular graph as in 14.17, with vertex setI×J . Then a Γ -grading of V , also called a rectangular grading, is a decomposition

V =⊕

(i,j)∈I×J

V(i,j)

such that for all (i, j) and (l,m) ∈ I × J and σ = ±,

Q(V σ(i,j))V−σ(i,j) ⊂ V

σ(i,j), V σ(i,j) V

−σ(i,j) V

σ(i,m) ⊂ V

σ(i,m), (1)

V σ(i,j) V−σ(i,j) V

σ(l,j) ⊂ V

σ(l,j), V σ(i,j) V

−σ(l,j) V

σ(l,m) ⊂ V

σ(i,m), (2)

§20] Root gradings 253

and all other types of products vanish. The details are left to the reader.For a concrete example, let A be a unital associative k-algebra. Generalizing

the example Mpq(A) of the Jordan pair of rectangular matrices in 6.6(a) we letI and J be arbitrary non-empty sets and denote by MatIJ(A) the k-module ofI×J-matrices with entries from A, which are finitary in the sense that only finitelymany entries of a matrix in MatIJ(A) are non-zero. Then it is easily seen thatV := MIJ(A) = (MatIJ(A),MatJI(A)) is a Jordan pair with operations Qxy = xyx(matrix product). This Jordan pair embeds in the Morita context

M =

(k · 1I + MatII(A) MatIJ(A)

MatJI(A) k · 1J + MatJJ(A)

)(3)

where 1I is the I × I-matrix with 1A on the diagonal and zeros elsewhere.Let a, b ⊂ A be k-submodules with aba ⊂ a and bab ⊂ b. Then one checks that

V := MIJ(a, b) := (MatIJ(a),MatJI(b)) (4)

is a Jordan subpair of MIJ(A). Denote by Eij the usual matrix units. Then V hasa rectangular grading given by

V(i,j) =(aEij , bEji

). (5)

The proof of (1) and (2) is left to the reader.

(b) Let Γ be a collision, realized as Γ = 2ε0, ε0 + ε1, 2ε1 ⊂ Cher2 , see 11.1.

With the definition of (11.1.1) it is easily verified that a Γ -grading of V is the sameas a Peirce grading.

More generally, let Γ = TI ∼= G (CherI ) be the extended triangular graph with

vertex set P1(I) ∪ P2(I), see 14.5(c) and 14.18. To simplify notation, we writeVij = Vi,j = Vji, in particular, Vii = Vi. Then a Γ -grading of a Jordan pair V ,also called a hermitian grading, is the same as a direct sum decomposition

V =⊕i,j∈Γ

Vi,j (6)

satisfying the multiplication rules of the Peirce spaces of a Jordan pair with respectto a system of orthogonal idempotents as in 6.16:

D(V σij , V−σmp ) = 0 if i, j ∩ m, p = ∅, (7)

V σij , V −σjm , V σpq = 0 if m 6∈ p, q, (8)

Q(V σij )V−σmp = 0 if m, p 6⊂ i, j, (9)

V σij , V −σjm , V σmp ⊂ V σip, (10)

Q(V σij )V−σij ⊂ V

σij , (11)

Q(V σij )V−σii ⊂ V

σjj . (12)

For the proof, it is convenient to use the realization of Γ as the subset R1 = εij =εi + εj : i, j ∈ I of the 3-graded root system Cher

I . Then (20.1.2) is the same as(6). Since α = εij and β = εmp are orthogonal if and only if i, j ∩ m, p = ∅,


the condition (20.1.3) is equivalent to (7). Hence, to evaluate the first equation in(20.1.1) we only need to consider α, β and γ with α ∼ β and β ∼ γ. Because of thesymmetry of the indices, we can therefore assume α = εij , β = εjm and γ = εpqwith j,m ∩ p, q 6= ∅. Then α − β + γ = εi − εm + εp + εq 6∈ Γ if and onlyif m 6∈ p, q. This shows that the first equation in (20.1.1) is equivalent to (8)and (10). For the second equation in (20.1.1), say α = εij , β = εmp and thereforeζ = 2α − β = 2(εi + εj) − (εm + εp) we have ζ 6∈ Γ if and only if m, p 6⊂ i, jwhich is (9), while ζ ∈ Γ is equivalent to m, p = i, j or m = p = i or m = p = j,that is, to (11) and (12).

The prototype of a Jordan pair with a hermitian grading is the Jordan pair ofhermitian matrices over a form ring (A, J, ε, Λ) of 6.6(d). As in (a), we extend thedefinition there to the case of infinite matrices by considering I × I matrices withonly finitely many non-zero entries. Accordingly, we define

V + = x ∈ MatI(A) : xJ = −εx and xii ∈ Λ+ for all i,V − = y ∈ MatI(A) : yJ = −yε−1 and yii ∈ Λ− for all i,

(V +, V −) = HI(A, J, ε, Λ).

Let again Eij be the standard matrix units. For i 6= j and a, b ∈ A we define

f+ij (a) = aEij − ε−1aJEji, f−ij (b) = bEji − bJεEij . (13)

Then one shows by a straightforward verification that

f+ij (a) = f+

ji (−ε−1aJ), f−ij (b) = f−ji (−b

Jε). (14)

Since the maps a 7→ −ε−1aJ and b 7→ −bJε are bijections of period two of A, wehave fσij(A) = fσji(A). Now we define

V +ii = Λ+Eii = ε−1ΛEii, V −ii = Λ−Eii = ΛEii, Vii = (V +

ii , V−ii ), (15)

V σij = fσij(A) and Vij = (V +ij , V

−ij ) for i 6= j. (16)

Then the Vij are subpairs and define a hermitian grading of V . This can beshown directly or follows from (20.8.1): V is a subpair of the Γ -graded Jordanpair V = MI,I(A) with root spaces Vij , and Vij = V ∩ Vij . If Λ = 0 then we obtainexamples of Jordan pairs graded by the triangular graph TI .

In the examples so far, there is no assumption on the existence of idempotents.On the other hand, in view of the multiplication rules (7) – (12) it is not surprisingthat a system of orthogonal idempotents yields a hermitian grading, as shown inthe following example.

(c) Let I be an index set, let O = (ei)i∈I be a family of orthogonal idempotentsof V , and assume that V has a Peirce decomposition with respect to O (which isalways the case if I is finite). Denoting by Vij for i, j ∈ I ′ = I ∪ 0 the Peirce

spaces of the ei as in 6.16 and putting Γ ′ = TI′ , we have the decomposition

V =⊕

i,j∈Γ ′Vij . (17)

Then it follows from the multiplication rules (6.16.4) – (6.16.6) and Example (b)that (17) is a Γ ′-grading with root spaces V σi,j = V σij .

(d) Examples of Jordan pairs graded by the octahedral graphs OI and OI willbe given in 23.25.


20.3. The category of Γ -graded Jordan pairs. Let Γ be a Jordan graph.We define the category gradjpΓ of Γ -graded Jordan pairs as follows. Its objectsare pairs (V,R) consisting of a Jordan pair V and a Γ -grading R = (Vγ)γ∈Γ of V .A morphism from (V,R) to (V ′,R′ = (V ′γ)γ∈Γ ) is a homomorphism h: V → V ′ ofJordan pairs satisfying

h(Vγ) ⊂ V ′γ for all γ ∈ Γ . (1)

It is immediately verified that in this way one obtains a category as asserted. Wedenote by AutΓ (V,R) the automorphism group of (V,R) as an object of gradjpΓ .Then

AutΓ (V,R) = h ∈ Aut(V ) : ∀γ ∈ Γ : h(Vγ) = Vγ. (2)

20.4. Lemma. Let f : Γ → ∆ be a morphism of Jordan graphs.

(a) Let R = (Vγ)γ∈Γ be a Γ -grading of a Jordan pair V . For δ ∈ ∆ put

Wδ =∑

f(γ)=δ

Vγ . (1)

Then (Wδ)δ∈∆ is a ∆-grading of V , said to be induced by f and denoted f•(R).

(b) Let h: (V,R) → (V ′,R′) be a morphism of Γ -graded Jordan pairs and putf•(R

′) = (W ′δ)δ∈∆. For zδ =∑f(γ)=δ xγ ∈Wδ define

f•(h)(zδ) =∑

f(γ)=δ

h(xγ) ∈W ′δ. (2)

Then f•(h): (V, f•(R))→ (V ′, f•(R′) is a morphism of ∆-graded Jordan pairs, and

the assignments (V,R) 7→ (V, f•(R)) and h 7→ f•(h) yield a functor

f•: gradjpΓ → gradjp∆. (3)

Proof. (a) It is clear that V is the direct sum of the Wδ. For δ, ε ∈ ∆ we have

Q(Wσδ )W−σε =

∑α,β

Q(V σα )V −σβ +∑α,β,γ

V σα , V −σβ , V σγ

where the sums are over all α, γ ∈ f−1(δ) and β ∈ f−1(ε). Hence the relations(20.1.1) imply Q(Wσ

δ )W−σε ⊂ Wσ2δ−ε. A similar proof shows Wσ

δ ,W−σε ,Wσ

ζ ⊂Wσ

δ−ε+ζ , so that (20.1.1) holds for ∆. If f(α) ⊥ f(β) then α ⊥ β by (i) of 14.12.This easily implies (20.1.3) for ∆.

(b) From (2) and (20.3.1) it follows that f•(h) is a morphism of gradjp∆, andit is easily checked that f• has the required functorial properties.

20.5. The category of root graded Jordan pairs. We now define a cat-egory gradjp containing all categories gradjpΓ introduced in 20.3. The objectsof gradjp are pairs (Γ, (V,R)) where Γ is a Jordan graph and (V,R) ∈ gradjpΓis a Γ -graded Jordan pair. A morphism from (Γ, (V,R)) to (∆, (W,S)) is a pair(f, h) where f : Γ → ∆ is a morphism of Jordan graphs, and h: V → W is ahomomorphism of Jordan pairs satisfying


h(Vγ) ⊂Wf(γ) for all γ ∈ Γ . (1)

Clearly, the automorphism group Aut(Γ, (V,R)) of an object of gradjp is thefollowing subgroup of Aut(Γ )×Aut(V ):

Aut(Γ, (V,R)) = (f, h) ∈ Aut(Γ )×Aut(V ) : ∀γ ∈ Γ : h(Vγ) = Vf(γ). (2)

There is a functorΠ: gradjp→ jgraph (3)

given by projection onto the first factor: Π(Γ, (V,R)) = Γ on objects and Π(f, h) =f on morphisms. For Γ ∈ jgraph, the fibre over Γ is by definition the (non-full)subcategory of gradjp whose objects are the objects of gradjp mapping onto Γ ,and whose morphisms are the morphisms of gradjp mapping onto IdΓ . It is easilyseen that the fibre over Γ is canonically isomorphic with the category gradjpΓdefined in 20.3.

20.6. Proposition. (a) The assignments Γ 7→ gradjpΓ on objects and f 7→f• on morphisms of jgraph, as defined in Lemma 20.4, define a functor

Φ: jgraph→ Cat

from the category of Jordan graphs to the category Cat of categories.

(b) The category gradjp is the category obtained from Φ by the Grothendieckconstruction, see 4.21. In particular, the projection Π: gradjp → jgraph of(20.5.3) is a split opfibration.

Proof. (a) We must show that (IdΓ )• = IdgradjpΓ and (f ′ f)• = f ′• f• formorphisms f : Γ → Γ ′ and f ′: Γ ′ → ∆ of jgraph. If f = IdΓ then (20.4.1) showsthat f•(V,R) = (V,R), and from (20.4.2) it is clear that for a morphism h: (V,R)→(V ′,R′) in gradjpΓ we have f•(h) = h. This proves (IdΓ )• = IdgradjpΓ .

Keep in mind that the equality (f ′f)• = f ′•f• is an equality between functorsdefined on gradjpΓ , so both sides must agree on objects and on morphisms ofgradjpΓ . First, let R = (Vγ)γ∈Γ , and put (f ′ f)•(V,R) = (V,Xδ)δ∈∆ andf ′•(f•(R)) = (V, Yδ)δ∈∆. Then by (20.4.1),

Xδ =∑

(f ′f)(γ)=δ

Vγ , Yδ =∑

f ′(γ′)=δ

Wγ′ =∑

f ′(γ′)=δ

∑f(γ)=γ′

Vγ ,

which proves Xδ = Yδ. Finally, it follows easily from (20.4.2) that (f ′ f)•(h) =f ′•(f•(h)) for a morphism h from (V,R) to (V ′,R′).

(b) Refer to 4.21 for the definition of the category X obtained from Φ bythe Grothendieck construction. The objects of X are the pairs (Γ, (V,R)) whereΓ ∈ jgraph and (V,R) ∈ Φ(Γ ) = gradjpΓ , so gradjp and X have the sameobjects. A morphism from (Γ, (V,R = (Vγ)γ∈Γ )) to (∆, (W,S = (Wδ)δ∈∆)) in X isa pair (f, h) where f : Γ → ∆ is a morphism of jgraph and h: (V, f•(R))→ (W,S)is a morphism of gradjp∆. By (20.3.1) and (20.4.1) this means

h( ∑f(γ)=δ

Vγ

)⊂Wδ, for all δ ∈ ∆. (1)

On the other hand, by 20.5, a morphism from (Γ, (V,R)) to (∆, (W,S)) in gradjpis a pair (f, h) where f : Γ → ∆ and h satisfy (20.5.1), and this is clearly equivalentto (1). Hence gradjp = X, so gradjp is a split opfibration over jgraph.


20.7. Peirce gradings associated with root gradings. Let R be a Γ -grading of V and fix α ∈ Γ . By Lemma 20.4, the morphism gα: Γ → T2

∼= G (Cher2 )

of (14.16.1) yields a T2-grading Rα := (gα)•(R) of V which, by Example (b) of20.2, may be identified with a Peirce grading. We denote the corresponding Peircespaces by Vi(α). Then

V = V0(α)⊕ V1(α)⊕ V2(α) (1)

with Peirce spaces given by

Vi(α) =∑

β∈Γi(α)

Vβ , (2)

where Γi(α) is defined in 16.7. The results of §10 can thus be applied to this Peircegrading. In particular, we have: if α 6= β then every (xα, yβ) ∈ V σα × V −σβ isquasi-invertible and

xyβα =

xα +Q(xα)yβ if β → αxα otherwise

, (3)

B(xα, yβ)−1 = B(−xα, yβ) = B(xα,−yβ). (4)

Indeed, if α 6= β then either α ⊥ β or α β or α→ β or β → α. In the first case,xα ∈ V σ0 (β), in the second and third case, yβ ∈ V −σ1 (α) while in the fourth case,xα ∈ V σ1 (β). Hence the assertions follow from Corollary 10.7(a).

Suppose α ⊥ β. Then Vα ⊂ V2(α) and Vβ ⊂ V0(α). This implies, by the Peircerelations of 10.1,

α ⊥ β =⇒ D(V σα , V−σβ ) = Q(V σα )V −σβ = Q(V σα )Q(V −σβ ) = 0, (5)

hence also B(V σα , V−σβ ) = Id.

Let G be a group over V and let again α 6= β. Then (9.10.7) and (3) imply(((((((x−σ(−yβ), xσ(xα)

)))))))= x−σ

(Q(yβ)xα

)· bσ(xα, yβ) · xσ

(Q(xα)yβ

)(6)

where Q(yβ)xα = 0 or Q(xα)yβ = 0.For a root graded Jordan pair (V,R), we define a subgroup of Aut(V ) by

EA(V,R) =⟨β(V +

γ , V−δ ) : γ, δ ∈ Γ, γ 6= δ

⟩, (7)

called the group of R-elementary automorphisms of V . This is consistent withthe definition given in 10.10 in case R is a Peirce grading. Indeed, we have seenin Example 20.2(b) that a Peirce grading is the same as a Γ -grading for Γ acollision. The reader should be warned that the elements of EA(V,R), unlike theautomorphisms of R, do not permute the root spaces Vα in any sensible way.

20.8. Hermitian gradings associated with root gradings. Let again Rbe a Γ -grading of V . We generalize 20.7 by replacing α ∈ Γ by an orthogonalsystem Ω = ωi : i ∈ I ⊂ Γ and by applying Lemma 20.4 to the morphism

f : Γ → Γ ′ = TI′ , I ′ = I ∪ 0, of Proposition 16.9. Then V is TI′ -graded: as in(20.2.6) we have


V =⊕

p,q∈Γ ′Vpq

where for p, q ∈ I ′ and Γpq as in (16.9.3) – (16.9.6),

Vpq =⊕γ∈Γpq

Vγ .

As a special case we consider the matrix pair V = MII(A) of 20.2(a), rootgraded by Γ = KI KI . It is immediate that Ω = (i, i) : i ∈ I ⊂ Γ is an

orthogonal system in Γ . Hence V has a TI -grading

MII(A) = V =⊕

i,j∈TI

Vij (1)

with root spaces Vii = V(i,i) = (AEii, AEii) and

Vij = V(i,j) ⊕ V(j,i) = (AEij , AEji)⊕ (AEji, AEij)

for i 6= j.

§21. Groups defined by root gradings

21.1. Definition of st(V,R). Let Γ a Jordan graph with associated 3-gradedroot system (R,R1) and let R be a Γ -grading of a Jordan pair V as in 20.1. LetG be a group over V . Here and in the sequel we often follow the abuse of languageof 9.1 and write simply G for an object (G, x+, x−, π) of st(V ). Define a familyU = (Uα)α∈R of subgroups of G by U0 = 1 and

Uσα = xσ(V σα ) for α ∈ Γ = R1, σ ∈ +,−, (1)

Uµ =⟨⋃

b(V +α , V

−β ) : α− β = µ, α, β ∈ R1

⟩for µ ∈ R×0 . (2)

As remarked in 20.7, (xα, yβ) ∈ V +α ×V −β is quasi-invertible for α 6= β, so (2) makes

sense. Define the full subcategory st(V,R) of st(V ) by

G ∈ st(V,R) ⇐⇒G ∈ st(V ) and G has R-commutator relationswith root subgroups defined by (1) and (2)

. (3)

Remarks. (a) Since a Peirce grading P is the same as a root grading of type

T2 by Example (b) of 20.2, the present definition is consistent with the definitionof st(V,P) in (11.1.7).

(b) As in case of st(V,P), it is clear from 3.3(b) that if there exists a groupG ∈ st(V,R) then G = PE(V ) must have R-commutator relations. We will show inCorollary 21.12 that this is the case, and thus conclude that st(V,R) is not empty.

(c) Let G be a group over V with root subgroups Uσ. From 20.1 and (1) it isclear that

§21] Groups defined by root gradings 259

Uσ ∼=⊕α∈Γ

Uσα (4)

where⊕

α∈Γ Uσα is the restricted direct product, i.e., the subgroup of the directproduct of the Uσα consisting of all elements with only finitely many components6= 1. Since G ∈ st(V ) is generated by U+ and U−, it follows that G is generatedby all U±α, α ∈ Γ . Following the convention of 3.2, we have

UR0=⟨Uµ : µ ∈ R0

⟩and observe that (9.7.3) implies

UR0⊂ G0 = π−1

(PE0(V )

). (5)

It is also worth noting that for any G ∈ st(V,R) the subgroup UR0 has R0-commutator relations with root groups Uµ (µ ∈ R0).

(d) Let G ∈ st(V,R) with root groups Uα as defined above, let π: G → G =PE(V ) be the canonical projection, and suppose that K ⊂ Ker(π) is a normalsubgroup of G. We have seen in Lemma 9.2(e) that then G = G/K is a groupover V with respect to U± = can(U±) where can: G → G is the canonical map.It is immediate that the subgroups can(Uα), α ∈ R, are the root groups definedin (1) and (2) for G and G has R-commutator relations with respect to the family(can(Uα))α∈R.

(e) DefineΓ×(R) = γ ∈ Γ : Vγ 6= 0, (6)

and let Γ ′ be the Jordan subgraph of Γ generated by Γ×(R), which exists by 15.16.Then Vγ = 0 for γ ∈ Γ Γ ′, and it follows immediately from the definitions thatR′ = (Vγ)γ∈Γ ′ is a Γ ′-grading of V , and st(V,R) = st(V,R′). More generally, forany Jordan subgraph Γ ′′ ⊂ Γ containing Γ×(R) the Jordan pair V is Γ ′′-gradedby R′′ = (Vγ)γ∈Γ ′′ and st(V,R) = st(V,R′′).

In general, Γ×(R) itself is not a Jordan subgraph. For example, let Γ be acollision α → β ← γ, thus R = Cher

2 , and let V = V2 ⊕ V1 ⊕ V0 = Vα ⊕ Vβ ⊕ Vγbe a Γ -grading with V2 6= 0 6= V1 but V0 = 0. Then Γ×(R) = α, β is not aJordan subgraph because γ = 2β−α belongs to Γ but not to Γ×(R). An exampleof such a Jordan pair can be constructed as follows. Let ∆ = α β = K2 bethe complete graph on two vertices and assume V has a ∆-grading R = (Vα, Vβ),for example, let V = M12(A) as in Example 20.2(a). Consider the morphism

f = gβ : ∆ → T2 of 20.7 sending (α, β) to (α, β). Then (V,R) = (V, f•(R)) as in20.4. It is straightforward to verify that any Γ -graded Jordan pair with Vγ = 0arises in this way from a ∆-grading of V . In (23.23.2) we will present anotherexample where Γ×(R) is not a Jordan subgraph.

While the definition of the root groups Uα for α ∈ R±1 in (1) is quite canonical,this is not so for the Uµ. The next lemma shows that the Uµ as defined in (2) arethe minimal ones if G is to have R-commutator relations and (5) is to hold.

21.2. Lemma. Let V be a Γ -graded Jordan pair and let G be a group over V .Let (U%)%∈R be a family of subgroups of G such that

(i) G has R-commutator relations with respect to the U%,


(ii) Uσα = xσ(V σα ) for all α ∈ R1, σ ∈ +,−,(iii) Uµ ⊂ G0 for all µ ∈ R×0 .

Then Uµ ⊃ Uµ for all µ ∈ R0 where Uµ is as in (21.1.2).

We will show in Corollary 21.15 that G belongs to st(V,R).

Proof. Let α 6= β, let (xα, yβ) ∈ V +α × V −β and put h = b(xα, yβ). We claim

h ∈ Uµ.Since R is 3-graded, A :=

(((((((α,−β

)))))))is a nilpotent subset containing µ, and it is

either contained in µ, µ+α or in µ, µ−β. We deal with the first case, and leavethe second case, which is analogous, to the reader. By (20.7.3), x

yβα = xα + zµ+α

where zµ+α = Q(xα)yβ ∈ V +2α−β = V +

µ+α. By (20.7.6),

b(xα, yβ) =(((((((

x−(−yβ), x+(xα)))))))· x+(−zµ+α).

Since A is nilpotent, Proposition 3.12(b) and the commutator relations for the U%show (((((((

x−(−yβ), x+(xα)))))))∈(((((((U−β , Uα

)))))))⊂ U(((((−β,α))))) = UA = Uµ · Uµ+α,

so that(((((((

x−(−yβ), x+(xα)))))))

= gµ · gµ+α where gµ ∈ Uµ and gµ+α ∈ Uµ+α ⊂ U+.Hence

h = gµ · gµ+α · x+(−zµ+α) ∈ Uµ · Uµ+α,

so h = gµ ·g+ where gµ ∈ Uµ and g+ ∈ U+. On the other hand, π(h) = β(xα, yβ) ∈PE0(V ) by (9.7.3) and therefore h ∈ G0. By assumption (iii), Uµ ⊂ G0 as well.

Now Lemma 9.2(a) implies g+ = 1, so h = gµ ∈ Uµ.

21.3. Definitions. Let Γ be a Jordan graph with associated 3-graded rootsystem (R,R1). We define a full subcategory gcjΓ of the category gcR introducedin 4.1 as follows. The objects of gcjΓ are groups with R-commutator relations withrespect to a family U = (Uα)α∈R of subgroups as in 4.1, but with the additionalrequirement that

G =⟨Uα : α ∈ R1 ∪R−1

⟩. (1)

Let (V,R) be a Γ -graded Jordan pair and let G = (G,U±, π) ∈ st(V,R), withroot groups U = (Uα)α∈R defined in (21.1.1) and (21.1.2). Since G is a group overV , it is generated by U+ and U−. From (21.1.4) it is clear that G satisfies (1), thus(G,U) ∈ gcjΓ .

Let G and G′ be in st(V,R), and let ϕ: G→ G′ be a morphism in st(V ), thatis, a group homomorphism ϕ: G → G′ such that the diagrams (9.1.4) commute.This immediately implies that ϕ: Uα → U ′α is an isomorphism for all α ∈ R1∪R−1.Moreover, ϕ: Uµ → U ′µ is surjective (but in general not an isomorphism), becauseϕ(b(xα, yβ)) = b′(xα, yβ) by (9.7.1). This shows that ϕ is a morphism from (G,U)to (G′,U′) in gcjΓ . Thus, the assignments G = (G,U±, π) 7→ (G,U) on objects andϕ 7→ ϕ on morphisms define a functor

LR : st(V,R)→ gcjΓ . (2)

This functor is faithful (injective on morphism sets) but not injective on objects,because it loses the information contained in the projection π: G→ G = PE(V ) ofan object (G,U±, π) of st(V,R).


Let f : Γ → ∆ be a morphism of Jordan graphs, and let (S, S1) be the 3-gradedroot system associated with ∆. To simplify the notation, we denote the inducedmorphism R(f): (R,R1)→ (S, S1) of Theorem 15.11 again by f .

For (G,U) ∈ gcjΓ let f•(G,U) = (G′,U′) ∈ gcS be defined as in Proposi-tion 4.22. Then

U ′ξ =⟨Uα : f(α) = ξ

⟩for ξ ∈ S×, (3)

f•(G,U) = (G,U′) ∈ gcj∆. (4)

Indeed, let ξ ∈ S×. Since 3-graded root systems are reduced, (4.18.1) becomesR[ξ] = f−1(ξ), and U ′ξ = UR[ξ] by (4.18.2), so we have (3). Again by (4.18.2), G′

is the subgroup of G generated by all Uα where f(α) 6= 0. For α ∈ R±1 we havef(α) ∈ S±1, in particular, f(α) 6= 0. Hence (1) shows G′ = G. From f(Ri) ⊂ Si fori ∈ 0,±1 follows f−1(S±1) = R±1. Now (1) shows that G is generated by all U ′ξ,ξ ∈ S1∪S−1, so that (4) holds. Thus the functor f•: gcR → gcS of Proposition 4.22maps the subcategory gcjΓ of gc to gcj∆. Hence, its restriction to gcjΓ is a functor,again denoted f•:

f•: gcjΓ → gcj∆. (5)

We now show that passing to an induced root grading is compatible with thecommutator relations in the following sense.

21.4. Proposition. Let Γ and ∆ be Jordan graphs with associated 3-gradedroot systems (R,R1) and (S, S1), let f : Γ → ∆ be a morphism of Jordan graphs anddenote the corresponding morphism (R,R1)→ (S, S1) also by f . Let R = (Vγ)γ∈Γbe a Γ -grading of the Jordan pair V , and let f•(R) be the induced ∆-grading ofV as in Lemma 20.4. Then st(V,R) is a full subcategory of st(V, f•(R)), and thediagram

st(V,R)i //

LR

st(V, f•(R))

Lf•(R)

gcjΓ f•

// gcj∆

(1)

is commutative, where i denotes the inclusion and the vertical arrows are definedin (21.3.2).

Proof. Let G ∈ st(V,R) and let LR(G) = (G,U) ∈ gcjΓ as in (21.3.2). By(21.3.4), f•(G,U) = (G,U′) ∈ gcj∆ has S-commutator relations with root groups(21.3.3). On the other hand, let U′′ = (U ′′ξ )ξ∈S be the root groups of G defined in21.1, but for the root grading f•(R) of V instead of R. We claim that U′′ = U′.From (20.4.1) it follows easily that U ′′ξ = U ′ξ for ξ ∈ S±1. Also, (21.1.5) and

(21.3.3) imply U ′ξ ⊂ G0 for all ξ ∈ S×0 . Therefore, the hypotheses of Lemma 21.2

are satisfied for U′, so we conclude U ′′ξ ⊂ U ′ξ for all ξ ∈ S×0 .

To prove the reverse inclusion, let µ ∈ R∩ f−1(ξ). Then µ ∈ R×0 because f is amorphism of 3-graded root systems, so µ = α − β for suitable α, β ∈ R1. Puttingζ := f(α), η := f(β), we have ζ, η ∈ S1 and ξ = ζ − η. From the definition ofthe ∆-grading in (20.4.1), it follows that V +

α ⊂ W+ζ and V −β ⊂ W−η , and therefore

b(V +α , V

−β ) ⊂ b(W+

ζ ,W−η ) ⊂ U ′′ξ . By definition of Uµ in (21.1.2), this implies


Uµ ⊂ U ′′ξ for all µ ∈ R∩ f−1(ξ) and therefore U ′ξ ⊂ U ′′ξ by (21.3.3). This completesthe proof that U′ = U′′, and shows that G ∈ st(V, f•(R)). Since st(V,R) is afull subcategory of the category st(V ), a morphism in st(V,R) is the same as amorphism in st(V ), whence st(V,R) is a full subcategory of st(V, f•(R)). We alsohave shown

Lf•(R)(i(G)) = (G,U′′) = (G,U′) = f•(G,U) = f•(LR(G)).

Hence (1) is commutative on objects. It also commutative on morphisms since forany morphism ϕ in st(V,R) the definitions in 21.3 and 4.22 yield

Lf•(R)

(i(ϕ)

)= Lf•(R)(ϕ) = ϕ = LR(ϕ) = f•

(LR(ϕ)

), (2)

as desired.

21.5. Corollary. Fix α ∈ Γ and let Rα be the Peirce grading of V associatedwith α as in (20.7.1). Then st(V,R) ⊂ st(V,Rα).

Proof. This follows immediately from the proposition applied to the homomor-phism fα: (R,R1)→ Cher

2 of (14.9.1).

21.6. Corollary. With the notations of (20.7.7), 21.1 and Proposition 21.4,we have

UR0=⟨ ⋃µ∈R0

Uµ

⟩= EA(V,R) and EA(V, f•(R)) ⊂ EA(V,R).

Proof. The first formula is an immediate consequence of the definitions. As tothe second, specialize (21.3.3) to the case where G = G and ξ ∈ S×0 . Since f mapsRi to Si, we have

U ′ξ =⟨ ⋃µ∈R0∩f−1(ξ)

Uµ⟩

for ξ 6= 0 which implies the inclusion EA(V, f•(R)) ⊂ EA(V,R).

21.7. Theorem. Let Γ be a Jordan graph with associated 3-graded root system(R,R1), and let R = (Vα)α∈Γ be a Γ -grading of a Jordan pair V . Let G be a groupover V and define subgroups (Uα)α∈R of G as in 21.1. Then the following conditionsare equivalent:

(i) G has R-commutator relations, i.e., G ∈ st(V,R),

(ii) the commutator relations(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) hold for all α ∈ R×0 , β ∈

R1 ∪R−1, and for all (α, β) ∈ R1 ×R−1 with α ⊥ β,

(iii) the group UR0normalizes U+ and U−, and

(((((((Uα, U−β

)))))))= 1 for all

α ⊥ β in R1,

(iv) for all α, β ∈ Γ and all xα ∈ V +α , yβ ∈ V −β , the relations

B(xα, yβ) if α ∼ β 6= α, (StR1)

b(xα, yβ) = 1 if α ⊥ β (StR2)

hold in G.


Proof. The implication (i) =⇒ (ii) is obvious.

(ii) =⇒ (iii): For µ ∈ R×0 and α ∈ R1 we have(((((((µ, α

)))))))⊂ R1 since R is 3-graded.

Hence(((((((Uµ, Uα

)))))))⊂ U(((((µ,α))))) ⊂ UR1 = U+, which implies that Uµ normalizes U+.

One shows in the same way that Uµ normalizes U−. For α ⊥ β in R1 we have(((((((α,−β

)))))))= ∅ by (14.8.9) and hence

(((((((Uα, U−β

)))))))⊂ U∅ = 1.

(iii) =⇒ (iv): If α − β = µ ∈ R×0 then b(xα, yβ) ∈ Uµ by (21.1.2). Henceb(xα, yβ) normalizes U+ and U−, so B(xα, yβ) holds by (9.8.5). If α ⊥ β then(20.7.5) and (20.7.6) show that b(xα, yβ) =

(((((((x−(−yβ), x+(xα)

)))))))= 1.

The proof of (iv) =⇒ (i) will be given in 21.11. It requires the followingpreliminary results.

21.8. Proposition. Let f : Γ → ∆ be a morphism of Jordan graphs, let R =(Vγ)γ∈Γ be a Γ -grading of the Jordan pair V , and let S = f•(R) = (Wδ)δ∈∆ bethe induced ∆-grading as in Lemma 20.4. Also, let G be a group over V whichsatisfies the relations (StR1) and (StR2). Then G satisfies the relations (StS1)and (StS2).

Proof. We begin by proving (StS2). Let γ ⊥ δ in ∆. We must show that x+(x)and x−(y) commute, for all x ∈ W+

γ and y ∈ W−δ . By definition, x =∑α xα

is a finite sum of elements xα ∈ V +α with f(α) = γ and, similarly, y =

∑β yβ

with yβ ∈ V −β and f(β) = δ. By 14.12(i), γ ⊥ δ implies α ⊥ β. Hence (StR2)

shows that(((((((

x−(yβ), x+(xα))))))))

= b(xα, yβ) = 1. Since x+(x) =∏α x+(xα) and

x−(y) =∏β x−(yβ), it follows that x+(x) and x−(y) commute.

We are left with proving (StS1). By (9.8.5), this means that b(x, y) normalizesU+ and U− for x ∈ W+

γ , y ∈ W−δ and γ 6= δ ∼ γ. This will follow from the moreprecise claim:

Let x =∑α∈f−1(γ) xα ∈W+

γ and y =∑β∈f−1(δ) yβ ∈W

−δ .

Then b(x, y) is a product with factors b(xα, yβ) in a suitable order.(1)

Indeed, assuming (1), each factor b(xα, yβ) normalizes both U+ and U−, hence sodoes the product b(x, y).

We will prove (1) by induction on the number of non-zero summands in thedecomposition of x and y. Although we have to consider the cases γ → δ, γ δand γ ← δ, we will only prove the case γ → δ, leave the easier case γ δ to thereader and note that the case γ ← δ follows from γ → δ by replacing V by V op.We have γ → δ ← 2δ − γ by (C1) and (15.3.1), in particular γ ⊥ 2δ − γ ∈ ∆.But 2γ − δ /∈ ∆, since 2γ − δ ∈ ∆ implies 〈2γ − δ, γ∨〉 = 4 − 1 = 3, contradicting(14.10.1).

First suppose x = xα ∈ V +α . There is nothing to prove if y = yβ ∈ V −β . So, by

induction, we can assume that y has the form y = yβ + t with yβ ∈ V −β , t ∈ W−δand that (1) holds for b(xα, t). Since B(xα, yβ) holds by (StR1), we can applythe formula (9.9.2) and get b(xα, y) = b(xα, yδ + t) = b(xα, yβ) · b(x

yβα , t). By

(20.7.3), xyβα = xα + Q(xα)yβ = xα since Q(xα)yβ ∈ Q(W+

γ )W−δ ⊂ W+2γ−δ = 0

because 2γ − δ 6∈ ∆. Hence b(xα, y) = b(xα, yβ) · b(xα, t) and (1) holds for (xα, y)by induction.


Let now x ∈W+γ and y ∈W−δ be arbitrary. We write x in the form x = xα + s

with x+α ∈ V +

α and s ∈ W+γ . We have just shown that (1) holds for (xα, y).

In particular, the relation B(xα, y) holds in G. Hence, by (9.9.1) we now getb(x, y) = b(xα+s, y) = b(s, yxα)·b(xα, y) where yxα = y+Q(y)xα. By induction wecan assume that (1) is true for (s, y), hence the relation B(s, y) holds inG. So (9.9.2)yields b(s, yxα) = b(s, y) · b(s,Q(y)xα). But Q(y)xα ∈ Q(W−δ )W+

γ ⊂ W−2δ−γ andγ ⊥ 2δ − γ, so b(s,Q(y)xα) = 1 by the relation (StS2) already proved. Altogetherwe have b(x, y) = b(s, y)b(xα, y), where now the induction hypothesis applies toboth (s, y) and (xα, y). This finishes the proof of (1).

21.9. Corollary. Let R be a Γ -grading of the Jordan pair V , fix a root α ∈ Γ ,and let Rα be the induced Peirce grading of V as in (20.7.1). Also, let G be agroup over V which satisfies the relations (StR1) and (StR2). Then G satisfies therelations (StRα1) and (StRα2) and hence belongs to st(V,Rα) by Theorem 11.2.

Proof. This follows from Proposition 21.8, applied to the morphism fα of(14.9.1) mapping Γ to Cher

2 .

21.10. Lemma. Let R be a Γ -grading of V , let α 6= β, γ 6= δ be in Γ and let(xα, yβ) ∈ V +

α × V −β and (uγ , vδ) ∈ V +γ × V −δ . If a group G over V satisfies the

relations B(xα, yβ) for all α, β ∈ Γ and all xα ∈ V +α and yβ ∈ V −β then also the

formulas(((((((b(xα, yβ), x+(uγ)

)))))))= x+

(− xαyβuγ+Q(xα)Q(yβ)uγ

), (1)(((((((

b(xα, yβ), x−(vδ))))))))

= x−(yβxαvδ+Q(yβ)Q(xα)vδ

), (2)(((((((

b(xα, yβ),b(uγ , vδ))))))))

= b(B(xα, yβ)uγ , B(−yβ , xα)vδ

)· b(uγ , vδ)

−1 (3)

hold in G.

Proof. Formulas (1) and (2) are (9.8.3) and (9.8.4), taking into account that(9.9.3) and (20.7.3) yield

b(xα, yβ)−1 =

b(−xα, yβ) if β → αb(xα,−yβ) otherwise

.

Also, (3) is a consequence of (9.9.5) and (20.7.4).

21.11. Proof of (iv) =⇒ (i) of Theorem 21.7. We set U% = 1 if % is a linearcombination of roots which is itself not a root. This is consistent with (21.1.1) sinceV ±% = 0, and we also have

b(V +α , V

−β ) ⊂ Uα−β for all α 6= β in R1. (1)

Indeed, if α−β ∈ R×0 this holds by (21.1.2). Otherwise, α ⊥ β so b(V +α , V

−β ) = 1

by (StR2).By definition of the commutator relations in 3.2 we must show (3.2.1), (3.2.2)

and (((((((Uµ, Uν

)))))))⊂ U(((((µ,ν))))) (2)

for all nilpotent pairs (µ, ν) in R. Since R is reduced by Corollary 15.15, (3.2.2)is vacuous, while (3.2.1) holds by definition. Again since R is reduced, it follows


from (2.17.1) that (µ, ν) is a nilpotent pair, i.e., µ, ν is prenilpotent, if and onlyif µ+ ν 6= 0.

We will prove (2) by invoking Lemma 3.8 for the sets Xµ = Uµ where µ ∈R1 ∪R−1, and for

Xµ =⋃b(V +

α , V−β ) : α− β = µ

where α, β ∈ R1 and µ ∈ R×0 . Then obviously Xµ = X−1µ in the first case,

while Xµ = X−1µ for µ ∈ R0 follows from (9.9.3) and (20.7.3): b(xα, yβ)−1 =

b(−xα, yxαβ ) = b(−xα, yβ) unless β → α, but then b(xα, yβ)−1 = b(xyβα ,−yβ) =

b(xα,−yβ). Therefore, (2) follows as soon as we have shown the assumption (3.8.1)of loc. cit., namely (((((((

Xµ, Xν

)))))))⊂ U(((((µ,ν))))), (3)

for all nilpotent pairs (µ, ν) in R.

We first deal with the case that µ and ν are linearly dependent (and of courseµ+ν 6= 0). Then necessarily µ = ν, so we must show that Uµ is abelian. If µ ∈ R±1

then Uµ ⊂ U± is clearly abelian. Let µ ∈ R×0 and fix a representation µ = β − αas in (15.9.2), thus 〈β, α∨〉 = 1. Let µ = δ − γ be another representation of µ as adifference of two roots in R1. Then

〈γ, α∨〉 − 〈δ, α∨〉 = 〈γ − δ, α∨〉 = 〈α− β, α∨〉 = 2− 1 = 1,

whence either 〈γ, α∨〉 = 2, 〈δ, α∨〉 = 1 or 〈γ, α∨〉 = 1, 〈δ, α∨〉 = 0. It follows that

b(V +γ , V

−δ ) ⊂ b

(V +

2 (α), V −1 (α))∪ b(V +

1 (α), V −0 (α)), (4)

where the V σi (α) are the spaces of the Peirce grading Rα as in 20.7. Let U ′ξ (ξ ∈ C×2 )be the root subgroups of G defined as in 11.1 with respect to the Peirce gradingRα. Then (4) and (11.1.4) show that b(V +

γ , V−δ ) ⊂ U ′ε1−ε0 , and since this is so for

all representations of µ as a difference of two roots in R1, we have Uµ ⊂ U ′ε1−ε0by (21.1.2). On the other hand, G ∈ st(V,Rα) by Corollary 21.9, so the C2-commutator relations for the U ′ξ hold in G. In particular, U ′ε1−ε0 is abelian, andhence so is Uµ.

From now on, we assume that µ and ν are linearly independent. Since R =R1 ∪R0 ∪R−1,

(((((((Uµ, Uν

)))))))=(((((((Uν , Uµ

)))))))and

(((((((Uµ, Uν

)))))))⊂(((((((Uσ, Uσ

)))))))= 1 for µ, ν ∈ Rσ1,

the following cases remain to be dealt with:

(a) µ ∈ R1, ν ∈ R−1,(b) µ ∈ R×0 , ν ∈ R±1,(c) µ, ν ∈ R×0 .

In Case (a), let µ = α and −ν = β ∈ R1. Then α 6= β since µ and ν arelinearly independent. Let xα ∈ V +

α and yβ ∈ V −β . Then Q(xα)yβ ∈ V +2α−β and

Q(yβ)xα ∈ V −2β−α. Hence (20.7.6) and (20.1.1) yield(((((((x−(−yβ), x+(xα)

)))))))= x−(Q(yβ)xα) · b(xα, yβ) · x+(Q(xα)yβ)

∈ Uα−2β · Uα−β · U2α−β = Uµ+2ν · Uµ+ν · U2µ+ν ⊂ U(((((µ,ν))))),

as desired.


In Case (b), we let µ = α−β where α, β ∈ R1, and let again (xα, yβ) ∈ V +α ×V −β .

Assume first ν = γ ∈ R1 and choose zγ ∈ V +γ . Then by (21.10.1) and (20.1.1),(((((((

b(xα, yβ), x+(zγ))))))))

= x+

(− xαyβzγ

)· x+

(Q(xα)Q(yβ)zγ

)∈ Uα−β+γ · U2α−2β+γ = Uµ+ν · U2µ+ν ⊂ U(((((µ,ν))))).

The case where ν = −δ ∈ R−1 follows in the same way from (21.10.2).

We come to Case (c) which is the most involved. By (3), it is enough to provethe commutator relations for the elements of the sets Xµ and Xν generating Uµand Uν respectively. Thus, let µ = α − β and ν = γ − δ where α, β, γ, δ ∈ R1.Let x = xα, y = yβ be as before and (u, v) = (uγ , vδ) ∈ V +

γ × V −δ . Then with theabbreviations

a = aγ+µ = −xyu ∈ V +γ+µ, a′ = a′γ+2µ = QxQyu ∈ V +

γ+2µ,

c = cδ−µ = yxv ∈ V −δ−µ, c′ = c′δ−2µ = QyQxv ∈ V −δ−2µ,

(21.10.3) yields the formula

g :=(((((((

b(x, y), b(u, v))))))))

= b(u+ a+ a′, v + c+ c′) · b(u, v)−1,

and we must show g ∈ U(((((µ,ν))))). We distinguish the following four cases:

1. γ + µ 6∈ R, δ − µ 6∈ R,

2. γ + µ ∈ R, δ − µ ∈ R,

3. γ + µ ∈ R, δ − µ 6∈ R,

4. γ + µ 6∈ R, δ − µ ∈ R.

Case 4 is of course analogous to Case 3 after interchanging the roles of α, β andγ, δ. Therefore, it suffices to consider the first three cases.

Case 1: γ + µ /∈ R and δ − µ /∈ R.

Since the µ-string through γ contains no gaps by [18, VI, §1.3, Proposition 9] or[63, A.5], it follows that also γ + 2µ /∈ R and similarly δ − 2µ /∈ R. Hence we havea = a′ = c = c′ = 0 and therefore g = 1.

Case 2: γ + µ ∈ R and δ − µ ∈ R.

Since R is 3-graded these roots are in R1 = Γ , and we set

τ = γ + µ = α− β + γ, % = δ − µ = δ − α+ β.

Then τ − γ + % − δ = µ − µ = 0. From the linear independence of µ and ν itfollows easily that τ, γ, %, δ are distinct. Also γ ∼ δ since ν = γ − δ ∈ R×0 . ByProposition 16.1(b2) either (τ, γ, %, δ) or (%, γ, τ, δ) is a kite:

Case 2.1:

τ

????

γ

???? δ

%

Case 2.2:

%

????

γ

???? δ

τ


In either subcase, we have 〈γ, τ∨〉 = 1, so γ → τ does not hold. By Lemma 16.11,2τ − γ = γ + 2µ /∈ Γ , and in the same way, 2% − δ = δ − 2µ /∈ Γ . Therefore,a′ = c′ = 0 and the commutator simplifies to

g = b(uγ + aτ , vδ + c%) · b(uγ , vδ)−1.

Continuing with Case 2.1, we have vδ, c% ∈ V −1 (γ) while uγ , aτ ∈ V +2 (γ). By

Corollary 21.9 and 11.5, the map b is bi-multiplicative on V +2 (γ) × V −1 (γ), so we

obtain g = b(aτ , vδ) · b(aτ , c%) · b(uγ , c%). Here the second factor is 1 since τ ⊥ %,so that

g ∈ Uτ−δ · Uγ−% = Uµ+ν ⊂ U(((((µ,ν)))))

since τ − δ = γ − % = µ+ ν. In Case 2.2, an analogous argument shows uγ + aτ ∈V +

1 (δ) and vδ + c% ∈ V −2 (δ) and again g ∈ Uµ+ν . This completes the proof ofCase 2.

Case 3: τ = γ + µ ∈ R and δ − µ /∈ R.

The root string argument used in Case 1 shows δ − 2µ /∈ Γ whence c = c′ = 0. By(9.9.1) and (20.7.3),

g = b((a+ a′) + u, v

)· b(u, v)−1 = b(a+ a′, vu) = b(a+ a′, v +Q(v)u). (5)

Since sδ(µ) = µ− 〈µ, δ∨〉δ ∈ R, the assumption 〈µ, δ∨〉 > 0 together with the rootstring argument leads to the contradiction µ − δ ∈ R. Therefore 〈µ, δ∨〉 6 0. Itfollows that

〈τ, δ∨〉 = 〈µ+ γ, δ∨〉6 〈γ, δ∨〉. (6)

Now we distinguish the following two subcases 3.1 and 3.2:

Case 3.1: 〈γ, δ∨〉 = 1.

Then uγ ∈ V +1 (δ) and thereforeQ(vδ)uγ = 0. Moreover, by (6), aτ ∈ V +

1 (δ)⊕V +0 (δ)

which yields Q(vδ)aτ = 0 and consequently va = v. Now (5) and (9.9.1) imply

g = b(a′ + a, v) = b(a′, va) · b(a, v) = b(a′τ+µ, vδ) · b(aτ , vδ)

∈ Uτ+µ−δ · Uτ−δ = U2µ+ν · Uµ+ν ⊂ U(((((µ,ν))))).

Case 3.2: 〈γ, δ∨〉 = 2.

Then γ → δ, and τ ∼ γ since τ − γ = µ ∈ R×0 . Also, as noted before, τ, γ, δare distinct. By Lemma 15.6, the induced subgraph of Γ on γ, δ, τ is one of thefollowing:

Case 3.2.1:γ

????

τ δCase 3.2.2:

γ

???? τ

δ

In the first case, it follows from Lemma 15.8 that

ε := τ − γ + δ = τ − ν ∈ Γ


and that (γ, δ, ε, τ) is a kite which by Lemma 16.13 generates a hexagram:

γ

????

τ δ

2τ − γ

??// ε

????

2δ − γoo

__????

In particular, τ + µ = τ + (τ − γ) = 2τ − γ → τ , whence τ and τ + µ belong toΓ2(τ) and therefore aτ + a′τ+µ ∈ V +

2 (τ).

Moreover, 2δ−γ ⊥ τ implies Q(vδ)uγ ∈ V −0 (τ). By Corollary 21.9, (11.4.1) and(StR1) the element g of (5) becomes

g = b(aτ + a′τ+µ, vδ +Q(vδ)uγ) = b(aτ + a′τ+µ, vδ) b(aτ + a′τ+µ, Q(vδ)uγ)

= b(aτ + a′τ+µ, vδ) = b(aτ , vδ) b(a′τ+µ, vδ) = b(aτ , vδ) ∈ Uτ−δ = Uµ+ν ,

using τ + µ = 2τ − γ and therefore τ + µ ⊥ δ.

In Case 3.2.2, γ, δ, τ generate, by Corollary 15.7, a pyramid:

γ

???? τ

δ

2δ − τ

??2δ − γ

__????

In particular, τ + µ = 2τ − γ /∈ Γ , else it would follow from Lemma 16.11 thatγ → τ . Hence a′τ+µ = 0, and vδ + Q(vδ)uγ ∈ V −1 (τ) while of course aτ ∈ V +

2 (τ).Now Corollary 21.9 and (11.5.1) yield

g = b(aτ , vδ +Q(vδ)uγ) = b(aτ , vδ) · b(aτ , Q(vδ)uγ)

∈ Uτ−δ · Uτ−(2δ−γ) = Uµ+ν · Uµ+2ν ⊂ U(((((µ,ν))))).

This establishes Case 3.2.2, and completes the proof of Theorem 21.7.

21.12. Corollary. The projective elementary group G = PE(V ) of a Jordanpair V with a root grading R has R-commutator relations with root groups U =(Uα)α∈R given by

Uσα = expσ(V σα ) for α ∈ R1, σ ∈ +,−, (1)

Uµ =⟨⋃

β(V +α , V

−β ) : α− β = µ, α, β ∈ R1

⟩⊂ G0 for µ ∈ R×0 , (2)

and thus belongs to st(V,R); in particular, st(V,R) is not empty.

Proof. From (21.1.2) it follows that the root groups of G are given by theabove formulas. By (7.7.1) and (10.7.17), the relations (StR1) and (StR2) hold inPE(V ).


21.13. Corollary. Let A be a unital associative algebra, and let a, b ⊂ A as in20.2(a). Let V = MIJ(a, b) be the Jordan pair defined in (20.2.4), with the KIKJ -grading R defined in (20.2.5). Then the elementary group E(M, V ) defined in 9.3belongs to st(V,R).

Proof. It suffices to verify condition (iv) of Theorem 21.7. The relations (StR1)are clear by (9.8.6). To prove (StR2), let α ⊥ β and xα ∈ V +

α , yβ ∈ V −β . Thenα = (i, j), β = (k, l) where i 6= k and j 6= l, and xα = aEij and yβ = bElk, fora ∈ a and b ∈ b. Hence xαyβ = abEijElk = 0 since j 6= l, and in the same way,yβxα = 0. Now b(xα, yβ) = 1 follows from (9.7.7).

Example. In case a = b = A the group E(M, V ) is the elementary linear groupEI∪J(A) of Example 3.16(c). The root groups defined in 21.1 coincide with the rootgroups of loc. cit. The corollary therefore provides a proof of the well-known factstated there that EI∪J(A) has AI∪J -commutator relations.

21.14. Corollary. Let G ∈ st(V,R). The projection

π0 = π∣∣UR0

: UR0→ EA(V,R)

makes UR0a central extension of EA(V,R).

Proof. From the definitions and Corollary 21.6 it is clear that π0 is surjective. By(iii) of Theorem 21.7, UR0

normalizes U±, and by (21.1.5), UR0⊂ G0. Lemma 9.2(c)

shows that Ker(π0) = Ker(π) ∩ UR0⊂ Ker(π) ∩N is central in G.

21.15. Corollary. A group G ∈ st(V ) satisfying the assumptions of 21.2 be-longs to st(V,R).

Proof. To keep the root groups of 21.1 apart from the ones in 21.2 we denotethe latter by U% for % ∈ R. Thus Uα = Uα for α ∈ R±1 and Uµ ⊂ Uµ for µ ∈ R0.For µ ∈ R0 and α ∈ R1 we have

(((((((µ, α

)))))))⊂ R1 since R is 3-graded. Hence the

commutator relations (i) of 21.2 imply Uµ normalizes U+, whence so does Uµ. Inthe same way, one sees that Uµ normalizes U−. Also, for α ⊥ β in R1 we know(((((((α,−β

)))))))= ∅ from (14.8.9). Therefore

(((((((Uα, U−β

)))))))=(((((((Uα, U−β

)))))))= 1 by assumption

(ii) of 21.2. Hence G ∈ st(V,R) by Theorem 21.7(iii).

21.16. The embedding IR. We have seen in Corollary 21.12 that the projec-tive elementary group G = PE(V ) has R-commutator relations with root groupsU = (Uα)α∈R, so G belongs to st(V,R), and by (21.3.2), LR(G) = (G, U) is anobject of the category gcjΓ which, by 21.3, is a full subcategory of gcR.

Let C be the comma category(gcR ↓ (G, U)

). Thus the objects of C are

the morphisms π: (G,U) → (G, U) of gcR, and a morphism ϕ: π → π′ of C is amorphism ϕ: (G,U)→ (G′,U′) of gcR satisfying π′ ϕ = π:

(G,U)ϕ //

π $$HHHHHHHHH(G′,U′)

π′zzuuuuuuuuu

(G, U)

(1)


Let (G,U±, π) ∈ st(V,R) with U as in (21.1.1) and (21.1.2). We define a functorIR: st(V,R)→ C by

IR(G,U±, π) = LR(π) =(π: (G,U)→ (G, U)

)(2)

on objects. If ϕ: (G,U±, π)→ (G′, U ′±, π′) is a morphism of st(V,R) then LR(ϕ) =ϕ: (G,U) → (G′,U′) is a morphism of gcR, and from the definitions and (21.3.2)it is clear that the diagram (1) is commutative. Hence IR(ϕ) := ϕ: π → π′ is amorphism of C . It is easily checked that this yields a functor IR from st(V,R) toC which is an embedding, that is, it is injective on objects and on morphisms.

In 4.4, we defined coverings for groups with commutator relations and intro-duced the notion of simply connected groups. Recall the category st(G, U) definedin 4.12, a full subcategory of the comma category C whose objects are the coveringsof (G, U). The following questions arise naturally:

(a) Which objects of st(V,R) are mapped under IR to a covering of (G, U)?

(b) Given a covering of (G, U), does it belong to the image of IR?

As noted in 21.3, the projection π of (G,U±, π) ∈ st(V,R) is surjective on allroot groups Uα, and even bijective for all α ∈ R±1. On the other hand, by 4.4, anecessary condition for π to be a covering is that it be bijective on all root groups.This condition is in general not fulfilled, as shown in 21.17 below.

We will show in Corollary 21.20 that bijectivity on all root groups is in fact suffi-cient for π to be a covering, thus answering Question (a). The proof rests on the factthat the projective elementary group, considered as a group with R-commutatorrelations as above, has unique factorization (Theorem 21.19). A positive answer toQuestion (b) will be given in Proposition 21.23.

21.17. Example. Let A be an associative unital ring and let a, b ⊂ A besubgroups ofA with aba ⊂ a and bab ⊂ b as in 20.2(a). Let n = p+q>3 and considerthe Morita context M defined in (20.2.3) where I = 1, . . . , p, J = 1, . . . , q. Wehave seen that V =

(Matpq(a), Matqp(b)

)has a rectangular grading R of type

Γ = Kp Kq. Let J ′ = p + 1, . . . , p + q and define the bijection J → J ′ byj 7→ j′ = p + j. Then the 3-graded root system (R,R1) determined by Γ is givenas follows:

R1 = εi − εj′ : (i, j) ∈ I × J, R−1 = εj′ − εi : (j, i) ∈ J × I,R0 = εi − εl : (i, l) ∈ I × I ∪ εj′ − εm′ : (j,m) ∈ J × J.

By Corollary 21.13, the elementary group G = E(V,R) belongs to st(V,R). Weclaim that its root subgroups are as follows:

Uεi−εj′ =

(1p aEij0 1q

), Uεj′−εi =

(1p 0bEji 1q

),

Uεi−εl =

(1p + (ab)Eil 0

0 1q

), Uεj′−εm′ =

(1p 00 1q + (ba)Ejm

).

Indeed, the first two cases follow immediately from (20.2.5) and (9.3.1). Now letµ = εi − εl ∈ R×0 where i 6= l in I. By (21.1.2), Uµ is generated by all b(xα, yβ)


where α, β ∈ R1 with µ = α− β, and xα ∈ V +α , yβ ∈ V −β . The only way of writing

µ in this form isµ = (εi − εj′)− (εl − εj′)

for some j ∈ J . Then xα = aEij and yβ = bEjl where a ∈ a and b ∈ b, soxαyβ = abEil while yβxα = 0. Now (9.7.7) shows b(xα, yβ) = 1n + abEil, whichyields the third formula. The last case is proved similarly.

The kernel of π restricted to a root subgroup Uµ, µ ∈ R×0 , is given by

Ker(π∣∣Uεi−εl) = 1n + cEil : c ∈ ab, ca = bc = 0 for i 6= l in I,

Ker(π∣∣Uεj′−εm′ ) = 1n + cEj′m′ : c ∈ ba, cb = ac = 0 for j 6= m in J .

Indeed, let g = 1n + cEil ∈ Uεi−εl where c ∈ ab. Then π(g) = (h+, h−) is theautomorphism of the Jordan pair V given by h+(z) = z + cEilz and h−(w) =w−wcEil, for z ∈ V + = Matpq(a) and w ∈ V − = Matqp(b). Hence π(g) = 1 if andonly if ca = 0 = bc. The second case is proved similarly. Thus, if for example a = bis an ideal with a2 6= 0 = a3 then all Uµ, µ ∈ R×0 , are non-trivial and belong to thekernel of π, whereas for a = b = A we see that π is injective on all root subgroups.

21.18. Lemma. Let Γ be a Jordan graph with associated 3-graded root system(R,R1), let (V,R) be a Γ -graded Jordan pair, and let 0 6= µ ∈ R0 and h =(h+, h−) ∈ Uµ as defined in 21.12. Then for all γ ∈ Γ = R1,

(h+ − IdV +) · V +γ ⊂

⊕i>1

V +γ+iµ, (h− − IdV −) · V −γ ⊂

⊕i>1

V −γ−iµ. (1)

Remark. The sums in (1) actually only run up to i = 2. Indeed, supposeV σγ+iµ 6= 0, so that γ + iµ ∈ R1. From (14.8.6) we know |〈µ, ν∨〉| 6 2 for any tworoots µ, ν in a 3-graded root system. Hence 〈γ + iµ, µ∨〉 = 〈γ, µ∨〉 + 2i 6 2 shows〈γ, µ∨〉6 2(1− i), so i6 2 follows.

Proof. By Corollary 21.12, h is a product of inner automorphisms of the formβ(xα, yβ) where µ = α−β and α ∼ β 6= α in Γ . Suppose first that h = g = β(xα, yβ)is one of the generators. Then g+ = B(xα, yβ), and putting Sγ :=

⊕i>1 V

+γ+iµ, we

have, by (20.1.1),

(g+− Id) ·V +γ ⊂ xα, yβ , V +

γ +Q(xα)Q(yβ)V +γ ⊂ V +

γ+α−β +Vγ+2(α−β) ⊂ Sγ . (2)

Furthermore, we claim that g+ · Sγ ⊂ Sγ . Indeed, replacing γ by γ + iµ in theabove computation and observing that Sγ+iµ ⊂ Sγ , we see

g+ · V +γ+iµ =

(Id + (g+ − Id)

)· V +

γ+iµ ⊂ V+γ+iµ + Sγ+iµ ⊂ Sγ + Sγ = Sγ .

Now let h = gh where h is a product of generators and assume by induction thath+ − Id maps V +

γ to Sγ . Then the identity

h+ − Id = g+(h+ − Id) + (g+ − Id)

together with (2) shows that also h+− Id maps V +γ to Sγ . Since the generating set

for Uµ is invariant under inversion by (20.7.4), this proves the first formula of (1).The proof of the second formula is analogous.


21.19. Theorem. Let (V,R) be a root graded Jordan pair and let G = PE(V )be the projective elementary group, with R-commutator relations and root groups Uas in Corollary 21.12. Then G has unique factorization in the sense of 3.15.

Proof. Let A ⊂ R be a finite nilpotent set and decompose A = A−1 ∪ A0 ∪ A1

where Ai = A ∩ Ri. It follows from Theorem 7.7(b) that it suffices to show: themultiplication map

∏µ∈A0

Uµ → G is injective. Since A0 is a closed subset of A,being the intersection of two closed subsets, it is again nilpotent. Thus we havereduced the problem to showing that the group PE0(V ), which has R0-commutatorrelations, has unique factorization. This will be done by verifying the hypothesesof Lemma 3.19.

Thus let A ⊂ R×0 be a finite nilpotent subset. We have to show that there existsα0 ∈ A such that (i) B = A α0 is closed and (ii) Uα0 ∩ UB = 1. Since Ris reduced (Corollary 15.15), we know the existence of α0 satisfying (i) from theRemark in 3.19. But in order to prove (ii) we will make a specific choice of α0.Namely, since A is finite and N-free by Corollary 1.14, we can apply Lemma 1.3.Thus, let VQ = spanQ(A), V = VQ ⊗Q R and C the pointed convex cone spannedby A ⊂ V . By loc. cit., there exists α0 ∈ A such that R+ · α0 is an extremal ray ofC. Then B = A α0 is closed. Indeed, assume to the contrary that there existβ1, . . . , βp ∈ B such that β1 + · · · + βp = α0. Since R+ · α0 is an extremal ray ofC, it follows that each βi is a unique positive real multiple of α0, say βi = riα0.Because R is reduced, this implies βi = α0 ∈ B, contradicting the definition of B.Thus condition (i) of Lemma 3.19 holds and it remains to verify condition (ii).

Enumerate B = α1, . . . , αn. Since B is nilpotent, being a closed subsetof the nilpotent set A, we have UB = Uα1 · · · Uαn by Proposition 3.12(b). Leth0 ∈ Uα0 ∩ UB , say, h0 = h1 · · ·hn where hi ∈ Uαi . Let hi = (h+

i , h−i ) and put

fi = h+i − Id. The relation h+

0 = Id + f0 = h+1 · · ·h+

n = (Id + f1) · · · (Id + fn) isequivalent to

f0 =

n∑p=1

∑16i1<···<ip6n

fi1 · · · fip . (1)

Let γ ∈ R1 be arbitrary, let xγ ∈ V +γ and apply (1) to xγ . By Lemma 21.18, we

havef0 · xγ =

∑l0>1

yγ+l0α0where yγ+l0α0

∈ V +γ+l0α0

.

Again by Lemma 21.18, fi · V +γ ⊂

∑l>1 V

+γ+lαi

. Hence (1) implies

f0 · xγ ∈n∑p=1

∑16i1<···<ip6n

∑l1,...,lp>1

V +γ+l1αi1+···+lpαip

. (2)

Assume that f0 · xγ 6= 0. Then at least one component yγ+l0α0 is non-zero. Henceat least one corresponding term must occur in (2). This implies that there exists arelation

l0α0 = l1αi1 + · · ·+ lnαin , (3)

where l0, li > 1 and iν > 1. Because R is reduced, this contradicts the fact thatR+ ·α0 is an extremal ray of C. Hence f0 ·xγ = 0. Since V + is the sum of the V +

γ ,

γ ∈ R1, it follows that f0 = 0, i.e., h+0 is the identity on V +. In the same way, one

shows h−0 = IdV − , whence h0 = 1, as desired.


21.20. Corollary. A morphism π: (G,U)→ (G, U) in gcR is a covering if andonly if π is bijective on all root groups. In this case, G has unique factorization,and for all nilpotent A ⊂ R the map π: UA → UA is an isomorphism.

In particular, for (G,U±, π) ∈ st(V,R) the morphism IR(G,U±, π) = π of(21.16.2) is a covering if and only if it is bijective on root groups.

Proof. Since G has unique factorization by Theorem 21.19, the first statementfollows from Lemma 4.5(c). The remaining statements follow from Lemma 4.3(d)and because IR(G,U±, π) is a morphism in gcR.

21.21. The category stbij(V,R). Let (G,U±, π) ∈ st(V,R) with root groupsU = (Uα)α∈R. We define the full subcategory stbij(V,R) of st(V,R) by

(G,U±, π) ∈ stbij(V,R) ⇐⇒ π: Uα → Uα is bijective, for all α ∈ R. (1)

In view of Corollary 21.20, this is equivalent to

(G,U±, π) ∈ stbij(V,R) ⇐⇒ π: (G,U)→ (G, U) is a covering. (2)

Hence by Corollary 21.20,

every G ∈ stbij(V,R) has unique factorization. (3)

For R = P(e) we have seen in Corollary 12.7 that st(V, e) ⊂ stbij(V,P(e)). For

arbitrary R we evidently have (G, U) ∈ stbij(V,R). More examples of groups instbij(V,R) will be given in the next section, in particular in Corollary 24.9(b) andCorollary 24.15.

21.22. Corollary. Let ϕ: G→ G′ be a morphism in stbij(V,R) and let A ⊂ Rbe nilpotent. Put UA =

⟨Uα : α ∈ A

⟩and define U ′A analogously. Then the induced

homomorphism ϕ: UA → U ′A is bijective.

Proof. G′ has unique factorization by (21.21.3). Hence Lemma 4.3(d) finishesthe proof.

Question (b) raised in 21.16 has now the following answer.

21.23. Proposition. We keep the notation introduced in 21.16. Then anycovering π: (G,U) → (G, U) is the image under IR of an object of stbij(V,R). Inmore detail, define subgroups U± of G by

U+ := UR1=⟨Uα : α ∈ R1

⟩and U− := UR−1

=⟨Uα : α ∈ R−1

⟩. (1)

Then G is generated by U+ ∪ U−, (G,U±, π) belongs to stbij(V,R) and is mappedto (G,U) under IR. The functor IR induces an isomorphism

stbij(V,R) ∼= st(G, U) (2)

of categories.

Proof. The sets R±1 are abelian, in particular, they are nilpotent. Henceπ: U± → U± is an isomorphism by by Corollary 21.20. Define isomorphismsxσ: V σ → Uσ by the commutativity of the diagram (9.1.3):


V σxσ∼=

//

expσ

∼=

!!CCCCCCCC Uσ

π

∼=

Uσ

(3)

Let G′ be the subgroup of G generated by U+ ∪ U−. Then (G′, U+, U−, π) is agroup over V , so the definitions of §9 apply to G′. In particular, if (x, y) ∈ V isquasi-invertible, (9.7.1) implies

b(x, y) = x−(−yx) x+(x) x−(y) x+(xy) ∈ G′.

Let us first show G = G′ as groups. Since G is generated by its root groupsU = (U%)%∈R, and Uα ⊂ G′ for all α ∈ R1 ∪ R−1, it suffices to show that Uµ ⊂ G′

for all µ ∈ R×0 . We define subgroups U ′µ of G′ by

U ′µ :=⟨⋃

b(V +α , V

−β ) : α− β = µ, α, β ∈ R1

⟩(4)

and claim thatU ′µ = Uµ. (5)

For the inclusion from left to right, write µ ∈ R×0 in the form µ = α − β whereα, β ∈ R1. Let (xα, yβ) ∈ V +

α × V −β , so

h′µ := b(xα, yβ)

is one of the generators of U ′µ. We know α ∼ β 6= α by (15.9.1). Thus either

α β or α ← β or α → β. Since R is 3-graded, the root interval A :=(((((((α,−β

)))))))equals A = µ if α β, it is A = µ, µ + α = 2α − β if α ← β and it isA = µ, µ − β = −(2β − α) if α → β. In any case, A is a nilpotent subset of R.We will deal with the first two cases, thus A ⊂ µ, µ + α, and leave the secondcase, which is analogous, to the reader.

By (20.7.3), xyβα = xα + zµ+α where zµ+α = Q(xα)yβ ∈ V +

2α−β = V +µ+α. By

(20.7.6),h′µ =

(((((((x−(−yβ), x+(xα

)))))))· x+(−zµ+α).

Since A is nilpotent, Proposition 3.12(b) and the commutator relations for G show(((((((x−(−yβ), x+(xα)

)))))))∈(((((((U−β , Uα

)))))))⊂ U(((((−β,α))))) = UA = Uµ · Uµ+α,

so that(((((((

x−(−yβ), x+(xα))))))))

= hµ · gµ+α where hµ ∈ Uµ and gµ+α ∈ Uµ+α. Hence

h′µ = hµ · gµ+α · x+(−zµ+α) ∈ Uµ · Uµ+α.

We put g′µ+α = gµ+α · x+(−zµ+α), apply π to this equation and obtain

β(xα, yβ) = π(h′µ) = π(hµ) · π(g′µ+α) ∈ Uµ · Uµ+α.

By (9.7.3) and (21.12.2), β(xα, yβ) ∈ Uµ ⊂ G0. Now it follows from Theorem 7.7(b)that π(g′µ+α) = 1, and since π is bijective on root groups, g′µ+α = 1. This provesh′µ = hµ ∈ Uµ.


It remains to prove the inclusion from right to left in (5). By (21.12.2) we haveπ(U ′µ) = Uµ. Since π: Uµ → Uµ is an isomorphism, the inclusion U ′µ ⊂ Uµ impliesU ′µ = Uµ, as desired.

So far, we have shown that (G,U+, U−, π) is a group over V . To prove that itbelongs to stbij(V,R) and is mapped to (G,U) under IR, it suffices to verify thatthe subgroups of G defined in (21.1.1) and (21.1.2) agree with the subgroups Uαof G as an object of gcR. For α ∈ R±1 this is clear in view of the definition ofthe isomorphism Uσ ∼= V σ in (3) and the definition (21.1.1), and for µ ∈ R0 itfollows from (21.1.2) and (5). Hence (G,U+, U−, π) belongs to st(V,R) and evento stbij(V,R) because π is bijective on root groups.

Finally, the isomorphism (2) follows easily from what we just proved, Corol-lary 21.20, and the fact that IR is an embedding. The details are left to thereader.

Our next aim is to show that stbij(V,R) is a reflective subcategory of st(V,R),see 5.11 for a review of this concept.

21.24. Proposition. Let A := (G,U±, π) ∈ A := st(V,R) with root groupsU = (Uα)α∈R as in 21.1. For µ ∈ R0 let Kµ = Ker(π

∣∣Uµ).

(a) Then K :=⟨Kµ : µ ∈ R×0

⟩is a central subgroup of G contained in Ker(π)

and normalizing U±.

(b) The quotient G := G/K, equipped with the root groups U± := can(U±) andthe induced map π: G→ G is an object B := (G, U±, π) of B := stbij(V,R).

(c) The canonical map can: A→ B is a B-reflection for A.

Proof. (a) By Theorem 21.7(iii), the groups Uµ, µ ∈ R0, normalize U+ andU−, and clearly Kµ ⊂ Ker(π). Now Lemma 9.2(c) shows that Kµ is central in G,so K is central as well.

(b) Since K is a normal subgroup of G contained in Ker(π) it follows fromLemma 9.2(e) that B ∈ st(V ), and the root subgroups of G are just the canonicalimages Uα of the Uα. Hence A ∈ A. To show that B ∈ B it remains to verifythat π is injective on all root groups Uµ, µ ∈ R0. This follows from the fact thatKer(π) ⊃ Ker(π: Uµ → Uµ) = Kµ by definition of K. Thus B ∈ B.

(c) Let B′ = (G′, U ′±, π′) ∈ B, let ϕ: A → B′ be a morphism of A and letµ ∈ R0. Then ϕ(Kµ) = 1, because for h ∈ Kµ we have π′(ϕ(h)) = π(h) = 1 andtherefore ϕ(h) = 1 since π′ is injective on root groups. Hence ϕ induces a uniquehomomorphism ψ: G→ G′ making the diagram

Gϕ //

can

G′

G

∃!ψ

99

commutative. Thus can: A→ B is a B-reflection for A.


§22. The Steinberg group of a root graded Jordan pair

22.1. Definition. Let R = (Vγ)γ∈Γ be a Γ -grading of a Jordan pair V , andlet (R,R1) be the 3-graded root system associated with Γ .

Generalizing 11.11, we now consider the quotient G of the free product Fr(V ) =V + ∗ V − of the additive groups V + and V − by the (normal subgroup generated

by the) relations (StR1) and (StR2) formulated in Theorem 21.7. Let U± =

can(V ±) ⊂ G. Since the projective elementary group G satisfies these relations by

Corollary 21.12, there is a canonical homomorphism π: G → G, and the diagramof isomorphisms

V σxσ=can //

expσ !!BBBBBBBB Uσ

π||||||||

Uσ

is commutative. Hence (G, U±, π) is a group over V , and in fact, by its verydefinition, has R-commutator relations with respect to the family of subgroupsU = (Uα)α∈R defined as in 21.1. Thus

St(V,R) := (G, U±, π) ∈ st(V,R),

called the Steinberg group of (V,R). From the definition it is clear that, for every(G,U±, π) ∈ st(V,R), there exists a unique morphism

κ: St(V,R)→ (G,U±, π) (1)

in st(V,R). Thus St(V,R) is an initial object in the category st(V,R), uniquelydetermined up to unique isomorphism by this universal property.

In order to simplify notation, let us for the moment identify the categoryst(V,R) with a subcategory of the comma category C =

(gcR ↓ (G, U)

)by means

of the embedding IR of 21.16. Under this identification, and by Proposition 21.23,we have

st(G, U) = stbij(V,R) ⊂ st(V,R),

the inclusion being in general strict. By 4.10, the category st(G, U) has also aninitial object

St(G, U) =(π: (G, U)→ (G, U)

),

which, by the unique factorization property of G shown in Theorem 21.19, may beconstructed as in Theorem 4.14. In general, St(G, U) will not be an initial objectof st(V,R), thus not isomorphic to St(V,R). However, the universal property ofSt(V,R) yields a commutative triangle

Gκ //

π >>>>>>>> G

π

G

§22] The Steinberg group of a root graded Jordan pair 277

and κ is surjective (as any morphism of st(V ), see 9.1). We claim that the kernel

of κ is the subgroup K of G defined in Proposition 21.24(a), thus

G ∼= G/K.

Indeed, let µ ∈ R×0 and let g ∈ Kµ ⊂ Uµ. Then 1 = π(g) = π(κ(g)) and κ(g) ∈ Uµ.

Since π is injective on Uµ, it follows that κ(g) = 1. As K is generated by the

Kµ, we have K ⊂ Ker(κ) and therefore an induced homomorphism η: G/K → G.By Proposition 21.24(b), St(V,R)/K ∈ stbij(V,R), and since St(G, U) is an initialobject of stbij(V,R), η is an isomorphism.

22.2. Lemma. (a) Let Γ be a Jordan graph with associated 3-graded rootsystem (R,R1) and let (V,R) and (V ′,R′) be Γ -graded Jordan pairs. Considera homomorphism h: (V,R) → (V ′,R′) of Γ -gradings as defined in 20.3, so thath(Vγ) ⊂ V ′γ for all γ ∈ Γ . Let

St(V,R) = (G, U±, π) and St(V ′,R′) = (G′, U ′±, π′)

be the respective Steinberg groups defined in 22.1. Then h lifts to a homomorphism

ϕh: G→ G′

in the sense of 9.5, thus satisfying

ϕh(xσ(x)) = x′σ(hσ(x)), (1)

for all x ∈ V σ.

(b) Let U = (Uα)α∈R be the family of root subgroups of G. Then there is afunctor

SΓ : gradjpΓ → gcjΓ , (2)

given on objects bySΓ (V,R) = LR(St(V,R)) = (G, U) (3)

(with LR as in (21.3.2)) and on morphisms by

SΓ (h) = ϕh.

Proof. (a) Let Fr(V ) = V + ∗ V − be the free product, and let η: Fr(V ) → G

and η′: Fr(V ′) → G′ be the canonical maps. We write the canonical injectionsV σ → Fr(V ) as x 7→ [x] and employ an analogous notation for Fr(V ′). By 9.5,h: V → V ′ lifts uniquely to a group homomorphism Fr(h): Fr(V ) → Fr(V ′) satis-fying Fr(h)([x]) = [hσ(x)] for all x ∈ V σ. Let K and K ′ be the kernels of η and η′,respectively. Once we have shown that Fr(h) maps K to K ′, the existence of ϕhwill follow from the commutative diagram with exact rows

1 // K

Fr(h)

inc // Fr(V )

Fr(h)

η // G

ϕh

// 1

1 // K ′inc

// Fr(V ′)η′

// G′ // 1

(4)


By the definition of G and G′ in 22.1, K and K ′ are the normal subgroups generatedby the relations (StR1) and (StR2), respectively.

For quasi-invertible (x, y) ∈ V , define, as in (9.7.1), bFr(x, y) = [−yx] · [x] ·[y] · [−xy] ∈ Fr(V ), and let b′Fr be similarly defined for V ′. By (9.7.5) wethen have Fr(h)

(bFr(x, y)

)= b′Fr

(h+(x), h−(y)

)∈ Fr(V ′). Let α 6= β ∼ α and

(xα, yβ) ∈ V +α × V −β . Since h is a morphism of Γ -graded Jordan pairs, we have

(h+(xα), h−(yβ)) ∈ V ′+α × V ′−β . Hence the relation B(xα, yβ) ∈ K is mapped byFr(h) to the relation B(h+(xα), h−(yβ)) ∈ K ′, so Fr(h) preserves the relations(StR1). Similarly, one sees that Fr(h) preserves the relations (StR2).

(b) The group G has R-commutator relations with respect to the family of

subgroups U. By 21.1(c), G is generated by the Uα, α ∈ R1 ∪ R−1, so we have

(G, U) ∈ gcΓ . Now let h: (V,R) → (V ′,R′) be a morphism of Γ -graded Jordan

pairs. Then ϕh is a morphism of gcΓ , thus satisfies ϕ(Uα) ⊂ U ′α for all α ∈ R,

where Uα and U ′α are the root subgroups of G and G′. This follows from (1)and the definition of the root groups in 21.1. Finally, it is easily checked thatSΓ preserves identities and satisfies SΓ (h′ h) = SΓ (h′) SΓ (h) for morphismsh: (V,R)→ (V ′,R′) and h′: (V ′,R′)→ (V ′′,R′′) in gradjpΓ .

22.3. The categories gcj and gcj′. Following the procedure of 4.16, we definea category gcj encompassing all gcjΓ as follows. The objects of gcj are pairs(Γ, (G,U)) where Γ ∈ jgraph and (G,U) ∈ gcjΓ . For a morphism f : Γ → ∆ ofjgraph, the induced morphism of the associated 3-graded root systems (R,R1) and(S, S1) is also denoted f . Then a morphism from (Γ, (G,U)) to (∆, (H,V)) of gcjis a pair (f, ϕ) where f : Γ → ∆ is a morphism of jgraph, and ϕ: G→ H is a grouphomomorphism satisfying ϕ(Uα) ⊂ Vf(α) for all α ∈ R. From these definitions, itis evident that projection onto the first factor, (Γ, (G,U)) 7→ Γ and (f, ϕ) 7→ f , isa functor

P : gcj→ jgraph.

Given (G,U) ∈ gcjΓ and f : Γ → ∆ in jgraph, we have by (21.3.5) a functorf•: gcjΓ → gcj∆. One shows as in Proposition 4.22(c) that the assignmentsΓ 7→ gcjΓ on objects and f 7→ f• on morphisms define a functor

Ψ : jgraph→ Cat. (1)

By the Grothendieck construction 4.21, Ψ gives rise to a split opfibration∫Ψ = P ′: gcj′ → jgraph.

Recall from loc. cit. that the objects of gcj′ are pairs (Γ, (G,U)), where Γ ∈ jgraphand (G,U) ∈ Ψ(Γ ) = gcjΓ , so that gcj′ and gcj have the same objects. However,a morphism (Γ, (G,U)) → (∆, (H,V)) of gcj′ is a pair (f, ϕ), with f : Γ → ∆ asbefore, but ϕ: f•(G,U)→ (H,V) a morphism in gcj∆, see (21.3.4).

We define a functorJ : gcj→ gcj′ (2)

analogously to the definition of E in Proposition 4.23: J is the identity on objectsand maps a morphism (f, ϕ) of gcj to the morphism (f, ϕ

∣∣f•(G,U)) of gcj′. Thenit is clear that P = P ′ J .


The connection between the present definitions and those of §4 may be visualizedin the following diagram:

gcj

J

##HHHHHHHHH

P

u′′ // gc

Π

E

~~||||||||

gcj′

P ′vvvvvvvvv

u′ // gc′

Π′ BBBBBBBB

jgraphu

// SF

(3)

On the right, SF and gc are defined in 1.5 and 4.16 respectively, Π in Proposi-tion 4.19, and gc′, E and Π ′ in Proposition 4.23.

It remains to define the horizontal functors. The bottom arrow u is the compo-sition of the functor R : jgraph→ RS3 of 15.11 and the obvious forgetful functorRS3 → SF. We define u′ and u′′ by (Γ, (G,U)) 7→ (u(Γ ), (G,U)) on objects and by(f, ϕ) 7→ (u(f), ϕ) on morphisms. Then it is clear that the three squares containedin (3) are commutative.

Let Φ be the functor defined in Proposition 4.22(c). Then Ψ is merely a sub-functor of Φ u in the sense that Ψ(Γ ) is a subcategory of (Φ u)(Γ ) = gcR(Γ ) andthat Ψ(ϕ) is the restriction of (Φ u)(ϕ) = f• to gcjΓ . In particular, Ψ 6= Φ u.

22.4. Proposition. The functor J : gcj→ gcj′ of (22.3.2) is an isomorphism.Hence gcj is a split opfibration over jgraph.

Proof. Clearly, J is bijective on objects, and one shows as in the proof ofProposition 4.23 that J is injective on morphisms. We show that J is also surjectiveon morphisms. Let f : Γ → ∆ be a morphism of Jordan graphs, and denote theinduced morphism of the associated root systems (R,R1) and (S, S1) also by f .Let (f, ϕ): (Γ, (G,U))→ (∆, (H,V)) be a morphism of gcj′, so that ϕ: f•(G,U)→(H,V) is a morphism of gcj∆. By (21.3.4), f•(G,U) = (G,U′) with U′ defined in(21.3.3). Thus ϕ: G → H is a group homomorphism satisfying ϕ(U ′ξ) ⊂ Vξ for allξ ∈ S. We wish to show that (f, ϕ): (Γ, (G,U)) → (∆, (H,V)) is a morphism ofgcj, so we must verify that ϕ(Uα) ⊂ Vf(α) for all α ∈ R. Let ξ = f(α) 6= 0. ThenUα ⊂ U ′ξ and therefore ϕ(Uα) ⊂ ϕ(U ′ξ) ⊂ Vξ = Vf(α).

22.5. By Proposition 20.6, the categories gradjpΓ appear as the fibres of thesplit opfibration Π: gradjp → jgraph, and by Proposition 22.4, the categoriesgcjΓ are the fibres of the split opfibration P : gcj → jgraph. It is an obviousquestion whether there exists a functor S: gradjp→ gcj making the diagram

gradjp∃? S //

Π ##FFFFFFFF gcj

P~~

jgraph

commutative, and whose restriction to the fibres gradjpΓ of Π are the functors SΓof Lemma 22.2(b). The answer is no, the reason being that a morphism f : Γ → ∆


of Jordan graphs need not be injective. Hence elements α 6= β in Γ can have thesame image under f , so the relations B(xα, yβ), defined for α 6= β, which occurin the definition of the groups St(V,R) in 22.1, and thus also implicitly in thedefinition of the functor SΓ in (22.2.3), are not preserved by a morphism (f, h) ofgradjp.

However, the covariant functors Φ and Ψ defined in Proposition 20.6 and (22.3.1)can be considered as contravariant functors

Φ∗: jgraphop → Cat and Ψ∗: jgraphop → Cat,

and the Grothendieck construction for such functors yields fibrations

Π∗: gradjp∗ → jgraphop and P ∗: gcj∗ → jgraphop,

with fibres gradjpΓ and gcjΓ over Γ ∈ jgraphop, so the same question can beasked in this context:

gradjp∗∃? S∗ //

Π∗ $$IIIIIIIII gcj∗

P∗||yyyyyyyy

jgraphop

Here the answer is yes, as we shall see in Proposition 22.11. For the convenience ofthe reader, we now explain the categories and functors involved in more detail.

22.6. The Grothendieck construction of fibrations. A functor P : X→ Bis called a fibration if P op: Xop → Bop is an opfibration, as defined in 4.17. Weleave it as an easy exercise to make this more explicit, by reversing all the arrowsin 4.17. In particular, an arrow u: Y → X in X with P (u) = f : b → a is calledcartesian if for all g: c→ b in B and all w: Z → X with P (w) = f g, there existsa unique v: Z → Y such that P (v) = g and w = u v:

Zw //

∃! v ))

X

Yu

::uuuuuu

c

g ))SSSSSSSSSSSfg // a

bf

::tttttt

(1)

We review next the contravariant version of the construction 4.21, and refer to [38,B.1.3.1] and [108, 3.1.3] for details.

Let F : B → Cat be a contravariant functor. Then a morphism f : b → a of Byields a morphism F (f) = f•: F (a)→ F (b) which satisfies (f g)• = g• f•. Withthese data, we associate the following fibration P =

∫F : X → B. The objects of

X are pairs (a,A) where a ∈ B and A ∈ F (a). A morphism from (b, B) to (a,A) isa pair (f, ϕ) where f : b → a and ϕ: B → f•(A) are morphisms of B and of F (b),respectively. The composition of (f, ϕ) and (g, ψ): (c, C)→ (b, B) is defined by


(f, ϕ) (g, ψ) = (f g, g•(ϕ) ψ). (2)

Projection onto the first factor yields a fibration P : X → B, and the fibre overa ∈ B is canonically isomorphic with F (a). A morphisms (f, ϕ): (b, B)→ (a,A) iscartesian if and only if ϕ: B → f•(A) is an isomorphism. A splitting is given bydefining

ω(f, (a,A)) = (f, Idf•(a)) (3)

for all f : b→ a and A ∈ F (a).A morphism (f, ϕ) is called horizontal resp. vertical if ϕ = IdB resp. f = Idb.

An arbitrary morphism factors uniquely as the composition of a vertical followedby a horizontal one:

(f, ϕ) = (f, Idf•(A)) (Idb, ϕ). (4)

Now let F ′: Bop → Cat be a second functor, and let∫F ′ = P ′: X′ → B be the

corresponding split fibration. We wish to describe the functors S: X → X′ suchthat P ′ S = P :

XS //

P @@@@@@@@ X′

P ′~~

B

(5)

For F ′ we also use the notation f• = F ′(f): F ′(a)→ F ′(b).Since S maps fibres to fibres, it induces functors Sa: F (a)→ F ′(a) for all a ∈ B,

given byS(a,A) =

(a, Sa(A)

), S(Ida, ϕ) =

(a, Sa(ϕ)

)(6)

on objects A ∈ F (A) and morphisms ϕ: F (a)→ F ′(a). For an arbitrary morphism(f, ϕ): (a,A)→ (b, B) of X, (5) implies that S(f, ϕ) has the form

S(f, ϕ) =(f, Σ(f, ϕ)

): (b, Sb(B))→ (a, Sa(A)), (7)

where Σ(f, ϕ): Sb(B)→ f•(Sa(A)).

22.7. Lemma. We use the notations introduced before. Then the functorsS: X→ X′ with P = P ′ S are in bijection with the following data:

(i) for all a ∈ B, a functor Sa: F (a)→ F ′(a),

(ii) for all f : b→ a in B and all A ∈ F (a), a morphism

χ(f,A): Sb(f•(A))→ f•(Sa(A)),

subject to the following conditions:

χ(Ida, A) = IdSa(A), (1)

χ(f g, A

)= g•

(χ(f,A)

) χ(g, f•(A)

), (2)

χ(g,B′

) Sc

(g•(ϕ)

)= g•

(Sb(ϕ)

) χ(g,B), (3)

for all morphisms g: c→ b of B, all ϕ: B → B′ of F (b).


The correspondence is given on objects and morphisms of X by the formulas

S(a, A) = (a, Sa(A)), S(f, ϕ) =(f, χ(f,A) Sb(ϕ)

), (4)

and S preserves cartesian arrows if and only if the χ(f,A) are isomorphisms.

Proof. (a) Let S: X → X′ making (22.6.5) commutative, let (f, ϕ): (b, B) →(a,A) be a morphism of X, factored as in (22.6.4), and define χ(f,A) by

S(f, Idf•(A)) = (f, χ(f,A)), (5)

for all A ∈ Xa. Then functoriality of S, (22.6.6) and (22.6.2) show that

S(f, ϕ) = S(f, Idf•(A)

) S(Idb, ϕ) =

(f, χ(f,A)

)(Idb, Sb(ϕ)

)=(f, χ(f,A) Sb(ϕ)

),

so (4) holds, and by (22.6.7) we have

Σ(f, ϕ) = χ(f,A) Sb(ϕ). (6)

The proof of (1) follows from

Id(a,Sa(A)) = (Ida, IdSa(A)) = S(Ida, IdA) =(Ida, χ(Ida, A)

).

Now we work out the conditions for S to be compatible with composition:

S((f, ϕ) (g, ψ)

)= S(f, ϕ) S(g, ψ), (7)

for morphisms (f, ϕ): (b, B)→ (a,A) and (g, ψ): (c, C)→ (b, B) of X. By (22.6.2)and (22.6.7), the left hand side of (7) is

S((f, ϕ) (g, ψ)

)= S(f g, g•(ϕ) ψ) =

(f g, Σ(f g, g•(ϕ) ψ)

),

while the right hand side is

S(f, ϕ) S(g, ψ) =(f, Σ(f, ϕ)

)(g, Σ(g, ψ)

)=(f g, g•(Σ(f, ϕ)) Σ(g, ψ)

).

Hence S is compatible with composition if and only if

Σ(f g, g•(ϕ) ψ) = g•(Σ(f, ϕ)) Σ(g, ψ).

By (6), this is equivalent to

χ(f g,A) Sc(g•(ϕ) ψ

)= g•

(χ(f,A) Sb(ϕ)

) χ(g,B) Sc(ψ). (8)

We obtain (2) by specializing ψ = Idg•(B) and ϕ = Idf•(A), and by specializingf = Idb and ψ = IdC and using (1) we get (3).

(b) Conversely, let the Sa and χ(f,A) satisfy (i) and (ii), and define S by(4). We must show that S preserves identities and is compatible with composition.The first property follows from (1) and the fact that the Sa are functors. By thecomputation in (a), the second is equivalent to (8) which we proceed to verify.

From functoriality of Sc and of g• it follows that (8) is equivalent to[χ(f g,A) Sc(g•(ϕ))

] Sc(ψ) =

[g•(χ(f,A)

) g•

(Sb(ϕ)

) χ(g,B)

] Sc(ψ),

so it suffices to show the equality of the expressions in brackets. Now

χ(f g,A) Sc(g•(ϕ)) = g•(χ(f,A)

) χ(g, f•(A)

) Sc(g•(ϕ))

holds by (2), and by (3), applied to B′ = f•(A), we have

χ(g, f•(A)

) Sc(g•(ϕ)) = g•

(Sb(ϕ)

) χ(g,B).

Finally, an arrow (f, ϕ) of X is cartesian if and only if ϕ is an isomorphism.Hence the last statement follows from the second formula of (4).


22.8. Let us return to the situation of 4.21, so Φ: B → Cat is a covariantfunctor. We define a contravariant functor Φ∗: Bop → Cat by Φ∗(a) = Φ(a) onobjects, and by

Φ∗(f : b→ a) = Φ(f : a→ b)

on morphisms f : b→ a of Bop. This makes sense, since Bop and B have the sameobjects, but the reversed composition of morphisms. We denote the compositionof morphisms in Bop by f ~ g = g f . To simplify notation, we write againΦ∗(f) = f• = Φ(f) = f•.

By applying the Grothendieck construction 22.6 to Φ∗, we obtain a split fibrationΠ∗ =

∫Φ∗ =: X∗ → Bop. The objects of X∗ are, as before, the pairs (a,A) with

a ∈ B and A ∈ Φ∗(a) = Φ(a). A morphism from (b, B) to (a,A) of X∗ is a pair(f, ϕ) where f : b→ a is a morphism of Bop, thus f : a→ b is a morphism of B, and

ϕ: B → f•(A) = f•(A) (1)

is a morphism of the category F (b). The composition of two morphisms in X∗, say(f, ϕ): (b, B)→ (a,A) and (g, ψ): (c, C)→ (b, B) is given by (22.6.2), with the soledifference that f g is replaced by f ~ g. The projection onto the first factor isa split fibration Π∗: X∗ → Bop. The fibre X∗b of Π∗ over b ∈ Bop is canonicallyisomorphic to the category F (b).

The category X∗ is is not, as one might suspect, the opposite of the categoryX =

∫F as defined in 4.21. While Xop and X∗ have the same objects, a morphism

from (b, B) to (a,A) in Xop is a morphism (f, ϕ): (a,A) → (b, B) in X. Thusf : a→ b is a morphism in B and

ϕ: f•(A)→ B (2)

is a morphism of F (b). From (1) and (2) we see that X∗ 6= Xop.

22.9. We now apply the construction just discussed to the functors Φ and Ψfrom jgraph to Cat defined in Proposition 20.6 and (22.3.1), thus

Π∗ =∫Φ∗: gradjp∗ → jgraphop and P ∗ =

∫Ψ∗: gcj∗ → jgraphop.

For a morphism f : ∆→ Γ of jgraphop, we accordingly write

Φ∗(f) = f• = f•: gradjpΓ → gradjp∆,

where f• is defined in Lemma 20.4, and

Ψ∗(f) = f• = f•: gcjΓ → gcj∆,

where f• is defined in (21.3.5).

We intend to apply Lemma 22.7 to the present situation. In order to easethe comparison of the notation there with the present one, we offer the followingdictionary:

B F F ′ X X′ P P ′ a b A B Sa χ(f,A)

jgraphop Φ∗ Ψ∗ gradjp∗ gcj∗ Π∗ P ∗ Γ ∆ (V,R) (W,S) SΓ χ(f,R)


22.10. Lemma. Let (V,R) be a Γ -graded Jordan pair and let f : ∆ → Γ be amorphism in jgraphop. We use the notation introduced in 22.9.

(a) By Proposition 21.4, st(V,R) is a full subcategory of st(V, f•(R)), withinclusion functor i: st(V,R)→ st(V, f•(R)). Since St(V, f•(R)) is an initial objectof st(V, f•(R)), there exists a unique homomorphism

κ: St(V, f•(R))→ i(St(V,R)) (1)

of the category st(V, f•(R)). Let LR and Lf•(R) be the functors defined in (21.3.2),taking their values in the categories gcjΓ and gcj∆, respectively. Then

χ(f,R) := Lf•(R)(κ): S∆(V, f•(R))→ f•(SΓ (V,R)) (2)

is a morphism of the category gcj∆ satisfying

χ(IdΓ ,R) = IdSΓ (V,R). (3)

If g: Θ → ∆ is another morphism of jgraphop then

χ(f ~ g,R) = g•(χ(f,R)) χ(g, f•(R)). (4)

(b) Let (W,S) and (W ′,S′) be in gradjp∆ and let k: (W,S)→ (W ′,S′) be amorphism of ∆-graded Jordan pairs. Then

χ(g,S′) SΘ(g•(k)) = g•(S∆(k)) χ(g,S). (5)

Proof. To achieve a more concise notation, we denote in this proof an object(Γ, (V,R)) of gradjp∗ simply by (Γ,R). We also write G = (G,U±, π) = St(V,R)

for the Steinberg group of st(V,R), and G = (G, U±, π) = St(V, f•(R)) for theSteinberg group of st(V, f•(R)).

(a) By (22.2.3), we have

SΓ (V,R) = LR(G) and S∆(V, f•(R)) = Lf•(R)(G).

By (21.4.1), the diagram

st(V,R)i //

LR

st(V, f•(R))

Lf•(R)

gcjΓ

f•//// gcj∆

(6)

is commutative, where the notation f•(R) and f• instead of f•(R) and f• is ex-plained in 22.8. Hence Lf•(R)(i(G)) = f•(LR(G)), so by (22.2.3) and functorialityof Lf•(R),

χ(f,R) = Lf•(R)(κ) : Lf•(R)(G) = S∆(V, f•(R))→Lf•(R)(i(G)) = f•(LR(G)) = f•(SΓ (V,R)),


as claimed in (2).For g as indicated, we have g•(f•(R)) = (g f)•(R) by functoriality of Φ in

Proposition 20.6(a), which becomes g•(f•(R)) = (f ~ g)•(R) in the notation usedhere. Thus, by Proposition 21.4, we get the inclusion

j: st(V, f•(R)) ⊂ st(V, g•(f•(R))) = st(V, (f ~ g)•(R))

as a full subcategory. Let≈

G ∈ st(V, g•(f•(R))) be the Steinberg group of

st(V, g•(f•(R))). By the universal property of≈

G there are unique morphisms

λ:≈

G → j(G) and µ:≈

G → j(i(G)) of the category st(V, g•(f•(R))). But also

j(κ) λ:≈

G→ G is such a morphism, so by uniqueness, µ = j(κ) λ: the diagram

≈

Gλ //

µ!!DDDDDDDDD j(G)

j(κ)wwwwwwwww

j(i(G))

is commutative. We apply the functor Lg•(f•(R)) to µ and use (21.4.2):

χ(f ~ g, R) = Lg•(f•(R))(j(κ) λ) = Lg•(f•(R))(j(κ)) Lg•(f•(R))(λ)

= g•(Lf•(R)(κ)

) Lg•(f•(R))(λ)

= g•(χ(f,R)

) χ((f ~ g)•,R).

(b) In order to be able to use notation introduced in the proof of (a), we prove(b) in the following equivalent form: Let (V,R) and (V ′,R′) be in gradjpΓ andlet h: (V,R)→ (V ′,R′) be a morphism of Γ -graded Jordan pairs. Then

χ(f,R′) S∆(f•(h)) = f•(SΓ (h)) χ(f,R). (7)

We also write again simply G = St(V,R), G′ = St(V ′,R′), G = St(V, f•(R)), and

G′ = St(V ′, f•(R′)). As before, i: st(V,R) → st(V, f•(R)) and i′: st(V ′,R′) →st(V ′, f•(R′)) are the inclusions. Let κ: G → i(G) and κ′: G′ → i′(G′) bethe canonical maps in st(V, f•(R)) and st(V ′, f•(R′)) respectively, coming from

the fact that G and G′ are initial objects in the respective categories. ByLemma 22.2(a), h lifts to a group homomorphism ϕh: G→ G′. Let

f•(h): (V, f•(R))→ (V ′, f•(R′))

be the induced morphism of ∆-graded Jordan pairs as in (20.4.2), which is simplygiven by f•(h) = h: V → V ′ on the underlying (non-graded) Jordan pairs. Thenthe diagram

Gκ //

ϕf•(h)

i(G)

ϕh

G′

κ′// i′(G′)

(8)


is commutative. Indeed, let xσ: V σ → G and xσ: V σ → G be the canonicalisomorphisms onto the subgroups Uσ and Uσ, and use the analogous notationfor G′ and G′. Since G is generated by U+ and U−, it suffices to show that thediagram commutes when restricted to Uσ. For κ and κ′ we have

κ(xσ(v)) = xσ(v) and κ′(x′σ(v′)) = x′σ(v′)

for all v ∈ V σ and v′ ∈ V ′σ. By 9.5,

ϕh(xσ(v)) = x′σ(h(v))

for all v ∈ V σ, and in the same way,

ϕf•(h)(xσ(v)) = x′σ(h(v)).

Now it is clear that

ϕh(κ(xσ(v))) = x′σ(h(v)) = κ′(ϕf•(h)(xσ(v)))

for all v ∈ V σ, so the diagram is commutative.Formula (7) is equivalent to the diagram

S∆(V, f•(R)

) χ(f,R) //

S∆(f•(h))

f•(SΓ (V,R)

)f•(SΓ (h))

S∆(V ′, f•(R′)

)χ(f,R′)

// f•(SΓ (V ′,R′)

) (9)

being commutative. For the proof, we relate it to the commutative diagram (8).

First, by definition in 22.2, S∆(V, f•(R)

)= Lf•(R)

(St(V, f•(R))

)= Lf•(R)(G),

S∆(V ′, f•(R′)

)= Lf•(R′)(G

′) and S∆(f•(h)) = ϕf•(h), SΓ (h) = ϕh. Also, by defi-nition in (a), χ(f,R) = Lf•(R)(κ) and χ(f,R′) = Lf•(R′)(κ

′), while commutativity

of (6) gives f•(SΓ (V,R)

)= f•

(LR(G)

)= Lf•(R)

(i(G)

)and f•

(SΓ (V ′,R′)

)=

Lf•(R′)(i(G′)

). Hence the diagram (9) is the same as the diagram

Lf•(R)(G)Lf•(R)(κ)

//

ϕf•(h)

Lf•(R)

(i(G)

)f•(ϕh)

Lf•(R′)(G

′)Lf•(R′)(κ

′)

// Lf•(R′)(i(G′)

)(10)

All groups occurring in this diagram belong to the category gcj∆, in particular, theyare groups with root groups and commutator relations. By definition of the functorsLf•(R) and Lf•(R′) in 21.3, the underlying groups and group homomorphisms of(10) are the same as those in (8), so the diagram commutes on the level of groups.But all morphisms in (8) respect the various root groups because f• has beenapplied to the groups in the right hand column. Thus (9) is a commutative diagramin gcj∆.


22.11. Proposition. We use the notation of 22.9 and Lemma 22.10. There ex-ists a functor S∗: gradjp∗ → gcj∗ as follows. For an object (Γ, (V,R)) ∈ gradjp∗,put

S∗(Γ, (V,R)) = (Γ, SΓ (V,R)). (1)

For a morphism (f, h): (∆, (W,S))→ (Γ, (V,R)) of gradjp∗, let

Σ(f, h) = χ(f,R) S∆(h) : S∆(W,S)→ S∆(V, f•(R))→ f•(SΓ (V,R)

)(2)

and define

S∗(f, h) = (f, Σ(f, h)) : S∗(∆, (W,S))→ S∗(Γ, (V,R)). (3)

Then S∗ is a functor making the diagram

gradjp∗S∗ //

Π∗ $$IIIIIIIII gcj∗

P∗||yyyyyyyy

jgraphop

(4)

commutative.

Proof. This follows from Lemma 22.10 and Lemma 22.7, by using the dictionaryat the end of 22.9.

Remark. The functor S∗ is in general not a cartesian functor in the sense thatit maps cartesian arrows to cartesian arrows. Indeed, by Lemma 22.7, S∗ cartesianwould imply that χ(f,R) is an isomorphism, for all f : ∆ → Γ . By (22.10.2),χ(f,R) is the image under Lf•(R) of the canonical homomorphism κ of (22.10.1).In general, κ is not an isomorphism, so neither will be χ(f,R). For example, let∆ = δ be Jordan graph consisting of a single vertex. Then (V, f•(R)) is theJordan pair V with the trivial grading V = Vδ. Hence the relations (StR1) and

(StR2) are vacuous, so the Steinberg group G = St(V, f•(R)) is the free productFr(V ) = V + ∗ V −. On the other hand, G = St(V,R) is, by 22.1, obtained as thequotient of Fr(V ) by these relations for the Γ -graded Jordan pair (V,R). Here κ isthe canonical map Fr(V )→ St(V,R) which is in general not an isomorphism.

22.12. Proposition. Let Γ =∐i Γ

(i) be a coproduct of Jordan graphs as in15.4(b), let R = (Vα)α∈Γ be a Γ -grading of V , and put V (i) =

∑α∈Γ (i) Vα.

(a) The V (i) are ideals of V with direct sum V , and R(i) = (V(i)α )α∈Γ (i) is a

Γ (i)-grading of V (i). In the following we identify PE(V ) with⊕

i∈I PE(V (i)) bymeans of the isomorphism (8.2.1).

(b) If G(i) ∈ st(V (i),R(i)) then the restricted direct product⊕

i∈I G(i), with

obvious definitions of the subgroups Uσ and the projection onto PE(V ), belongs tost(V,R).

(c) Conversely, let G ∈ st(V,R). For i ∈ I and σ ∈ +,− define U (i)σ =xσ(V (i)σ) and G(i) =

⟨U (i)+ ∪ U (i)−⟩. Then the groups G(i) with the subgroups

U (i)± and the projection πi = π|G(i) belong to st(V (i),R(i)). The groups G(i)


and G(j) commute for i 6= j, and there is a canonical surjective homomorphism⊕i∈I G

(i) → G.

(d) The group St(V,R) is isomorphic to the restricted direct product of theSt(V (i),R(i)):

St(V,R) ∼=⊕i∈I

St(V (i),R(i)). (1)

Proof. (a) is immediate. The easy verifications in (b) are left to the reader.

(c) From (21.1.4) we have Uσ ∼=⊕

i∈I xσ(V (i)σ) =⊕

i U(i)σ. Hence G(i) is a

group over V (i). Next observe(((((((xσ(V (i)σ), x−σ(V (j)−σ)

)))))))= 1 for all i 6= j. (2)

Indeed, any v ∈ V (i)σ and w ∈ V (j)−σ has a decomposition v =∑α vα and

w =∑β wβ where α ⊥ β for all occurring vα, wβ . Thus (2) follows from (StR2).

This relation implies that the root groups U(i)% , % ∈ R(i), of (21.1.1) and (21.1.2),

are the root groups U% of G. Hence G(i) has R(i)-commutator relations, provingG(i) ∈ st(V (i),R(i)).

The relation (2) implies that G(i) and G(j) commute for i 6= j. Hence we have acanonical homomorphism

⊕iG

(i) → G. It is surjective since the restricted directproduct of the U (i)σ is mapped onto the Uσ and the latter generate G.

(d) Let G(i) = St(V (i),R(i)) and G = St(V,R). One checks easily that the

injections V (i) → V induce group homomorphisms ψ(i): G(i) → G which commuteby (2), so there is a homomorphism ψ:

⊕G(i) → G. In the other direction,

let p(i): V → V (i) be the canonical projection and f (i): Γ (i) → Γ the canonicalembedding. Then (f (i), p(i)) is a morphism of the category gradjp∗ defined in 22.5

and thus induces a group homomorphism ϕ(i): G → G(i) by (22.11.2). Then the

homomorphism ϕ =⊕ϕi: G→

⊕G(i) has image

⊕G(i) and is inverse to ψ.

Remark. The surjective homomorphism⊕

iG(i) → G in (c) is in general not

an isomorphism: not every group in st(V,R) is a restricted direct product. Forexample, let R = A1 × A1, Γ = K1 q K1 and V = M1,1(C) ×M1,1(C) with thecanonical Γ -grading. The group (SL2(C)×SL2(C))/±(12, 12) lies in st(V,R) butis not a direct product.

§23. Cogs

23.1. Definition. Let ∆ be a mixed graph as in 14.10, let V be a Jordan pair,and denote the set of idempotents of V by Idp(V ).

A cog of type ∆ in V is a map E : ∆→ Idp(V ), written E (α) = eα, satisfying

eα 6= 0 for all α ∈ ∆, (1)

eα ∈ V〈α,β∨〉(eβ) for all α, β ∈ ∆, (2)

where 〈α, β∨〉 ∈ 0, 1, 2 is defined in (15.1.1) and (14.10.1). The case ∆ = ∅ isallowed and occurs naturally, see Remark (a) of 23.11.

§23] Cogs 289

The reader may wonder about the reason for (1). In fact, if ∆ is connected thena map satisfying (2) is zero or injective, more precisely:

(a) if eα 6= 0 for all α ∈ ∆ then E : ∆→ Idp(V ) is injective,

(b) if ∆ is connected and eα = 0 for one α ∈ ∆ then eβ = 0 for all β ∈ ∆,

(c) if ∆ is connected, has at least two vertices, and E is injective then eα 6= 0for all α ∈ ∆.

In particular, E is injective if E is a cog. Therefore, we will often identify E withits image Im(E ) = eα : α ∈ ∆ ⊂ Idp(V ).

Proof. (a) Since 〈α, α∨〉 = 2, we have eα ∈ V2(eα). Suppose eα = eβ forα 6= β. Then eβ = eα ∈ V2(eα) ∩ V〈β,α∨〉(eα) by (2) and similarly, eα = eβ ∈V2(eβ)∩V〈α,β∨〉(eβ). Because α 6= β, 〈α, β∨〉 6= 2 or 〈β, α∨〉 6= 2 by (14.10.1). Sincedistinct Peirce spaces have zero intersection, we get eα = eβ = 0 in both cases,contradiction.

(b) Let eα = 0 and β ∈ ∆. If β ∼ α then 〈β, α∨〉 6= 0. Then eβ ∈ V2(eβ) ∩V〈β,α∨〉(eα), but since eα = 0, V0(eα) = V and Vi(eα) = 0 for i = 1, 2, so eβ = 0.If β ⊥ α then α and β can be connected by a chain α = α1 ∼ α2 ∼ · · · ∼ αn = β,and by induction it follows again that eβ = 0.

(c) If eα = 0 for some α then by (b), eβ = 0 for all β, which contradictsinjectivity since ∆ has at least two vertices.

The terminology “cog” comes from the fact that the graph relations α β,α ⊥ β, α→ β in ∆ are translated into the relations eα > eβ (eα [c]ollinear to eβ),eα ⊥ eβ (eα [o]rthogonal to eβ), eα a eβ (eα [g]overned by eβ) of the correspondingidempotents, see (6.15.4). This is an immediate consequence of (2). In particular,any two idempotents of a cog are compatible.

23.2. Lemma. Let M be a k-module, let ∆ and L be index sets, and for everyα ∈ ∆ let M =

⊕l∈LM

αl be a direct sum decomposition. Let I = L∆ =

I: ∆ →

L

, and put, for I ∈ I,

MI :=⋂α∈∆

MαI(α). (1)

Then the sum of submodules∑I∈IMI is direct.

Proof. We must show: if I1, . . . , In ∈ I are n different elements and xν ∈ MIν

for ν = 1, . . . , n such that∑nν=1 xν = 0 then all xν = 0. The proof is by induction

on n, the case n = 1 being trivial.For the induction step, let n> 2 and suppose the assertion holds for all m < n.

Since I1 6= I2, there exists α0 ∈ ∆ such that l1 := I1(α0) 6= l2 := I2(α0).Now decompose the interval N := 1, . . . , n ⊂ N into subsets

N =⋃l∈L

Sl (2)

where Sl := ν ∈ N : Iν(α0) = l. Then 1 ∈ Sl1 and 2 ∈ Sl2 because I1(α0) = l1and I2(α0) = l2, so at least two of the Sl are not empty. Hence Card(Sl) < n forall l ∈ L. Put, for l ∈ L,


yl :=∑ν∈Sl

xν .

Since the xν ∈ MIν , they are in particular, by (1), in Mα0

Iν(α0). Hence yl ∈ Mα0

l

and 0 =∑nν=1 xν =

∑l∈L yl. But the sum Mα0 =

∑l∈LM

α0

l is direct, so we haveall yl = 0. Since Card(Sl) < n and yl =

∑ν∈Sl xν = 0 it follows by induction that

xν = 0 for all ν ∈ Sl and l ∈ L. Now (2) shows that xν = 0 for all ν ∈ N .

23.3. Peirce decomposition with respect to a cog. Let E : ∆ → Idp(V )be a cog of type ∆ in a Jordan pair V . For a function I: ∆ → Z we define theI-Peirce space of E as the intersection of all I(α)-Peirce spaces of the eα:

VI(E ) :=⋂α∈∆

VI(α)(eα). (1)

It follows from Lemma 23.2 (for L = Z) that the sum∑I∈Z∆

VI(E ) ⊂ V (2)

is direct. Since the idempotents eα are compatible, so are their Peirce decomposi-tions

P(eα) : V = V2(eα)⊕ V1(eα)⊕ V0(eα). (3)

If ∆ is finite, V is the sum of the VI(E ). If ∆ is infinite, this need not be so, seefor example 6.16. The VI(E ) are subpairs of V , being intersections of such, and themultiplication rules (10.1.1) hold mutatis mutandis:

V σI V −σJ V σL ⊂ V σI−J+L, Q(V σI )V −σJ ⊂ V σ2I−J . (4)

(Since Z is an abelian group so is Z∆, hence I−J+Lmakes sense.) If I(α), J(α) =0, 2 for some α ∈ ∆ then

D(V σI , V−σJ ) = 0. (5)

For α ∈ ∆ let Pα ∈ Z∆ be defined as in (15.1.1), thus

Pα(β) = 〈α, β∨〉. (6)

Then it follows from (23.1.2) that

eα ∈ VPα(E ) =⋂β∈∆

V〈α,β∨〉(eβ) ⊂ V〈α,α∨〉(eα) = V2(eα) for all α ∈ ∆. (7)

23.4. Associated cogs. Recall from 6.17 that two idempotents e, e′ of a Jor-dan pair are associated, written e ≈ e′, if their Peirce spaces coincide: Vi(e) = Vi(f),for i ∈ 0, 1, 2. Two cogs E ,E ′: ∆ → Idp(V ) are called associated , in symbolsE ≈ E ′, if eα ≈ e′α for all α ∈ ∆. The following conditions are equivalent:

(i) E and E ′ are associated,

(ii) VI(E ) = VI(E ′) for all I ∈ Z∆,

(iii) VPα(E ) = VPα(E ′) for all α ∈ ∆, with Pα defined in (23.3.6),

(iv) eα ∈ VPα(E ′) and e′α ∈ VPα(E ) for all α ∈ ∆.

The implication (i) =⇒ (ii) is true in view of the definition (23.3.1) of VI(E ) andthe assumption Vi(eα) = Vi(e

′α) for all α ∈ ∆ and i = 0, 1, 2. Further, (ii) =⇒ (iii)

is clear, and (iii) =⇒ (iv) =⇒ (i) follow from (23.3.7).

§23] Cogs 291

23.5. Definition. In 13.1 we introduced the category st(V,S ) for a set S ofidempotents. Now let E : ∆→ Idp(V ) be a cog. Then we define a full subcategoryof st(V ) by

st(V,E ) := st(V, Im(E )

)=⋂α∈∆

st(V, eα).

The following results complement those in §13.

23.6. Proposition. Let ∆ be a mixed graph without isolated vertices, see 17.4,and let E and E ′ be associated cogs defined on ∆. Then

st(V,E ) = st(V,E ′). (1)

Proof. By 23.4 we have eα = E (α) ≈ e′α = E ′(α) for all α ∈ ∆. Fix α ∈ ∆.By assumption, there exists β ∈ ∆ satisfying α ∼ β 6= α, that is, α βor α → β or α ← β. This implies eα > eβ or eα a eβ or eα ` eβ for thecorresponding idempotents. In the first two cases we get st(V, eα) = st(V, e′α)from Proposition 13.7(ii). In the third case, st(V, eβ) = st(V, e′β) follows againfrom Proposition 13.7(ii), so that (13.14.1) yields st(V, eα, eβ) = st(V, eβ) =st(V, e′β) = st(V, e′α, e′β).

Hence in all three cases st(V, eα, eβ) = st(V, e′α, e′β). Now write ∆ =⋃i∈I ∆i where each ∆i is a connected induced subgraph with two vertices. Then

st(V,E ) =⋂i∈I

st(V,E |∆i) =⋂i∈I

st(V,E ′|∆i) = st(V,E ′)

proves (1).

23.7. Proposition. Let V be a Jordan pair, Γ a Jordan graph, and E : Γ →Idp(V ) a cog. For every connected component of Γ pick a vertex which is not theendpoint of an arrow, and let ∆ ⊂ Γ be the subset thus obtained. Then

st(V,E ) = st(V,E∣∣∆).

Proof. The inclusion from left to right is clear. For the proof of the otherinclusion let β ∈ Γ and let α ∈ ∆ be in the same connected component as β. It isenough to show st(V, eα) = st(V,E

∣∣Ξ) for some Ξ ⊂ Γ with α, β ⊂ Ξ. We mayassume α 6= β. If α β then eα > eβ , so (13.11.1) shows that Ξ = α, β hasthe required property.

In case α → β, whence eα a eβ , our claim follows from Proposition 13.14.Since β → α is excluded by assumption, the only remaining possibility is α ⊥ β.By 15.6 there exists γ ∈ Γ such that the induced subgraph on α, β, γ is either(i) α γ β or (ii) α → γ β or (iii) α → γ ← β. In Case (i) wehave eα > eγ > eβ , so that st(V, eα) = st(V, eα, eγ) = st(V, eα, eγ , eβ) by(13.11.1). In Case (ii), where eα a eγ > eβ , we get st(V, eα) = st(V, eα, eγ) fromProposition 13.14 and st(V, eγ) = st(V, eβ) from (13.11.1). Finally, in Case (iii) wehave eα a eγ ` eβ so that again Proposition 13.14 implies our claim.


23.8. Idempotent root gradings. Let Γ be a Jordan graph and let R =(Vγ)γ∈Γ be a Γ -grading of a Jordan pair V as in 20.1. Let ∆ ⊂ Γ and letE : ∆ → Idp(V ) be a cog, written E (δ) = eδ. We say R and E are compatible,and the pair (R,E ) is an idempotent root grading of V , if

Vγ =⋂δ∈∆

V〈γ,δ∨〉(eδ) for all γ ∈ Γ . (1)

Idempotent root gradings (R,E ) form the objects of a category idgradjp, with amorphism from (R,E ) to (R′,E ′) being defined as a morphism (f, h): R → R′ inthe category gradjp (see 20.5) for which

f(dom(E )

)⊂ dom(E ′) and h E = E ′ (f

∣∣dom(E )). (2)

There is a forgetful functor F : idgradjp → gradjp, defined by (R,E ) 7→ R onobjects and (f, h) 7→ (f, h) on morphisms. The fibre of F at R is the (non-full)subcategory of idgradjp consisting of the objects mapping to R and the morphismsmapping to IdR. It is easily seen that the fibre of F at R can be identified withthe partially ordered set (considered as a category in the usual way)

cog(R) (3)

of all cogs E , defined on some subset ∆ = dom(E ) of Γ and compatible with R,partially ordered by

E 6 E ′ ⇐⇒ dom(E ) ⊂ dom(E ′) and E = E ′∣∣dom(E ). (4)

From (1) it is clear that the root spaces Vγ of R are uniquely determined bythe cog E and the matrix

(〈γ, δ∨〉

)(γ,δ)∈Γ×∆, which encodes the way in which ∆ is

embedded in Γ . The Peirce spaces of the set E (∆) of idempotents of V alone arein general not sufficient to determine the root grading R, as the following simpleexample shows. Let V = (k, k), let Γ = α β and let R be the root gradinggiven by Vα = V and Vβ = 0. Put ∆ = α, and define E (α) = (1, 1). Then theidempotent eα has but a single Peirce space V = V2(eα), so the root grading cannotbe recovered from the Peirce spaces of E (∆) alone.

It follows from (23.1.1) and (23.3.7) that

0 6= eδ ∈ Vδ for all δ ∈ ∆. (5)

Since V is the direct sum of the Vγ and since by (1) every Vγ is a Peirce space ofE , the sum of all Peirce spaces (23.3.2) is all of V , even when ∆ is infinite. By (5),Vδ 6= 0 for δ ∈ ∆, but Vγ = 0 for γ ∈ Γ ∆ is possible, see the examples in 23.10.

By 20.7, every α ∈ Γ determines a Peirce grading Rα of V given by (20.7.2):

Vi(α) =∑

γ∈Γi(α)

Vγ =∑

〈γ,α∨〉=i

Vγ , (6)

for i = 0, 1, 2. On the other hand, for α ∈ ∆ the idempotent eα determines a Peircedecomposition (23.3.3). These two decompositions agree: Rα = P(eα), explicitly:

§23] Cogs 293

α ∈ ∆ =⇒ Vi(α) = Vi(eα) for i = 0, 1, 2. (7)

Indeed, let 〈γ, α∨〉 = i. Then by (1), Vγ ⊂ V〈γ,α∨〉(eα) = Vi(eα) which implies,by summing over all γ ∈ Γi(α), that Vi(α) ⊂ Vi(eα) for i ∈ 0, 1, 2. SinceV = V2(α)⊕ V1(α)⊕ V0(α) = V2(eα)⊕ V1(eα)⊕ V0(eα), the assertion follows.

It will be useful to introduce the following definition: two vertices α 6= β inΓ are said to be separated by a subset ∆ of Γ if there exists δ ∈ ∆ such that〈α− β, δ∨〉 6= 0. We say ∆ separates Γ if this holds for all α 6= β in Γ , thus,

for all α 6= β in Γ there exists δ ∈ ∆ such that 〈α, δ∨〉 6= 〈β, δ∨〉. (8)

23.9. Lemma. Let Γ be a Jordan graph, let ∆ ⊂ Γ be a subset and let E : ∆→Idp(V ) be a cog of type ∆ in a Jordan pair V with a root grading R = (Vγ)γ∈Γ .

(a) Assume that ∆ separates Γ , and that

Vγ ⊂⋂δ∈∆

V〈γ,δ∨〉(eδ), for all γ ∈ Γ. (1)

Then E ∈ cog(R).

(b) Suppose E ∈ cog(R). Then

α 6= β in Γ not separated by ∆ =⇒ Vα = Vβ = 0. (2)

Remarks. (a) Condition (1) is weaker than (23.8.1): only inclusion insteadof equality is required. Instead, ∆ has to be “sufficiently big”.

(b) Condition (2) does not involve the cog E . It is a necessary condition for ∆to be the domain of definition of a cog compatible with R.

Proof. (a) Extend the definition of Pα, α ∈ ∆, in (23.3.6) in the obvious wayand define Iγ ∈ Z∆, γ ∈ Γ , by

Iγ(δ) = 〈γ, δ∨〉 for δ ∈ ∆.We first claim that the sum ∑

γ∈ΓVIγ (E ) (3)

is direct. Indeed, by (23.3.2), the sum∑I∈Z∆ VI(E ) is direct. By the injectivity of

the map γ 7→ Iγ , every space VIγ (E ) lies in (in fact, equals) a unique space VI(E ),namely the one where I = Iγ . Hence directness of the sum (23.3.2) (taken over allI ∈ Z∆) implies that of (3) (taken over all γ ∈ Γ ).

Our assumption (1) saysVγ ⊂ VIγ (E ),

for all γ ∈ Γ . Forming the sum over all γ ∈ Γ we see that

V =⊕γ∈Γ

Vγ ⊂⊕γ∈Γ

VIγ (E ) ⊂ V.

This implies

Vγ = VIγ (E ) =⋂α∈∆

V〈γ,α∨〉(eα),

for all γ ∈ Γ , so (R,E ) is an idempotent root grading.

(b) From (23.8.1) and 〈α, δ∨〉 = 〈β, δ∨〉 for all δ ∈ ∆ we have Vα = Vβ . On theother hand, α 6= β implies Vα ∩ Vβ = 0 since the sum of the (Vγ)γ∈Γ is direct.


23.10. First examples of idempotent root gradings. Let V be a Jordanpair with a root grading R = (Vγ)γ∈Γ .

(a) Suppose R is idempotent with respect to the empty cog. Then ∆ = ∅ doesnot separate any two different roots. Hence Lemma 23.9(b) shows V = 0 as soonas Γ has more than one element.

(b) Let Γ = γ be a singleton and let R be idempotent with respect to anon-empty cog E : ∆ → Idp(V ). Thus ∆ = Γ , and (23.8.1) says that V = V2(eγ),so eγ is simply an invertible idempotent of V .

(c) Let Γ be a collision α → β ← γ. A root grading R of type Γ is then thesame as a Peirce grading P, see 10.1 and 20.2(b): V = V2⊕V1⊕V0 where V2 = Vα,V1 = Vβ , and V0 = Vγ . This example occurred already in 11.1 where we used the

fact that a collision is isomorphic to G (Cher2 ) = T2 via the identification α = 2ε1,

β = ε1 + ε0, γ = 2ε0.A cog E of type α is the same as a non-zero idempotent e = E (α) ∈ Vα,

and (R,E ) is an idempotent root grading if and only if Vi = Vi(e) for i = 0, 1, 2;equivalently, if the Peirce grading is idempotent with respect to e.

Conversely, suppose that R is idempotent with respect to a cog E defined on asubset ∆ ⊂ Γ , and assume V 6= 0. Then α ∈ ∆ or γ ∈ ∆. Indeed, assume to thecontrary that ∆ ⊂ β. By (a), ∆ = ∅ would imply V = 0 which is excluded. Thus∆ = β. Since 〈α, β∨〉 = 〈β, β∨〉 = 〈γ, β∨〉 = 2, ∆ does not separate α and β norβ and γ. Hence Vα = Vβ = Vγ = 0 by Lemma 23.9(b), contradiction.

(d) Let I be a set of cardinality >2 and let V be a Jordan pair which has aPeirce decomposition with respect to an orthogonal system O = (ei)i∈I of non-

zero idempotents. Let I ′ = I ∪ 0 and Γ ′ = TI′ . We use the isomorphism

p, q 7→ εp + εq =: εpq of (14.18.5) to identify TI′ and G (CherI′ ). We have seen in

Example 20.2(c) that the decomposition (20.2.17)

V =⊕p,q∈I′

Vpq

is a hermitian grading R of type Γ ′.Let Σ be the induced subgraph of Γ ′ with vertex set P1(I) = i : i ∈ I.

Then the map i 7→ ei is a cog defined on Σ which we may identify with O. Weuse the definition of the Peirce spaces Vij in (6.16.2) and (6.16.7) to show that R isidempotent with respect to O. Thus we must verify that Vεpq =

⋂i∈I V〈εpq,ε∨ii〉(ei)

holds for all εpq ∈ Γ ′. Indeed, for distinct i, j ∈ I we have

Vεii = Vii = V2(ei) = V2(ei) ∩⋂k 6=i

V0(ek) = V〈εii,ε∨ii〉(ei) ∩⋂k 6=i

V〈εii,ε∨kk〉(ek),

Vεij = Vij = V1(ei) ∩ V1(ej) = V1(ei) ∩ V1(ej) ∩⋂i,j,k 6=

V0(ek)

= V〈εij ,ε∨ii〉(ei) ∩ V〈εij ,ε∨jj〉(ej) ∩⋂i,j,k 6=

V〈εij ,ε∨kk〉(ek),

Vεi0 = Vi0 = V1(ei) ∩⋂k 6=i

V0(ek) = V〈εi0,ε∨ii〉(ei) ∩⋂k 6=i

V〈εi0,ε∨kk〉(ek),

Vε00 = V00 =⋂k∈I

V0(ek) =⋂k∈I

V〈ε00,ε∨kk〉(ek).

§23] Cogs 295

Let ∆ ⊂ TI ⊂ TI′ = Γ ′ be a subset satisfying Σ ⊂ ∆. Generalizing the above,we claim that any cog E , defined on ∆ and extending O, is also compatible with R.Since we already know that O ∈ cog(R), it suffices to show that every Vγ , γ ∈ Γ ′,lies in a Peirce space of any idempotent eij = E (εij), i 6= j, and εij ∈ ∆. Becauseεii → εij ← εjj the idempotent eij governs the orthogonal idempotents ei and ej ,whence eij ≈ ei + ej by (13.14.3). Hence Vpq ⊂ Vl(eij) = Vl(ei + ej) for suitablel ∈ 0, 1, 2 follows from (6.16.7).

In the spirit of Proposition 23.7 we have

st(V,E ) = st(V,O). (1)

Indeed, for εij ∈ ∆, i 6= j, the idempotent eij governs the two orthogonal idempo-tents, whence

st(V, ei, eij , ei) = st(V, ei) = st(V, ej). (2)

by (13.14.2) and therefore

st(V,E ) =(⋂i∈I

st(V, ei))∩( ⋂εij∈∆, i 6=j

st(V, eij))

= st(V,O) ∩( ⋂εij∈∆, i 6=j

st(V, ei, eij , ej))

= st(V,O).

Moreover, if ∆ = TI then

st(V,E ) = st(V,O) = st(V, ei) (3)

for any i ∈ I. Indeed, we only need to prove the second equation. Fix i ∈ I, andlet j ∈ I, j 6= i. Since by assumption εij ∈ ∆, (2) can be applied and provesst(V, ei) = st(V, ej).

(e) The conditions of Lemma 23.9(a) are sufficient but not necessary for aroot grading to be idempotent. For example, let V = (k, k), let Γ = α, β, γ bea triangle, and let R be the root grading given by Vα = V and Vβ = Vγ = 0.Put ∆ = α, and define E (α) = (1, 1). Then it is immediate that (R,E ) is anidempotent root grading but α does not separate β and γ because β 6= γ but〈β, α∨〉 = 1 = 〈γ, α∨〉.

23.11. Lemma. Let Γ =∐i∈I Γ

(i) be a coproduct of Jordan graphs Γ (i), andlet V be a Jordan pair with a Γ -grading R. By 22.12, V is then the direct sum ofideals V (i), each of which has a Γ (i)-grading R(i).

(a) Let E ∈ cog(R), with ∆ = dom(E ). Then each E (i) = E∣∣(∆ ∩ Γ (i)) is a

cog in V (i), and E (i) ∈ cog(R(i)).

(b) Conversely, let E (i) ∈ cog(R(i)) for all i, with dom(E (i)) = ∆(i) ⊂ Γ (i), let

∆ =⋃i∈I ∆

(i) ⊂ Γ , and let E : ∆→ Idp(V ) be the obvious extension of the E (i) toa map E : ∆→ Idp(V ). Then E is a cog in V . Define

Γ0(∆) = γ ∈ Γ : 〈γ,∆∨〉 = 0


and assume that, for all i 6= j in I,

α, β ∈ Γ0(∆) and α ∈ Γ (i), β ∈ Γ (j) =⇒ Vα = Vβ = 0. (1)

Then E ∈ cog(R).

Proof. Let i 6= j and let e ∈ V (i) and f ∈ V (j) be idempotents. From thedescription of the Peirce projections in (6.14.3) and the fact that V (i) and V (j) aredisjoint ideals, one sees that

V (i) ⊂ V0(f), V (j) ⊂ V0(e), (2)

V2(e)⊕ V1(e) ⊂ V (i), (3)

and therefore Vn(e) = V(i)n (e) for n = 1, 2, while V0(e) = V

(i)0 (e) ⊕

⊕j 6=i V

(j).Hence

Vn(e) ∩ V (i) = V (i)n (e) for all n ∈ 0, 1, 2. (4)

(a) For γ ∈ Γ (i) we have Vγ ⊂ V (i). Let δ ∈ ∆(i) := ∆ ∩ Γ (i). Then E (δ) = eδis a non-zero idempotent in V (i), hence E (i) is a cog in V (i).

Since (V,R) is idempotent with respect to E , (23.8.1) holds for all γ ∈ Γ (i). Byintersecting (23.8.1) with V (i) we obtain

Vγ = Vγ ∩ V (i) =⋂δ∈∆

(V〈γ,δ∨〉(eδ) ∩ V (i)

). (5)

Suppose δ ∈ ∆(j) where i 6= j. Then f = eδ ∈ V (j), so V (i) ⊂ V0(eδ) by (2).Thus the corresponding term in (5) is just V (i) and can therefore be omitted in theintersection. Hence it follows from (5) and (4) that

Vγ =⋂

δ∈∆(i)

V(i)〈γ,δ∨〉(eδ), (6)

which shows that (V (i),R(i)) is idempotent with respect to E (i).

(b) Fix γ ∈ Γ (i). Then (6) holds, and we must show (23.8.1), thus

D :=⋂δ∈∆

V〈γ,δ∨〉(eδ) = Vγ . (7)

We distinguish two cases.

Case 1: γ /∈ Γ0(∆). Then we decompose ∆ with respect to γ as follows:

∆ = ∆2 ∪∆1 ∪∆0 ∪∆′

where ∆n = δ ∈ ∆(i) : 〈γ, δ∨〉 = n and ∆′ = ∆ ∆(i). Correspondingly,D = D2 ∩D1 ∩D0 ∩D′ is the intersection of four terms. Since γ /∈ Γ0(∆), we have∆2 ∪∆1 6= ∅. Hence by (3),

D2 ∩D1 =⋂

δ∈∆2∪∆1

V〈γ,δ∨〉(eδ) ⊂ V (i),

§23] Cogs 297

and therefore D ⊂ V (i). This implies

D = D ∩ V (i) =⋂δ∈∆

V〈γ,δ∨〉(eδ) ∩ V (i).

If δ /∈ ∆(i) then 〈γ, δ∨〉 = 0, so V0(eδ) ⊃ V (i) by (2) and the corresponding termscan be omitted in the intersection. Thus we obtain from (4),

D =⋂

δ∈∆(i)

V〈γ,δ∨〉(eδ) ∩ V (i) =⋂

δ∈∆(i)

V(i)〈γ,δ∨〉(eδ) = Vγ ,

because of our assumption (6).

Case 2: γ ∈ Γ0(∆). Here we must show⋂δ∈∆

V0(eδ) = Vγ . (8)

For every j ∈ I we have⋂δ∈∆(j)

V0(eδ) =( ⋂δ∈∆(j)

V(j)0 (eδ)

)⊕⊕k 6=j

V (k),

whence ⋂δ∈∆

V0(eδ) =⋂j∈I

⋂δ∈∆(j)

V0(eδ) =⊕j∈I

( ⋂δ∈∆(j)

V(j)0 (eδ)

). (9)

Define Γ(j)0 (∆(j)) in analogy to Γ0(∆) for Γ (j) and ∆(j). The Γ (j) are pairwise

orthogonal, whence Γ(j)0 (∆(j)) = Γ0(∆) ∩ Γ (j). Since (V (j),R(j)) is idempotent

with respect to E (j), (23.8.6) implies⋂δ∈∆(j)

V(j)0 (eδ) =

⊕β∈Γ (j)

0 (∆(j))

V(j)β =

⊕β∈Γ (j)

0 (∆(j))

Vβ . (10)

From (9) and (10) we obtain⋂δ∈∆

V0(eδ) =⊕j∈I

⊕β∈Γ (j)

0 (∆(j))

Vβ =⊕

β∈Γ0(∆)

Vβ . (11)

If Γ0(∆) = γ then (11) shows that (8) holds. If β 6= γ in Γ0(∆) and β andγ belong to the same component Γ (i) then Vβ = Vγ = 0 follows from (23.9.2),applied to V (i), R(i), E (i). If β and γ lie in different components Γ (i) and Γ (j),then Vβ = Vγ = 0 holds by (1). Hence (8) holds as well since by (11) both sidesvanish.

Remarks. (a) In Lemma 23.11(a), empty cogs occur naturally. For example,let Γ = α, β with α ⊥ β, and let V = Vα ⊕ Vβ be the direct sum of ideals whereVα = (k, k). Then V is Γ -graded, and it is idempotent with respect to the cogE : ∆ = α → Idp(V ) given by eα = (1, 1). Also Γ = Γ (1) q Γ (2) = α q β,and ∆(2) = ∅, so E (2) is empty.


(b) Let Γ (i) = αi → βi ← γi be collisions for i ∈ I = 1, 2, let Γ = Γ (1)qΓ (2),let V = V (1) ⊕ V (2) be a direct sum of ideals, and let e(i) ∈ V (i) be non-zeroidempotents. By 23.10(c), V (i) is Γ (i)-graded and idempotent with respect to thecog E (i): ∆(i) = αi → V (i), E (i)(αi) = e(i). Assume that the extension E of E (1)

and E (2) makes V an idempotent root grading. Then

Vγ1 = V〈γ1,α∨1 〉(eα1) ∩ V〈γ1,α∨2 〉(eα2

) = V0(e(1)) ∩ V0(e(2))

=(V

(1)0 (e(1))⊕ V (2)

)∩(V (1) ⊕ V (2)

0 (e(2)))

= Vγ1 ⊕ Vγ2 ,

so (1) is not fulfilled as soon as Vγ2 = V(2)0 (e(2)) 6= 0.

23.12. Group actions on cogs. Let Γ be a Jordan graph and V a Jordanpair, and consider cogs E for V defined on subsets ∆ of Γ . Let h ∈ Aut(V ) andt ∈ Aut(Γ ). For a cog E : ∆ → Idp(V ) define E ′ := h E t−1 on ∆′ = t(∆),explicitly,

E ′(α′) = h(E (t−1(α′))), for α′ ∈ ∆′.Then E ′ is a cog. Indeed, write eα = E (α) and e′α′ = E ′(α′), for α ∈ ∆ and α′ ∈ ∆′.It is clear that e′α′ is a non-zero idempotent in V . Moreover, for all α′ = t(α) andβ′ = t(β) in ∆′, since h and t are automorphisms of V and Γ , respectively,

e′α′ = h(eα) ∈ h(V〈α,β∨〉(eβ

)= V〈α,β∨〉(h(eβ) = V〈α′,β′∨〉(e

′β′),

so E ′ is a cog defined on ∆′. Clearly, this defines an action of the group Aut(V )×Aut(Γ ) on the set of cogs defined on subsets of Γ . For E and E ′ = h E t−1 asabove, the Peirce spaces of E and E ′ are related by

h(VI(E )

)= VIt−1(E ′). (1)

This follows from

h(VI(E )

)=⋂α∈∆

h(VI(t−1(α))(et−1(α))

)=⋂α∈∆

VI(t−1(α))(e′α)) = VIt−1(E ′).

Now let R be a Γ -grading of V and let E ∈ cog(R). Recall from 20.5 the groupAut(Γ, (V,R)), consisting of all pairs (t, h) ∈ Aut(Γ )×Aut(V ) such that

h(Vγ) = Vt(γ) for all γ ∈ Γ . (2)

Then Aut(Γ, (V,R)) acts on cog(R) by the action defined above.Indeed, as shown before, E ′ is a cog. Since E ∈ cog(R), we have by (23.8.1)

and (23.3.7) thatVγ = VPγ (E ) for all γ ∈ Γ .

From the fact that t is an automorphism of Γ it follows that Pγ t−1 = Pt(γ). Wenow apply the automorphism h and use (2) and (1):

Vt(γ) = h(Vγ) = h(VPγ (E )

)= VPγt−1(E ′) = Vt(γ)(E

′).

Hence E ′ ∈ cog(R). It is immediately verified that this action respects the partialorder on cog(R) defined in (23.8.4).

The next lemma gives some explicit examples of elements of Aut(Γ, (V,R)),constructed from cogs compatible with R.

§23] Cogs 299

23.13. Lemma. Let R be a Γ -grading of V and let E ∈ cog(R), defined on∆ ⊂ Γ . Suppose α, β ∈ ∆ satisfy 〈β, α∨〉 = 1.

(a) Then t := tα,β = sαsβsα ∈ T (∆) ⊂ Inn(Γ ) as in Lemma 15.18, and puttingh := ωeβ ,eα as in Lemma 10.11 we have (t, h) ∈ Aut(Γ, (V,R)). Hence

T (∆) ⊂ Π(

Aut(Γ, (V,R))), (1)

where Π: gradjp→ jgraph is the functor of 20.5.

(b) Let R = R−1 ∪R0 ∪R1 be the 3-graded root system associated with Γ andlet G ∈ st(V,R), viewed as a group in gcR via the forgetful functor LR of (21.3.2).Then the element

weβ ,eα = Int(weα) · weβ ∈ G

of Lemma 13.10 is a Weyl element for the root β − α ∈ R×0 . It is contained in thesubgroup G0 ∩N where G0 and N are defined in Lemma 9.2.

Proof. (a) For γ ∈ Γ we put i = 〈γ, β∨〉, j = 〈γ, α∨〉, k = 〈α, β∨〉. Then

t(γ) = γ − (i− jk)(β − α).

by (15.18.1). Let e = eα and f = eβ . Then h = ωf,e = Int(ωe)·ωf by Lemma 10.11.Put V(ij) = Vi(f) ∩ Vj(e). Then (23.8.1) shows Vγ ⊂ V(ij). Now use the formulasof Propositions 10.12 and 10.13 to compute h · xγ for xγ ∈ V σγ .

We present the details for k = 1, that is, α β and leave the easier case k = 2to the reader. The proof proceeds by distinguishing the following cases where, forbetter readability, the superscripts ±σ have been suppressed.

Case i = j. Then t(γ) = γ. On the other hand, for i = 2 or i = 0, h · xγ =h · x(ii) = x(ii) = xγ ∈ Vγ . For i = 1, we have h · xγ = h · x(11) = x(11) −f e e f x(11) = xγ−eβ eα eα eβ xγ and eβ eα eα eβ xγ ∈ Vβ−α+α−β+γ =Vγ , whence again h · xγ ∈ Vγ = Vt(γ).

Case i − j = −1. Then t(γ) = γ − α + β, and h · xγ = h · x(ij) = f e x(ij) =eβ eα xγ ∈ Vβ−α+γ = Vt(γ).

Case i − j = 1. Then t(γ) = γ + α − β and h · xγ = h · x(ij) = −eα eβ xγ ∈Vα−β+γ = Vt(γ).

Case i− j = 2. Then i = 2, j = 0, t(γ) = γ + 2α− 2β, while h · xγ = h · x(02) =QeβQeαxγ ∈ V2β−2α+γ = Vt(γ).

Case i− j = −2. Then i = 0, j = 2, t(γ) = γ−2α−2β, while h ·xγ = h ·x(20) =QeαQeβxγ ∈ V2α−2β+γ = Vt(γ).

We have now shown h(Vγ) ⊂ Vt(γ). To prove equality, observe that h−1 = ω−f,eby (10.11.3). Hence the calculations above show, after replacing f by −f , thath−1(Vγ) ⊂ Vt(γ) for all γ ∈ Γ , and therefore h−1(Vt(γ)) ⊂ Vt2(γ) = Vγ , sincet2 = sαs

2βsα = s2

α = IdΓ . Thus also Vt(γ) ⊂ h(Vγ) which proves (23.12.2). Finally,(1) follows immediately from the definitions.

(b) Put µ = β − α. Since eβ ∈ V1(eα) and G ∈ st(V,P(eα)) by Corollary 21.5,Lemma 13.10 yields


w := weβ ,eα = b(e+α , e−β ) b(−e+

β , e−α ) b(e+

α , e−β ) ∈ U−µUµU−µ,

by definition of the root groups U±µ in (21.1.2).To complete the proof that w is a Weyl element for µ we must verify, by the

definition in 5.1, that Int(w) · U% = Ut(%), for all % ∈ R. First, let γ ∈ R1 and% = σγ where σ = ±. Then by (21.1.1) and (13.10.4),

Int(w) · Uσγ = Int(w) · xσ(V σγ ) = xσ(h · V σγ ) = xσ(V σt(γ)) = Uσt(γ).

Next let % = ν ∈ R×0 . By (21.1.2), Uν is generated by all b(xγ , yδ) where γ, δ ∈ R1

and ν = γ−δ. Since w ∈ N∩G0 by Lemma 13.10, formula (9.7.6) applies and yieldsInt(w) · b(xγ , yδ) = b

(h+(xγ), h−(yδ)

)with h+(xγ) ∈ V +

t(γ) and h−(yδ) ∈ V −t(δ) by

(a). Therefore b(h+(xγ), h−(yδ)

)∈ Ut(ν), so we have

Int(w) · Uν ⊂ Ut(ν). (2)

The same argument applies to w−1 = Int(weα) · w−1eβ

= Int(weα) · w−eβ = w−eβ ,eαand shows Int(w)−1 ·Uν ⊂ Ut(v). Together with (2) this proves Int(w) ·Uν = Ut(ν),as desired.

23.14. Invertible idempotents. Let V be a Jordan pair. It follows from 6.13and 6.14 that the following conditions are equivalent for (x, y) ∈ V :

(i) x is invertible in V and y = x−1,

(ii) y is invertible in V and x = y−1,

(iii) e = (x, y) is an idempotent with V = V2(e).

Such a pair will be called an invertible idempotent of V . We denote by V × theset of invertible idempotents of V . In particular, when V + = V − = 0, the pair(0, 0) ∈ V is an invertible idempotent. This agrees with the usual convention inring theory.

Example. As in Example (a) of 6.6 let V = Mp,q(A) be the Jordan pairof rectangular matrices over a unital associative k-algebra A. Suppose that A hasinvariant basis number. We have seen in the Example discussed in 6.13 that V × 6= ∅if and only if p = q.

Let U be a subpair of V . For an idempotent e of U we have

e ∈ U× ⇐⇒ U ⊂ V2(e), (1)

which follows from U2(e) = U ∩ V2(e). In particular, e ∈ V2(e)×. The equivalence(1) implies

U ∩ V × ⊂ U×,

an inclusion which is in general not an equality: in V = M2(K) for K a field theidempotent e = (E11, E11) is invertible in U = V2(e) = (K · E11,K · E11) but notin V .

Let e ∈ U×. For an idempotent f ∈ U we have

f ∈ U× ⇐⇒ e ≈ f. (2)

§23] Cogs 301

Indeed, if f ∈ U× then U ⊂ V2(e)∩V2(f) by (1), whence e ∈ V2(f) and f ∈ V2(e) sothat e ≈ f follows from the definition of associated idempotents in 6.15. Conversely,if e ≈ f then V2(e) = V2(f) by 6.17 and therefore U ⊂ V2(f) because U ⊂ V2(e)again by (1).

In the following lemma we will apply this to U = Vδ, a homogeneous space of aΓ -grading of V .

23.15. Lemma. Let V be a Jordan pair with a Γ -grading R = (Vα)α∈Γ , andlet E ∈ cog(R). Then 0 6= eδ = E (δ) ∈ V ×δ for all δ ∈ ∆ = dom(E ). If alsoE ′ ∈ cog(R) then

eδ ≈ e′δ for all δ ∈ dom(E ) ∩ dom(E ′).

Proof. Since E is a cog, 0 6= eδ ∈ Vδ by (23.8.5) and then eδ ∈ V ×δ by (23.14.1)and (23.8.6). The second assertion then follows from (23.14.2).

23.16. Lemma. Let E ∈ cog(R) and let E ′: ∆ = dom(E )→ Idp(V ), α 7→ e′α,be a map satisfying e′α ∈ Vα and e′α ≈ eα, for all α ∈ ∆. Then E ′ is a cog,associated with E , and E ′ ∈ cog(R).

Proof. We first show that E ′ is a cog. Obviously, only (23.1.2) has to be shown,that is, e′α ∈ V〈α,β∨〉(e′β) for all α, β ∈ ∆. Since associated idempotents have thesame Peirce spaces we have V〈α,β∨〉(e

′β) = V〈α,β∨〉(eβ) for all β ∈ ∆. Hence, using

(23.8.1) for E , our claim becomes e′α ∈ Vα for all α ∈ ∆ which holds by assumption.Thus E ′ is a cog associated with E . Since E and E ′ have the same Peirce spaces,(23.8.1) shows that E ′ is compatible with R.

23.17. Lemma. Let R be a Γ -grading of a Jordan pair V and let E and E ′ bein cog(R), with dom(E ) = ∆ and dom(E ′) = ∆′. Assume that E and E ′ agree on∆ ∩∆′ and define F : ∆ ∪∆′ → Idp(V ) by

F (α) =

E (α) if α ∈ ∆E ′(α) if α ∈ ∆′

.

Then F ∈ cog(R).

Proof. As E and E ′ agree on ∆∩∆′, it causes no confusion to write F (α) = eαfor all α ∈ ∆ ∪ ∆′. To show that F is a cog, it suffices by 23.1 to verify thateα ∈ V〈α,β∨〉(eβ) for all α, β ∈ ∆ ∪ ∆′. If both α and β belong to ∆ this is clearfrom (23.1.2), and likewise if both α and β belong to ∆′. Now let α ∈ ∆ andβ ∈ ∆′, and put i = 〈α, β∨〉. Since E ′ is compatible with R, we have, by (23.8.6)and (23.8.7),

Vi(eβ) = Vi(β) =∑

〈γ,β∨〉=i

Vγ .

In particular, 〈α, β∨〉 = i, hence Vα is one of the summands on the right hand side,so by (23.8.5),

eα ∈ Vα ⊂∑

〈γ,β∨〉=i

Vγ = Vi(eβ) = V〈α,β∨〉(eβ).


By symmetry, it follows in the same way that eα ∈ V〈α,β∨〉(eβ) if α ∈ ∆′ and β ∈ ∆.Thus F is a cog.

It remains to show that (23.8.1) holds for F . Since E and E ′ are compatiblewith R, we have

Vγ =⋂α∈∆

V〈γ,α∨〉(eα) =⋂β∈∆′

V〈γ,β∨〉(eβ)

for all γ ∈ Γ . As E and E ′ agree on ∆ ∩∆′, it follows

Vγ = Vγ ∩ Vγ =⋂α∈∆

V〈γ,α∨〉(eα) ∩⋂β∈∆′


=⋂

α∈∆∩∆′V〈γ,α∨〉(eα) ∩

⋂α∈∆ ∆′

V〈γ,α∨〉(eα) ∩⋂

α∈∆′ ∆V〈γ,α∨〉(eα)

=⋂

δ∈∆∪∆′V〈γ,δ∨〉(eδ),

as required.

23.18. Proposition. Let R = (Vγ)γ∈Γ be a root grading of V and let cog(R)be the partially ordered set of cogs compatible with R defined in 23.8. Recall theprojection functor Π: gradjp→ jgraph of 20.5.

(a) cog(R) is inductively ordered, so by Zorn’s Lemma, it contains maximalelements.

(b) Let E ′ be a maximal element of cog(R) and let E ∈ cog(R) be arbitrary.Then dom(E ) ⊂ dom(E ′).

(c) The maximal elements of cog(R) have the same domain of definition∆max and are associated. They are permuted by the group Aut(Γ, (V,R)) act-ing as in 23.12, and ∆max is a Jordan subgraph of Γ , stable under the subgroupΠ(

Aut(Γ, (V,R)))

of Aut(Γ ).

(d) Every E ∈ cog(R) can be extended to a cog E ′ ∈ cog(R) defined on ∆max.

Proof. (a) Clearly, (23.8.4) defines a partial order on cog(R). Now let C ⊂cog(R) be a chain. Let ∆ =

⋃E∈C dom(E ) and define E on ∆ in the obvious way.

Then it follows easily that E is a cog, and clearly E 6 E for all E ∈ C. For E to bein cog(R), it remains to show (23.8.1) for E . Since all E ∈ C are in cog(R), wehave

Vγ =⋂

δ∈dom(E )

V〈γ,δ∨〉(eδ),

for all γ ∈ Γ . By taking the intersection over all E ∈ C, it follows that

Vγ =⋂

E∈C

⋂δ∈dom(E )

V〈γ,δ∨〉(eδ) =⋂δ∈∆

V〈γ,δ∨〉(eδ),

as required.

(b) Assume that ∆ = dom(E ) is not contained in ∆′ = dom(E ′), so that∆ ∆′ 6= ∅. Define E on ∆ by

§23] Cogs 303

E (δ) =

E (δ) if δ ∈ ∆ ∆′

E ′(δ) if δ ∈ ∆ ∩∆′. (1)

Then Lemma 23.15 shows that E (δ) ≈ E (δ) for all δ ∈ ∆, and by (23.8.5) for Eand E ′, we get E (δ) ∈ Vδ for all δ ∈ ∆. Hence E ∈ cog(R) by Lemma 23.16, anddom(E ) = ∆.

By (1), E and E ′ agree on ∆∩∆′. Hence Lemma 23.17 shows that there existsF ∈ cog(R) with dom(F ) = ∆ ∪∆′ ' ∆′ and F

∣∣∆′ = E ′, contradicting the factthat E ′ is maximal.

(c) From (b) it follows immediately that any two maximal elements of cog(R)have the same domain of definition, and Lemma 23.15 shows that any two maximalcogs are associated.

As remarked in 23.12, the action of Aut(Γ, (V,R)) is compatible with the partialorder. Hence maximal elements of cog(R) are permuted by Aut(Γ, (V,R)), andtheir common domain of definition ∆max is stable under Π

(Aut(Γ, (V,R))

). In

particular, (23.13.1) shows that ∆max is invariant under T (∆max). Hence ∆max isa Jordan subgraph by Lemma 15.18(b).

(d) Choose F ∈ cog(R) with dom(F ) = ∆max. By (b), ∆ = dom(E ) ⊂ ∆max.Define

E ′(δ) =

E (δ) if δ ∈ ∆F (δ) if δ ∈ ∆max ∆

.

Then E ′(δ) ∈ Vδ for all δ ∈ ∆max, by (23.8.5), applied to E in case δ ∈ ∆, and toF in case δ ∈ ∆max ∆, and E ′(δ) ≈ F (δ) for all δ ∈ ∆max, by Lemma 23.15.Hence E ′ ∈ cog(R) by Lemma 23.16.

23.19. Lemma. We use the notation of Proposition 23.18, and recall from(21.1.6) the definition

Γ×(R) = γ ∈ Γ : Vγ 6= 0.

(a) Then ∆max ⊂ Γ×(R) and Γ×(R) is stable under the action of the subgroupΠ(

Aut(Γ, (V,R)))

of Aut(Γ ).

(b) Suppose Γiso ∪ Γ 0 ⊂ ∆max. Then for all γ ∈ Γ

V +γ = 0 ⇐⇒ V −γ = 0, (1)

and Γ×(R) is a Jordan subgraph. For a connected component Σ of Γ precisely oneof the following alternatives occurs:

(i) Σ ⊂ ∆max,

(ii) Σ ∩∆max = Σ0 & Σ ⊂ Γ×(R),

(iii) Σ ∩∆max = Σ ∩ Γ×(R) = Σ0 & Σ.

Proof. (a) The inclusion ∆max ⊂ Γ×(R) follows from 0 6= eδ ∈ Vδ for allδ ∈ ∆max. If (f, h) ∈ Aut(Γ, (V,R)) then, by definition, h(Vγ) = Vf(γ) whence

f(Γ×(R)

)= Γ×(R).

(b) If γ ∈ Γiso ∪ Γ 0 then (1) is clear because we have the non-zero idempotenteγ ∈ Vγ . Otherwise, γ is either the initial point of an arrow γ = δ → α of hermitiantype, or the endpoint of an arrow α→ γ = ε of orthogonal type.


In the first case, embed δ → α in a hexagram (17.2.2). Then α, β, γ are inΓ 0. Assume V σδ = 0. Since eα is invertible in V2(eα) = Vδ ⊕ Vα ⊕ Vδ′ , it followsthat Q(e−σα )V σδ = V −σδ′ = 0 and similarly Q(eσβ)V −σδ′ = V σδ′′ = 0 which implies

Q(e−σγ )V σδ′′ = V −σδ = 0.In the second case, α → ε embeds in a pyramid (17.2.4) where α and α′ are

in Γ 0. Then V ′ = Vα ⊕ Vε ⊕ Vα′ is a Jordan subpair, e1 = eα and e2 = eα′ is acomplete orthogonal system of idempotents in V ′ and Vε = V ′12 in the notation of6.16. Hence Q(eσ1 + eσ2 )V −σε = V σε , proving (1) in this case.

Let Inn(Γ 0) be the subgroup of Aut(Γ ) generated by all tα,β for edges α βin Γ 0. By Lemma 17.13, Inn(Γ 0) acts transitively on Σ0 and on ∂Σ. This fact andthe stability of ∆max and Γ×(R) under T (Γ 0) imply easily that (i) – (iii) holds.By Proposition 17.9(a), Σ0 is a Jordan subgraph of Σ, hence of Γ . Therefore,Γ×(R)∩Σ is a Jordan subgraph for all Σ. Since Γ×(R) is the orthogonal disjointunion of its intersections with the connected components of Γ , Γ×(R) is a Jordansubgraph as well.

Remark. The induced graph Γ×(R) is in general not a Jordan subgraph, seethe example in 23.23. Moreover, (23.23.2) shows that all inclusions in the chain∆max ⊂ Γ×(R) ⊂ Γ can be proper.

23.20. Lemma. Let R be a Γ -grading of a Jordan pair V , let ∆ ⊂ Γ , and letE : ∆→ Idp(V ) be a map satisfying

(i) E (δ) ∈ V ×δ for all δ ∈ ∆, and

(ii) there exist E ′, E ′′ ∈ cog(R) such that dom(E ′) ⊂ ∆ ⊂ dom(E ′′).

Then E ∈ cog(R).

Proof. We put ∆′ = dom(E ′), ∆′′ = dom(E ′′) and e′δ = E ′(δ), eδ = E (δ),e′′δ = E ′′(δ) for δ in the appropriate domains. Since 0 6= e′α ∈ V ×α for α ∈ ∆′

by Lemma 23.15, it follows from (23.14.2) and assumption (i) that e′α ≈ eα, inparticular 0 6= eα. The same reasoning shows eβ ≈ e′′β for β ∈ ∆.

For E to be a cog, it remains to verify (23.1.2). Since eα ∈ Vα by assumption,this is a special case of (23.8.1) defining compatibility of E and R. The followingsequence of equalities and inclusions and the fact that (23.8.1) holds for E ′ and E ′′

proves our assertion:

Vγ =⋂α∈∆′

V〈γ,α∨〉(e′α) =

⋂α∈∆′

V〈γ,α∨〉(eα) ⊃⋂β∈∆


=⋂β∈∆

V〈γ,β∨〉(e′′β) ⊃

⋂δ∈∆′′

V〈γ,δ∨〉(e′′δ ) = Vγ .

23.21. Corollary. For e ∈ Idp(V ) and δ ∈ ∆max the following conditions areequivalent.

(i) e ∈ V ×δ ,

(ii) there exists E ∈ cog(R) with δ ∈ dom(E ) and e = E (δ),

(iii) there exists E ∈ cog(R) with dom(E ) = ∆max and e = E (δ).

§23] Cogs 305

Proof. The implication (ii) =⇒ (i) follows from Lemma 23.15, and (iii) =⇒ (ii) istrivial. For the proof of (i) =⇒ (iii) we choose F ∈ cog(R) with dom(F ) = ∆max,and define E : ∆max → Idp(V ) by

E (α) =

e if α = δF (α) if α 6= δ

.

Then E (α) ≈ F (α) by (23.14.2), so Lemma 23.20 with E ′ = E ′′ shows thatE ∈ cog(R).

23.22. Lemma. Let (R,E ) be an idempotent Γ -grading of V with dom(E ) =∆, and assume Γ 0 ⊂ ∆ ⊂ Γ .

(a) Let α β in Γ so that in particular α, β ∈ Γ 0 ⊂ ∆. Then

Vα = Vαeβeβ = eαeβVβ. (1)

(b) Let δ → α be a hermitian arrow in Γ and let δ → α ← δ′ be the collisionit generates; thus δ′ = 2α− δ and α ∈ Γ 0 ⊂ ∆ by Proposition 17.7(a). Then

Vδ = QeαVδ′ . (2)

(c) Let α→ ε be an arrow of orthogonal type and let α→ ε← γ be the collisionit generates; thus γ = 2ε− α and α, γ ∈ Γ 0 by Proposition 17.7(b). Then

Vε = eαVεeγ. (3)

(d) Let (α, β, γ, δ) be a square in Γ . Then

Vα = eβ eγ Vδ = eβ Vγ eδ. (4)

(e) Let Γiso = Γher = ∅, so Γ = Γlin ∪ Γorth by (17.5.1). Then V has trivialextreme radical. Hence PE(V ) = FPE(V ), see 8.8, and PE(V ) has trivial centre.

Proof. (a) We have Vα ⊂ V1(eβ) because of (23.8.1) and 〈α, β∨〉 = 1. But

V σ1 (eβ) =x ∈ V σ : eσβ e−σβ x = x

by (6.14.7). Hence x = eσβ e

−σβ x for x ∈ V σα , which proves the first equality in

(1).For the second equality, observe eσα e−σβ V σβ ⊂ V σα by (20.1.1). Since Vα ⊂

V2(eα) ∩ V1(eβ) by (23.8.1), we get xσα = eσα e−σβ eσβ e−σα xσα from (10.12.5) with

f = eα and e = eβ . But eσβ e−σα xσα ∈ V σβ by (20.1.1) again, so that the secondequation in (1) also holds.

(b) From 〈δ, α∨〉 = 〈δ′, α∨〉 = 2 it follows that Vδ ⊂ V2(α) and Vδ′ ⊂ V2(α). Infact, V2(α) = V0⊕ V1⊕ V2 is a Peirce grading with V0 = Vδ, V1 = Vα and V2 = Vδ′ .Since eα ∈ V2(α) is an invertible idempotent, (2) is a special case of (10.3.1).


(c) Let e1 = eα and e2 = eγ . Then α ⊥ γ implies that e1 and e2 areorthogonal idempotents. Let V =

⊕Vij be the Peirce decomposition of V with

respect to e1 and e2 as in (6.16.3). From 〈ε, α∨〉 = 〈ε, γ∨〉 = 1 we concludeVε ⊂ V1(e1) ∩ V1(e2) = V12. Thus, the subpair U = Vα ⊕ Vε ⊕ Vγ of V satisfies theassumptions of Lemma 10.3(b), and (3) is a special case of (10.3.2).

(d) From α = β − γ + δ by (15.3.2) and (20.1.1) we have Vβ Vγ Vδ ⊂ Vα.For the inclusion Vα ⊂ eβ eγ Vδ, let f = eβ and e = eγ . Then e and f arecollinear, and Vα ⊂ V1(f) ∩ V0(e). Let xα ∈ V σα . Then (10.12.5) shows thatxα = fσ e−σ eσ f−σ xα, and zδ := eσ f−σ xα ∈ V σγ V −σβ V σα ⊂ V σδ .

For the inclusion Vα ⊂ eβ Vγ eδ, let xα ∈ V σα and put yγ = e−σβ xα e−σδ ∈

V −σγ . Then by (JP15), (1) and (20.1.3),

eσβ yγ eσδ = eσβ e−σβ xα e−σδ e

σδ

= xα e−σβ eσβ e−σδ eσδ − xα e−σβ eσβ e−σδ eσδ + eσβ e−σδ xα e

−σβ eσδ

= xα e−σβ eσβ − 0 + 0 = xα.

(e) Let z ∈ Extr(V σ), decomposed in its components z =∑α∈Γ zα in the

V σα . By (8.6.1), 0 = zV −σβ V σβ =∑α∈Γ zαV

−σβ V σβ . Here zαV −σβ V σβ ⊂ V σα

by (20.1.1), and the sum of the Vα being direct, we conclude zαV −σβ V σβ = 0 for

all β ∈ Γ . Since Γ = Γlin ∪ Γorth, there exists β ∈ Γ 0 with α β or β → α.Let eβ = E (β). Then zα ∈ V σ1 (eβ), so zα = eσβe

−σβ zα by (6.14.7) and therefore

zα = 0.

23.23. Cogs for rectangular matrices. Let A be a unital associative k-algebra, let a and b be as in 20.2(a), and let V = MIJ(a, b) be as in (20.2.4), withthe root grading R of type Γ = KI KJ defined in (20.2.5). This root grading Ris in general not idempotent.

Indeed, suppose R is compatible with some cog E . Then, by (23.1.1) and(23.8.5), some root space (a·Eij , b·Eji) contains a non-zero Jordan pair idempotente = (aEij , bEji). But this means 0 6= a = aba and 0 6= b = bab, in particularab ∈ a ∩ b is a non-zero idempotent of the algebra A. Hence, if for example a is anil ideal, the root grading is not idempotent.

On the other hand, if a = A = b, then R is idempotent, for example withrespect to the cog E = e(i,j) : i ∈ I, j ∈ J, e(i,j) := (Eij , Eji). Since ∆ = Γ inthis case, the condition (23.8.8) is clearly satisfied: if 〈α, δ∨〉 = 〈β, δ∨〉 for δ = αand δ = β, i.e., 〈α, β∨〉 = 2 = 〈β, α∨〉 then α = β follows. Moreover, one easilyverifies that Vγ ⊂ V〈γ,α∨〉(eα) for all α ∈ ∆, so that also (23.9.1) holds. Hence(V,R) is idempotent by Lemma 23.9.

Let B be the grid basis for KI KJ defined in Example (a) of 18.4. Then asimilar argument shows that R is also idempotent with respect to the smaller cogB = E

∣∣B.

Let in particular Γ = K2 K3, let a1, a2 be proper ideals of A satisfyinga1a2 = 0 = a2a1, and define submodules

U =⊕

16i62, 16j63

U(ij) (1)

§23] Cogs 307

of V = M23(A) by

U(11) = (AE11, AE11) = V(11), U(12) = (AE12, AE21) = V(12),

U(21) = (a2E21, a2E12) ⊂ V(21), U(22) = (a2E22, a2E22) ⊂ V(22),

U(13) = (a1E13, a1E31) ⊂ V(13), U(23) = (0, 0) ⊂ V(23).

Then it is straightforward to check that U is a subpair of V , whence (1) is a Γ -grading S of U . Let ∆ = (1, 1), (1, 2) ⊂ Γ and put F = E |∆. Then F (∆)consists of two collinear idempotents e1 = F (1, 1) and e2 = F (1, 2) whose jointPeirce spaces U[ij] = Ui(e1) ∩ Uj(e2) are

U[21] = U(11), U[12] = U(12), U[11] = U(13),

U[10] = U(21), U[01] = U(22), U[00] = U(23).

This implies that F ∈ cog(S). While U(13), U(21) and U(22) may contain non-zeroidempotents, the condition ai & A prevents F from being extendable to a largercog in cog(S). Hence ∆ = ∆max in the notation of Proposition 23.18. As statedthere in general, ∆ ∼= K2 is a Jordan subgraph of Γ . For Γ×(S) = γ ∈ Γ : Uγ 6= 0we get

∆max & Γ×(S) & Γ, (2)

and Γ×(S) is not a Jordan subgraph of Γ .

23.24. Cogs for hermitian matrices. Let (A, J, ε, Λ) be a form ring as inExample (d) of 6.6 and let HI(A, J, ε, Λ) be the Jordan pair of Λ-hermitian matrices

of size I defined in 20.2(b). Let Γ = TI and consider the Γ -grading defined by(20.2.15) and (20.2.16).

We now exhibit cogs defined on Γ 0 = TI compatible with the Γ -grading. Fori 6= j let u ∈ A× be a unit of A and let fij be as in (20.2.13). Then

fij(u) :=(f+ij (u), f−ij (u−1)

)∈ Vij

is an idempotent of V . This follows easily from the fact that, for all a, b ∈ A,

f+ij (a)f−ij (b) = abEii + (ba)JEjj , f−ij (b)f+

ij (a) = baEjj + (ab)JEii.

We obtain a cog E defined on Γ 0 by choosing a total order on I and putting

E (i, j) = fij(u) for i < j.

By (20.2.14), these idempotents satisfy fij(u) = fji(−εu−1) and will therefore besymmetric in i and j if and only if u2 = −ε, for example if ε = −1 and u = 1. Ingeneral, different choices of u and of the order on I will yield different cogs.

Let |I| > 3. Then a straightforward verification, using Lemma 23.9(a), showsthat (R,E ) is an idempotent root grading.


23.25. Cogs for Jordan pairs of quadratic forms. Let K be a commu-tative ring, let M be a K-module with a quadratic form q: M → K and letV = J(M, q) be the Jordan pair defined in Example (f) of 6.6, with quadraticoperators Qxy = b(x, y)x − q(x)y, where b is the polar form of q. There are twomain examples of idempotents in V . First, let (e+, e−) ∈ M ×M be a hyperbolicpair, thus q(e+) = q(e−) = 0 and b(e+, e−) = 1. Then it is immediate that (e+, e−)and (e−, e+) are idempotents of V . Second, if x ∈M satisfies q(x) = u ∈ K× then(x, xu−1) is an idempotent of V as well.

We now construct quadratic forms whose associated Jordan pairs have rootgradings of type OI and OI . Let Ph = K · e+⊕K · e− be the free hyperbolic planeover K, with quadratic form q(λe+ + µe−) = λµ. Let I be a non-empty (possiblyinfinite) index set, and let Mh be the orthogonal sum of I copies of Ph. Thus Mh

is the free K-module with basis e+i , e−i : i ∈ I, and the quadratic form

qh

(∑i∈I

λie+i + µie

−i

)=∑i∈I

λiµi. (1)

Recall from 14.19 that the graph OI has vertex set I × +,− and is simply lacedwith edges (i, σ) (j, τ) ⇐⇒ i 6= j. Then V = J(Mh, qh) has a root grading Rof type OI given by

V σ(i,+) = K · eσi , V σ(i,−) = K · e−σi . (2)

Define a cog E : OI → Idp(V ) by

E ((i,+)) = (e+i , e−i ), E ((i,−)) = (e−i , e

+i ). (3)

Using Lemma 23.9(a), it is easily checked that E is compatible with R.

Let OI = OI ∪ ω as in 14.19. Recall that α → ω for all α ∈ OI and that

OI is the induced subgraph of OI . We construct examples of root gradings of typeOI as follows. Let (M0, q0) be an arbitrary quadratic form on a K-module M0, letM = M0 ⊕Mh with the quadratic form q = q0 ⊥ qh (orthogonal sum), and let

W = J(M, q). Then W has a OI -grading given by

Wσω = M0, Wσ

α = V σα for α ∈ OI , (4)

which is still idempotent with respect to the cog E defined in (23.10.1). If M0

contains an element x0 with q0(x0) = 1 then e0 = (x0, x0) is an idempotent of

J(M0, q0) and hence of W , and the cog E can be extended to a cog E defined on

all of OI by E (ω) = e0.

23.26. Proposition. Let V be a Jordan pair over k with an idempotent rootgrading of type Γ = OI , |I|>3, which is idempotent with respect to a cog defined on∆ ⊃ OI . Then there exists K ∈ k-alg and a quadratic form q0: M0 → K such thatV is isomorphic as a Γ -graded Jordan pair to the base ring restriction kW whereW is as in 23.25.

Proof. This is shown in [76, III, Theorem 2.8] in the setting of Jordan triplesystems. The translation to Jordan pairs is immediate and is left to the reader, see[79, §5].

§24] Weyl elements for idempotent root gradings 309

§24. Weyl elements for idempotent root gradings

24.1. Definition. Let V be a Jordan pair, let R = (Vγ)γ∈Γ be a root grading oftype Γ , let E ∈ cog(R) be a cog compatible with R as 23.8, defined on∆ = dom(E ).

Let (R,R1) be the 3-graded root system associated with Γ as in Theorem 15.11,thus R1

∼= Γ . Consider a group G ∈ st(V,R) with R-commutator relations androot groups (U%)%∈R defined in 21.1. As in 23.1, we write E (α) = eα for α ∈ ∆, sothat eα ∈ Vα by (23.8.5). Let

wα := weα = x−(e−α ) x+(e+α ) x−(e−α ) ∈ U−α Uα U−α (1)

be defined as in 9.11.Recall from 5.1 the notion of a Weyl element for a group (G, (U%)%∈R) with

R-commutator relations: an element w ∈ G is a Weyl element for a root α ∈ R ifw ∈ U−α Uα U−α and

wUβ w−1 = Usα(β) for all β ∈ R. (2)

Although wα as defined above clearly satisfies the first requirement, this will ingeneral not be the so for (2). Therefore, we define, in analogy to 12.1, a fullsubcategory of st(V,R) by

G ∈ st(V,R,E ) :⇐⇒G ∈ st(V,R) and wα is a Weyl elementfor (G, (U%)%∈R

), for all α ∈ ∆.

(3)

We will show in Corollary 24.8 that the projective elementary group PE(V )belongs to st(V,R,E ); in particular, this category is not empty.

Similarly to the definitions given in 13.1 and 22.1, one sees that the categoryst(V,R,E ) has an initial object

St(V,R,E ), (4)

called the Steinberg group of (V,R,E ).

Let us discuss these definitions for the simplest cases of idempotent root gradingslisted in 23.10.

(a) Γ = α a singleton and ∆ = ∅. Then V = Vα is simply a Jordanpair with no further structure, and there are no Weyl elements to consider, sost(V,R,E ) = st(V,R) = st(V ) and St(V,R,E ) = Fr(V ).

(b) Γ = α = ∆. Then V = V2(eα) where e = eα is an invertible idempotent,and wα is a Weyl element if and only if the relations W(e) of 9.17 hold in G,so st(V,R,E ) = st(V, e) as in 12.1, in the special case where V = V2(e). HenceSt(V,R,E ) = St(V, e) as in 13.1.

(c) Γ = α→ β ← γ a collision and ∆ = α. By 23.10(c), E ∈ cog(R) if andonly if R = P(e) is the Peirce grading induced by a non-zero idempotent e = eα.After identifying α → β ← γ with 2ε1 → ε1 + ε0 ← 2ε0, we see from (12.1.2) thatG ∈ st(V,R,E ) if and only if G ∈ st(V, e), and hence St(V,R,E ) = St(V, e).

The main result of this section is the following Theorem 24.2, the analogue ofTheorem 12.5, which characterizes these groups by weaker conditions. Since thecase where the root system is of type A1 or C2 has been dealt with in 12.5, we nowexclude the cases where R contains a connected component of these types.


24.2. Theorem. Let Γ be a Jordan graph with Γiso = ∅. Let V be a Jordanpair with a Γ -grading R = (Vγ)γ∈Γ , let E ∈ cog(R) be a cog compatible with Ras in 23.8, defined on dom(E ) = ∆ and assume Γ 0 ⊂ ∆. For a group G ∈ st(V )consider the following conditions:

(i) G ∈ st(V,R,E ) (as defined in (24.1.3)),

(ii) G ∈ st(V, eδ) (as defined in (12.1.2)), for all δ ∈ ∆,

(iii) G ∈ st(V, eδ) for all δ ∈ Γ 0,

(iv) G ∈ st(V,R).

Then (i) ⇐⇒ (ii) ⇐⇒ (iii) =⇒ (iv). If Γ contains no connected component

of type T3 then all four conditions are equivalent.

We leave it to the reader to formulate the corresponding statements for therespective Steinberg groups.

Remarks. Let (R,R1) be the 3-graded root system associated with Γ . ThenΓiso = ∅ if and only if R has no irreducible component of type A1 or C2, and theabsence of components of type T3 means R contains no irreducible component oftype C3.

At first glance, the implication (iv) =⇒ (i) (under the additional assumption

on the absence of irreducible components of type T3) might seem too good to betrue: how can the mere fact that G belongs to st(V,R) imply the seemingly muchstronger condition (i)? There are two remarks to be made.

First, the root grading R is not arbitrary, but must admit a cog E making R anidempotent root grading. This is a strong restriction on R; for example, the rootgrading considered in 21.17 is not always of this type, as shown in 23.23.

Second, the assumptions Γiso = ∅ and Γ 0 ⊂ ∆ are essential. In the anal-ogous Theorem 12.5, the root system is of type C2, and there the conditionG ∈ st(V,P(e)) is not sufficient to guarantee G ∈ st(V, e), as Example 12.6 shows.

For Γ ∼= TI , |I|>2, a description of st(V,R,E ) in case ∆ = Γ is given in (24.6.15).

The proof of Theorem 24.2 will occupy most of this section. In the course of theproof, a number of partial results under weaker assumptions occur naturally whichmay be of independent interest, for example, Propositions 24.6, 24.10 and 24.13.The final step of the proof is presented in 24.14.

24.3. Notations and assumptions. The following conventions will be inforce throughout this section:

V is a Jordan pair, Γ is a Jordan graph with associated 3-graded root system(R,R1), R = (Vγ)γ∈Γ is a Γ -grading of V , and E ∈ cog(R) is a cog, compatiblewith R and defined on ∆ = dom(E ). By G we denote a group over V , and for all% ∈ R, the subgroups U% of G are defined as in 21.1. Unless mentioned otherwise,we do not assume G ∈ st(V,R), so the R-commutator relations for the U% need nothold. For all α ∈ ∆, the elements wα ∈ G are defined as in (24.1.1).

We also recall the definition

st(V,E ) :=⋂δ∈∆

st(V, eδ) (1)


from 23.5, and the notationInt(a) · b = aba−1

for the inner automorphism of a group G defined by a ∈ G.

24.4. Lemma. Let G ∈ st(V,E ). Then for all α ∈ ∆, β ∈ Γ and σ = ±,

Int(wα) · Uσβ = Uσsα(β), (1)

and w2α normalizes Uσβ. In more detail, the action of Int(wα) on U±β = x±(V ±β )

is given as follows.

(a) If 〈β, α∨〉 = 0 then sα(β) = β and for all t ∈ V σβ ,

Int(wα) · xσ(t) = xσ(t). (2)

(b) If 〈β, α∨〉 = 1 then sα(β) = β − α =: µ ∈ R×0 . For all u ∈ V +β , v ∈ V −β ,

Int(wα) · x+(u) = b(−u, e−α ), Int(wα) · x−(v)· = b(e+α , v), (3)

and thereforeUµ = b(V +

β , e−α ), U−µ = b(e+

α , V−β ). (4)

(c) If 〈β, α∨〉 = 2 then sα(β) = β − 2α ∈ −Γ = R−1 and for all t ∈ V σβ ,

Int(wα) · xσ(t) = x−σ(Q(e−σα )t). (5)

Finally, Int(w2α) acts on u ∈ Uσβ by

Int(w2α) · xσ(u) = xσ((−1)〈β,α

∨〉u). (6)

Proof. By (23.8.1), Vβ ⊂ V〈β,α∨〉(eα) = Vi(eα) and by (12.1.2), st(V, eα) ⊂st(V,P(eα)). In particular, Lemma 12.2 is applicable to e = eα.

(a) Here Vβ ⊂ V0(eα), so (2) and hence (1) follows from (12.2.1).

(b) Recall the definition of Uµ in (21.1.2):

Uµ =⟨

b(V +γ , V

−δ ) : γ − δ = µ, γ, δ ∈ Γ

⟩. (7)

We prove the first equality of (4). Since eα ∈ Vα by (23.8.5), it is clear thatb(V +

β , e−α ) ⊂ Uµ. To prove the reverse inclusion, let µ = γ − δ for γ, δ ∈ Γ . Then

〈γ, α∨〉 − 〈δ, α∨〉 = 〈γ − δ, α∨〉 = 〈β − α, α∨〉 = 1− 2 = −1.

Hence j := 〈γ, α∨〉 = 〈δ, α∨〉 − 1 satisfies 0 6 j 6 1, and V +γ ⊂ V +

j (eα) and

V −δ ⊂ V−j+1(eα) by (23.8.1). Since G ∈ st(V, eα), Theorem 12.5 shows that the shift

relations S01 and S12 hold for eα. They imply b(V +γ , V

−δ ) = b(V +

γ V −δ e+α, e−α ),

and V +γ V −δ e+

α ⊂ V +µ+α = V +

β by (20.1.1). This completes the proof of Uµ =

b(V +β , e

−α ). The analogous proof of the second equation in (4) is left to the reader.


Since Vβ ⊂ V1(eα), the formulas in (3) follow from (12.2.2) and (12.2.4), and (4)and (3) together show that (1) holds. Finally,

Int(w2α) · x+(u) = wα · b(−u, e−α ) · wα = x+(−u)

(by (12.2.3)) shows that w2α normalizes Uβ , and in the same way it follows from

(12.2.5) that w2α normalizes U−β .

(c) Here we have Vβ ⊂ V2(eα). Since G ∈ st(V, eα), Theorem 12.5(ii) impliesthat the relation W(eα) holds, so (9.16.1) shows that (5) holds for all t ∈ V σ2 (eα),in particular, for all t ∈ V σβ . To prove (1), let β′ = 2α − β = −sα(β) ∈ Γ . Then

Uσsα(β) = x−σ(V −σβ′ ), so (1) is equivalent to wα · xσ(V σβ ) · w−1α = x−σ(V −σβ′ ). In

view of (5), the left hand side is x−σ(Q(e−σα )V σβ ), which reduces us to showing

Q(e−σα )V σβ = V −σβ′ . (8)

From (20.1.1) we have

Q(e−σα )V σβ ⊂ V −σ2α−β = V −σβ′ , Q(eσα)V −σβ′ ⊂ Vσ2α−β′ = V σβ . (9)

Observe that 〈β′, α∨〉 = 〈2α − β, α∨〉 = 4 − 2 = 2, and therefore Vβ′ ⊂ V2(eα) aswell. By (6.14.7), Q(e−σα )Q(eσα) acts like the identity on V −σ2 (eα) and hence also onV −σβ′ . We apply Q(e−σα ) to the second inclusion of (9) and obtain V −σβ′ ⊂ Q(e−σα )V σβ .Together with the first inclusion of (9) this proves (8). Finally,

Int(wα)2 · xσ(t) = xσ(Q(eσα)Q(e−σα )t) = xσ(t)

for all t ∈ V σβ ⊂ V σ2 (eα) shows that w2α centralizes Uσβ .

Remark. The observant reader will have noticed that the proof of this lemmadoes not use the full force of the assumptions. In fact, the lemma holds, with thesame proof, for a Jordan pair with a root grading R of type Γ , a single α ∈ Γand an idempotent eα ∈ Vα satisfying Vγ ⊂ V〈γ,α∨〉(eα) for all γ ∈ Γ , and a groupG ∈ st(V, eα).

24.5. Lemma. If µ ∈ R0 satisfies 〈µ,∆∨〉 = 0 then Uµ = 1.

Proof. We may assume µ 6= 0. Then µ = α − β where α, β ∈ Γ are notseparated by ∆. Hence it follows from (23.9.2) that Vα = Vβ = 0, and thereforeb(V +

α , V−β ) = 1 by (9.7.4). This implies Uµ = 1 by (21.1.2).

24.6. Proposition. We have

st(V,R,E ) = st(V,R) ∩ st(V,E ). (1)

Proof. Since we are dealing with full subcategories, it suffices to prove

G ∈ st(V,R,E ) ⇐⇒ G ∈ st(V,R) ∩ st(V,E ).

“=⇒”: Let G ∈ st(V,R,E ). Then G ∈ st(V,R) by definition, and Corol-lary 21.5 shows that G ∈ st(V,Rα) for all α ∈ Γ . By Theorem 12.5 it remains toshow that W(eα) holds for the Cher

2 -grading defined by α ∈ ∆, that is,


wα · U±2ε1 · w−1α = U∓2ε1 , (2)

where now U±2ε1 = x±(V ±2 (eα)). By (23.8.7), V2(eα) =⊕

β∈Γ2(α) Vβ which implies

U±2ε1 =⊕

β∈Γ2(α)

U±β , (3)

the restricted direct product of the U±β . If β ∈ Γ2(α) then 〈−sα(β), α∨〉 =〈2α − β, α∨〉 = 4 − 2 = 2, so the map β 7→ f(β) := −sα(β) is a bijection of Γ2(α)onto itself, equal to its own inverse. Since wα is a Weyl element for (G, (U%)%∈R),

wα · U±β · w−1α = Usα(±β) = U∓f(β)

for all β ∈ Γ2(α). Together with (3) this implies (2).

“⇐=”: It remains to show that, for all δ ∈ ∆,

wδ · U% · w−1δ = Usδ(%) for all % ∈ R. (4)

The case % ∈ R±1 having been dealt with in Lemma 24.4, we now prove

wδ · Uµ · w−1δ = Usδ(µ) for all µ ∈ R×0 . (5)

By Lemma 15.10 the possible values of 〈µ, δ∨〉 are 0 and ±1. Assume first 〈µ, δ∨〉 =±1. Then τ := sδ(µ) = µ∓δ ∈ R∓1. By Lemma 24.4 w2

δ normalizes Uτ , so we havew−1δ ·Uτ ·wδ = wδ ·Uτ ·w−1

δ = Usδ(τ) = Uµ and hence wδ ·Uµ ·w−1δ = Uτ = Usδ(µ),

as desired. Therefore, we assume 〈µ, δ∨〉 = 0 from now on. Then sδ(µ) = µ, andwe must show:

wδ · Uµ · w−1δ = Uµ. (6)

If 〈µ,∆∨〉 = 0 then (6) follows from Lemma 24.5. In the following we maytherefore assume that there exists α ∈ ∆ such that 〈µ, α∨〉 6= 0 and therefore〈µ, α∨〉 = ±1. We will do the case 〈µ, α∨〉 = −1 and leave the other case to thereader.

Let β := sα(µ) = µ+ α ∈ Γ . Then

〈β, α∨〉 = 〈µ+ α, α∨〉 = −1 + 2 = 1, (7)

sα(β) = s2α(µ) = µ and by Lemma 24.4,

Uµ = b(V +β , e

−α ) = wα · Uβ · w−1

α = w−1α · Uβ · wα. (8)

Also, 0 = 〈µ, δ∨〉 = 〈β−α, δ∨〉 implies 〈β, δ∨〉 = 〈α, δ∨〉 and this can take the values0, 1, 2. Accordingly, we distinguish the following three cases.

Case 〈β, δ∨〉 = 〈α, δ∨〉 = 0. Then Vα + Vβ ⊂ V0(eδ) by (23.8.1). Since G ∈st(V, eδ) ⊂ st(V,P(eδ)), Theorem 11.2(iii) shows

(((((((wδ,wα

)))))))=(((((((

wδ, Uβ)))))))

= 1. Hence(8) implies

Int(wδ) · Uµ = Int(wδ) Int(wα) · Uβ = Int(wα) Int(wδ) · Uβ = Int(wα) · Uβ = Uµ,


as desired.

Case 〈β, δ∨〉 = 1 = 〈α, δ∨〉. The first equation implies β δ or β ← δ. Sincealso 〈β, α∨〉 = 1, the same holds for the pairs (α, δ) and (β, α). By Lemma 15.6,the induced subgraph on α, β, δ is one of the following:

(a)δ

????

α β(b)

α

???? δ

β(c)

δ

????

α β

Using (8) we rewrite (6) in the form

Int(wδwαw−1δ ) · Int(wδ) · Uβ = Int(wα) · Uβ . (9)

Put e = eδ and f = eα. Then 〈α, δ∨〉 = 1 implies f ∈ V1(e), so Lemma 13.10 isapplicable. Also, 〈β, δ∨〉 = 1 implies, by Lemma 24.4, that

Int(wδ) · Uβ = b(V +β , e

−δ ) = b(V +

β , e−),

and by (8) we have Int(wα) · Uβ = b(V +β , e

−α ) = b(V +

β , f−). Finally, wδ wα w−1δ =

wf,e in the notation of Lemma 13.10. Hence (9) is equivalent to

Int(wf,e) · b(V +β , e−) = b(V +

β , f−).

From (13.10.4) and (9.7.6) it follows that the left hand side is b(ωf,e ·V +β , ωf,e ·e−),

so that finally (6) is equivalent to

b(ωf,e · V +β , ωf,e · e−) = b(V +

β , f−). (10)

For the proof, we will use Propositions 10.12 und 10.13. Let V(ij) = Vi(f) ∩ Vj(e)as in 10.11. Then

Vβ ⊂ V〈β,α∨〉(f) ∩ V〈β,δ∨〉(e) = V1(f) ∩ V1(e) = V(11). (11)

We first treat the cases (a) and (b). Then 〈δ, α∨〉 = 1 implies e = eδ ∈V〈δ,α∨〉(eα) = V1(f), so that e and f are collinear idempotents, and the actionof ωf,e is given in Proposition 10.12. In particular, ωf,e · e− = f− e+ e− = f−since f ∈ V1(e), and therefore (10) is equivalent to ωf,e · V +

β = V +β . By (11),

Proposition 10.12 shows that, for all uβ ∈ V +β ,

ωf,e · uβ = uβ − f+ e− e+ f− uβ.

By (20.1.1),

f+ e− e+ f− uβ = e+α e−δ e

+δ e−α uβ ∈ V +

α−δ+δ−α+β = V +β ,

so that ωf,e · uβ ∈ V +β . Again by Proposition 10.12, ωf,e maps V +

(11) bijectively to

itself and is its own inverse. Hence ωf,e · V +β = V +

β , as desired.

Now assume we are in case (c). Then δ, α, β generate a kite (δ, α, γ, β) by (C4):


δ

????

α

???? β

γ

In order to work out the left hand side of (10), we observe first that α← δ impliesf ` e. To preserve the notation of Proposition 10.13 let us write eα = g instead off . By (11) we still have Vβ ⊂ V1(g) ∩ V1(e) = V(11). Furthermore,

e = eδ ∈ V〈δ,α∨〉(eα) ∩ V〈δ,δ∨〉(eδ) = V2(g) ∩ V2(e) = V(22).

By (15.3.2), α − δ + β = γ. Hence Proposition 10.13 shows that for an elementuβ ∈ V +

β ⊂ V+(11),

ωg,e · uβ = Qg+e−, g−, uβ ∈ V +2α−δ−α+β = V +

γ ⊂ V +(10),

ωg,e · e−δ = Qg−Qe+e− = Qe−α e+δ ∈ V

−2α−δ ⊂ V

−(20).

Put zγ := ωg,e · uβ ∈ V +γ . Then the left hand side of (10) is

b(ωg,e · uβ , ωg,e · e−δ ) = b(zγ , Q(e−α )e+

δ

). (12)

With respect to g = eα we have Vγ ⊂ V〈γ,α∨〉(eα) = V1(g). Hence the right handside of (12) is of the form b(x1, Qy2z2) with respect to g. Since G ∈ st(V, eα), theshift formula S′12 of 12.3 is applicable and yields

b(zγ , Qe−α e+δ ) = b

(zγ e−α e+

δ , e−α ),

where zγ e−α e+δ ∈ V

+γ−α+δ = V +

β . As zγ ∈ V +(10), we have

zγ e−α e+δ = e+ g− zγ = −ωg,e · zγ = −ω2

g,e · uβ = uβ

by (10.13.2). Altogether, we have shown that

b(ωg,e · uβ , ωg,e · e−) = b(zγ , Q(e−α )e+δ ) = b(uβ , g−)

which proves (10).

Case 〈β, δ∨〉 = 〈α, δ∨〉 = 2. Thus β = δ or β → δ, and α = δ or α → δ. As〈β, α∨〉 = 1, Lemma 15.6 shows that the induced subgraph on α, β, δ is

α // β = δ orα

???? β

δ

Using (8), we rewrite (6) in the form

Int(wα wδ w−1α ) · Uβ = Uβ . (13)

Let us put e := eα and g := eδ. Then the relation δ ← α implies g ` e, in particular,g ∈ V1(e). Hence by Lemma 13.10, (13) is equivalent to

ωg,e · V +β = V +

β . (14)

The proof follows the pattern of the previous cases. First,

Vβ ⊂ V〈β,δ∨〉(eδ) ∩ V〈β,α∨〉(eα) = V2(g) ∩ V1(e) = V(21).

For uβ ∈ V σβ ⊂ V σ(21) we have by Proposition 10.13,

ωg,e · uβ = eσα, Q(e−σδ ) · uβ , Q(eσδ )e−σα ∈ V σα−(2δ−β)+2δ−α = V σβ ,

and ωg,e maps V(21) bijectively onto itself and is its own inverse, again by 10.13.This proves (14).


Remarks. (a) The attentive reader may have noticed that the proof of thedirection from right to left above in fact shows that (4) holds for a group G ∈st(V,E ) which not necessarily satisfies the R-commutator relations.

(b) In the setting of Example 23.10(d) it follows from (1) that

st(V,R,E ) = st(V,R,O) (15)

since st(V,E ) = st(V,O) by (23.10.1).

24.7. Corollary. Let E ′ ∈ cog(R) with dom(E ′) = dom(E ) = ∆, and suppose

e′ξ = eξ for all isolated vertices ξ ∈ ∆. (1)

Then

st(V,E ) = st(V,E ′) and st(V,R,E ) = st(V,R,E ′). (2)

Proof. By Lemma 23.15(a), E and E ′ are associated. By Proposition 24.6 itsuffices to show the first equation in (2). Let ∆1 be the union of the isolated verticesof ∆, let ∆2 = ∆ ∆1 and denote by Ei and E ′i the restrictions of E and E ′ to ∆i,respectively. Then (1) implies st(V,E1) = st(V,E ′1), and Proposition 23.6 applies toE2 and E ′2, and yields st(V,E2) = st(V,E ′2). Hence st(V,E ) = st(V,E1)∩st(V,E2) =st(V,E ′1) ∩ st(V,E ′2) = st(V,E ′).

24.8. Corollary. The projective elementary group PE(V ) of a Jordan pair Vwith an idempotent root grading (R,E ) belongs to st(V,R,E ).

Proof. We know PE(V ) ∈ st(V,R) by 21.12 and PE(V ) ∈ st(V, e) for allidempotents e ∈ V by 12.8. Thus PE(V ) ∈ st(V,E ) by (24.3.1), so that PE(V ) ∈st(V,R,E ) follows from 24.6.

24.9. Corollary. Let ϕ: G→ G′ be a morphism of groups in st(V,R,E ). Forδ ∈ ∆, let wδ ∈ G and w′δ ∈ G′ be the Weyl elements defined as in (24.1.1). Then:

(a) ϕ(wδ) = w′δ for all δ ∈ ∆.

(b) ϕ: U% → U ′% is bijective, for all % ∈ R. Hence G ∈ stbij(V,R) as defined in(21.21.1), and therefore st(V,R,E ) ⊂ stbij(V,R).

(c) G has unique factorization.

(d) If A ⊂ R is a nilpotent subset and UA =⟨Uα : α ∈ A

⟩then ϕ: UA → U ′A is

bijective.

Proof. (a) This is clear from the definition (24.1.1) and the fact that ϕ(xσ(v)) =x′σ(v), for all v ∈ V σ.

(b) We know from 21.1 that ϕ(U%) = U ′% for all % ∈ R and that ϕ∣∣U% is bijective

for % ∈ R1∪R−1. Let µ ∈ R×0 . By Lemma 24.5, we may assume that there exists δ ∈∆ with 〈µ, δ∨〉 6= 0, thus c := 〈µ, δ∨〉 = ±1. Then % := sδ(µ) = µ− cδ ∈ R1 ∪R−1.Since wδ and w′δ are Weyl elements for G and G′, we have Int(wδ) · Uµ = U% andInt(w′δ) · U ′µ = U ′%. Hence the diagram


Uµ∼= //

ϕ∣∣Uµ

U%

ϕ∣∣U%

U ′µ ∼=// U ′%

is commutative, where the horizontal arrows are given by conjugation with wδ andw′δ, respectively. Since ϕ

∣∣U% is an isomorphism, so is ϕ∣∣Uµ.

To prove the second statement, let G′ = PE(V ) and ϕ = π. This is possible byCorollary 24.8. Then π is bijective on all root groups, whence G ∈ stbij(V,R).

(c) and (d): Since st(V,R,E ) ⊂ stbij(V,R), these are special cases of (21.21.3)and Corollary 21.22 respectively.

24.10. Proposition. Let ∆ satisfy the following condition: for all α, β ∈ Γ ,

if α ⊥ β then there exists δ ∈ ∆ such that α ∼ δ ⊥ β or α ⊥ δ ∼ β. (1)

Then st(V,E ) ⊂ st(V,R) and hence, by Proposition 24.6, st(V,R,E ) = st(V,E ).

Proof. Let G ∈ st(V,E ), so by definition, G ∈ st(V, eγ) for all γ ∈ ∆. To proveG ∈ st(V,R) we apply Theorem 21.7(iv) and show that for all α, β in Γ and allxα ∈ V +

α and yβ ∈ V −β the relations

B(xα, yβ) if α ∼ β 6= α, (2)

b(xα, yβ) = 1 if α ⊥ β (3)

hold in G.

Let α ∼ β 6= α in Γ . Then µ := α − β ∈ R×0 by 15.9. If 〈µ,∆∨〉 = 0 thenUµ = 1 by Lemma 24.5, so b(xα, yβ) ∈ Uµ = 1 and B(xα, yβ) holds trivially.Otherwise, there exists an element γ ∈ ∆ such that i := 〈α, γ∨〉 6= 〈β, γ∨〉 =: j.Then 〈α− β, γ∨〉 = i− j 6= 0 together with Lemma 15.10 shows |i− j| = 1.

Let e = eγ ∈ Vγ and let P(e) be the Peirce grading defined by e. Then G ∈st(V, e) ⊂ st(V,P(e)) by (12.1.2). We have Vα =

⋂δ∈∆ V〈α,δ∨〉(eδ) ⊂ V〈α,γ∨〉(eγ) =

Vi(e) and in the same way, Vβ ⊂ Vj(e). Hence (2) holds by Theorem 11.2.

Now let α ⊥ β in Γ and choose δ ∈ ∆ as in (1). If α ∼ δ ⊥ β, let e = eδ ∈ Vδ bethe corresponding idempotent. Then i := 〈α, δ∨〉 ∈ 1, 2 and 〈β, δ∨〉 = 0. If i = 2then Vα ⊂ V2(e) while Vβ ⊂ V0(e), so b(V +

α , V−β ) ⊂ b(V +

2 , V −0 ) = 1 follows fromTheorem 11.2. If i = 1 then Vα ⊂ V1(e) (and still Vβ ⊂ V0(e)). Since G ∈ st(V, e),Theorem 12.5(iii) shows that all shift formulas of 12.3 hold. Let x1 ∈ V +

1 (e) andy0 ∈ V −0 (e). By the shift formula S10 we have

b(e+, e− x1 y0

)= b(x1, y0). (4)

In particular, let xα ∈ V +α ⊂ V +

1 (e) and yβ ∈ V −β ⊂ V −0 (e). Then e− xα yβ = 0by (20.1.3), so (3) follows from (4). Finally, if α ⊥ δ ∼ β, then a similar proof,using the shift formula S01, shows that (3) holds.


24.11. Example. Condition (24.10.1) is not fulfilled in general. For example,let V = H3(k) be the Jordan pair of symmetric 3×3 matrices over k as in 6.6(c). It

has a canonical root grading R of type Γ = T3, see 20.2(b). Thus Γ is a hexagram:

11

????

12 13

22

??// 23

???? 33

__????oo

=

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

where ij stands for i, j. Let E be the cog defined on Γ 0 = α, β, γ as in23.24 for the form ring (A = k, J = Id, ε = −1, Λ = k) and u = 1, thusE (i, j) = (Eij + Eji, Eij + Eji), and let ∆ = α, γ. We claim that R is anidempotent root grading with respect to the cog D = E

∣∣∆. This will follow fromLemma 23.9 once we have verified the assumptions there. The first of them is(23.8.8):

〈µ,∆∨〉 6= 0 for all µ ∈ R×0 . (1)

Indeed, let Γi(α) = ζ ∈ Γ : 〈ζ, α∨〉 = i as in 16.7; explicitly:

Γ2(α) = δ, α, δ′, Γ1(α) = β, γ, Γ0(α) = δ′′.

Observe that the subsets Γi(α) are separated by α in the sense that 〈ζ, α∨〉 6= 〈η, α∨〉for ζ ∈ Γi(α), η ∈ Γj(α) and i 6= j. Similarly, Γ2(α) and Γ1(α) are separated by γ.This implies (1). It is easily checked that D satisfies (23.9.1). Hence (R,D) is anidempotent root grading of V . But ∆ does not satisfy (24.10.1) for the orthogonalroots δ and β, so the proof above for b(V +

β , V−δ ) = 1 is not applicable.

On the other hand, (24.10.1) holds for E , but also for the cog B defined onB := δ, α, γ by B(δ) = (E11, E11) and B|∆ = D . In the first case, the validityof (24.10.1) is a general fact and follows from Lemma 24.12 below.

24.12. Lemma. Let Γiso = ∅ and Γ 0 ⊂ ∆. Then ∆ satisfies (24.10.1), andhence st(V,R,E ) = st(V,E ).

Proof. If α ∈ Γ 0 or β ∈ Γ 0 it suffices to put δ = α resp. δ = β. Thus we nowassume α, β ∈ ∂Γ . By Proposition 17.7, there are the following cases.

Case 1: α is the initial point of an arrow α → ξ of hermitian type, so ξ ∈ Γ 0.If ξ ⊥ β we put γ = ξ. If ξ ∼ β then ξ → β is impossible by (15.6.5) and ξ βwould imply β ∈ Γ 0. Hence we have α→ ξ ← β. Since α→ ξ is a hermitian arrow,it embeds by (17.2.2) into a hexagram

α

????

ξ γ

β

??//

????

__????oo

Case 2: α is the end point of an arrow γ → α of orthogonal type, so γ ∈ Γ 0. Wethen have γ ⊥ β. Indeed, γ β would imply β ∈ Γ 0 and β → γ → α contradicts(15.6.5). If γ → β then it follows from Lemma 15.6 that α β, contradictingα ⊥ β.


Remark. As we have already seen in 24.11, the conditions Γiso = ∅ and Γ 0 ⊂ ∆are by no means necessary for ∆ ⊂ Γ to satisfy (24.10.1) and (24.11.1). Forexample, any grid basis ∆ ⊂ Γ as defined in 18.3 has these properties. The setδ, α, γ used in 24.11 is a grid basis by 18.4(c). The details are left to the reader.

24.13. Proposition. Assume that Γiso = ∅ and Γ 0 ⊂ ∆ and put E 0 = E∣∣Γ 0.

Thenst(V,E 0) = st(V,E ) ⊂ st(V,R).

Proof. Clearly, a group G ∈ st(V,E ) belongs to st(V,E 0). Conversely, letG ∈ st(V,E 0). By Lemma 24.12, Γ 0 satisfies (24.10.1), so by Proposition 24.10,st(V,E 0) ⊂ st(V,R). In particular, G satisfies the R-commutator relations.

Let ξ ∈ ∆ Γ 0. Then G ∈ st(V,P(eξ)) by Corollary 21.5. Hence, to proveG ∈ st(V, eξ), it remains by Theorem 12.5 to show that the Weyl relations W(eξ)hold. Now the vertex ξ ∈ ∂Γ is external, and since Γ has no isolated vertices orarrows, ξ is either of hermitian or orthogonal type (by (17.7.1)), so either ξ = δ fora hermitian arrow δ → α, or ξ = ε for an arrow α→ ε of orthogonal type.

Case (a): δ → α is a hermitian arrow, embedded in a hexagram (17.2.2), whereα, β, γ ∈ Γ 0. The Weyl relation W(eδ) says

weδ · xσ(z) · w−1eδ

= x−σ(Q(e−σδ )z), (1)

for all z ∈ V σ2 (eδ). We claim that V2(eδ) = Vδ. Indeed, by (23.8.7), V2(eδ) =V2(δ) =

∑ε∈Γ2(δ) Vε, and by 16.7, Γ2(δ) consists of δ and all initial points of arrows

λ → δ. Since δ → α, (15.6.5) shows there are no such arrows, so Γ2(δ) = δ andtherefore Vδ = V2(δ) = V2(eδ).

We now show (1) for σ = + and leave the analogous proof for σ = − to thereader. Since δ → γ is a hermitian arrow as well, (23.22.2) shows that V2(δ) =QeγVδ′′ . Thus (1) is equivalent to

weδ · x+(Qe+γ yδ′′) · w−1eδ

= x−(Qe−δQe+γ yδ′′), (2)

for all yδ′′ ∈ V −δ′′ . Here eγ ∈ V1(eδ) and yδ′′ ∈ V0(eδ). Hence (2) can be reduced asfollows:

(2) ⇐⇒ d(eδ)+ (Qe+γ yδ′′) = 1 (by (12.1.7))

⇐⇒ zeδ(e+γ , yδ′′) = 1 (by (12.2.10))

⇐⇒ b(e+γ , yδ′′) = b(e+

δ , e−δ e

+γ yδ′′) (by (12.1.5)). (3)

The proof of (3) is now obtained by conjugating both sides with weα and using theshift relations S′′10 and S′′21 of 12.3 for eα (which is admissible by Theorem 12.5because G ∈ st(V, eα)). On the left, we have

weα · b(e+γ , yδ′′) · w−1

eα = x−(−e−α , e+γ , yδ′′),

since e+γ ∈ V +

1 (eα) and yδ′′ ∈ V −0 (eα). On the right, we use S′′21 and obtain, since

eδ ∈ V2(eα) and e−δ , e+γ , yδ′′ ∈ V −δ−γ+δ′′ = V −γ ⊂ V −1 (eα), that


weα · b(e+δ , e

−δ e

+γ yδ′′) · w−1

eα = x−(−e−α , e+δ , e

−δ e

+γ yδ′′).

Hence it remains to show that e−α , e+γ , yδ′′ = e−α , e+

δ , e−δ e−γ yδ′′, equivalently

D(yδ′′ , e+γ )e−α = D(yδ′′ e+

γ e−δ , e

+δ )e−α . But this follows from the second formula of

(10.8.2) for j = 0 with respect to the idempotent eδ.

Case (b): ξ = ε for an arrow α → ε of orthogonal type. Let e = eα ∈ Vα andg = eε ∈ Vε. Then G ∈ st(V, e) ∩ st(V,P(g)) and g governs e. Hence G ∈ st(V, g)by Lemma 13.13.

Remark. The astute reader will have noticed that Proposition 24.13 actuallyholds under the following weaker assumptions, with the same proof:

(i) ∆0 := ∆ ∩ Γ 0 satisfies (24.10.1),

(ii) ∆ ∩ Γiso = ∅,(iii) (Γher)

0 ⊂ ∆.

24.14. Proof of Theorem 24.2. By Lemma 24.12, ∆ satisfies (24.10.1).Hence the assumptions made in Propositions 24.10 and 24.13 hold. The follow-ing implications have already been shown:

(i) =⇒ (ii) follows from 24.6.

(ii) =⇒ (iii) and (i) =⇒ (iv) are trivial.

(iii) =⇒ (iv) follows from 24.10.

(iii) =⇒ (ii) follows from 24.13.

(ii) =⇒ (i): since we already have (ii) =⇒ (iii) =⇒ (iv), this follows from 24.6.

It remains to show (iv) =⇒ (iii) under the assumption that Γ have no connected

component of type T3∼= Cher

3 , a hexagram.

Let G ∈ st(V,R) and let α ∈ Γ 0. By 20.7, α induces a Peirce grading Rα,and G ∈ st(V,Rα) by 21.5. By (23.8.7), the Peirce grading Rα is the same as theidempotent Peirce grading P(eα) defined by the idempotent eα. By Theorem 12.5(and since G ∈ st(V,P(eα))), we have G ∈ st(V, eα) if and only if the Weyl relationW(eα) holds in G, which in the form (9.16.1) is

weα xσ(z) w−1eα = x−σ

(Q(e−σα )z

)(1)

for all z ∈ V σ2 (eα) and σ ∈ +,−. We will show (1) for σ = +. The case σ = −then follows by interpreting the case σ = + for V op.

Again by (23.8.7), V2(eα) = V2(α) =∑δ∈Γ2(α) Vδ, and δ ∈ Γ2(α) = δ ∈ Γ :

〈δ, α∨〉 = 2 if and only if δ = α or δ → α by 16.7. Since (1) is multiplicative in z,it suffices to prove it for all z ∈ V +

α and all δ → α and z ∈ V +δ .

First, let z ∈ V +α . Since α ∈ Γ 0, there exists β ∈ Γ 0 such that α β. Hence

Vα = Vα Vβ Vβ by (23.22.1). Then (12.2.12) (for the case j = 1) implies that (1)holds for z ∈ V +

α .Next, let δ → α. Since Γiso = ∅, it follows from Lemma 17.3(b) that δ →

α is of hermitian or orthogonal type. The second alternative would imply, by


Proposition 17.7, that α ∈ ∂Γ which contradicts α ∈ Γ 0. Hence δ → α is an arrowof hermitian type. By our assumption on Γ , δ → α is contained in a subgraphΣ ∼= T4 of Γ . It will be helpful to picture Σ as in (14.18.4), where ij stands

symbolically for i, j resp. εij , after identifying T4 with G (Cher4 ) by (14.18.5):

T4 =

11

444444444444444

14

ooooooooOOOOOOOO

444444444444444

12

444444444444444 13

44

OO

wwoooooooo

''OOOOOOOO

24 34

22 //

77oooooooo

DD23

OOOOOOOOoooooooo

33oo

ggOOOOOOOO

ZZ444444444444444

We may assume δ = ε11 and α = ε12. Let Vij = Vεij . Since εij ∈ Γ 0 for i 6= j, theVij , i 6= j, contain the idempotents eij := E (εij). In particular,

e := eα = e12.

Recall from (23.1.2) that the relations between the eij are determined by thecorresponding relations between the εij :

eα > eβ ⇐⇒ α β, eα a eβ ⇐⇒ α→ β, eα ⊥ eβ ⇐⇒ α ⊥ β.

By (23.22.2) applied to the arrow δ → γ := ε13, we have V +δ = V +

11 = Qe+13V −33 so

that it is enough to prove (1) for z = Qe+13y33 with y33 ∈ V −33 , i.e.,

we · x+(Qe+13y33) · w−1

e = x−(Qe−Qe+13y33). (2)

Observe that x1 = e+13 ∈ V1(e) while y0 = y33 ∈ V −0 (e). Hence (2) is, by the

equivalence of the shift relations S10 and S′10 (see 12.3), equivalent to

b(e+13, y33) = b(e+

12, e−12 e

+13 y33). (3)

Now the proof of (3) proceeds along the following lines, the justification for thevarious steps being supplied below:

A := b(e+13, y33) =

(((((((b(e+

13, e−23), b(−e+

23, y33))))))))

=: A1 (Step 1)

=(((((((

b(e+13, e

−23), b(−e+

23 y33 e+34, e

−34))))))))

=: A2 (Step 2)

= b(e+

13 e−23 e

+23 y33 e

+34, e

−34

)=: A3 (Step 3)

= b(e+

13 y33 e+34, e

−34

)=: A4 (Step 4)

= b(e+

12, e−12 e

+13 y33 e

+34 e

−34)

=: A5 (Step 5)

= b(e+

12, e−12 e

+13 y33

). (Step 6)


In the following computations, we constantly use the rules (20.2.7) – (20.2.12) of20.2(b).

Step 1: we work out the commutator A1. Using that G ∈ st(V,R) and thussatisfies the relation B(e+

13, e−23) we apply (9.9.5) together with B(e−23, e

+13)−1 =

B(−e−23, e+13) by (10.7.4) and (6.9.1), and get

A1 = b(−B(e+

13, e−23)e+

23, B(−e−23, e+13)y33

)· b(− e+

23, y33

)−1.

HereB(e+

13, e−23)e+

23 = e+23 − e

+13 e−23 e

+23+Qe+13

Qe−23e+

23

= e+23 − e

+13 +Qe+13

e−23 = e+23 − e

+13,

because e13 ∈ V1(e23). Moreover,

B(−e−23, e+13)y33 = y33 + e−23 e

+13 y33+Qe−23

Qe+13y33 = y33,

since e−23 e+13 y33 = 0 = Q(e−23)Q(e+

13)y33. By (11.5.2) we have b(−e+23, y33)−1 =

b(e+23, y33). Hence, by (11.5.1) for the Peirce grading induced by e12,

A1 = b(−e+23 + e+

13, y33) · b(e+23, y33) = b(e+

13, y33) = A.

Step 2: let f = e34. Then −e+23 = u1 ∈ V1(f) and y33 = v2 ∈ V −2 (f). Hence by

(12.1.6) (applied for j = 1 and f instead of e) we have

b(−e23, y33) = b(u1, v2) = b(u1 v2 f+, f−) · zf (u1, v2)

= b(−e+23 y33 e

+34, e

−34) · zf (−e+

23, y33)

and, by Lemma 12.2, zf (u1, v2) = zf (−e+23, y33) belongs to the centre of G since

G ∈ st(V,P(f)). Substituting this in A1 we obtain A2 because a central elementmay be omitted in a commutator.

Step 3 consists of working out the commutator A2. Put a24 = e+23 y33 e

+34.

Then, using again (9.9.5), we obtain

A2 =(((((((

b(e+13, e

−23), b(−a24, e

−34))))))))

= b(−B(e+

13, e−23)a24, B(−e−23, e

+13)e−34

)· b(−a24, e

−34)−1

= b(−a24 + e+13 e−23 a24, e−34) · b(a24, e

−34) = b

(e+

13 e−23 a24, e−34

)= A3.

Step 4: by the second formula of (10.8.2) for the idempotent e23 we have, sincee+

34 ∈ V+1 (e23) and y33 ∈ V −2 (e23):

e+13 y33 e

+34 = D(e+

34, y33)e+13 = D(e+

34 y33 e+23, e

−23)e+

13 = e+13 e−23 e

+23 y33 e

+34,

which proves A3 = A4.

Step 5: let us put u14 = e+13 y33 e

+34 ∈ V +

14 . Then Qu14e−34 = 0 by (20.2.9).

On the other hand, for e = e12 we have V14 ⊂ V1(e) and V34 ⊂ V0(e). Putx1 = u14 ∈ V +

1 (e) and y0 = e−34 ∈ V−0 (e). Then the relation S′10 holds trivially,

whence also S10 (by 12.3). This shows

b(x1, y0) = b(e+, e−, x1, y0) = b(e+12, e

−12 u14 e

−34).

Finally, Step 6 follows from the first formula of (10.8.2) for the idempotentf = e34 since y33 ∈ V −2 (f) and e+

13 ∈ V+1 (f):

e−12 e+13 y33 = D(y33, e

+13)e−12 = D(e−34, e

+34 y33 e

+13)e

−12 = e−12 e

+13 y33 e

+34 e

−34.

This completes the proof.


24.15. Corollary. Under the assumptions of Theorem 24.2,

st(V,E ) = st(V,R,E ) = st(V,E |Γ 0) ⊂ stbij(V,R) ⊂ st(V,R),

and all five categories coincide if Γ contains no component of type T3.

Proof. The first two equalities follow from Theorem 24.2, and the inclusionst(V,R,E ) ⊂ stbij(V,R) follows from Corollary 24.9. The inclusion stbij(V,R)⊂ st(V,R) holds by definition. The last claim follows again from Theorem 24.2.

Most low rank cases, to which Theorem 24.2 and Corollary 24.15 do not apply,can be dealt with at the expense of requiring ∆ = Γ .

24.16. Corollary. Let Γ be a Jordan graph without isolated vertices. Let Vbe a Jordan pair with a Γ -grading R which is idempotent with respect to a cog Edefined on all of Γ . Then

st(V,E ) = st(V,R,E ) = stbij(V,R) ⊂ st(V,R).

Proof. Since dom(E ) = Γ , the condition (24.10.1) is obviously fulfilled (forδ = α) which proves the first equality. The last inclusion holds by definitionof stbij(V,R). This leaves us with the second equality, for which the inclusionst(V,R,E ) ⊂ stbij(V,R) was shown in Corollary 24.9. So it remains to provestbij(V,R) ⊂ st(V,R,E ).

Let G ∈ stbij(V,R). As in the beginning of 24.14, the proof that G ∈ st(V,R,E )is reduced to showing (24.14.1) for all α ∈ Γ . By Lemma 9.16 and 9.17, and usingthe abbreviation wα = weα , this is equivalent to

Int(wα) · U% = Usα(%) for all % ∈ Γ2(α) ∪ (−Γ2(α)). (1)

To this end we will first establish,

Int(wα) · U% = Usα(%) for all % ∈ R with % 6= ±α, (2)

where as usual (R,R1) is the 3-graded root system associated with Γ . For the proofof (2) recall Corollary 21.22: if A ⊂ R is nilpotent then the canonical morphismπ: G→ PE(V ) = G induces an isomorphism π: UA → UA.

Let % be as in (2) and consider the set A = R∩(Zα+N+%). Then A is nilpotent.Indeed, A is clearly closed. By local finiteness of R and Proposition 1.13(iv), itsuffices to show that 0 /∈ A. Assuming nα + p% = 0 for n ∈ Z and p ∈ N+ implies% = ±α because R is reduced, contradiction. By using the commutator formula(3.2.3) repeatedly one sees Int(wα) · U% ⊂ UA. From sα(%) = % − 〈%, α∨〉α ∈ Awe have Usα(%) ⊂ UA. Since π: UA → UA is an isomorphism, it suffices to prove(2) after applying π. Now π(wα) = wα and π(Uβ) = Uβ for all β ∈ A. Henceit suffices to show Int(wα) · U% = Usα(%), which holds because G ∈ st(V,R,E ) byCorollary 24.8. This completes the proof of (2).

We come to the proof of (1). By (2), this equality holds for % = ±δ where δ ∈ Γsatisfies δ → α. As Γ2(α) = δ ∈ Γ : δ → α ∪ α by (16.7.2), we are left withproving (1) for % = ±α. Since α is not an isolated vertex of Γ , there exists β ∈ Γsatisfying α β or α ← β or α → β. In the first two cases 〈α, β∨〉 = 1, whence


Vα ⊂ V1(eβ) by (23.8.1) and thus V σα = V σα e−σβ eσβ by (6.14.7). Now (1) followsfrom (12.2.12) and (12.2.13) with j = 1.

This leaves us with the case α→ β. We complete this configuration to a collisionα → β ← γ in Γ . Let xβ ∈ V +

β and (zγ , yγ) ∈ Vγ . Since the relation B(xβ , yγ)holds in G, it follows from (9.8.3) that

x+(QxβQyγzγ) = x+(xβ yγ zγ) ·(((((((

b(xβ , yγ), x+(zγ)))))))). (3)

Recall 2β = α+γ from (15.3.1). Then we have uα := QxβQyγzγ ∈ V+2β−γ = V +

α and

xβ yγ zγ ∈ V +β . Let us put aα := x+(uα) ∈ Uα and bβ := x+(xβ yγ zγ) ∈ Uβ

as well as cγ := x+(xγ) ∈ Uγ . Also, let µ = β − γ = α − β ∈ R0. Thendµ := b(xβ , yγ) ∈ Uµ by (21.1.2). With these abbreviations, (3) becomes

aα = bβ ·(((((((dµ, cγ

))))))). (4)

By (2), the conjugation formula Int(wα) · U% = Usα(%) holds for all % 6= ±α; inparticular, it holds for % = γ, % = β and % = µ. Furthermore, sα(γ) = γ sinceα ⊥ γ, and sα(β) = β−α = −µ which implies sα(µ) = −s2

α(β) = −β. By applyingInt(wα) to (4), we therefore obtain

Int(wα) · aα = b′−µ ·(((((((d′−β , c

′γ

))))))), (5)

where b′−µ = Int(wα) · bβ ∈ U−µ, d′−β = Int(wα) · dµ ∈ U−β , and c′γ ∈ Uγ . Considerthe closed root interval

C :=[[[[− β, γ,

]]]]= −β, γ, −β + γ = −µ, −2β + γ = −α.

Then (5) shows that Int(wα) · aα ∈ UC .

Since C is nilpotent, π: UC → UC is an isomorphism. By the argument usedbefore, we find that

Int(wα) · aα = Int(wα) · x+(uα) = x−(Q(e−α )uα

)holds in G for all uα ∈ V +

α of the form uα = QxβQyγzγ . In particular, for xβ = e+β

and yγ = e−γ we have V −γ = Q(e−γ )V +γ and V +

α = Q(e+β )V −γ = Q(e+

β )Q(e−γ )V −γ ,which proves (1) for % = α. The same proof for the idempotent −eα yields (1) for% = −α.

Remark. The projective elementary group G = PE(V ) is a group with R-commutator relations (Corollary 21.12). For such a group we have defined theSteinberg category st(G) in 4.12, and we have shown in Proposition 21.23 that thereexists an isomorphism st(G) ∼= stbij(V,R) of categories. Hence Corollaries 24.15and 24.16 also contain information about PE(V ).

We now give an application of Proposition 24.6 and Corollary 24.9 to specialJordan pairs. It complements Corollary 21.13.


24.17. Proposition. Let R be a Γ -grading of a Jordan pair V , let E ∈ cog(R)and assume Γiso ∪ Γ 0 ⊂ ∆ = dom(E ). Assume further that V is a specialJordan pair, embedded in a Morita context M as in 6.4. Then the elementarygroup E(M, V ) belongs to the categories st(V,E ), stbij(V,R) ∼= st(PE(V )) andst(V,R,E ):

E(M, V ) ∈ st(V,E ) ∩ stbij(V,R) ∩ st(V,R,E ).

Proof. By Corollary 12.9 we have G = E(M, V ) ∈ st(V, eδ) for all δ ∈ ∆,so G ∈ st(V,E ). We show next that G ∈ st(V,R). The relations (StR1) ofTheorem 21.7 hold in G since (x+

α , y−β ) ∈ V +

α × V −β , α 6= β, is quasi-invertibleby (20.7.3) so that (9.8.6) applies. Hence, in view of Theorem 21.7, it remains toprove the relations (StR2), that is: b(xα, yβ) = 1 for (xα, yβ) ∈ V +

α ×V −β , providedα ⊥ β. We distinguish three cases.

(i) Suppose there exists γ ∈ ∆ with Vα ⊂ V2(eγ) and Vβ ⊂ V0(eγ). Thenb(xα, yβ) ∈ b(V +

2 (eγ), V −0 (eγ)) = 1 since G ∈ st(V, eγ). In particular, this is soif α ∈ ∆, for then Vα ⊂ V2(eα) and Vβ ⊂ V0(eα). Therefore, we may assumeα 6∈ ∆ and, by symmetry, also β 6∈ ∆ from now on. By (17.7.1), we then have α,β ∈ ∂(Γher) ∪ ∂(Γorth).

(ii) Assume α = ε ∈ ∂(Γorth). Then ε is part of a pyramid as in (17.2.4):

γ

???? δ

ε

δ′

??γ′

__????

which is contained in the connected component Γε of Γ containing ε. By Propo-sition 17.12, Γε ∼= OI , so every vertex of Γε is connected to ε. As β ⊥ ε thisshows β /∈ Γε, and therefore β ⊥ Γε. By (23.22.3), V +

ε = V +γ V −ε V +

γ′ . SinceG ∈ st(V, eγ) and Vε ⊂ V1(eγ), Vγ′ ⊂ V0(eγ), it follows from (12.4.1) and γ′ ⊥ βthat

b(V +ε , V

−β ) = b(V +

γ V −ε V −γ′ , V−β ) = b(V +

γ , V −ε V +γ′ V

−β ) = 1.

(iii) We are left with the case α, β ∈ ∂(Γher). It suffices to show that there existsγ ∈ Γ such that α→ γ ⊥ β, because then γ ∈ Γ 0 ⊂ ∆ so that an application of (i)finishes the proof. The existence of γ is clear if α and β lie in different connectedcomponents of Γ . If Γ is connected we have Γ = Γher by Proposition 17.5, soΓ ∼= TI , |I|> 3 by Proposition 17.10, from which the existence of γ is immediate.

We now know that G ∈ st(V,E ) ∩ st(V,R). From Proposition 24.6 we thenobtain G ∈ st(V,R,E ) and from Corollary 24.9 that in fact G ∈ stbij(V,R).

24.18. Example. Let A be a unital associative k-algebra, let I and J be non-empty sets, put N = I ∪ J and suppose |N |> 3. Let

V = MIJ(A)

be the rectangular matrix pair defined in 20.2(a). There we have shown that V hasa root grading R of type Γ = KIKJ , given by V(i,j) = (AEij , AEji). By 23.23, R


is idempotent with respect to the cog E : Γ → Idp(V ), E (i, j) = e(i,j) = (Eij , Eji).

Since the root system R associated with Γ is R = AN , thus irreducible, andR 6= A2 = A1, and since Γ = Γ 0 = ∆, Corollary 24.15 applies in its strongform:

st(V,R,E ) = st(V,E ) = stbij(V,R) = st(V,R) ∼= st(PE(V )). (1)

The Jordan pair V is special, embedded in the Morita context M of (20.2.3). ByProposition 24.17, the elementary group E(M, V ) belongs to st(V,R,E ), whence toall categories (1). One arrives at the same conclusion by invoking Corollary 21.13,where we proved that E(M, V ) belongs to st(V,R). To elaborate on this, we observethat

E(M, V ) = EN (A), (2)

the elementary linear group EN (A) of 3.16(c). It is generated by all transvectionsemn(a) = 1N +Emn(a) for m,n ∈ N , m 6= n, and a ∈ A, and satisfies the relations

emn(a) emn(b) = emn(a+ b), (3)(((((((emn(a), enp(b)

)))))))= emp(ab) for m 6= n 6= p, (4)(((((((

emn(a), epq(b))))))))

= 1 for n 6= p, m 6= q (5)

where a, b ∈ A. These relations show that the group EN (A) has R-commutatorrelations with root groups Uα = emn(A), for α = εm− εn ∈ R×. These root groupscoincide with the root groups for E(M, V ) defined in (21.1.1) and (21.1.2), as wehave seen in Example 21.17.

Extending the definition of the Steinberg group Stn(A) in 9.18, we define theSteinberg group StN (A) as the group presented by generators xmn(a), a ∈ A andm,n ∈ N , m 6= n, satisfying the relations (3) – (5) above with emn(a) replacedby xmn(a). Since EN (A) has unique factorization for nilpotent pairs (3.16(c)),Theorem 4.14 shows that StN (A) is the Steinberg group of EN (A), i.e., an initialobject in the Steinberg category st(EN (A)). As EN (A) belongs to the Steinbergcategory of PE(V ), it follows from Corollary 4.12 that StN (A) is also an initialobject in st(PE(V )). On the other hand, by its very definition in 22.1, St(V,R) isan initial object in st(V,R) ∼= st(PE(V )) by (1). Hence, both groups are isomorphicunder an isomorphism preserving root groups:

StN (A) ∼= St(V,R). (6)

As for the elementary groups, this can also be seen directly by verifying that therelations defining StN (A) and St(V,R) respectively hold in the other group.

§25. The monomial group

25.1. Preliminaries. In this section, Γ is a Jordan graph with associated 3-graded root system (R,R1) and R is a Γ -grading of a Jordan pair V . Recall from(21.1.6) the definition

Γ×(R) = γ ∈ Γ : Vγ 6= 0.

We assume throughout that

§25] The monomial group 327

Γ×(R) generates Γ as a Jordan graph. (1)

By 21.1(c), this is no restriction of generality.Let X = X•(Γ ) as in Lemma 14.11 be the free abelian group defined by Γ .

Then (1) and (15.21.1) imply

X = spanZ(Γ×(R)). (2)

25.2. The category stw(V,R). Let us assume that R is an idempotent rootgrading as in 23.8, so the set cog(R) of all cogs compatible with R is not empty.By Proposition 23.18, cog(R) contains maximal elements, all defined on the samedomain ∆max ⊂ Γ . Recall from 23.14 that U× denotes the set of invertible idem-potents of a subpair U of V . We introduce the set

P = (δ, e) ∈ ∆max × Idp(V ) : e ∈ V ×δ (1)

and define a full subcategory

stw(V,R) ⊂ st(V,R)

as follows. For a group G ∈ st(V,R), let U = (Uα)α∈R be the root subgroupsdefined in 21.1, so that (G,U) is a group with R-commutator relations. Recall from(9.11.2) the notation

we = x−(e−) · x+(e+) · x−(e−)

for an idempotent e of V . Now we define

G ∈ stw(V,R) ⇐⇒ ∀ (δ, e) ∈P : we is a Weyl element for δ (2)

in the sense of 5.1. Explicitly, this means that

Int(we) · Uα = Usδ(α) for all α ∈ R. (3)

Let E ∈ cog(V,R) be defined on some subset ∆ of Γ . We relate stw(V,R) tothe subcategories st(V,R,E ) of st(V,R) defined in 24.1 and show:

G ∈ stw(V,R) ⇐⇒ ∀E ∈ cog(R) : G ∈ st(V,R,E ). (4)

For the implication from left to right, let E ∈ cog(R) and δ ∈ ∆ = dom(E ).Then e = E (δ) ∈ V ×δ by Lemma 23.15, so (δ, e) ∈ P by Corollary 23.21. SinceG ∈ stw(V,R), we is a Weyl element for G, so G ∈ st(V,R,E ) by (24.1.3).

Conversely, let (δ, e) ∈ P. Again by Corollary 23.21, there exists a cog Edefined on δ with e = E (δ). Since G ∈ st(V,R,E ), we is a Weyl element for δ, asdesired.

As a consequence of (4), we see that stw(V,R) is not empty in the mostimportant cases. For example, let Γ be a Jordan graph with Γiso = ∅ and let (V,R)be Γ -graded Jordan pair admitting a cog defined on Γ 0. If Γ has no connectedcomponents of type T3 then Theorem 24.2 and (4) show that st(V,R) = stw(V,R).

More generally, assume only that Γ does not have isolated vertices. Thenstw(V,R) = st(V,R,F ) for any cog F defined on ∆max. Indeed, stw(V,R) ⊂st(V,R,F ) by (4). Conversely, by Proposition 23.18(d), any E ∈ cog(R) canbe extended to a cog E ′ defined on ∆max. Then st(V,R,E ) ⊃ st(V,R,E ′) =st(V,R,F ) by definition in (24.1.3) and Corollary 24.7.


25.3. The monomial group. Let G ∈ stw(V,R). The monomial group of Gis the subgroup

M = M(G) =⟨

we : (δ, e) ∈P⟩. (1)

We also define a subgroup K of M by

K =⟨

wewf : (δ, e) ∈P and (δ, f) ∈P⟩. (2)

The following example explains our choice of terminology. Let A be a ringand let V = Mp,q(A) be the Jordan pair of rectangular matrices of 6.6(a). LetΓ = Kp Kq. By 20.2(a), V has a Γ -grading R, and by 23.23, R is idempotent.For δ = (i, j) ∈ Γ we have Vδ = V(i,j) = (A · Ei,j , A · Eji) by (20.2.5). Hence

V ×δ = (aEij , a−1Eji) : a ∈ A×. By 21.17, the 3-graded root system (R,R1)

associated with Γ is Apn where n = p+ q, and the isomorphism Γ ∼= R1 is given by

(i, j) ∈ Γ 7→ εi − εp+j ∈ R1.

The Weyl group of R is the symmetric group Sn, and under this isomorphism, sδcorresponds to the transposition τi,p+j interchanging i and p+ j.

Consider the elementary group G = En(A) of 6.1, and let M be the Moritacontext defined there. Then En(A) = E(M, V ), and Proposition 24.17 impliesEn(A) ∈ st(V,R,E ) for any cog defined on Γ , for example the one exhibited in23.23. But then En(A) ∈ stw(V,R) by the example discussed in 25.2.

Let (δ, e) ∈ P, so δ = (i, j) and e = (aEij , a−1Eji) where a ∈ A×. Then

x+(e+) = 1n + aEi,p+j and x−(e−) = 1n − a−1Ep+j,i, and a matrix computationshows that

we =

(1p 0

−a−1Eji 1q

)(1p aEij0 1q

)(1p 0

−a−1Eji 1q

)=

(1p − Eii aEij−a−1Eji 1q − Ejj

)= DP, (3)

where D = diag(d1, . . . , dn) is the diagonal matrix with entries di = a, dp+j =−a−1, and dk = 1 otherwise, and P is the permutation matrix corresponding tothe transposition τi,p+j .

A matrix in GLn(A) is called monomial [32, p. 29] if it is the product ofa diagonal matrix and a permutation matrix, equivalently, if each row and eachcolumn contains exactly one non-zero entry, and this entry is a unit of A. From(3) it is clear that we is a monomial matrix, so the monomial group M of Gconsists of monomial matrices which are, moreover, elementary. The factors D andP appearing in (3) are in general not elementary. We leave it to the reader to showthat, conversely, the elementary monomial matrices are generated by matrices ofthe form (3).

Let also (δ, f) ∈ P, so that f = (bEij , b−1Eji) for some b ∈ A×. Then a

computation shows that wewf = diag(h1, . . . , hn) is a diagonal matrix with entrieshi = −ab−1, hp+j = −a−1b and hk = 1 otherwise.

Let us return to the general situation of a group G ∈ stw(V,R). By Proposi-tion 23.18(c), ∆max is a Jordan subgraph of Γ , in particular, it is a Jordan graph in


its own right. Let (S, Y, S1) be the 3-graded root system associated with ∆max asin Theorem 15.11. Thus we have S1 = ∆max, and by Proposition 15.20, Y may beidentified with span(∆max) ⊂ X = span(Γ ). As well, we identify the Weyl groupW(S) with a subgroup of W(R), and by (15.17.4) we have

W(S) =⟨sδ : δ ∈ ∆max

⟩. (4)

The example (23.23.2) shows that ∆max & Γ and hence S & R may very wellhappen, even in the simply laced case.

The main result of this section is the following theorem.

25.4. Theorem. With the notation and conventions introduced in 25.2 and25.3, consider a group G ∈ stw(V,R). Then there is unique homomorphismθ: M(G)→W(S) satisfying

θ(we) = sδ (1)

for all (δ, e) ∈P, and hence

Int(g) · Uα = Uθ(g)·α (2)

for all g ∈ M(G) and α ∈ R. The sequence

1 // Kinc // M(G)

θ //W(S) // 1 (3)

is exact.

Proof. To show that θ exists, we first prove:

if α ∈ R±1 and β ∈ R satisfy Uα = Uβ 6= 1 then α = β. (4)

By symmetry, we may assume α ∈ R1. Let G0 ⊂ G be the subgroup definedin Lemma 9.2. By definition of the root groups in 21.1, we have Uβ ⊂ U± orUβ ⊂ G0, depending on whether β ∈ R±1 or β ∈ R0. From Lemma 9.2(a) it followsthat U+∩G0 = U+∩U− = 1. Hence our assumption Uα = Uβ 6= 1 implies β ∈ R1.By (21.1.4), U+ is the direct sum of the abelian groups (Uγ)γ∈R1

. Hence α 6= βwould imply Uα ∩ Uβ = 1, contradicting Uα = Uβ 6= 1, so we have α = β.

To show the existence of θ, it suffices to verify:

if (δ1, e1), . . . , (δn, en) ∈P and we1 · · ·wen = 1 then sδ1 · · · sδn = 1.

Let g = we1 · · ·wen and w = sδ1 · · · sδn . Then (25.2.3) applied several times shows

Uα = Int(g) · Uα = Uw(α)

for all α ∈ R. In particular, let α ∈ Γ×(R), so that V +α 6= 0 or V −α 6= 0. We

use (4) with α ∈ R1 if V +α 6= 0 and α ∈ R−1 in case V −α 6= 0 and β = w(α), and

obtain α = w(α). Hence w fixes Γ×(R), so by (25.1.2), w = IdX . Thus θ existsand is obviously uniquely determined by (1). Moreover, (2) follows immediatelyfrom the definition of M and from (25.2.3), while surjectivity of θ is a consequenceof (25.3.4). The proof of the exactness of (3), i.e., that K = Ker(θ), requires somepreparation and will be given in 25.10.


Example. The sequence (3) is in general not split, as the following exampleshows. Let V = (F3,F3) be the Jordan pair defined by the field with three elements,with the obvious Γ -grading R given by Γ = γ and Vγ = V as in 23.10(b). The

associated root system is R = A1 = A2. By the example treated in 25.3, theelementary group G = E2(F3) belongs to stw(V,R). We have P = (γ,±e),where e = (1, 1) ∈ V , so w−e = w−1

e and (25.3.3) shows that M is generated by thematrix

we =

(0 1−1 0

)of order 4. Therefore, M(G) is cyclic of order 4, and (3) becomes the non-splitsequence of abelian groups

0 // Z/2Z // Z/4Z // Z/2Z // 0 . (5)

In the following subsections until 25.10, V is an arbitrary Jordan pair.

25.5. Quadrangles of idempotents. Recall the symbols > and ⊥ for idem-potents defined in 6.15. A quadruple (e1, e2, e3, e4) of idempotents in V is called aquadrangle if

ei−1 > ei ⊥ ei+2 and eσi e−σi+1 eσi+2 = eσi+3, (1)

for σ ∈ +,− and i taken modulo 4. The shift formulas 10.8 applied for ei+3 give

D(eσi , e−σi+1) = D(eσi+3, e

−σi+2), (2)

Q(eσi )Q(e−σi+1) = Q(eσi+3)Q(e−σi+2), (3)

B(eσi , e−σi+1) = B(eσi+3, e

−σi+2). (4)

We also have

D(eσi , e−σi+1)D(eσi+2, e

−σi+3)x = x, for x ∈ V σ2 (ei) ∩ V σ1 (ei+1). (5)

Indeed, by (2), D(eσi , e−σi+1)D(eσi+2, e

−σi+3)x = D(eσi , e

−σi+1)D(eσi+1, e

−σi )x. Let f = ei

and e = ei+1. Then x ∈ V σ(21) = V σ2 (f) ∩ V σ1 (e). Hence the assertion follows from

(10.12.5).If one of the idempotents in a quadrangle is zero then they all are by the second

condition in (1). On the other hand, let ∆ = K2 K2 = (α1, α2, α3, α4) be asquare. If all ei 6= 0 then the first condition in (1) says that αi 7→ ei defines a cogof type ∆. The following lemma describes the relation between cogs of this typeand quadrangles. For easier notation we use the abbreviation

ei ej ek =(e+i e−j e

+k , e

−i e

+j e−k )

(6)

for idempotents of V .

25.6. Lemma. Let V be a Jordan pair.

(a) Let e1, e2, e3 be idempotents in V satisfying e1 > e2 > e3 ⊥ e1. Thene4 = e1 e2 e3 is an idempotent and (e1, e2, e3, e4) is a quadrangle.


(b) Let ∆ = (α1, α2, α3, α4) ∼= K2 K2 be a square, let Q: ∆ → Idp(V ) be acog and put ei = Q(αi). Then e′4 := e1 e2 e3 is an idempotent of V , associatedwith e4. Moreover, (e1, e2, e3, e

′4) is a quadrangle, and Q′: ∆→ Idp(V ), defined by

Q′(αi) =

ei for i = 1, 2, 3e′4 for i = 4

, (1)

is a cog associated with Q.

Proof. (a) Let f = e1 and e = e2, so that e and f are collinear idempotents.Let h = ωf,e as in (10.11.1). Then e3 ∈ V0(f) ∩ V1(e) = V(01), so

h · e3 = e1 e2 e3 = e4

by Proposition 10.12. Since h is an automorphism of V , e4 is an idempotent. Fromthe Peirce relations (6.14.8) we obtain e4 ∈ V1(e1) ∩ V0(e2) ∩ V1(e3). Since by(10.12.2) the automorphism h maps e2 to e1 we get

e4 e4 e1 = h · e3, h · e3, h · e2 = h · e3, e3, e2 = h · e2 = e1,

whence e1 ∈ V1(e4) by (6.14.7), proving e1 > e4. By symmetry, e3 > e4. Thus(e1, e2, e3, e4) satisfies the first part of (25.5.1), and for (e1, e2, e3, e4) to be a quad-rangle it remains to show

(i) e2 e3 e4 = e1, (ii) e3 e4 e1 = e2, (iii) e4 e1 e2 = e3.

Observe that e1 ∈ V1(e2) ∩ V0(e3), so (i) follows from (10.12.5) by putting theref = e2 and e = e3. The same method yields (iii), this time putting f = e2

and e = e1 in (10.12.5). For (ii) we observe that e1 + e3 is an idempotent withe2, e4 ∈ V1(e1) ∩ V1(e3) ⊂ V2(e1 + e3), whence e4 = Q(e3 + e1)e2 and so

e3 e4 e1 = Q(e3 + e1)e4 = Q(e3 + e1)Q(e3 + e1)e2 = e2

by (6.14.7) again.

(b) We know from (a) that (e1, e2, e3, e′4) is a quadrangle of idempotents.

Moreover, by (23.1.2), e′4 = e1 e2 e3 ∈ V1−0+1(e4) = V2(e4). We claim thatalso e4 ∈ V2(e′4). Indeed, letting h = ωf,e as in (a) and using (1), this holds ifand only if h−1 · e4 ∈ V2(e3). By (10.11.3), h−1 = ωf,−e, and e4 ∈ V1(f) ∩ V0(e)since α1 α4 ⊥ α2. By (6.14.5), V0(e) = V0(−e). Hence Proposition 10.12 isapplicable to ωf,−e, so that h−1 · e4 = −−e2 e1 e4 ∈ V1−0+1(e3) = V2(e3), asdesired. We have shown e′4 ∈ V2(e4) and e4 ∈ V2(e′4), so e4 and e′4 are associatedby 6.15. The last statement follows from 23.4.

We next establish the analogue of 25.5 and 25.6 for kites instead of squares.

25.7. Diamonds of idempotents. Besides the symbols > and ⊥ we will alsouse the symbols a and `, all of them defined in 6.15. A quadruple (e0, e1, e2, e3) ofidempotents ei in a Jordan pair V is called a diamond if

e1 ` e0 a e3, e0 ⊥ e2, e1 > e2 > e3 > e1, (1)

e0 e1 e2 = e3, e2 e3 e0 = e1, e3 e0 e1 = e2, (2)

e1 e2 e3 = 2e0, (3)

where here and again in Lemma 25.8 we use the abbreviation (25.5.6).

If e0 = 0 then the other ei = 0 by (2). On the other hand, let ∆ =(α0, α1, α2, α3) be a kite. If all ei 6= 0 then the map αi 7→ ei defines a cog oftype ∆.


25.8. Lemma. Let V be a Jordan pair.

(a) Let e0, e1, e2 be idempotents in V satisfying e0 a e1 > e2 ⊥ e0. Thene3 = e0 e1 e2 is an idempotent and (e0, e1, e2, e3) is a diamond.

(b) Let e0, e1, e3 be idempotents satisfying e0 a e1 > e3 ` e0. Then e2 =e3 e0 e1 is an idempotent and (e0, e1, e2, e3) is a diamond.

(c) In any diamond (e0, e1, e2, e3),

D(eσi , e−σi+1) = D(eσi+3, e

−σi+2) (1)

holds for i modulo 4 and σ ∈ +,−.

(d) Let ∆ = (α0, α1, α2, α3) be a kite, let D : ∆ → Idp(V ) be a cog and putei = D(ei). Then e′3 = e0 e1 e2 and e′′2 = e3 e0 e1 are idempotents of V ,associated with e3 and e2, respectively. Moreover, (e0, e1, e2, e

′3) and (e0, e1, e

′′2 , e3)

are diamonds, and D ′: ∆→ Idp(V ) and D ′′: ∆→ Idp(V ), defined by

D ′(αi) =

ei for i = 0, 1, 2e′3 for i = 3

and D ′′(αi) =

ei for i = 0, 1, 3e′′2 for i = 2

, (2)

are cogs associated with D .

Proof. (a) The element c = e0 + e2 is an idempotent and e1 ∈ V1(e0) ∩V1(e2) ⊂ V2(c) by 6.16. Since Q(cσ)e−σ1 = eσ0 e−σ1 eσ2 = eσ3 , it follows that e3

is an idempotent. By (10.8.2) we have the equations

D(e1, e0) = D(e2, e2 e1 e0) = D(e2, e3),

D(e2, e1) = D(e2 e1 e0, e0) = D(e3, e0),

D(e0, e1) = D(e0 e1 e2, e2) = D(e3, e2),

which imply (25.7.2) and (25.7.3): e2 e3 e0 = e1 e0 e0 = e1, e3 e0 e1 =e2 e1 e1 = e2 and e1 e2 e3 = e3 e2 e1 = e0 e1 e1 = 2e0. It then fol-lows that e3 e3 e2 = e2 e3 e3 = e1 e0 e3 = e2, i.e., e2 ∈ V1(e3). Sincee3 ∈ V0−1+2(e2) = V1(e2) by the Peirce multiplication rules, we get e2 > e3.

For the proof of e1 > e3 we use the automorphism h = ωg,e of Proposition 10.13for g = e1 and e = e0. By (10.13.1) and since e2 ∈ V1(g) ∩ V0(e) we obtainh(e1) = −e1, h(e2) = −e0 e1 e2 = −e3. From e1 > e2 we then infer h(e1) > h(e2),i.e., (−e1) > (−e3), from which e1 > e3 follows by definition of collinearity.

We also know h(Qge

)= e by (10.13.1), whence h−1(e0) = Q(e1)e0 which, by

the Peirce multiplication rules, is an element of V2(e2). Hence e0 ∈ h(V2(e2)

)=

V2

(h(e2)

)= V2(−e3) = V2(e3). Since clearly e3 = e0 e1 e2 ∈ V2−1+0(e0) = V1(e0)

we get e0 a e3, the last missing relation in (25.7.1).

(b) Let again h = ωg,e be the automorphism of the proof of (a). Since e3 ∈V1(g)∩V1(e), Lemma 10.13 and (10.8.1) show that h(e3) = Qge g e3 = g e e3 =e2 is an idempotent satisfying h(g) > h(e3), i.e., (−g) > e2 which is equivalent tog = e1 > e2. Moreover, e2 ∈ V1−2+1(e0) = V0(e0). Thus e0 ⊥ e2. Hence by (a),(e0, e1, e2, e0 e1 e2) is a diamond. But e0 e1 e2 = −ωg,e(e2) = −ω2

g,e(e3) = e3

by (10.13.2), finishing the proof of (b).


(c) With the exception of i = 1, these equations have been established in theproof of (a). We have also seen D(e2, e1) = D(e3, e0). Interpreting this in V op

shows D(e1, e2) = D(e0, e3) which is the missing case i = 1 in (1).

(d) We have already proved in (a) and (b) that e′3 and e′′2 are idempotents andthat (e0, e1, e2, e

′3) and (e0, e1, e

′′2 , e3) are diamonds. It follows that D ′ and D ′′ are

cogs of type ∆. By 23.4 they are associated with D if e3 ≈ e′3 and e2 ≈ e′′2 .For the proof of e3 ≈ e′3, it is immediate from the Peirce multiplication rules

that e′3 ∈ V2−1+1(e3) = V2(e3). For the proof that also e3 ∈ V2(e′3) we use asin (a) the idempotent c = e0 + e2. We have seen there that e′3 = Qc(e1). Sincee1 ∈ V1(e0)∩V1(e2) it follows that e1 ∈ V2(c) and thus V2(e1) ⊂ V2(c). Analogously,e3 ∈ V1(e0) ∩ V1(e2) ⊂ V2(c) implying Qce3 = e0 e3 e2 ∈ V2−1+1(e1) = V2(e1).Suppressing the superscripts ±σ, we then calculate

Q(e′3)Q(e′3)e3 = Q(Qce1

)Q(Qce1

)e3 = QcQe1Q

2cQe1Qce3 (by (JP3))

= QcQ2e1Qce3 (since Qe1Qce3 ∈ V2(e1) ⊂ V2(c))

= Q2ce3 = e3 (since Qce3 ∈ V2(e1) and e3 ∈ V2(c)).

Thus e3 ∈ V2(e′3), so that indeed e3 ≈ e′3.It remains to prove e2 ≈ e′′2 . We use again the automorphism h = ωg,e where

(g, e) = (e1, e0). We have seen in the proof of (b) that h(e3) = e1 e0 e3 = e′′2 .Hence, replacing e3 by e′3, we get h(e′3) = e1 e0 e

′3 = e2 since (e0, e1, e2, e

′3) is a

diamond. Because e3 ≈ e′3, this implies e′′2 ≈ e2.

25.9. Lemma. Let S be a set of idempotents in a Jordan pair V and letG ∈ st(V,S ) as defined in 13.1. For an idempotent c ∈ V we put wc = x−(c−) ·x+(c+) · x−(c−) ∈ G as in (9.11.2), so that wc is a Weyl element for G in casec ∈ S .

(a) For e ∈ S and any c ∈ Idp(V ) with e ⊥ c we have

Int(we) · wc = wc. (1)

(b) Let e, f ∈ S with e > f . Then

we wf we = wf we wf . (2)

(c) For d, e ∈ S with d ` e the element f = Qd(e) is an idempotent of Vsatisfying G ∈ st(V, f) and

Int(wd) · we = wf . (3)

If f ′ ∈ Idp(V ) with d ` f ′ ⊥ e then f ′ ≈ f .

(d) Assume e, f, d ∈ S satisfy e > f > d ⊥ e or e ` f a d > e. Then

(i) c = e f d ∈ Idp(V ) and (f, d, c, e) is a quadrangle in the first anda diamond in the second case.

(ii) Moreover, G ∈ st(V, c) and

Int(wf ) · we = Int(wd) · wc. (4)


(e) Suppose e, f, d ∈ S satisfy e > f and either e a d ` f or e > d > f .Then c = d − f e e f d is an idempotent associated with d, and G ∈ st(V, c).Moreover,

(we wf w−1e ) · wd = wc · (we wf w−1

e ). (5)

Proof. (a) follows from (12.2.1) taking into account that G ∈ st(V, e) ⊂st(V,P(e)), and (b) is a restatement of (13.11.2) for easier reference.

(c) It was shown in Lemma 10.13 that f is an idempotent of V satisfyingd ` f ⊥ e. Hence it follows from Proposition 13.14, by replacing g there by d here,that G ∈ st(V, f). By (13.13.1) and (9.11.4) we get

Int(wd) · we = Int(wd) ·(x−(e−) x+(e+) x−(e−)

)= x+(f+) x−(f−) x+(f+) = wfop = wf ,

since the Weyl relation W(f) holds in G ∈ st(V, f) and therefore wf = wfop byLemma 9.16. The final claim was already shown in Proposition 13.14.

(d) Part (i) follows from Lemma 25.6(a) in the first and from Lemma 25.8(b)in the second case. For part (ii), we have G ∈ st(V, c) since in both cases d > c, sothat Proposition 13.11 applies: st(V, d) = st(V, d, c) = st(V, c). To prove (4), weobserve e ∈ V1(f) in both cases, whence by (13.10.2)

Int(wf ) we = b(f+, e−) b(−e+, f−) b(f+, e−).

Recall st(V, d) = st(V,−d) by (12.1.3). Since f ∈ Vj+1(−d) and e ∈ Vj(−d) withj = 0 in the first case and j = 1 in the second case, the formulas (12.2.6) and(12.2.7) yield

Int(w−d) · b(f+, e−) = x−(d− f+ e−

)= x−(c−),

Int(w−d) · b(−e+, f−) = x+

(e+ f− d+

)= x+(c+).

Therefore,

Int(w−d) ·((Int(wf ) · we)

)= wc

which is equivalent to (4) because w−d = w−1d .

(e) In both cases e > f . Hence Lemma 13.10 shows Int(we) · wf = wf,e ∈N ∩ G0 and π(wf,e) = ωf,e. Let c = ωf,e(d) ∈ Idp(V ), which by Lemma 10.12 isc = d− f e e f d. Now it follows from (13.10.4) that

Int(wf,e) · wd = x−(ωf,e · d−) x+(ωf,e · d+) x−(ωf,e · d−) = wc.

This proves (5). By the Peirce rules, f e e f d ∈ V2(d), whence also c ∈ V2(d).This implies

ω−1f,e(c) = d ∈ ω−1

f,e

(V2(d)

)= V2

(ω−1f,e(d)

).

But ω−1f,ed = ωf,−e(d) = c, again by Lemma 10.12, and therefore d ≈ c. Thus

st(V,P(d)) = st(V,P(c)), and then (9.17.1) shows st(V, d) = st(V, c).


25.10. Proof of the exactness of (25.4.3). Let (δ, e) and (δ, f) be in P.From (25.4.1) we see θ(wewf ) = s2

δ = Id, so K ⊂ Ker(θ) by (25.3.2). We show nextthat K is a normal subgroup of M . By the definition of M and K, it suffices that

Int(wc) · (wewf ) ∈ K (1)

for all (γ, c) ∈P. Let k = we wf ∈ K. Then

Int(wc) · k = wc(

Int(k) · w−1c

)k,

so (1) will follow if we can show Int(k) · w−1c = wd for some (γ, d) ∈P.

Since e ≈ f by (23.14.2), Proposition 13.2 says that k = b(e+ + f+, f−) andthat the relation B(e+ + f+, f−) holds in G. Putting h = π(k) = β(e+ + f+, f−) ∈Aut(V ), we therefore have

Int(k) · xσ(v) = xσ(hσ(v))

for all v ∈ V σ, which implies, by (9.11.2) and (9.11.5),

Int(k) · w−1c = Int(k) · w−c = wh(−c).

Since both e and f belong to Vδ, it follows from the composition rules for a rootgrading (20.1.1) that h(Vγ) = Vγ , and h induces an automorphism of Vγ . Henced := h(−c) ∈ V ×γ , as desired.

Now the homomorphism θ factors via M/K, so we obtain the commutativediagram

Mcan //

θ ""DDDDDDDDD M/K

θ′vvvvvvvvv

W(S)

of surjective group homomorphisms, and it remains to show that θ′ is injective.This will be done by defining a homomorphism τ : W(S) → M/K satisfying

τ θ′ = IdM/K as follows. By (25.3.4), W(S) is generated by sδ : δ ∈ ∆max. Let

δ ∈ ∆max and choose e ∈ V ×δ , so that (δ, e) ∈P. Then

tδ := can(we) ∈M/K (2)

is independent of the choice of e. Indeed, if also (δ, f) ∈P, then wf = we (w−ewf )and −e ∈ V ×δ , so w−ewf ∈ K. We claim that there exists a group homomorphismτ : W(S)→M/K satisfying

τ(sδ) = tδ for all δ ∈ ∆max.

Assuming the existence of τ , we then have, for all (δ, e) ∈P, that τ(θ′(can(we))) =τ(θ(we)) = τ(sδ) = tδ = can(we). Hence τ θ′ = IdM/K , so θ′ is injective.

To prove the existence of τ , we use the presentation of W(S) given in [63,Corollary 18.12], and therefore must show that the tδ satisfy the following sixrelations (i) – (vi), where always α, β, γ, δ ∈ ∆max:

(i) t2α = 1,


(ii) tαtβ = tβtα if α ⊥ β,

(iii) tαtβtα = tβtαtβ if α β,

(iv) tαtβtα = t2α−β if α← β,

(v) tαtβtα = tδtγtδ if (δ, α, β, γ) is a square or a kite,

(vi) tαtβtα · tγ = tγ · tαtβtα if α β and 〈γ, α∨〉 = 1 = 〈γ, β∨〉.

Relation (i) follows from (2) and the definition of K in (25.3.2): t2α = can(w2e) =

1M/K . Before continuing with the remaining five relations let us remark thatLemma 25.9 is applicable by putting

S := pr2(P) =⋃

δ∈∆max

V ×δ .

In particular, (25.9.1) and (25.9.2) imply (ii) and (iii).

Proof of (iv): Let (α, d) ∈ P and (β, e) ∈ P, so that tα = can(wd) andtβ = can(we). Then d ` e and f = Qde ∈ V ×2α−β . Hence

tαtβtα = tαtβt−1α (by (i)) = can(wdwew

−1d ) = can(wf ) (by (25.9.3)) = t2α−β .

Proof of (v): We represent tα, . . . , tδ as

tα = can(wd), tβ = can(wc), tγ = can(we), tδ = can(wf ),

and get, using (i),

tαtβtα = can(wdwcw−1d ), tδtγtδ = can(wfwew

−1f ).

Since α δ γ ⊥ α in the case of a square and α ← δ → γ α in the caseof a kite, we have the corresponding relations among the idempotents d, f, e, thatis d > f > e ⊥ d and d ` f a e > d in the respective cases. Now Lemma 25.9(d)shows that c′ = d f e is an idempotent. Moreover, c′ ∈ Vα−δ+γ = Vβ , andc′ ≈ c by Lemma 25.6(b) and Lemma 25.8(d). Hence c′ ∈ V ×β by (23.14.2), so thattβ = can(wc′). Without loss of generality we may therefore assume c = c′, and then(v) follows from (25.9.4).

Proof of (vi): The stated conditions and Lemma 15.6 imply that the configura-

tion of α, β, γ is eitherγ

α

β

<< orγ

α

@@β

^^<< . Let

tγ = can(wd), tα = can(we), tβ = can(wf ).

The relations between α, β, γ imply e > f and either e a d ` f or e > d > f .By Lemma 25.9(e), c = d − f e e f d is an idempotent associated with d.

Since f e e f d ∈ Vβ−α+α−β+γ = Vγ it follows that c ∈ Vγ , and therefore c ∈ V ×γby (23.14.2). Hence tγ = can(wc) and by (25.9.5) and (i) we obtain

tαtβtα · tγ = can(wewfw−1e · wd) = can(wc · wewfw−1

e ) = tγ · tαtβtα.



25.11. The subgroup M0(G) of the monomial group. We work in thesetting of 25.3: V is a Jordan pair with a Γ -grading R and G ∈ stw(V,R). Recallthe subgroup N = NormG(U+) ∩NormG(U−) ⊂ G of Lemma 9.2. We define

M0 = M0(G) = M(G) ∩N, (1)

and denote the restriction of the homomorphism θ: M → W(S) of Theorem 25.4to M0 by

θ0: M0 →W(S).

By Proposition 15.20(c), we identify the Weyl group W(S0) with a subgroup ofW(R0).

25.12. Proposition. (a) Let G ∈ stw(V,R) and θ as in (25.4.1). Then theexact sequence (25.4.3) restricts to an exact sequence

1 // Kinc // M0(G)

θ0 //W(S0) // 1 . (1)

(b) M0(G) ⊂ G0 where G0 is defined in Lemma 9.2.

Proof. (a) For k ∈ K = Ker(θ) we have, by (25.4.2), Int(k)·Uα = Uθ(k)(α) = Uαfor all α ∈ R. This implies Int(k) · Uσ = Uσ by (21.1.4), so k ∈M ∩N = M0.

To show that W(S0) ⊂ θ(M0), let µ ∈ S×0 and write µ in the form µ = β−α withα, β ∈ S1 = ∆max and 〈β, α∨〉 = 1, as in Proposition 17.9(c). Let (α, e) and (β, f)be in P, and choose any cog E ∈ cog(R) defined on ∆max. Then e ≈ E (α) = eαand f ≈ E (β) = eβ . After modifying E as in Lemma 23.16, we may assume thate = eα and f = eβ . Hence it follows from (23.1.2) that f ∈ V〈β,α∨〉(e) = V1(e). ByLemma 13.10 and 25.3,

wf,e = Int(we) · wf ∈M ∩N ∩G0 = M0 ∩G0, (2)

and by (25.4.1) and (15.17.3),

θ(wf,e) = θ(

Int(we) · wf)

= sαsβsα = sµ. (3)

Conversely, let g ∈ M0. Then w := θ(g) ∈ W(S) by Theorem 25.4, and weclaim that w ∈W(S0). Let us first show that w(Γ×(R)) ⊂ R1.

Thus consider α ∈ Γ×(R) so that Vα 6= 0. If V +α 6= 0, then 1 6= x+(V +

α ) =Uα ⊂ U+ and 1 6= Int(g) · Uα = Uβ where β = w(α) ∈ R, by (25.4.2). Sinceg ∈ M0 and Uα ⊂ U+, also Uβ ⊂ U+. This implies β ∈ R1. Indeed, assume tothe contrary that β ∈ R−1 ∪ R0. The definition of the root groups for µ ∈ R0 in(21.1.2) shows Uµ ⊂ G0. Hence β ∈ R−1 ∪R0 implies Uβ ⊂ U− ∪G0 and thereforeUβ ⊂ U+ ∩

(U− ·G0

)= 1 by Lemma 9.2(a), contradiction. The same argument

works in case V −α 6= 0, so we have shown w(Γ×(R)) ⊂ R1 = Γ . But then (25.1.1)together with (15.21.2) proves w ∈W(S0).

(b) Let g ∈ M0. By (1), θ(g) = sµ1 · · · sµn where µi ∈ S0. As in the proof of(a), write µi = βi − αi as differences of elements of S1 with 〈β, α∨〉 = 1, andchoose (αi, ei) and (βi, fi) in P. Then gi = wfi,ei ∈ M0 ∩ G0 by (2). Putg′ = g1 · · · gn. Then (3) implies θ(g′) = θ(g), so g = g′ k for some k ∈ K =Ker(θ). Let (δ, e) and (δ, f) ∈ P. Then e and f are associated by (23.14.2),so it follows from Proposition 13.2 that wewf = b(e+ + f+, f−) and thereforeπ(wewf ) = β(e+ + f+, f−) ∈ Aut(V ), so that wewf ∈ G0. Hence K ⊂ G0, andtherefore g ∈ G0.


25.13. Corollary. We keep the assumptions and notation of Theorem 25.4 andProposition 25.12. Recall the group Aut(Γ, (V,R)) defined in (20.5.2) and the groupAutΓ (V,R) defined in (20.3.2) which we identify with the subgroup of all (IdΓ , h)of Aut(Γ, (V,R)). Let G ∈ stw(V,R) and let π: G → PE(V ) be the canonicalprojection. Then there is a homomorphism

ψ: M0(G)→ Aut(Γ, (V,R)), g 7→ (θ(g), π(g)),

and ψ maps K to AutΓ (V,R).

Proof. We know M0(G) ⊂ N ∩G0 by (25.11.1) and Proposition 25.12(b). Henceπ(g) = (h+, h−) ∈ PE(V ) ∩Aut(V ) by Lemma 9.2, and it follows that

Int(g) · xσ(v) = xσ((hσ(v)

)for all v ∈ V σ. On the other hand, by (25.4.2),

Int(g) · xσ(V σγ ) = Int(g) · Uσγ = Uθ(g)·σγ = xσ(V σθ(g)·γ).

Hence h(V σγ ) = V σθ(g)·γ , i.e., (θ(g), π(g)) ∈ Aut(Γ, (V,R)). Obviously, ψ is a homo-

morphism, and if g ∈ K then θ(g) = Id, and therefore ψ(g) ∈ AutΓ (V,R).

§26. Centrality results

26.1. Lemma. Let (R′, X ′) be a subsystem of of a finite root system (R,X)satisfying R∩ spanZ(R′) = R′ and rankR > 4 rank(R′). Then there exists 0 6= α ∈R such that α ⊥ R′.

Proof. We use the categorical equivalence between RS and RSR of Proposi-tion 2.9(c) and view R and hence also R′ embedded in XR = spanR(R). As asubgroup of the free group X, spanZ(R′) is a free. This implies spanZ(R′)⊗Z R =spanR(R′), whence R ∩ spanR(R′) = R′. By [18, VI, §1.7, Proposition 24] thereexist root bases B′ of R′ and B of R such that B′ ⊂ B. For β′ ∈ B′ define its linkset

L(β′) = β ∈ B B′ : 〈β′, β∨〉 6= 0.

By [18, VI, §4.1, Lemme 2], |L(β′)|63. Assume no element of B B′ is orthogonal toB′. Then B = B′∪

(⋃β′∈B′ L(β′)

), whence |B|6|B′|+3|B′| = 4|B′|, contradiction.

Hence some α ∈ B B′ is orthogonal to B′ and therefore to R′.

26.2. Lemma. Let R =⊕

i∈I R(i) be the decomposition of a root system

(R,X) into irreducible components.

(a) For each i ∈ I, choose 0 6= δi ∈ R(i). Then there is a well-defined linear formf : X → Z given by f(x) =

∑i∈I〈x, δ∨i 〉. The set A := A(f) := α ∈ R : f(α) > 0

is a nilpotent subset of class 6 4.

(b) Suppose that F is a finite subset of R and that each R(i) has infinite rank.Then the δi may be chosen in such a way that 〈F, δ∨i 〉 = 0 and hence F ⊂ Ker(f).

Proof. (a) Since X is spanned by R, every x ∈ X belongs to the Z-span offinitely many R(i). Hence there are only finitely many i ∈ I such that 〈x, δ∨i 〉 6= 0.

§26] Centrality results 339

This shows that f is well-defined. It is clear that f is a positive functional for A asdefined at the end of 1.3. If α ∈ A then α ∈ R(i) for some i, hence f(α) = 〈α, δ∨i 〉64by well-known properties of the Cartan numbers, see [18, VI, §1.3] or [63, 3.4]for locally finite root systems. Since A is obviously closed in R, it follows fromLemma 1.10 that A is nilpotent of class at most 4.

(b) Let F (i) = F ∩ R(i). Since R(i) has infinite rank and F (i) is finite, itfollows from Lemma 26.1, applied to a sufficiently large subsystem S of R(i) withF (i) ⊂ S, that there exists 0 6= δi ∈ R(i) orthogonal to F (i). For j 6= i it is clearthat F (j) ⊥ δi, so we have F ⊥ δi, and the assertion follows.

26.3. Proposition. Let V be a Jordan pair with a Γ -grading R, let E ∈cog(R) defined on ∆ = dom(E ), and assume that ∆ contains Γ 0 and the endpoints of all isolated arrows. Let G ∈ st(V,R,E ) and let f and A ⊂ R be as inLemma 26.2(a). Then, in the notation of 3.2, G is generated by UA ∪ U−A.

Proof. In a first step, we reduce this to the case where Γ is connected. LetΓ (i) be the connected components of Γ , let (R,R1) be the 3-graded root system

associated with Γ and let (R(i), R(i)1 ) be the irreducible components of (R,R1),

corresponding to the Γ (i). By Proposition 22.12, V is the direct sum of ideals V (i),each of which is R(i)-graded, and G is generated by groups G(i) ∈ st(V (i),R(i)). ByLemma 23.11(a), (V (i),R(i)) is idempotent with respect to the cog E (i) = E

∣∣(∆ ∩Γ (i)). One checks easily that G(i) ∈ st(V (i),R(i),E (i)). Put A(i) = A ∩R(i). Then

A =⋃i∈I A

(i), and it suffices to show that each G(i) is generated by UA(i) ∪U−A(i) .This reduces us to the case where R irreducible and A has the form

A = A(δ) = α ∈ R : 〈α, δ∨〉 > 0,

and we must show that G is generated by the set

M(δ) := UA(δ) ∪ U−A(δ).

Let R = R−1 ∪ R0 ∪ R1 be the 3-grading of R. In a second step, we reduceto the case where δ ∈ R1 = Γ . Since A(−δ) = −A(δ) and R−i = −Ri, it isclear that we may assume, replacing δ by −δ if necessary, that δ ∈ R0 ∪ R1. Ifδ ∈ R0 we write δ = β − α with α, β ∈ Γ , 〈β, α∨〉 = 1 and α ∈ Γ 0 ∪ Γiso ⊂ ∆as in Proposition 17.9(c). In fact, because of 〈β, α∨〉 = 1 we either have β α,whence α ∈ Γ 0, or β ← α, in which case α is the end point of an isolated arrow.Then δ = sα(β) and since sα is an automorphism of the root system R, we haveA(δ) = sα(A(β)). By definition of st(V,R,E ) in 24.1, the element wα of (24.1.1)is a Weyl element for (G, (U%)%∈R). Hence

M(δ) = Int(wα) ·M(β),

so M(δ) generates G if and only if M(β) does. Therefore, we may replace δ by βand assume that δ ∈ Γ .

Let G′ be the subgroup of G generated by M(δ). Since G is generated by allU±γ , γ ∈ Γ , it suffices to show that U±γ ⊂ G′, for all γ ∈ Γ .

Let Γi = Γi(δ) = γ ∈ Γ : 〈γ, δ∨〉 = i as in 16.7. Then Γ = Γ2 ∪ Γ1 ∪ Γ0 andΓ2 ∪ Γ1 ⊂ A(δ), so that U±α ⊂ G′ for all α ∈ Γ2 ∪ Γ1. It remains to show thatU±β ⊂ G′ for all β ∈ Γ0. For this, it suffices that


∃ α1, . . . , αn ∈ ∆ ∩ (Γ2 ∪ Γ1) such that % := sα1 · · · sαn(β) ∈ A(δ) ∪A(−δ). (1)

Indeed, assuming (1), we have Weyl elements wαi ∈ G because αi ∈ ∆, and evenwαi ∈ G′ because αi ∈ Γ2 ∪ Γ1. Since also U% ∈ G′ the proof follows from theobservation

U±β = Usαn ···sα1 (±%) = Int(wαn) · · · Int(wα1) · U±% ⊂ G′.

Now let us prove (1). Since R is irreducible, Γ is connected, so Lemma 16.5 showsthat β ∈ Γ0 is connected to δ ∈ Γ2 by a chain of length 2. Hence there exists α ∈ Γsuch that δ ∼ α ∼ β ⊥ δ. For the induced subgraph on these three roots, there areby Lemma 15.6 the following possibilities:

(a) δ α β (b) δ // α β (c) δ α βoo (d) δ // α βoo

We now discuss these in turn.

Case (a). By (C2) of 15.3(b), δ, α, β generate a square

α β

δ γ

This shows α, γ ∈ Γ1 ∩ Γ 0, and by (15.3.2),

sγsα(β) = sγ(β − α) = β − γ − α = −δ ∈ Γ2 ⊂ A(δ).

Case (b). Complete δ → α β to a kite, using condition (C3) of a Jordangraph:

δ

????

α???? γ

β

Then α ∈ Γ1 ∩ Γ 0 and sα(β) = β − α ∈ A(−δ), because 〈β − α, δ∨〉 = 0− 1 = −1.

Case (c). Complete δ α← β to a kite and further to a hexagram:

β

????

α γ

β′

??// δ

????

β′′

__???oo

Then α ∈ Γ1 ∩ Γ 0 and sα(β) = β − 2α = −β′ ∈ −Γ2 ⊂ A(−δ).

Case (d). Since (δ, α, β) is a collision, we have sα(β) = −δ as in case (c), and(1) holds as soon as α ∈ ∆. By our assumption on ∆ this is so whenever δ → α

§26] Centrality results 341

is an isolated arrow, and it is also the case when δ → α is an arrow of hermitiantype since then α ∈ Γ 0 by Proposition 17.7(a). In the remaining case, δ → α is anarrow of orthogonal type by Lemma 17.3. By (17.2.4) δ → α embeds in a pyramid

δ

???? γ

α

γ′

??β

__????

Then δ, γ ∈ Γ 0 ∩ (Γ2 ∪ Γ1) and

sδsγ(β) = sδ(β − γ) = β − γ + δ = γ′ ∈ Γ1 ⊂ A(δ),

again by (15.3.2).

26.4. Lemma. Let V,R,E be as in Proposition 26.3 and let ϕ: G → G′ bea morphism of groups in st(V,R,E ). Let f and A be as in Lemma 26.2(a), letS = R ∩ Ker(f) and put H = US =

⟨Uβ : β ∈ S

⟩. Then H normalizes UA and

U−A, and H ∩Ker(ϕ) ⊂ Z (G).

Proof. We show that(((((((A,S

)))))))⊂ A. Indeed, by (1.6.2) an element of

(((((((A,S

)))))))has

the form γ = α1 + · · ·+αp+β1 + · · ·+βq where αi ∈ A, βj ∈ S, and p, q>1. Hencef(γ) = f(α1) + · · · + f(αp) > p > 0. In particular, for all α ∈ A and β ∈ S theopen root intervals

(((((((α, β

)))))))⊂ A and

(((((((− α, β

)))))))⊂ −A. Since G has R-commutator

relations, it follows from (3.4.4) that all Uβ , β ∈ S, normalize UA and U−A, andtherefore so does H.

Let h ∈ H ∩ Kerϕ and u ∈ UA. Then huh−1 = u′ ∈ UA and thereforeϕ(huh−1) = ϕ(u) = ϕ(u′). By Lemma 26.2, A is nilpotent. Hence Corollary 24.9(d)implies that ϕ

∣∣UA is injective. It follows that u = u′ = huh−1, so h centralizes UA.One proves in the same way that h centralizes U−A. But UA ∪U−A generates G by26.3 whence h ∈ Z (G).

26.5. Theorem. Let V be a Jordan pair with a Γ -grading R, let E ∈ cog(R)and assume Γ 0 ⊂ dom(E ). Assume furthermore that every irreducible component ofthe root system R associated with Γ has infinite rank. Then st(V,R) = st(V,R,E )and every morphism ϕ: G→ G′ of groups in st(V,R,E ) has central kernel.

Proof. Since G is generated by the root groups (Uα)α∈R, an element h ∈ Kerϕis contained in UF for a finite subset F of R. By Lemma 26.2(b), there exists alinear form f with the properties of 26.2(a) and such that F ⊂ S = R∩Ker(f). LetH = US . The assumption that R have no component of finite rank implies that Γcontains no isolated arrows, i.e., Γiso = ∅, because such an arrow would generatea component of type C2 of R. The rank assumption also implies that Γ doesnot contain a connected component of type T3, so that st(V,R) = st(V,R,E ) byTheorem 24.2. Hence Lemma 26.4, which relies on Proposition 26.3, is applicable inthe present situation. Now h ∈ H ∩Ker(ϕ) ⊂ Z (G) follows from Lemma 26.4.

For the next result, recall the Steinberg group St(V,R,E ) defined in 24.1, aninitial object of the category st(V,R,E ). In the setting of Theorem 26.5, we havest(V,R,E ) = st(V,R) and therefore St(V,R,E ) = St(V,R).


26.6. Corollary. (a) Under the assumptions of Theorem 26.5 the kernel ofthe canonical homomorphism π: St(V,R)→ PE(V ) is central.

(b) Moreover, if none of the irreducible components of R has type CI , thenKer(π) = Z (St(V,R)) and PE(V ) = FPE(V ).

Proof. (a) This is the special case G = St(V,R) and G′ = PE(V ) of Theo-rem 26.5.

(b) Here Γher = ∅, so PE(V ) = FPE(V ) is centreless by Lemma 23.22(e). Hencethe centre of St(V,R) is contained in Ker(π).

Remark. We will show in Corollary 27.6 that π: St(V,R) → PE(V ) is auniversal central extension of PE(V ).

Notes

§23. The terminology “cog” was first introduced in [76] for a family of tripotents whose

members are either [c]ollinear, [o]rthogonal, or [g]overning.Corollary 24.16 of §24 is an adaptation of [94, Lemma 37(a)] to our setting.

§25. Special cases of Theorem 25.4 are known. For Γ = Kp Kq , V = Mpq(A), A anassociative algebra, R the idempotent Γ -grading of 23.23 and G = Stn(A), n = p+ q, our groups

M(G) and K are the groups Wn(A) and Hn(A) of [32, 1.4E]. In this case our Main Theorem 25.4

is proved in [32, 1.4.18]. Similarly, in the unitary case (Γ = Tn or Γ = Tn) our groups M(G) andK become the groups introduced in [32, 5.5F*], and 25.4 becomes [32, 5.5.13 and 5.5.14], stated

there without proof.

If V is a finite-dimensional Jordan pair over a field and all V ±γ , γ ∈ Γ , are one-dimensional,the exact sequence (25.4.5) becomes the exact sequence 1→ Tk → Nk →W → 1 studied in [99].

The notion of quadrangle goes back to K. McCrimmon [70] where this concept was introduced

for tripotents in Jordan triple systems. The formulas (25.5.2) and (25.5.3) are proved in [70, 3.3],and Lemma 25.6 is stated as [70, 3.2] with a hint how to prove it. Diamonds in Jordan triple

systems were introduced in [76, I.2.4] and in Jordan pairs in [79, 1.4].

§26. In the setting of 26.5 assume that the Jordan pair V is special (it can be shown that thisis in fact always the case), embedded into a Morita context M. Then the canonical homomorphism

π: St(V,R)→ PE(V ) factors as π = τ ψ: St(V,R)→ E(M, V )→ PE(V ). Since Ker(π) is central,so is Ker(ψ), a classical result [32, 1.4.6 and 5.5.6].

The assumption in 26.5 and 26.6 that all irreducible components have infinite rank can be

disposed of at the expense of imposing stronger conditions on St(V,R) [59, 1.12] or on R, see

for example [32, 1.4.15] and [105] in case R is finite of type A. Centrality has recently also

been shown for the other irreducible finite types in case all V ±γ , γ ∈ Γ , are free of rank 1 over

a commutative ring and some E ∈ cog(R) is defined on all of Γ [48, 49, 89]. All these papers

require that the rank of R be not too small. Centrality is known to be false for R = A2 [111].

CHAPTER VI

CENTRAL CLOSEDNESS

Summary. This chapter contains our main results. Let R = (Vγ)γ∈Γ be a root grading

of a Jordan pair V , idempotent with respect to a sufficiently large cog, and assume that everyconnected component of Γ has rank at least 5. Then Theorem 27.4 states that the Steinberg

group G = St(V,R) defined in 22.1 is centrally closed. In fact, we prove a more precise resultwhich also allows certain components of rank 4. As a consequence, we show in Corollary 27.6: if

G′ ∈ st(V,R) and the canonical homomorphism G→ G′ has central kernel then G is the universal

central extension of G′. This is always the case if all connected components of Γ have infiniterank.

To help the reader in following the somewhat long and technical proofs, we sketch the main

steps in §27 and carry them out in the following sections. In short, we have to prove that for anycentral extension p: E → G there exists a unique section s: G→ E of p. Now G is in particular a

group over V , thus generated by the abelian subgroups U± ∼= V ±. If s exists then the pre-images

p−1(U±) are abelian subgroups of E. Therefore, we first prove this necessary condition in §28and §29. Next, we show, in §30 and §31, that the induced central extensions p−1(U±) → U±

admit sections s±. Finally, it remains to show that the s± satisfy the defining relations of G and

thus yield a section s: G→ E. The proof of this last and final step is contained in §32.Here is a more detailed outline of the proof. An essential tool are the so-called binary and

ternary symbols. By §9, G is generated by two subgroups U+ and U−, endowed with canonical

isomorphisms xσ : V σ → Uσ where σ = ±. Let x, z ∈ V σ . Then the commutator of p−1(xσ(x))and p−1(xσ(z)) consists of a single element

[[x, z ]

]∈ Ker(p), called the binary symbol defined

by x and z. The vanishing of these symbols is equivalent to the commutativity of p−1(U±). It

is easily seen that[[x, z ]

]is an alternating function of x and z, invariant under the elementary

automorphism group of (V,R). Such functions are studied, independently of central extensions,

in §28. They turn out to be trivial in many cases, but not in all (28.13). However, if they arise

from a central extension, we show in §29 that they do vanish under the assumptions of our mainresult and even in somewhat greater generality (Theorem 29.12).

The ternary symbols⟨〈x, y, z〉

⟩are defined for a quasi-invertible pair (x, y) ∈ V σ × V −σ and

an arbitrary z ∈ V σ by a similar process. They show a much more complicated behaviour than

the binary ones. The main result of §30 is Theorem 30.14, a vanishing theorem for certain ternary

symbols (see the first part of the formula below).This paves the way for the proof of the second step, the existence of the partial sections s±,

in §31. We first define sections on the root spaces V ±γ in Propositions 31.4, 31.6, 31.9, dependingon the way in which γ is embedded in the graph Γ , and then multiply them together to obtainthe partial sections.

The relations which the partial sections must satisfy correspond to the relations (StR1) and

(StR2) of Theorem 21.7. While the second set of relations is relatively easy to verify, the first setamounts to the following. For α, β, γ ∈ Γ , α 6= β let δ = α − β + γ and ε = 2(α − β) + γ. Then

the relations corresponding to (StR1) hold in E if and only if

⟨〈xα, yβ , zγ〉

⟩=

1 if δ 6∈ Γ ,

sσ(xσ(− xα yβ zγ

))if δ ∈ Γ , ε 6∈ Γ ,

sσ(xσ(− xα yβ zγ+QxαQyβ zγ

))if δ, ε ∈ Γ ,

for all xα ∈ V σα , yβ ∈ V −σβ and zγ ∈ V σγ . The first case is the content of Theorem 30.14. In theremaining cases the proof consists of a detailed discussion of the possible configurations of α, β, γ,

carried out in §32.

343

344 CENTRAL CLOSEDNESS [Ch. VI

§27. Statement of the main result and outline of the proof

27.1. Extensions and central extensions. Let G be a group. An extensionof G is a surjective group homomorphism p: E → G. If p is clear from the context,we sometimes speak simply of E as an extension of G. A section or splitting of Eis a homomorphism s: G→ E such that p s = IdG. Then E ∼= Ker(p) o s(G) is asemidirect product.

The extension is called central if Ker(p) ⊂ Z (E). If a central extension admitsa splitting then clearly E ∼= Ker(p)× s(G) is a direct product.

A morphism between extensions q: X → G and p: E → G is a group homomor-phism f : X → E such that q = pf . With this notion of morphisms the extensionsof G form a category, and the central extensions form a full subcategory. An initialobject of the latter is called a universal central extension of G. Thus a centralextension q: X → G is universal if and only if for all central extensions p: E → Gthere exists a unique f : X → E such that q = p f :

X∃! f //

q @@@@@@@@ E

p~~~~~~~~

G

(1)

The question of which groups admit universal central extensions is answered by thefollowing classical result. Recall that a group G is said to be perfect if equals itscommutator group D(G) =

(((((((G,G

))))))).

27.2. Theorem. A group has a universal central extension if and only if it isperfect.

Proof. See, for example, [32, 1.4.11], [83, Theorem 4.1.3] or [109, Theo-rem 6.9.5].

27.3. Centrally closed groups. A group X is called centrally closed ifIdX : X → X is a universal central extension. An equivalent condition is: ev-ery central extension p: E → X splits uniquely. Indeed, X is centrally closedif and only if for every central extension p: E → X there exists a unique grouphomomorphism f : X → E such that p f = IdX , which is the second condition.

By Theorem 27.2, a centrally closed group is perfect. It is easy to give a directproof as follows. Let A = X/D(X) and consider the central extension p: X×A→ Xgiven by projection onto the first factor. Then we have sections s and s′ of p givenby s(x) = (x, 1A) and s′(x) = (x, can(x)), where can(x) is the canonical image ofx in A. Since X is centrally closed, s = s′ which means X = D(X).

We can now state our main result.

27.4. Theorem. Let Γ be a Jordan graph with associated 3-graded root system(R,R1). Let V be a Jordan pair with a Γ -grading R which is idempotent with respectto a cog defined on a subset ∆ of Γ containing Γ 0. Assume that

(i) every connected component of Γ has rank > 4,

(ii) if Σ is a connected component of type O3 = G (Bqf4 ) or T4 = G (Cher

4 )then Σ ⊂ ∆,

§27] Statement of the main result and outline of the proof 345

(iii) Γ contains no connected components of type T4 = G (Dalt4 ).

Then the Steinberg group St(V,R) defined in 22.1 is centrally closed.

We discuss the case where Γ is connected, equivalently, where (R,R1) is irre-ducible, in more detail. It is immediate from the classification of 3-graded rootsystems in 14.7 and the description of the associated Jordan graphs in 14.17 –14.20 (see (14.20.12) for a comprehensive list) that the conditions listed above areequivalent to Γ and ∆ being one of the following:

Γ = KI KJ with 16 |I|6 |J | and |I|+ |J |> 5, (1)

Γ = TI or Γ = TI with |I|> 5, (2)

Γ = OI or Γ = OI with |I|> 4, (3)

Γ = T4, (4)

Γ = O3, (5)

Γ = Cl or Γ = Sch, (6)

and ∆ = Γ in all cases, except possibly for Γ = TI with |I| > 5 and Γ = OI with|I|> 4 where we only require Γ 0 ⊂ ∆ ⊂ Γ .

The proof of this theorem will occupy the sections §28 – §32. A preliminaryreduction and an outline of the proof is given in 27.7 and 27.10.

27.5. Lemma. Let X be a centrally closed group, let K be a normal subgroupof X and let q: X → G := X/K be the canonical projection. Let p: E → G be acentral extension. Then there exists a unique homomorphism f : X → E such thatq = p f . In particular, if K is central then q is a universal central extension of G.

Proof. Let E′ be the pull-back of q and p:

E′ = X ×G E = (x, e) ∈ X × E : q(x) = p(e).

Denoting by p′ = pr1: E′ → X and q′ = pr2: E′ → E the canonical projections, wehave the commutative diagram

E′q′ //

p′

E

p

X

∃! s

OO

q//

f

88qqqqqqqqqqqqqqqqqG

where p′ is a central extension. Indeed, p′ is surjective because p is surjective. AlsoKer(p′) = 1X ×Ker(p) shows that Ker(p′) is central.

Since X is centrally closed, there exists a unique section s: X → E′ of p′. Thishomomorphism has the form s(x) = (x, f(x)) where f : X → E is a homomorphism.Since s(x) ∈ E′, we have q(pr1(s(x))) = p(pr2(s(x))), equivalently, q(x) = p(f(x)),for all x ∈ X, and f is unique with this property by the uniqueness of s.


27.6. Corollary. With the notation and assumptions of Theorem 27.4, let G ∈st(V,R) and assume that the canonical homomorphism κ: St(V,R)→ G of (22.1.1)is a central extension. Then St(V,R) is a universal central extension of G. This isalways so if every irreducible component of R has infinite rank.

Proof. The first statement follows from Lemma 27.5, and the second fromTheorem 26.5.

27.7. Towards the proof of Theorem 27.4. Our aim is to show that theSteinberg group St(V,R) is centrally closed, equivalently, that every central exten-sion of it splits uniquely. We will show below in Proposition 27.8 that St(V,R) isperfect. This obviates proving uniqueness of a splitting, because of the followingwell-known fact:

A perfect group X with the property that everycentral extension of X splits is centrally closed.

(1)

Indeed, let p: E → X be a central extension and let s and s′ be splittings of p. Thenf(x) := s′(x)s(x)−1 defines a homomorphism f : X → Ker(p) because f(xy) =s′(xy)s(xy)−1 = s′(x)s′(y)s(y)−1s(x)−1 = s′(x)f(y)s′(x)−1 = s′(x)s′(x)−1 · f(y)(since f(y) is central in X) = f(x)f(y). But a homomorphism from a perfect toan abelian group is trivial, which proves s′ = s.

27.8. Proposition. Let V be a Jordan pair with a Γ -grading R, let E ∈cog(R) and assume Γ 0 ⊂ dom(E ) and Γiso = ∅. Then the groups in st(V,R)are perfect.

Proof. Let G ∈ st(V,R) and define, as in 11.12,

Kσ = x−1σ (D(G)) =

a ∈ V σ : xσ(a) ∈ D(G)

, (1)

a subgroup of the additive group of V σ. SinceG is generated by x+(V +)∪x−(V −) =U+ ∪ U− and Uσ =

⊕ξ∈Γ xσ(V σξ ) by (21.1.4), it suffices to show that V σξ ⊂ Kσ

for all ξ ∈ Γ .Let Rξ : V = V2(ξ)⊕ V1(ξ)⊕ V0(ξ) be the Peirce grading of V associated with

ξ as in (20.7.1). By 21.5, a group G ∈ st(V,R) also belongs to st(V,Rξ). Thisallows us to apply the results of 11.12 on the commutator subgroup of G. For theconvenience of the reader, we list here the following three special cases of (11.12.1)– (11.12.3) which will be used in the sequel. Let V = V2 ⊕ V1 ⊕ V0 be a Peircegrading of V and let x2 ∈ V σ2 etc. Then

V σ2 V −σ1 V σ1 + V σ1 V −σ0 V σ1 ⊂ Kσ, (2)

Qx2Qy1z0 + x2y1z0 ∈ Kσ, (3)

Qx1Qy0z0 + x1y0z0 ∈ Kσ. (4)

We now distinguish the following cases.

Case 1: ξ = α ∈ Γ 0 is an internal vertex, whence α β for some β ∈ Γ 0.Then, by (23.22.1) and (2), V σα = V σα e−σβ eσβ ⊂ V σ2 (α), V −σ1 (α), V σ1 (α) ⊂ Kσ.

Case 2: ξ ∈ ∂Γ is an external vertex. Because of our assumption Γiso = ∅, itfollows from Proposition 17.7 that ξ is the initial point of an arrow of hermitiantype or the endpoint of an arrow of orthogonal type.

§27] Statement of the main result and outline of the proof 347

Case 2.1: ξ = ε is the end point of an arrow α → ε of orthogonal type.Let γ = 2ε − α. By (23.22.3), V σε = V σα V −σε V σγ . Let xα ∈ V σα ⊂ V σ2 (α),

yε ∈ V −σε ⊂ V −σ1 (α) and zγ ∈ V σγ ⊂ V σ0 (α). Then by (3),

xαyεzγ+QxαQyεzγ ∈ Kσ.

Here QxαQyεzγ ∈ V σ2α−(2ε−γ) = V σα because α + γ = 2ε. Since α ∈ Γ 0, we

have V σα ⊂ Kσ by Case 1. Hence also xαyεzγ ∈ Kσ. It follows that V σε =V σα V −σε V σγ ⊂ Kσ.

Case 2.2: ξ = β is the initial point of an arrow of hermitian type which weembed in a hexagram as in (17.2.2):

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

Consider the Peirce grading determined by δ. Then δ ∈ Γ2(δ), α ∈ Γ1(δ) andδ′′ ∈ Γ0(δ). Put Vi = Vi(δ) for short. By (23.22.2),

V σδ = Q(V σα )Q(V −σβ )V σδ′′ ⊂ Q(V σ1 )Q(V −σ0 )V σ0 .

We have VαVβVδ′′ ⊂ Vα−β+δ′′ (by (20.1.1)) = Vγ (since α − β + γ = δ), and byCase 1, Vγ ⊂ K. Hence (4) implies that V σδ = QVαQVβVδ′′ ⊂ QV1

QV0V0 ⊂ K.

Remark. Proposition 27.8 applies in particular to the cases listed in (27.4.1)– (27.4.3), but also to others. For example, it follows from the classification of3-graded root systems in 14.7 and Proposition 17.7 that the only connected Jordangraphs Γ with Γiso 6= ∅ are the cases where Γ is a single vertex or a collision. Thusfor a connected Γ , Proposition 27.8 applies also to the following eight cases whichdo not fall under Theorem 27.4 (for O3 and T4, the reason being that E need notbe defined on all of Γ ):

(R,R1) A13 A1

4 A24 Bqf

3 Cher3 Bqf

4 Cher4 Dqf

4

Γ K2 K3 K2 K2 O2 T3 O3 T4 T4

For Γ = K1 or Γ = O1∼= T2 one needs additional assumptions on V to deduce

perfectness of a group in the various subcategories of st(V ). For example, it followsfrom [59, Theorem 2.6] that for a nondegenerate simple Jordan pair V with dcc onprincipal inner ideals the group PE(V ) is perfect if and only if V is not isomorphic

to V ∼= (F2,F2), (F3,F3) (where Γ = K1) or H2(F2) (where Γ = O1). In thesethree exceptional cases PE(V ) is isomorphic to, respectively, S3, A4 and S6 [59,Lemma 2.2].


27.9. Steinberg’s Central Trick. Let p: E → G be a central extension,giving rise to an exact sequence of groups

0 // Ai // E

p // G // 1 (1)

where the abelian group A is written additively. For an element g ∈ G and a subsetU ⊂ G, we use the notation

Eg := p−1(g), EU := p−1(U) ⊂ E.

By surjectivity, the fibres of p are not empty and the product of two fibres is a fibre:

Eg 6= ∅ and EgEh = Egh (2)

for g, h ∈ G. Since K = Ker(p) ⊂ Z (E), the action of E on itself by innerautomorphisms induces an action of G on E by automorphisms via

Int(g) · b := Int(a) · b = aba−1, (3)

for any a ∈ Eg. This action leaves Z (E) pointwise fixed and maps fibres ontofibres:

Int(g) · Eh = Eghg−1 . (4)

A crucial fact for the theory of central extensions is Steinberg’s Central Trick :

the commutator of two fibres of p consists of a single element of E. (5)

Indeed, let Eg = aK and Eh = bK. Since K ⊂ Z (E) we have, for all z, w ∈ K,that

(((((((az, bw

)))))))= azbwz−1a−1w−1b−1 = aba−1b−1 =

(((((((a, b)))))))

. By abuse of notation,

we will therefore identify the one-element subset(((((((Eg, Eh

)))))))= (((((((a, b))))))) of E with its

unique element(((((((a, b)))))))

.

27.10. Outline of the proof of Theorem 27.4. Let G = St(V,R) and letp: E → G be a central extension as in (27.9.1). Our approach to proving theexistence of a section of p is the obvious one: since G is defined by generators andrelations, we define a map s from the generators of G to E such that their imagesin E satisfy the defining relations of G and p s is the identity on the generatorsof G.

The group G is in particular a group over V as in 9.1, thus it is generated byabelian subgroups Uσ equipped with isomorphisms xσ: V σ → Uσ, for σ ∈ +,−.If a section s of p exists then E ∼= A×s(G) (direct product) with central A = Ker(p),and hence also p−1(Uσ) ∼= A× s(Uσ), in particular:

if s exists then p−1(Uσ) is abelian. (1)

The proof of Theorem 27.4 proceeds now in three steps. First, we verify thenecessary condition (1):

Step 1: the groups p−1(Uσ) are abelian.

Let x, z ∈ V σ. By Steinberg’s Central Trick,

§27] Statement of the main result and outline of the proof 349[[x, z ]

]:=(((((((p−1(xσ(x)), p−1(xσ(z))

)))))))(2)

is a well-defined single element of E and satisfies p([

[x, z ]])

=(((((((

xσ(x), xσ(z))))))))

= 1

since Uσ is abelian. We call[[x, z ]

]the binary symbol defined by x and z. Thus[

[ , ]]: V σ × V σ → Ker(p) is a map which measures the lack of commutativity of

p−1(Uσ):

p−1(Uσ) is abelian ⇐⇒ the binary symbols vanish. (3)

Using the group isomorphism i: A → Ker(p), we define a map F : V σ × V σ → Aby i(F (x, z)) =

[[x, z ]

]. Then F is bi-additive, alternating and invariant under

the group EA(V,R) (Lemma 29.2). In §28 we study, independently of centralextensions, arbitrary alternating maps on V σ with values in an abelian group whichare invariant under EA(V,R) and show that in the cases (27.4.1), (27.4.2), (27.4.4)and (27.4.6) of 27.4 such maps are indeed trivial (28.13). This is in general notso for the cases (27.4.3) and (27.4.5), see Example 28.16(iii). However, if F arisesfrom a central extension E of G as above, then we show in Theorem 29.12 thatF = 0, finishing the proof of Step 1.

Step 2: The extensions p: p−1(Uσ)→ Uσ admit sections sσ: Uσ → E.

An important role is played by the ternary symbols, defined as follows. Suppose(x, y) ∈ V σ×V −σ is quasi-invertible, and let bσ(x, y) ∈ G be defined as in 9.7. Letz ∈ V σ. Then we put ⟨

〈x, y, z〉⟩

=(((((((p−1(bσ(x, y)), p−1(z)

))))))). (4)

Our interest in §30 concerns the case where x ∈ V σα and y ∈ V −σβ for α 6= β inΓ . The main result is Theorem 30.14: if α, β, γ ∈ Γ satisfy α − β + γ /∈ Γ then⟨〈V σα , V −σβ , V σγ 〉

⟩= 1.

Having assembled the necessary tools, we show in §31 that there exist grouphomomorphisms ϕσ: V σ → E satisfying p(ϕσ(u)) = xσ(u) for all u ∈ V σ andtherefore sections sσ: Uσ → E given by sσ(xσ(u)) = ϕσ(u).

Step 3: the sections sσ of Step 2 extend to a section s: G→ E of p.

Since G is generated by U+ and U−, it remains to show that the sσ preserve thedefining relations (StR1) and (StR2) of G. This is done in §32.

27.11. Comparison with known results. We indicate how our Main Theo-rem 27.4 compares to previously known results. Theorems about central closednessof appropriately defined Steinberg groups are usually only proved in the case ofgroups associated with irreducible root systems. We therefore restrict our discussionto this case as well. It will be seen that Theorem 27.4 is new for Γ = OI = G (Bqf

I )

and TI = G (CherI ), although special cases were known before. But we stress that

in all cases our approach is completely different from the previously known proofs,requires fewer relations and avoids case-by-case considerations to a large extent.The following survey does not follow the historical development and does not claimto be complete.

(i) Γ = Cl = G (Ebi6 ) or Γ = Sch = G (Ealb

7 ). By [79, 7.2, 7.3], up to base ringrestriction a Jordan pair with such an idempotent Γ -grading is isomorphic to the


exceptional Jordan pair M12(C) or H3(C) for C a split octonion algebra, see 6.6(e),(h).

These two cases are examples of split root gradings, defined by the requirementthat ∆ = Γ and that all root spaces V ±γ , γ ∈ Γ , are free of rank 1. One canshow that the Steinberg group of a split root grading is isomorphic to the Steinberggroup considered by Stein in [91] who proved its central closedness. Examples ofsplit root graded Jordan pairs exist for all Γ . Hence, in the following discussionthe interest is in the non-split case.

(ii) Γ = KI KJ , i.e., (R,R1) ∼= AII∪J , with 16 |I|6 |J | and |I|+ |J |> 5. By

[72, 3.4] and [79, 3.2.3], a Jordan pair with an idempotent Γ -grading of this type isisomorphic to the Jordan pair of rectangular matrices V = MIJ(A) over a ring A,discussed in 20.2(a) and 23.23. By (24.18.6), St(V,R) ∼= StN (A) with N = I ∪ J ,the linear Steinberg group, and by (24.18.2) the elementary linear group EN (A)belongs to st(V,R). Hence, we recover the known facts that StN (A) is centrallyclosed for |N |>5 and is the universal central extension of EN (A) for infinite N , theso-called Kervaire-Milnor Theorem [44, 74], [94, Theorem 14] or [32, 1.4.12 and1.4.13].

(iii) Γ = TI or Γ = TI , |I| > 5, so (R,R1) is isomorphic to DaltI or to Cher

I .By [72, 4.4] and [79, 6.1], the Jordan pairs with an idempotent root grading oftype TI are precisely the Jordan pairs AI(k) of alternating I × I-matrices oversome commutative ring k, see 6.6(b). The root grading is split, so that (i) applies.

Alternatively, one can view these Jordan pairs as examples of TI -graded Jordanpairs with ∆ & Γ .

An example of a Jordan pair with an idempotent root grading of type TI isthe Jordan pair HI(A, J, ε, Λ) of hermitian matrices over a form ring, discussed in23.24. If ∆ = Γ and under a mild additional condition, valid if 2 is invertible in thebase ring, these are all examples of Jordan pairs with an idempotent TI -grading[72, 5.6], [79, 4.1.2]. We recall from 6.6(d) that HI(A, J, ε, 0) = AI(A) sinceΛ = 0 forces ε = 1, J = IdA and A commutative.

The elementary group E(M, V ) of the Morita context associated with V =HI(A, J, ε, Λ) is the elementary unitary group EU2I(R,Λ), cf. [9, II, (5.1a)], [32,5.3.18], [59, (4.6b]. It then follows from [58, 2.8] that PE(V ) is a central quotientof EU2I(R,Λ). We leave it to the reader to verify that EU2I(R,Λ) ∈ st(V,R).

For countable I and V = HI(A, J, ε, Λ) our Theorem 27.4 follows from resultsof Sharpe and Bak [32, 5.5.10]. For |I|> 5 central closedness of St(V,R) is statedin a special case in [32, 5.5.11]. It has recently been proved in general by Lavrenov[50]. We point out that our results apply to all Jordan pairs V with an idempotent

TI -grading which need not be isomorphic to HI(A, J, ε, Λ).

(iv) Γ = OI or Γ = OI with |I|> 4, thus (R,R1) is isomorphic to DqfI+1 or to

BqfI+1. Up to base ring restriction, a Jordan pair with such an idempotent Γ -grading

is isomorphic to the Jordan pair of a quadratic form, described in Proposition 23.26.In particular, Jordan pairs with an OI -grading are split, so that (i) applies. (Thiscould also be deduced from the case Γ = TI since the corresponding elementarygroups are isomorphic.)

The Steinberg groups associated with Jordan pairs with an OI -grading andM0 6= 0 in the notation of 23.26 do not seem to have been considered before, apart

§28] Invariant alternating maps 351

from Boge’s paper [12] which deals with nondegenerate quadratic forms over infinitefields of characteristic 6= 2 and Witt index > 6, i.e., |I|> 6.

§28. Invariant alternating maps

28.1. Invariant alternating maps. Let X and A be additive abelian groups.An alternating map on X with values in A is a bi-additive map F : X × X → Asuch that

F (x, x) = 0

for all x ∈ X. Then F is in particular skew-symmetric:

F (z, x) = −F (x, z), for all x, z ∈ X,

which follows by expanding F (x + z, x + z) = 0. We denote by Alt(X,A) theabelian group of A-valued alternating maps on X. Let H be a group acting on Xby group automorphisms. For h ∈ H and F ∈ Alt(X,A) we define

Fh(x, y) = F (h · x, h · y).

and say F is H-invariant if Fh = F for all h ∈ H. The group of H-invariantalternating maps is denoted Alt(X,A)H .

Let V be a Jordan pair over k and let H be a subgroup of Aut(V ), acting onV + by projection onto the first factor:

h · x = h+(x),

for h = (h+, h−) ∈ H and x ∈ V +. We consider H-invariant alternating maps onX = ZV

+, the abelian group obtained from V + by restricting scalars from k to Z,with values in some abelian group A. By abuse of notation, we define

Alt(V,A)H := Alt(ZV+, A)H .

The preference of V + over V − is not essential: one can always replace V by V op.

Unless mentioned otherwise, we assume for the remainder of this section that Vis a Jordan pair with a Γ -grading R = (Vγ)γ∈Γ which is idempotent with respectto a cog E ∈ cog(R) whose domain of definition ∆ = dom(E ) contains Γ 0. We putH = EA(V,R), as defined in (20.7.7), and let F ∈ Alt(V,A)H .

As a matter of notation, we will frequently write F (α, β) instead of F (V +α , V

+β ),

for α, β ∈ Γ .

28.2. Lemma. Let β, γ ∈ Γ satisfy β γ or β → γ. Then 2β − γ /∈ Γ and3β − 2γ /∈ Γ .

Proof. In both cases, 〈γ, β∨〉 = 1. Hence 〈2β − γ, β∨〉 = 4 − 1 = 3 and〈3β − 2γ, β∨〉 = 6− 2 = 4, so the assertion follows from (15.1.1).


28.3. Lemma. Let α, β, γ, δ in Γ and assume β 6= γ.

(a) Let xα, uβ etc. be in the respective root spaces and assume

B(uβ , vγ)xα = xα. (1)

Then for all zδ ∈ V +δ ,

F (xα, uβ vγ zδ) = F (xα, QuβQvγzδ). (2)

In particular, this is the case when γ ⊥ α.

(b) If β ∈ Γ 0 thenF (β, β) = 0. (3)

(c) Let α, β ∈ Γ and suppose that there exists γ ∈ Γ such that one of thefollowing conditions holds:

(i) β γ ⊥ α,

(ii) β γ α and β − γ + α /∈ Γ .

ThenF (α, β) = 0. (4)

Proof. (a) From β 6= γ it follows that (uβ , vγ) is quasi-invertible by 20.7.The inner automorphism β(uβ , vγ) belongs to H, so since F is bi-additive andH-invariant,

F (xα, zδ) = F (B(uβ , vγ)xα, B(uβ , vγ)zδ)

= F (xα, zδ − uβ vγ zδ+QuβQvγzδ)

= F (xα, zδ)− F (xα, uβ vγ zδ) + F (xα, QuβQvγzδ),

which proves (2). If γ ⊥ α then (1) follows from (20.7.5).

(b) Since β ∈ Γ 0 there exists γ such that β γ. Then 2β − γ and 3β − 2γare not in Γ by Lemma 28.2, and

B(uβ , vγ)xβ = xβ − uβ vγ xβ+QuβQvγxβ = xβ

because uβ vγ xβ ∈ V +2β−γ = 0 and QuβQvγxβ ∈ V +

3β−2γ = 0. Hence (1) and

therefore (2) holds for α = β. Let δ = γ. Then also QuβQvγzγ ∈ V+2β−γ = 0, so

we have F (xβ , uβ , vγ , zγ) = 0 by (2). Specialize vγ = e−γ and zγ = e+γ . Then

uβ = uβ , vγ , zγ by (6.14.7) and uβ ∈ V +1 (eγ) which proves (3).

(c) In all cases, it suffices to show that

F (V +α , V +

β V −γ V +γ ) = 0 (5)

because V +β = V +

β e−γ e+γ by (23.22.1).

Case (i): By (a), (1) follows from γ ⊥ α so (2) holds. For δ = γ we haveQuβQvγzγ ∈ V

+2β−γ = 0, since β γ and therefore 2β − γ /∈ Γ by Lemma 28.2.

Thus the right hand side of (2) vanishes, which proves (5).

Case (ii): From β − γ + α /∈ Γ it follows that uβ vγ xα ∈ V +β−γ+α = 0. Since

γ α we have 〈α, γ∨〉 = 1 and therefore 〈2β − 2γ + α, γ∨〉 = 2− 4 + 1 = −1, so2β − 2γ + α /∈ Γ . This implies QuβQvγxα ∈ V2β−2γ+α = 0, so that B(uβ , vγ)xα =xα − uβ vγ xα + QuβQvγxα = xα. Thus (1) and therefore (2) holds. Finally,

let δ = γ in (2). Then QuβQvγzγ ∈ V +2β−γ = 0 because β γ and therefore

2β − γ /∈ Γ by Lemma 28.2. Now (5) follows from (2).


28.4. Proposition. If Γ be isomorphic to a complete graph KI with |I| > 3then Alt(V,A)H = 0.

Proof. Let α, β ∈ Γ . By (28.3.3), we have F (α, α) = 0. Since |I| > 3, anedge α β embeds in a triangle α, β, γ which is closed, because non-closedtriangles occur only in the non-simply laced (hermitian) case, see 19.8. HenceLemma 28.3(c)(ii) shows that F (α, β) = 0. Now F = 0 follows from V =

⊕γ∈Γ Vγ

and the fact that F is bi-additive.

28.5. Lemma. Let (α1, . . . , α4) be a square in Γ .

(a) Then F (αi, αi+1) = 0 and 2F (αi, αi+2) = 0 for all i = 1, . . . , 4, with indicestaken mod 4.

(b) If F (e+αi , V

−αi+2

) = 0 for one i then F (αi, αj) = 0 for all i, j.

Proof. By symmetry, we may assume i = 1, and then write (α1, . . . , α4) =(α, β, γ, δ) for simpler notation.

(a) By Lemma 28.3(c)(i), we have F (α, β) = 0. From γ β and δα it follows from Lemma 28.2 that 2γ − β /∈ Γ and 2δ − α /∈ Γ . Hence also2γ−2β+α = 2δ−α (by (15.3.2)) is not in Γ . Let xα and uγ be in the respective rootspaces. Since γ 6= β, (uγ , e

−β ) is quasi-invertible. Hence QuγQe−

βxα ∈ V +

2γ−2β+α = 0

and QuγQe−βe+β ∈ V +

2γ−β = 0. This implies B(uγ , e−β )xα = xα − uγ e−β xα and

B(uγ , e−β )e+

β = e+β − uγ e

−β e

+β = e+

β − uγ by (6.14.7). Since F is invariant under

B(uγ , e−β ), we obtain

0 = F (xα, e+β )− F (B(uγ , e

−β )xα, B(uγ , e

−β )e+

β )

= F (uγ e−β xα, e+β ) + F (xα, uγ)− F (uγ e−β xα, uγ). (1)

Here uγ e−β xα ∈ V+γ−β+α = V +

δ . As γ δ, the third term in (1) vanishes byLemma 28.3(c)(i) (applied to δ, γ, β instead of α, β, γ), so

F (xα, uγ) = −F (uγ e−β xα, e+β ) = F (e+

β , uγ e−β xα), (2)

since F is alternating. The right hand side being symmetric in xα and uγ , so is theleft hand side. Again since F is alternating, 2F (xα, uγ) = 0.

(b) Assume F (e+α , V

+γ ) = 0. Since (δ, α, β, γ) is a square as well, (2) shows

F (xδ, uβ) = F (e+α , uβ e−α xδ). (3)

Hence

F (V +δ , V

+β ) ⊂ F (V +

δ , V+β e−α V

+δ ) ⊂ F (e+

α , V+β−α+δ) = F (e+

α , V+γ ) = 0, (4)

and then F (V +α , V

+γ ) = 0 follows from (2) and (4).

We will show in 28.16 that the statement of (a) can in general not be improved.


28.6. Lemma. Let α → ε be an arrow of orthogonal type, embedded in apyramid

Π =

α

???? β

ε

β′

??α′

__????=

α1

???? α2

ε

α4

??α3

__????(1)

as in (17.2.4).

(a) For all i = 1, . . . , 4, F (αi, αi) = F (αi, αi+1) = F (ε, αi) = 0.

(b) If F (V +α , e

+α′) = 0 then F (γ, δ) = 0 for all γ, δ ∈ Π. In any case, 2F (γ, δ) =

0 for all γ, δ ∈ Π.

(c) If ζ ∈ Γ and ζ ⊥ α, β, ε then F (ζ, γ) = 0 for all γ ∈ Π.

Proof. (a) The first two statements follow from Lemma 28.5(a). For the thirdstatement, we may by symmetry assume i = 1. We claim that ξ := α− β + ε /∈ Γ .Assume to the contrary that ξ ∈ Γ . Then 〈ξ, α∨〉 = 〈α−β+ ε, α∨〉 = 2− 1 + 1 = 2,which implies ξ → α → ε and contradicts (15.6.4). We also have 2α − 2β + ε /∈ Γbecause 〈2α−2β+ ε, α′∨〉 = 0−2 + 1 = −1. This implies xα yβ zε ∈ V +

α−β+ε = 0and QxαQyβzε ∈ V2α−2β+ε = 0 and shows

B(xα, yβ)zε = zε − xα yβ zε+QxαQyβzε = zε.

Hence (28.3.1) (with β, γ, α replaced by α, β, ε) holds, so by replacing α, β, γ, δ in(28.3.2) by ε, α, β, β, we conclude

F (xε, uα vβ zβ) = F (xε, QuαQvβzβ).

Here the right hand side vanishes because α β and Lemma 28.2 implyQuαQvβzβ ∈ V +

2α−β = 0. Now specialize (uβ , vβ) = eβ . Then (23.22.1) shows

F (xε, uα) = 0. Since xε ∈ V +ε and uα ∈ V +

α were arbitrary we have F (ε, α) = 0.

(b) By Lemma 28.5(b), F (α, α′) = 0 implies F (αi, αj) = 0 for all i, j. Wehave 2α − ε /∈ Γ since 〈2α − ε, α∨〉 = 4 − 1 = 3, and 2ε − α′ = α by (15.3.1). Let(uε, vε) ∈ Vε, and put zα = e+

α vε uε ∈ V +α for short. Then

B(e+α , vε)uε = uε − e+

α vε uε = uε − zα.

Further,B(e+

α , vε)e+α′ = e+

α′ − e+α vε e

+α′+Qe+αQvεe

+α′

where the second term on the right is in V +α−ε+α′ = V +

ε and the last term is in

V +2α−(2ε−α′) = V +

α . Writing these as xε and yα, respectively, we obtain from H-

invariance of F that

F (uε, e+α′) = F (B(e+

α , vε) · uε, B(e+α , vε) · e+

α′)

= F (uε − zα, e+α′ − xε + yα)

= F (uε, e+α′)− F (uε, xε) + F (uε, yα)− F (zα, e

+α′)

+ F (zα, xε)− F (zα, yα).


Recalling the definition of xε, we conclude

F (uε, e+α vε e

+α′) = F (uε, xε)

= F (uε, yα)− F (zα, e+α′) + F (zα, xε)− F (zα, yα) = 0

since F (α, ε) = 0 and F (α, α) = 0 by (a), and F (zα, e+α′) = 0 by assumption. Now

(23.22.3) shows F (ε, ε) = 0, as desired.

(c) From α β ⊥ ζ and β α ⊥ ζ it follows by Lemma 28.3(c)(i) thatF (ζ, α) = F (ζ, β) = 0, and in the same way, from β′ α ⊥ ζ and α′ β ⊥ ζthat F (ζ, β′) = F (ζ, α′) = 0. It remains to show F (ζ, ε) = 0. Since ζ ⊥ ε, (28.3.2)is applicable and yields

F (V +ζ , e

+α , Vε, e

+α′) = F (V +

ζ , Qe+αQV −ε e+α′).

Here V +ε = e+

α , V−ε , e

+α′ by (23.22.3), and

Qe+αQV −ε e+α′ ⊂ Qe+αV

−2ε−α′ = Qe+αV

−α ⊂ V +

α .

Hence F (ζ, ε) ⊂ F (ζ, α) = 0.

28.7. Lemma. Let

Ξ =

δ1

????

α3 α2

δ2

??// α1

???? δ3

__???oo

(1)

be a hexagram in Γ .

(a) Let i, j, k = 1, 2, 3, and let ζ ∈ Γ satisfy ζ ⊥ αi. Then F (αi, αi) = 0and F (ζ, δi) = F (ζ, αj) = F (ζ, αk) = 0. In particular, F (δi, δi) = F (δi, αj) = 0.

(b) If Ξ ⊂ ∆, thus E is defined on all of Ξ, then F (ξ, η) = 0 for all ξ ∼ η inΞ, and 2F (ξ, η) = 0 for all ξ ⊥ η in Ξ.

Proof. (a) By symmetry, it suffices to show this for (i, j, k) = (1, 2, 3). Tosimplify notation, we write Ξ in the form

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

Thus αi = α1 = β, and we claim F (β, β) = F (ζ, δ) = F (ζ, α) = F (ζ, γ) forζ ∈ Γ , ζ ⊥ β, in particular F (δ, α) = F (δ, γ) = 0. From α β we see β ∈ Γ 0,and therefore F (β, β) = 0 by Lemma 28.3(b). We also have α β ⊥ ζ andγ β ⊥ ζ. Hence it follows from Lemma 28.3(c)(i) that F (ζ, α) = F (ζ, γ) = 0.In particular, δ ⊥ β whence F (δ, α) = F (δ, γ) = 0.


By (23.22.2), Qe−βV +δ′′ = V −δ′ and Qe+αV

−δ′ = V +

δ , so that V +δ = Qe+αQe−β

V +δ′′ .

Also, (δ′′;β, α, γ) is a kite, so by (15.3.2), α−β+δ′′ = γ and therefore e+α , e−β , V

+δ′′

⊂ V +γ . Now we use (28.3.2) (with α, β, γ, δ replaced by ζ, α, β, δ′′) and obtain

F (V +ζ , V

+δ ) = F (V +

ζ , Qe+αQe−βV +δ′′) = F (V +

ζ , e+α , e−β , V

+δ′′) ⊂ F (V +

ζ , V+γ ) = 0.

In particular, for ζ = δ this proves F (δ, δ) = 0.

(b) Now assume Ξ ⊂ ∆, the domain of definition of the cog E . We show firstthat

F (αi, αj) = 0 for i 6= j. (2)

By symmetry, and with the notation used in the proof of (a), it suffices thatF (β, α) = 0. Since α 6= δ ⊥ β, it follows from (28.3.2) (applied to β, α, δ, δ insteadof α, β, γ, δ) that

F (xβ , uα e−δ e+δ ) = F (xβ , QuαQe−

δe+δ ).

Here uα ∈ V +〈α,δ∨〉(eδ) = V +

1 (eδ), so uα e−δ e+δ = uα. Also, QuαQe−

δe+δ = Quαe

−δ ∈

V +2α−δ = V +

δ′ , and F (V +β , V

+δ′ ) = 0 by (a). Thus F (xβ , uα) = 0, as desired.

We show next 2F (V +δi, V +δj

) = 0 for i 6= j. By symmetry, it suffices that

2F (V +δ , V

+δ′ ) = 0. Let xδ ∈ V +

δ and uδ′ ∈ V +δ′ . Then xδ e−α e+

α = 2xδ becauseVδ ⊂ V〈δ,α∨〉(eα) = V2(eα). Also, δ → α implies 2δ − α /∈ Γ by 28.2. Hence

Qxδe−α ∈ V +

2δ−α = 0 and

B(xδ, e−α )e+

α = e+α − xδ e−α e+

α+QxδQe−α e+α = e+

α − 2xδ +Qxδe−α = e+

α − 2xδ.

Also, B(xδ, e−α )uδ′ = uδ′ − xδ e−α uδ′ + QxδQe−α uδ′ . Here vα := xδ e−α uδ′ ∈

V +δ−α+δ′ = V +

α since δ + δ′ = 2α, and yδ := QxδQe−α uδ′ ∈ V+2δ−2α+δ′ = V +

δ , againsince δ + δ′ = 2α. Thus

B(xδ, e−α )uδ′ = uδ′ − vα + yδ.

By invariance of F under B(xδ, e−α ), we now get

F (uδ′ , e+α ) = F (B(xδ, e

−α )uδ′ , B(xδ, e

−α )e+

α ) = F (uδ′ − vα + yδ, e+α − 2xδ)

= F (uδ′ , e+α )− 2F (uδ′ , xδ)− F (vα, e

+α ) + 2F (vα, xδ)

+ F (yδ, e+α )− 2F (yδ, xδ).

From what we proved in (a) it follows that 2F (uδ′ , xδ) = 0.

It remains to show 2F (V +δi, V +αi) = 0. Again by symmetry, it suffices to do

this for i = 1. We compute B(xδ, e−α )uβ = uβ − xδ e−α uβ + QxδQe−α uβ . Here

zγ := xδ e−α uβ ∈ V +δ−α+β = V +

γ , and QxδQe−α uβ ∈ V +2δ−2α+β = 0, because

2δ − 2α + β = 2γ − β by (15.3.2), and 2γ − β /∈ Γ by γ β and 28.2. Thus (2)shows

0 = F (e+α , uβ) = F (B(xδ, e

−α )eα, B(xδ, e

−α )uβ)

= F (e+α − 2xδ, uβ − zγ) = F (e+

α , uβ)− F (e+α , zγ)− 2F (xδ, uβ) + 2F (xδ, zγ).

Now 2F (xδ, uβ) = 0 follows from (2) and (a).


28.8. Lemma. Suppose Γiso = ∅ and assume that F vanishes on each con-nected component Σ of Γ in the sense that F (α, β) = 0 for all α, β ∈ Σ. ThenF = 0.

Proof. (a) We extend the notation F (α, β) to subsets of Γ in the obviousway. Since F is bi-additive, it suffices to show: if Σ and Θ are different connectedcomponents of Γ then F (Σ,Θ) = 0. We first show F (Σ,Θ0) = 0. Indeed, let α ∈ Σand β ∈ Θ0. Then there exists γ ∈ Θ with β γ. Since different components areorthogonal, we have β γ ⊥ α and therefore F (α, β) = 0 by Lemma 28.3(c)(i).Thus we are reduced to proving F (Σ, ∂Θ) = 0.

Since Θiso = ∅, Proposition 17.7(d) shows ∂Θ = ∂(Θher) ∪ ∂(Θorth). Now Θ isconnected, so ∂Θ 6= ∅ implies Θ = Θorth or Θ = Θher is either of orthogonal orof hermitian type. In the first case, Θ ∼= OI , |I| > 2, by Proposition 17.12. Then∂Θ = ε is a singleton, contained in a pyramid as in (28.6.1). Let ζ ∈ Σ. SinceΣ ⊥ Θ, it follows from Lemma 28.6(c) that F (ζ, ε) = 0.

Now let Θ = Θher∼= TI , and let δ ∈ ∂Θ. Since |I| > 3, δ is contained in a

hexagram Ξ as in (28.7.1), say, δ = δi. If ζ ∈ Σ then ζ ⊥ Ξ, in particular, ζ ⊥ αi.Hence F (ζ, δ) = 0 by Lemma 28.7(a).

28.9. Lemma. Let Υ be an induced subgraph of type T4 in Γ .

(a) Then F (α, β) = 0 for all α, β ∈ Υ , except possibly when α, β ∈ Υ 0 andα ⊥ β in which case 2F (α, β) = 0.

(b) If Υ ⊂ ∆, that is, if E is defined on all of Υ , then F (α, β) = 0 for allα, β ∈ Υ .

Proof. We picture Υ as in (14.18.4), where ij stands for the subset i, j of I:

Υ =

11

444444444444444

14

ooooooooOOOOOOOO

444444444444444

12

444444444444444 13

44

OO

wwoooooooo

''OOOOOOOO

24 34

22 //

77oooooooo

DD23

OOOOOOOOoooooooo

33oo

ggOOOOOOOO

ZZ444444444444444

(1)

Then Υ 0 consists of the six vertices ij (i 6= j) and ∂Υ of the four vertices ii.

(a) By Lemma 28.3(b), F (α, α) = 0 for all α ∈ Υ 0. From (1) and (14.18.1) itis clear that Υ 0 ∼= T4 is an octahedron. This shows that any edge α β of Υembeds in a square, and any pair of orthogonal vertices α ⊥ γ in Υ 0 is part of asquare. Hence F (α, β) = 0 and 2F (α, γ) = 0 by Lemma 28.5(a).

It remains to prove F (δ, ξ) = 0 if δ ∈ ∂Υ and ξ ∈ Υ . There are four possibilities:

(i) ξ = δ,


(ii) ξ = α ∈ Υ 0 and δ → α,

(iii) ξ = β ∈ Υ 0 and β ⊥ δ,(iv) ξ = δ′ ∈ ∂Υ and δ′ 6= δ.

The first two cases follow by embedding ξ and δ in a subgraph Υ ′ ∼= T3 of Υ andinvoking Lemma 28.7(a). In the remaining cases, we may assume, after a relabeling,that δ = 1, 1, β = 2, 3, and δ′ = 2, 2. As in 20.2(b), we put Vij = Vi,j forshort, and for i 6= j we let eij = E (i, j) be the idempotent belonging to i, j.Since 2, 3 3, 4 ⊥ 1, 1, Lemma 28.3(c)(i) yields

F (δ, β) = 0. (2)

By (28.3.2), we have

F (x11, u23 v34 z44) = F (x11, Qu23Qv34z44) (3)

for elements in the respective root spaces. Here u23 v34 z44 ∈ V +24 , and since

ε24 ⊥ δ = ε11, the left hand side of (3) vanishes by (2), which shows

F (V +δ , Qe+23

Qe−34V +

44) = 0.

By (23.22.2), Qe+23Qe−34

V +44 = Qe+23

V −33 = V +22 = V +

δ′ , and therefore F (δ, δ′) = 0, as

desired.

(b) By what we proved in (a), it remains to show F (α, β) = 0 for α ⊥ β in Υ 0.By symmetry, we may assume α = 1, 4 and β = 2, 3. From 2, 2 → 2, 3we have V23 ⊂ V1(e22) and therefore V +

23 = V +23 e−22 e

+22. We apply Lemma 28.3(a)

with α and β as before and γ = δ = 2, 2. Specializing (28.3.2) yields

F (α, β) = F (V +14 , V

+23) = F

(V +

14 , V+23 e−22 e

+22)

= F(V +

14 , QV +23Qe−22

e+22

)⊂ F (V +

14 , V+33) = 0,

by Case (iii) of (a).

28.10. Proposition. Let Γ be connected, recall the invariant c(Γ ) of 17.14,and assume c(Γ ) = 1.

(a) Then F (γ, δ) = 0 for all γ, δ ∈ Γ , except possibly in the following cases:

(i) γ ⊥ δ in Γ 0,

(ii) Γ = Γorth and γ = δ ∈ ∂Γ .

(b) If there exist β ⊥ β′ in Γ 0 such that F (e+β , V

+β′ ) = 0 then F = 0.

(c) In any case, 2F = 0.

Proof. (a) By (17.14.10), Γ 0 ∼= OI for |I|>2, so by Proposition 17.11, for everyα ∈ Γ 0 there exist a unique α′ ∈ Γ 0 such that α ⊥ α′. Let us first establish thatevery α ∈ Γ 0 embeds in a square. Indeed, since α ∈ Γ 0 there exists β ∈ Γ 0 withα β. We have β α′, else α, β would both be orthogonal to α′, contradictingc(Γ 0) = c(Γ ) = 1 by (17.14.2). A similar argument shows α β′ α′. Hence(α, β, α′, β′) is a square:


α β

β′ α′

(1)

By Lemma 28.5(a), we have

F (α, α) = F (α, β) = 2F (α, α′) = 0. (2)

In particular, this proves (a) in case Γ = Γ 0 is simply laced.We are left with considering F (δ, γ) for δ ∈ ∂Γ and arbitrary γ ∈ Γ . Then

δ is part of an arrow which, by the classification of arrows in 17.3, is of isolated,hermitian or orthogonal type. The first case is excluded, else Γ = Γiso

∼= T2, acollision, which has c(Γ ) = 0 by (17.14.8).

In the second case, where δ is part (in fact, the starting point) of a hermitian

arrow, we have Γ = Γher since Γ is connected, and then Γ ∼= TI by 17.10. Fromc(Γ ) = 1 follows |I| = 4 by (17.14.4) or (17.14.9), so Lemma 28.9(a) shows F (δ, γ) =0.

It remains to deal with the case that δ is part (in fact, the end point) of an arrowof orthogonal type. Then Γ = Γorth by connectedness of Γ . By Proposition 17.12,∂Γ = ε is a singleton and α → ε for all α ∈ Γ 0. Embedding α in a square as in(1), we obtain a pyramid (28.6.1), so Lemma 28.6(a) shows F (ε, α) = 0.

(b) In view of (a), it remains to show that F (α, α′) = 0 for all α ∈ Γ 0, andF (ε, ε) = 0, where ε is the uniqe vertex in ∂Γ , in case Γ = Γorth is of orthogonaltype.

Let α ∈ Γ 0. If β 6= α 6= β′ then α β, and as in the proof of (a), thisgenerates a square (1). Hence Lemma 28.5(b) shows F (α, α′) = 0. If α = β orα = β′ then, again embedding β in a square, F (β, β′) = 0 by Lemma 28.5(b).

Finally, let Γ = Γorth and let ∂Γ = ε. Then ε is the tip of a pyramid (28.6.1),so F (ε, ε) = 0 by Lemma 28.6(b).

(c) This follows from (2) and by applying (b) to 2F which is also an alternatingH-invariant bilinear form.

28.11. Lemma. Let Σ ⊂ Γ be an induced subgraph isomorphic to K2 K3, aprism. Then F (α, β) = 0 for all α, β ∈ Σ.

Proof. The case α = β follows from (28.3.3). If α 6= β then there are two cases.If α β then α and β even embed in a square (because any edge of a prism ispart of a square), so F (α, β) = 0 by Lemma 28.5(a).

Now let α ⊥ β. The set of vertices of Σ orthogonal to some α ∈ Σ consists oftwo vertices connected by an edge, as one sees by inspecting (17.15.2). Hence thereexists γ such that β γ ⊥ α, so F (α, β) = 0 follows from Lemma 28.3(c)(i).

28.12. Proposition. If Γ is connected and c(Γ )> 2 then F = 0.

Proof. We show first:

F (α, β) = 0 for all α, β ∈ Γ 0. (1)


Indeed, the case α = β follows from (28.3.3). If α 6= β then by Lemma 17.16, αand β embed in a prism, say Σ. There are two cases. If α β then α and β evenembed in a square (because any edge of a prism is part of a square), so F (α, β) = 0by Lemma 28.5(a).

Now let α ⊥ β. The set of vertices of Σ orthogonal to α consists of two verticesconnected by an edge, as one sees by inspecting (17.15.2). Hence there exists γsuch that β γ ⊥ α, so F (α, β) = 0 follows from Lemma 28.3(c)(i).

By (1), we are done if Γ = Γ 0, that is, if Γ is simply laced. It remains to considerthe non-simply laced cases. By Proposition 17.5, either Γ = Γorth or Γ = Γher. Inthe first case, Γ ∼= OI by Proposition 17.12 which has c(Γ ) = 1 by (17.14.9), so

this case is excluded. If Γ = Γher∼= TI then (17.14.4) and our assumption c(Γ )> 2

imply |I|> 5.It remains to show that

F (δ, ξ) = 0 for all δ ∈ ∂Γ and all ξ ∈ Γ . (2)

This follows from Lemma 28.9 by embedding δ and ξ in an induced subgraphΥ ∼= T4.

Remark. In the hermitian case, the assumption c(Γ ) > 2 and its implication|I| > 5 is used in an essential way in order to prove F (α, β) = 0 for α ⊥ β in Γ 0,which relies on Lemma 17.16. The proof of (2) only required |I|> 4.

28.13. Summary. Let V be a Jordan pair with a Γ -grading R, idempotentwith respect to a cog defined on ∆ containing Γ 0, and let H = EA(V,R). If Γis connected then Propositions 28.4, 28.10 and 28.12 together with (17.14.8) and(17.14.9) show that Alt(V,A)H 6= 0 is only possible in the following cases:

(i) Γ = Kn for n6 2,

(ii) Γ = Tn for n = 2 or 3,

(iii) Γ = OI or OI for |I|> 2,

(iv) Γ = T4 and ∆ & Γ .

In case (i) and (ii), we have c(Γ ) = 0 while c(Γ ) = 1 in case (iii) and (iv). We nowshow by example that in these cases non-trivial alternating invariant bilinear formsdo exist.

28.14. Case (i): Γ = Kn, n 6 2. If Γ = K1 consists of a single vertex, then(by (20.7.7)) H is trivial. On the other hand, V is a Jordan pair with an invertibleidempotent, thus essentially a unital Jordan algebra J by 23.10(b). Hence obviously0 6= Alt(V,A)H is possible.

For Γ = K2 = α, β with α β, there is the following example, see 23.23.Let V = M12(k) be the rectangular matrix pair of 1× 2 and 2× 1 matrices over acommutative ring k. We write the matrices in V + as row vectors x = (x1, x2) and

those in V − as column vectors y =(y1y2

)= (y1, y2)T. Then V has a K2-grading R

given by V +α = (k, 0), V −α = (k, 0)T, and V +

β = (0, k), V −β = (0, k)T, idempotent

with respect to the cog eα =(

(1, 0), (1, 0)T)

and eβ =(

(0, 1), (0, 1)T)

.


Let A =∧2

V + be the second exterior power of V +, a free k-module of rank 1with basis a = (1, 0) ∧ (0, 1), and define F : V + × V + → A by

F (x, z) = x ∧ z = (x1z2 − x2z1) · a.

This is evidently alternating and satisfies F (e+α , e

+β ) = a. It is also invariant under

H = EA(V,R). Indeed, the V +-component of an element h ∈ EA(V,R) is aproduct of maps B(λe+

α , µe−β ) and B(λe+

β , µe−α ) (where λ, µ ∈ k) which act on

x = (x1, x2) ∈ V + by

B(λe+α , µe

−β )x = (x1 − λµx2, x2) = (x1, x2)

(1 0−λµ 1

),

B(λe+β , µe

−α )x = (x1, x2 − λµx1) = (x1, x2)

(1 −λµ0 1

).

Thus the action of H on V + is induced from the action of the elementary groupE2(k) by multiplication on the right on V +. For g ∈ E2(k) we have det(g) = 1,hence F (xg, zg) = (xg) ∧ (zg) = det(g) · (x ∧ z) (by standard properties of theexterior product) = x ∧ z = F (x, z), so F is H-invariant.

28.15. Case (ii): Γ = Tn, n = 2, 3. Let V = Hn(K) be the Jordan pair of n×nsymmetric matrices over a commutative ring K. Let Eij be the standard matrixunits, and let

hii = Eii, hij = Eij + Eji for i 6= j

be the usual basis of V +. We have seen in 23.24 that V has a Γ -grading R givenby V σij = K ·hij , which is idempotent with respect to the cog E (i, j) = (hij , hij).

Let u ∈ V +, v ∈ V −. Since V is special, we have B(u, v)x = (1− uv)x(1− vu).The group H = EA(V,R) is generated by all B(uα, vβ) where uα ∈ V +

α , vβ ∈ V −βand α 6= β in Γ and α not orthogonal to β. For α = i, i, β = i, j we haveuα = λhii, vβ = µhij (λ, µ ∈ K), hence

B(λhii, µhij)x = (1− λµEii(Eij + Eji))x(1− λµ(Eij + Eji)Eii)

= (1− λµEij)x(1− λµEji).

Similarly, one computes

B(µhij , λhii)x = (1− λµEji)x(1− λµEij),

and for β as before and γ = j, l where i, j, l 6=, one obtains

B(λhij , µhjl)x = (1− λµEil)x(1− λµEli).

This shows that H is generated by the transformations x 7→ gxgT where g =glm(r) = 1n + rElm, with r ∈ K and l 6= m in 1, . . . , n, so H consists of thetransformations x 7→ gxgT where g ∈ En(K), the elementary subgroup of GLn(K).

Now let n = 2 and assume 2K = 0. Then an alternating H-invariant bilinearform is given by F (x, y) = tr(xy]), where y] denotes the adjoint of the matrix y.


Indeed, tr(xx]) = tr(det(x)12) = 2 det(x) = 0. An element g ∈ E2(K) has g] = g−1

since det(g) = 1, from which it follows that F is H-invariant:

F (g · x, g · z) = F (gxgT, gzgT) = tr(gxgTgT]z]g]) = tr

(gxgT(gT)−1z]g−1)

= tr(g−1gxz]) = F (x, z).

For n = 3, define a quadratic form q on V + = H3(K) by

q(x) =∑cyc

xii(xjj + xjk),

where∑

cyc means the sum over the cyclic permutations of (1, 2, 3), and x = (xij) ∈H3(K). Then q vanishes on the basis (hij), and the polar form F of q is given by

F (hij , hkl) =

0 if i, j ∩ k, l 6= ∅1 if i, j ∩ k, l = ∅

.

One can show by a lengthy but straightforward computation that

F is H-invariant ⇐⇒ K is a Boolean ring,

that is, r2 = r for all r ∈ K. Since 2 = 0 in a Boolean ring, F is alternating in thiscase, being the polar form of a quadratic form over a ring in which 2 = 0. However,it is not true that q is invariant under H, even in case K = F2.

28.16. Case (iii): Γ = OI or OI . Let V = J(M, q) be the Jordan pair of aquadratic form q on a k-module M as in Example 6.6(f). A lengthy computationshows that the Bergmann operators B(x, y) satisfy

q(B(x, y)z

)=(1− b(x, y) + q(x)q(y)

)2q(z) (1)

for all x, y, z ∈ M . Hence if b(x, y) = 0 and either q(x) or q(y) = 0, then B(x, y)belongs to the orthogonal group of q. These transformations are special cases ofEichler transformations [32, 5.2.9],

We consider 23.25 in the special case where K = k and 2k = 0. Let (Mh, qh)be the orthogonal sum of I copies of the hyperbolic plane, and let V = J(Mh, qh).Then V has a ∆ = OI -grading R given by (23.25.2), which is idempotent withrespect to the cog (23.25.3). The generators of the group H = EA(V,R) are the(B(uα, vβ), B(vβ , uα)−1

)where α 6= β in ∆ and (uα, vβ) = (λe+

α , µe−β ) ∈ V +

α ×V −β ,

for λ, µ ∈ k. From (23.25.1) and (23.25.3) it is clear that q(e+α ) = 0 and b(e+

α , e−β ) =

0, for α 6= β in ∆. Hence (1) shows that q and therefore b is invariant under H.Since b(x, x) = 2q(x) = 0, we have b ∈ Alt(V, k)H .

An example for Γ = OI = ω ∪ ∆ is obtained by putting M = M0 ⊕ Mh

and q = qh ⊥ q0 (orthogonal sum) where M0 = k · x0 is free of rank one and

q0(x0) = 1. By (23.25.4), the Jordan pair W = J(M, q) has a Γ -grading R givenby Wα = Vα for α ∈ ∆ and Wω = (M0,M0). It is idempotent with respect to the

cog E extending E by ω 7→ e0 = (x0, x0). One checks that the polar form of q isinvariant under EA(W, R), and it is alternating because 2 = 0 in k.

§29] Vanishing of the binary symbols 363

28.17. Case (iv): Γ = T4. By Lemma 28.9, there can be no example with a cogdefined on all of Γ . On the other hand, Γ 0 ∼= T4

∼= O3 by 14.19, so a rather trivialexample with ∆ = Γ 0 is obtained by letting V be the Jordan pair of a hyperbolicquadratic form of rank 6 as in 28.16, and defining Vi = 0 for the roots i ∈ ∂Γ .

§29. Vanishing of the binary symbols

29.1. Notations and conventions. Let V be a Jordan pair. From now on,G = (G, x+, x−, π) ∈ st(V ) is a group over V as in 9.1 and π: G → PE(V ) is thecanonical homomorphism. As usual, we write Uσ = xσ(V σ). Elements of V σ willoften be denoted by u, x, z while v, y is our preferred notation for elements in V −σ.

As in (27.9.1), let

0 // Ai // E

p // G // 1 (1)

be a central extension; thus i is injective, p is surjective and K := Ker(p) = Im(i) ⊂Z (E), the centre of E. We do not assume that E is a group over V . For σ ∈ +,−and x ∈ V σ we write, simplifying the notation introduced in 27.9,

Eσ(x) := Exσ(x) = p−1(xσ(x)

). (2)

The index σ at Eσ(x) will often be dropped when it is clear from the context orunimportant. Since xσ(x)xσ(z) = xσ(x+ z) in G, formula (27.9.2) shows

E(x)E(z) = E(x+ z) = E(z)E(x) for x, z ∈ V σ. (3)

As in Lemma 9.2 we put

G0 = π−1(PE0(V )) and N = NormG(U+) ∩NormG(U−).

For h ∈ N∩G0 we have π(h) = (h+, h−) ∈ PE0(V ). This yields an action of N∩G0

on V σ by defining

h · x = hσ(x)

for h ∈ N ∩G0 and x ∈ V σ. As special cases of the action of G on E described in(27.9.3) we then have, for h ∈ N ∩G0 and x ∈ V σ,

Int(h) · E(x) = E(h · x). (4)

Let x, z ∈ V σ. Recall from (27.10.2) the binary symbols[[x, z ]

]:=(((((((E(x), E(z)

)))))))∈ Ker(p) ⊂ Z (E). (5)

It would be more precise to write[[x, z ]

]σ. But the simpler notation should not

lead to difficulties if the reader keeps in mind that the arguments of the binarysymbols come from the same space V σ. We collect their main properties in thefollowing lemma.


29.2. Lemma. Let p: E → G be a central extension as in (29.1.1).

(a) The binary symbols[[x, z ]

]are alternating, bi-multiplicative, and invariant

under the action of N ∩G0:[[x, x ]

]= 1,

[[x, z ]

]=[[ z, x ]

]−1, (1)[

[x+ u, z ]]

=[[x, z ]

]·[[u, z ]

],

[[x, u+ z ]

]=[[x, u ]

]·[[x, z ]

], (2)[

[h · x, h · z ]]

=[[x, z ]

], (3)

for all x, z, u ∈ V σ and h ∈ N ∩G0.

(b) We use the isomorphism i: A → K of (29.1.1) to define a map F : V σ ×V σ → A by

i(F (x, z)

)=[[x, z ]

]. (4)

Then F is alternating, bi-additive and invariant under the action of π(N ∩ G0),and p−1(Uσ) is abelian if and only if F = 0. In particular,

(i) if V has a root grading R and G ∈ st(V,R) then F is EA(V,R)-invariant,

(ii) if G = PE(V ) then F is invariant under PE0(V ), in particular, under allinner automorphisms of V .

Proof. (a) The properties (1) are evident from the definition (27.10.2). Leta ∈ E(x), b ∈ E(u), c ∈ E(z). Then ab ∈ E(x+ u) by (29.1.3), hence[

[x+ u, z ]]

=(((((((ab, c

)))))))=(((((((a,(((((((b, c))))))))))))))·(((((((b, c)))))))·(((((((a, c)))))))

(by (3.6.2))

=(((((((a,[[u, z ]

])))))))·[[u, z ]

]·[[x, z ]

]=[[x, z ]

]·[[u, z ]

]since

[[u, z ]

]∈ K ⊂ Z (E) and therefore

(((((((a,[[u, z ]

])))))))= 1. This shows

[[x, z ]

]is

multiplicative in x. The second formula of (2) follows from the first and (1) bytaking inverses. For h ∈ N ∩G0 we have, by (29.1.4),[

[h · x, h · z ]]

=(((((((E(h · x), E(h · z)

)))))))=(((((((

Int(h) · E(x), Int(h) · E(z))))))))

= Int(h) ·(((((((E(x), E(z)

)))))))= Int(h) ·

[[x, z ]

]=[[x, z ]

],

because[[x, z ]

]is central.

(b) The first statements follow immediately from the definition of F and (a).If R = (Vγ)γ∈Γ is a root grading of V , then b(V +

γ , V−δ ) ∈ N ∩G0 for all γ 6= δ by

Theorem 21.7(iv), whence F is invariant under all π(b(V +

γ , V−δ ))

= β(V +γ , V

−δ ).

This implies (i) by (20.7.7). For G = PE(V ) we know PE0(V ) ⊂ N by (7.7.1), so Fis PE0(V )-invariant. Since Inn(V ) ⊂ PE0(V ) by Theorem 7.7(c), F is in particularinvariant under all inner automorphisms.

29.3. The ternary symbols. We continue to assume the situation of 29.1.Suppose (x, y) ∈ V σ × V −σ is quasi-invertible, recall from (9.10.4) and (9.10.5)

bσ(x, y) =

b(x, y) for σ = +b(−y,−x)−1 for σ = −

= b−σ(−y,−x)−1, (1)


and define

Eσ(x, y) := Ebσ(x,y) = p−1(bσ(x, y)

)∈ E. (2)

From (1) we see

Eσ(x, y) = E−σ(−y,−x)−1. (3)

As before, the index σ at Eσ(x, y) will often be omitted. Suppose (x, y) ∈ V σ×V −σis quasi-invertible and G satisfies the relations Bσ(x, y) (see (9.10.9)) so that inparticular (((((((

bσ(x, y), xσ(z))))))))

= xσ(− xyz+QxQyz

)(4)

holds by (9.10.10) for all z ∈ V σ. Recall from (27.10.4) the ternary symbols⟨〈x, y, z〉

⟩=(((((((E(x, y), E(z)

))))))). (5)

It would be more precise to write⟨〈x, y, z〉

⟩σ, but the simpler notation should not

lead to difficulties as long as, similarly to the Jordan triple product, care is takenthat the parities of the variables x, y, z alternate; that is, x, z ∈ V σ and y ∈ V −σ.By applying p to (5) and observing (4), we obtain

p(⟨〈x, y, z〉

⟩)= xσ

(− xyz+QxQyz

), (6)

whence⟨〈x, y, z〉

⟩∈ E

(− xyz+QxQyz

). From (1) we see

⟨〈x, y, z〉

⟩=(((((((E(−y,−x)−1, E(z)

))))))), (7)

Similarly to (29.1.4), we have, for h ∈ N ∩ G0, x, z ∈ V σ resp. quasi-invertible(x, y) ∈ V σ × V −σ,

Int(h) · E(x, y) = E(h · x, h · y), (8)

Int(h) ·⟨〈x, y, z〉

⟩=⟨〈h · x, h · y, h · z〉

⟩. (9)

29.4. General assumptions. In addition to 29.1, we make from now on thefollowing assumptions which will be in force for the remainder of this section:

(i) V is a Jordan pair with a root grading R = (Vα)α∈Γ of type Γ , andE : ∆→ Idp(V ) is a cog compatible with R and defined on a subset ∆ ofΓ containing Γ 0.

(ii) G is a group in st(V,R,E ), as defined in (24.1.3).

Thus we are working in a more general situation than in Step 1 of 27.10 inasmuchas we do not assume G to be the Steinberg group St(V,R), nor do we impose therestrictions of Theorem 27.4 on Γ . This greater generality does not cause any addeddifficulties.

As in 23.1 we write eα = E (α) for the idempotent defined by E and α ∈ ∆.The notation introduced in 29.1 and 29.2 will be used without further reference.


29.5. Lemma. (a) Suppose α, β ∈ Γ satisfy α β or α→ β. Then for xαand yβ , zβ in the appropriate root spaces,⟨

〈xα, yβ , zβ〉⟩∈ E(−xα, yβ , zβ). (1)

(b) If α− β + γ /∈ Γ then⟨〈V σα , V −σβ , V σγ 〉

⟩⊂ Z (E). (2)

Proof. (a) Lemma 28.2 shows that 2α − β /∈ Γ . This implies QxαQyβzβ ∈V σ2α−β = 0. Now (1) follows from (29.3.5).

(b) We have

p(⟨〈xα, yβ , zγ〉

⟩)= xσ

(− xα, yβ , zγ+QxαQyβzγ

)= 1

because xα, yβ , zγ ∈ V σα−β+γ = 0 and QxαQyβzγ ∈ V σ2α−2β+γ = 0 in view of2(α− β) + γ /∈ Γ by Lemma 19.9(a).

29.6. Lemma. Let α, β, γ ∈ Γ be three distinct roots satisfying⟨〈V σα , V −σγ , V σβ 〉

⟩= 1. (1)

(a) Assume α, β, γ form a closed triangle. Then[[V σα , V

σβ ]]

= 1, (2)

and for xα ∈ V σα , yβ ∈ V −σβ etc. in the respective root spaces,⟨〈xα, yβ , uβ vγ zγ〉

⟩=⟨〈xα yβ uβ, vγ , zγ〉

⟩, (3)⟨

〈xα, yβ , uβ〉⟩

=⟨〈xα yβ uβ, e−σβ , eσβ〉

⟩, (4)⟨

〈xα, e−σβ , eσβ〉⟩

=⟨〈xα, e−σγ , eσγ 〉

⟩. (5)

(b) Suppose α, β, γ embed in a pyramid

α

???? γ

β

γ′

??α′

__???(6)

and assume, in addition to (1), that⟨〈V σα , V −σβ , V σγ′〉

⟩= 1 =

⟨〈V σα , V −σγ , V σγ′〉

⟩. (7)

Then (3) holds as well.

Proof. We first prove (3) for (a) and (b) simultaneously. In Case (b), we haveγ′ = 2β − γ by (15.3.1). In Case (a), we define formally γ′ = 2β − γ ∈ X•(Γ ).


Then γ′ /∈ Γ by Lemma 28.2, so Vγ′ = 0 by (20.1.2) and therefore E(V σγ′) ⊂ Z (E).This implies (7) also in Case (a) because any commutator with one factor a centralelement is trivial.

In the remainder, we give the proof for σ = +, the case σ = − then follows bypassing to V op. From the assumptions on α, β, γ made in (a) and (b) one obtainsthe following relations:

2α− β /∈ Γ, 2α− γ /∈ Γ, (8)

α− β + γ /∈ Γ, β − γ + α /∈ Γ, γ − α+ β /∈ Γ, (9)

2(α− β) + γ /∈ Γ, 2(β − α) + γ /∈ Γ (10)

Indeed, assume 2α − β = δ ∈ Γ . Then by Proposition 16.1(b1), β → α, contra-diction. In the same way, one shows 2α − γ /∈ Γ . For Case (a), (9) follows from(19.9.2), and for Case (b) this is an immediate consequence of Proposition 16.1(b2).Finally, (10) is a consequence of (9) and Lemma 19.9(a).

Recall the commutator formula (3.6.9):(((((((a,(((((((b, c))))))))))))))

=((((((( (((((((

a, b))))))),(((((((b, c))))))) )))))))·((((((( (((((((

b, c))))))),((((((((((((((a, b))))))), c))))))) )))))))·((((((((((((((a, b))))))), c))))))), (11)

which holds provided(((((((a, c)))))))

is central. We will show that (11) becomes (3) with

a ∈ E(xα, yβ), b ∈ E(−uβ , vγ), c ∈ E(zγ).

By (9) and (10), xα yβ zγ ∈ V +α−β+γ = 0 and QxαQyβzγ ∈ V

+2(α−β)+γ = 0. Hence

(29.3.5) shows(((((((a, c)))))))

=(((((((E(xα, yβ), E(zγ)

)))))))∈ E(−xα yβ zγ+QxαQyβzγ) = E(0) ⊂ Z (E).

Thus (11) is applicable. Let us abbreviate

u′β = uβ vγ zγ ∈ V +β , wγ′ = QuβQvγzγ ∈ V

+γ′ .

Then (((((((b, c)))))))

=(((((((E(−uβ , vγ), E(zγ)

)))))))∈ E(uβ vγ zγ+QuβQvγzγ)

= E(u′β) · E(wγ′) (12)

by (27.9.2). Choose b′ ∈ E(u′β) and c′ ∈ E(wγ′). Then(((((((b, c)))))))≡ b′c′ mod Z (E) and

(((((((a, c′

)))))))=⟨〈xα, yβ , wγ′〉

⟩= 1 (13)

by (12) and (7). Hence the left hand side of (11) is, using the commutator formula(3.6.3),(((((((

a,(((((((b, c))))))))))))))

=(((((((a, b′c′

)))))))=(((((((a, b′

)))))))· Int(b′)

(((((((a, c′

)))))))=(((((((a, b′

)))))))=⟨〈xα, yβ , u′β〉

⟩.

Thus the left hand side of (11) is the left hand side of (3).For the evaluation of the right hand side of (11) we have, using (9.9.5),


p((((((((a, b)))))))

) =(((((((

b(xα, yβ), b(−uβ , vγ))))))))

= b(−B(xα, yβ)uβ , B(yβ , xα)−1vγ

)· b(−uβ , vγ)−1.

With the abbreviationx′α = xα yβ uβ ∈ V +

α

we get

−B(xα, yβ)uβ = −uβ + xα yβ uβ −QxαQyβuβ = x′α − uβ ,B(yβ , xα)vγ = vγ − yβ xα vγ+QyβQxαvγ = vγ ,

since 2α − β /∈ Γ , β − α + γ /∈ Γ and 2(β − α) + γ /∈ Γ by (9) and (10).By Corollary 21.5, G ∈ st(V,R) ⊂ st(V,Rγ) for Rγ defined in 20.7. Because〈α, γ∨〉 = 1 = 〈β, γ∨〉 we can apply (11.5.1) for Rγ and obtain from the above

p((((((((a, b)))))))

) = b(x′α − uβ , vγ) · b(−uβ , vγ)−1 = b(x′α, vγ).

Thus(((((((a, b)))))))∈ E(x′α, vγ) and so((((((((((((((

a, b))))))), c)))))))

=⟨〈x′α, vγ , zγ〉

⟩,((((((((((((((a, b))))))), b′)))))))

=⟨〈x′α, vγ , u′β〉

⟩. (14)

Two more factors in (11) remain to be computed. As((((((((((((((a, b))))))), c′)))))))

=(((((((E(x′α, vγ), E(wγ′)

)))))))=⟨〈x′α, vγ , wγ′〉

⟩= 1

by (7), we get from (13), then (3.6.3), (14) and finally (1),((((((((((((((a, b))))))),(((((((b, c))))))))))))))

=((((((((((((((a, b))))))), b′c′

)))))))=((((((((((((((a, b))))))), b′)))))))· Int(b′)

((((((((((((((a, b))))))), c′)))))))

=((((((((((((((a, b))))))), b′)))))))

=⟨〈x′α, vγ , u′β〉

⟩= 1.

For the calculation of((((((((((((((b, c))))))),((((((((((((((a, b))))))), c))))))))))))))

in (11) we first show[[V +β , V

+α ]]

= 1 =[[V +γ′ , V

+α ]]. (15)

Indeed, in Case (a) the assumption (ii) of Lemma 28.3(c) holds by (9). HenceF (V +

α , V+β ) = F (V +

β , V+α ) = 0 by (28.3.4) and anticommutativity of F which

implies (15) by definition of F in (29.2.4). The second equality in (15) holds triviallybecause V +

γ′ = 0. In Case (b), F (V +β , V

+α ) = F (V +

γ′ , V+α ) = 0 and therefore (15)

follows from Lemma 28.6(a). Let x′′α = −x′α vγ zγ ∈ V +α . Then

((((((((((((((a, b))))))), c)))))))∈

E(x′′α) by (14), (29.5.1) and (8). Choose a′′ ∈ E(x′′α) and recall from (3.6.2) that(((((((b′c′, a′′

)))))))=(

Int(b′) ·(((((((c′, a′′

))))))))·(((((((b′, a′′

))))))).

Hence by (13), (29.2.4) and (15),((((((((((((((b, c))))))),((((((((((((((a, b))))))), c))))))))))))))

=(((((((b′c′, a′′

)))))))=(

Int(b′)(((((((c′, a′′

))))))))·(((((((b′, a′′

)))))))=(

Int(b′)[[wγ′ , x

′′α ]])·[[u′β , x

′′α ]]

= 1.

This finishes the proof of (1) for (a) and (b).

Regarding the remaining assertions in (a), we have already established (2) in(15). For (4) we compute as follows, using (3):⟨

〈xα, yβ , uβ〉⟩

=⟨〈xα, yβ , uβ e−γ e+

γ 〉⟩

=⟨〈xα yβ uβ, e−γ , e+

γ 〉⟩

=⟨〈x′α, e−γ , e+

γ 〉⟩

=⟨〈x′α, e−β , e

+β , e

−γ , e

+γ 〉⟩

=⟨〈x′α, e−β , e

+β , e−γ , e

+γ 〉⟩

=⟨〈x′α, e−β , e

+β 〉⟩.

This proves (4). Specializing (uβ , yβ) = eβ and (zγ , vγ) = eγ in (3) yields (5).


29.7. Lemma. Let α β be an edge in Γ .

(a) Fix σ ∈ +,− and suppose[[V −σα , V −σβ ]

]= 1. Then the map

ϕ: V σα → E, ϕ(xα) :=⟨〈−xα, e−σβ , eσβ〉

⟩, (1)

is a group homomorphism satisfying p(ϕ(xα)

)= x+(xα).

(b) Assume that α β embeds in a closed triangle α, β, γ satisfying(29.6.1). Then for xα, yβ , zβ in the respective root spaces,⟨

〈xα, yβ , zβ〉⟩

= ϕ(−xα yβ zβ). (2)

Hence⟨〈xα, yβ , zβ〉

⟩is multiplicative in each variable. Moreover, ϕ is independent

of β in the sense that also

ϕ(xα) =⟨〈−xα, e−γ , e+

γ 〉⟩. (3)

Proof. (a) For simpler notation we assume σ = +. The case σ = − follows asusual by passing to V op. From α β and 20.7 it follows that (xα, e

−β ) is quasi-

invertible, so ϕ is well-defined. By (29.5.1) we have p(ϕ(xα)) = x+(xα, e−β , e+β ) =

x+(xα). To prove that ϕ is a homomorphism, let also uα ∈ V +α . In G we have

b(−xα − uα, e−β ) = b(−xα, e−β )b(−uα, e−β )

by (11.5.1) which implies

ϕ(xα + uα) =(((((((E(−xα, e−β )E(−uα, e−β ), E(e+

β )))))))).

Let a ∈ E(−xα, e−β ), b ∈ E(−uα, e−β ) and c ∈ E(e+β ), so that ϕ(xα + uα) =

(((((((ab, c

))))))).

Then (((((((a, c)))))))

=(((((((E(−xα, e−β ), E(e+

β ))))))))

=⟨〈−xα, e−β , e

+β 〉⟩

= ϕ(xα)

by (29.3.5), and similarly(((((((b, c)))))))

= ϕ(uα). The commutator formula (3.6.2) shows

ϕ(xα + uα) =(((((((ab, c

)))))))=(((((((a,(((((((b, c))))))))))))))·(((((((b, c)))))))·(((((((a, c)))))))

=(((((((a,(((((((b, c))))))))))))))

ϕ(uα)ϕ(xα).

Hence it remains to show(((((((a,(((((((b, c))))))))))))))

=(((((((E(−xα, e−β ), E(uα)

)))))))=⟨〈−xα, e−β , uα〉

⟩= 1.

Let wα = weα be the Weyl element defined by eα. Since G ∈ st(V,R,E ) ⊂ st(V, eα)by 24.6 and β α, we have

Int(wα) · b(−xα, e−β ) = x−(e−α xα e−β ) ∈ x−(V −β )

by the relation S′′21 of 12.3, and Int(wα) · x+(uα) = x−(Qe−α uα) ∈ x−(V −α ), by theWeyl relation W(eα), see 9.17. This implies the corresponding relations for thefibres,

Int(wα) · E(−xα, e−β ) = E(e−α xα e−β ) ⊂ E(V −β ),

Int(wα) · E(uα) = E(Qe−α uα) ⊂ E(V −α ).

It follows that

Int(wα) ·(((((((a,(((((((b, c))))))))))))))∈(((((((E(V −β ), E(V −α )

)))))))=[[V −β , V

−α ]]

= 1,

and therefore also(((((((a,(((((((b, c))))))))))))))

= 1.

(b) The formulas (2) and (3) are (29.6.4) and (29.6.5) respectively. Nowmultiplicativity of

⟨〈xα, yβ , zβ〉

⟩in each variable follows from (2), (a), and the fact

that the Jordan triple product is trilinear.


29.8. Proposition. Let α and α′ be orthogonal vertices in Γ which embed inan octahedron

α

MMMMMMMMMMMMMM β

pppppppppppppp

4444444

γ′

555555

MMMMMMMMMMMMMM γ

β′

qqqqqqqqqqqqqqα′

(1)

Assume further that

all triangles α1, α2, α3 in (1) are closed and satisfy⟨〈V σα1

, V −σα2, V σα3〉⟩

= 1. (2)

Then F (V σα′ , Vσα ) = 0, equivalently,

[[V σα′ , V

σα ]]

= 1.

Proof. We prove this for σ = +, the case σ = − being obtained by passingto V op. Since α and α′ are contained in the square (α, β, α′, β′), Lemma 28.5(b)shows that it suffices to prove F (e+

α′ , x+α ) = 0 for all xα ∈ V +

α , equivalently,[[ e+α′ , x

+α ]]

= 1. (3)

We first give an outline of the proof.Consider the Weyl element w := weα′ ∈ G defined by eα′ and let gα ∈ E(xα).

We will compute Int(w) · gα in two different ways: first directly, using the fact that

w = xσ(eσα′) x−σ(e−σα′ ) xσ(eσα′) (4)

for σ ∈ +,−, and then by using the formula (29.7.3). Comparing the results willshow (3).

(a) First computation of Int(w) · gα. In E, we have, by (27.9.3),

Int(x+(e+α′)) · gα =

(((((((E(e+

α′), E(xα))))))))· gα =

[[ eα′ , xα ]

]· gα,

Int(x−(e−α′)) · gα =(((((((E(e−α′), E(xα)

)))))))· gα,

and the factors

k :=[[ e+α′ , xα ]

]and l :=

(((((((E(e−α′), E(xα)

)))))))belong to K = Ker(p). Indeed, for k this is clear from (29.1.5). For l, it followsfrom

(((((((x−(e−α′), x+(xα)

)))))))= 1 since α ⊥ α′ and Uα and U−α′ commute by Theo-

rem 21.7(iii). Thus an application of Int(xσ(e±α′)) to gα changes the latter by acentral factor k or l, so from (4) we conclude

Int(w) · gα = k2 l gα = k l2gα.

This implies k = l and hence Int(w) · gα = k3 · gα. Since k = i(F (e+α′ , xα)), we have

k2 = i(2F (eα′ , xα)

)= 1 by Lemma 28.5(a). Hence we obtain


Int(w) · gα = k · gα, (5)

independent of the choice of gα ∈ E(xα).

(b) Second computation of Int(w) · gα. By Lemma 25.6, e′β′ = eα eβ eα′ isan idempotent associated with eβ′ , and by (20.1.1) we know e′β′ ∈ Vα−β+α′ = Vβ′ .Hence Corollary 23.16 allows us to replace eβ′ by e′β′ and thus assume that Q =(eα, eβ , eα′ , eβ′) is a quadrangle of idempotents. In particular, this implies

eα eβ′ eα′ = eβ . (6)

By (2), the triangles α1, α2, α3 contained in (1) satisfy the assumptions ofLemma 29.6. Hence

[[V +αi , V

+αj ]]

= 1 by (29.6.2), so the hypotheses of Lemma 29.7hold. Applying (29.7.3) to the triangles T = α, β, γ and T ′ = α, β′, γ weconclude

ϕ(xα) =⟨〈−xα, e−β , e

+β 〉⟩

=⟨〈−xα, e−γ , e+

γ 〉⟩

=⟨〈−xα, e−β′ , e

+β′〉⟩∈ E(xα).

Now put gα = ϕ(xα) and compute Int(w) · gα as follows:

Int(w) ·gα = Int(w) ·⟨〈−xα, e−β′ , e

+β′〉⟩

=(((((((

Int(w) ·E(−xα, e−β′), Int(w) ·E(e+β′)))))))). (7)

Let zβ := xα, e−β′ , e+α′. Then zβ ∈ Vα−β′+α′ = V +

β by (15.3.2), and since Q is aquadrangle, (25.5.5) implies

e+α , e−β , zβ = e+

α , e−β , e

+α′ , e

−β′ , xα = xα. (8)

In G we have

Int(w) · b(−xα, e−β′) = x+(−xα, e−β′ , e+α′) = x+(−zβ)

by the relation S′′01 of 12.3 for e = eα′ and since xα ∈ V +0 (eα′) and eβ′ ∈ V1(eα′).

Moreover,Int(w) · x+(e+

β′) = b(−e+β′ , e

−α′),

by (12.2.2) and since eβ′ ∈ V1(eα′). From this we conclude, using (7),

Int(w) · gα =(((((((E(−zβ), E(−e+

β′ , e−α′)))))))). (9)

We compute in G, using α β′, S10 for −eα and (6),

b(−e+β′ , e

−α′) = b(−e+

α , e−α e+β′ e−α′) = b(−e+

α , e−β ).

Taking pre-images with respect to p, we obtain E(−e+β′ , e

−α′) = E(−e+

α , e−β ), so from

(9) it follows

Int(w) · gα =(((((((E(−zβ), E(−e+

α , e−β ))))))))

=(((((((E(−e+

α , e−β ), E(−zβ)

)))))))−1 =⟨〈−e+

α , e−β ,−zβ〉

⟩−1

= ϕ(− e+

α , e−β , zβ

)−1(by (29.7.2))

= ϕ(−xα)−1 (by (8))

= ϕ(xα) = gα (by Lemma 29.7(a)).

Combined with (5) we have shown k =[[ e+α′ , xα ]

]= 1 for all xα ∈ V +

α , and theproof is complete.


29.9. Lemma. (a) Let U and H be subgroups of G with(((((((U,H

)))))))= 1. Then(((((((

p−1(U), p−1(H))))))))⊂ Z (E), (1)(((((((

p−1(U), p−1(D(H)

))))))))= 1. (2)

(b) For α ∈ R letZ(α) = β ∈ R : α+ β /∈ R, (3)

and let UZ(α) be the subgroup of G generated by Uβ : β ∈ Z(α). Then(((((((Uα, UZ(α)

)))))))= 1, and therefore (a) applies to U = Uα and H = UZ(α).

Proof. (a) (1) follows from(((((((U,H

)))))))= 1 since the kernel of p is central. For the

proof of (2), first observe

p−1(D(H)

)⊂ Z (E) ·D

(p−1(H)

),

which is an easy consequence of the fact that p: p−1(H)→ H is surjective and hascentral kernel. Hence it suffices to show that

(((((((p−1(U), D

(p−1(H)

))))))))= 1. For fixed

a ∈ p−1(U), define a map f : p−1(H) → E by f(b) =(((((((a, b)))))))

. Then (1) shows thatf takes values in the abelian group Z (E). Moreover, f is a homomorphism: forb, c ∈ p−1(H),

f(bc) =(((((((a, bc

)))))))=(((((((a, b)))))))· Int(b)

(((((((a, c)))))))

= f(b) · Int(b)f(c) = f(b) f(c),

by (3.6.3) and (1). Hence f is vanishes on the commutator subgroup D(p−1(H)

).

Since a ∈ p−1(U) was arbitrary, this proves (2).

(b) Since R is a reduced root system, it follows from Proposition 2.17 that(α, β) is a nilpotent pair if and only if α+β 6= 0, and in this case

(((((((α, β

)))))))= ∅ if and

only if α+ β /∈ R. Hence(((((((Uα, Uβ

)))))))⊂ U(((((α,β))))) = 1 for all β ∈ Z(α) by (3.2.3).

29.10. Lemma. Let T = α, β, γ be a triangle in Γ which embeds in a tetra-hedron or in a prism as in (19.9.5). Then T is closed and⟨

〈V σα , V −σβ , V σγ 〉⟩

= 1. (1)

Proof. Closedness of T follows from Lemma 19.9(c). We prove (1) for σ = +,the case σ = − then follows by passing to V op. From α β and (15.9.1) itfollows that µ = α − β ∈ R×0 . For xα, yβ , zγ in the respective root spaces, wehave, by (29.3.2) and (29.1.2), E(xα, yβ) ⊂ p−1(Uµ) and E(zγ) ⊂ p−1(Uγ). As inLemma 29.9, let

Z(µ) = ν ∈ R : µ+ ν /∈ R and H = UZ(µ).

We claim that x+(zγ) ∈ D(H). Write T = α1, α2, α3 and choose a vertexδ = α4 resp. δ = δ3 as in (19.9.5). Then by (19.9.6) and (19.9.7), δ ∈ Z(µ) andγ− δ ∈ Z(µ) so that x+(e+

δ ) ∈ H. Since γ δ, we have γ− δ ∈ R×0 and thereforeb(−zγ , e−δ ) ∈ Uγ−δ (by (21.1.2)) ⊂ H, whence

(((((((b(−zγ , e−δ ), x+(e+

δ ))))))))∈ D(H). On

the other hand, by (21.10.1),

§29] Vanishing of the binary symbols 373(((((((b(−zγ , e−δ ), x+(e+

δ ))))))))

= x+

(zγ e−δ e

+δ +Q(zγ)Q(e−δ )e+

δ

).

Here zγ e−δ e+δ = zγ since γ δ, and Q(zγ)Q(e−δ )e+

δ = Q(zγ)e−δ ∈ V+2γ−δ = 0

because 2γ − δ /∈ Γ by Lemma 28.2. It follows that

x+(zγ) =(((((((

b(−zγ , e−δ ), x+(e+δ ))))))))∈ D(H).

Now (29.9.2) shows⟨〈xα, yβ , zγ〉

⟩=(((((((E(xα, yβ), E(zγ)

)))))))∈(((((((p−1(Uµ), p−1(Uγ)

)))))))⊂(((((((p−1(Uµ), p−1

(D(H)

))))))))= 1.

29.11. Lemma. Let T = α, β, γ be a triangle in Γ which embeds in an

induced subgraph of type O3 as in (14.19.5):

α

3333333

NNNNNNNNNNNNNNN β

ooooooooooooooo

4444444

γ′ //

555555

OOOOOOOOOOOOOO ε γoo

β′

EE

pppppppppppppppα′

YY3333333

(1)

Then T is closed and ⟨〈V σα , V −σβ , V σε 〉

⟩= 1. (2)

If ε belongs to the domain of definition ∆ of E then⟨〈V σα , V −σβ , V σγ 〉

⟩= 1. (3)

Proof. As before we will only prove this for σ = +. We show first that T isclosed. Assume δ = α−β+γ ∈ Γ . Then δ → α by (19.8.1) and therefore δ → α→ ε,contradicting (15.6.5). By symmetry, it follows that in fact all triangles containedin (1) are closed.

Let xα ∈ V +α and yβ ∈ V −β . By Lemma 23.22(d), applied to the square

(β, γ, β′, γ′), there exists uβ′ ∈ V +β′ such that

yβ = e−γ uβ′ e−γ′. (4)

By (9.9.5) we have(((((((b(uβ′ , e

−γ′), b(xα, e

−γ ))))))))

= b(B(uβ′ , e

−γ′)xα, B(e−γ′ , uβ′)

−1e−γ)· b(xα, e

−γ )−1.

We compute the arguments of the first factor b( , ) on the right. Since thetriangle β′, γ′, α is closed we have β′ − γ′ + α /∈ Γ , and by Lemma 19.9(a),also 2β′ − 2γ′ + α /∈ Γ . Hence uβ′ e−γ′ xα ∈ V

−β′−γ′+α = 0 and Q(uβ′)Q(e−γ′)xα ∈

V −2β′−2γ′+α = 0, which shows


B(uβ′ , e−γ′)xα = xα − uβ′ e−γ′ xα+Q(uβ′)Q(e−γ′)xα = xα.

Also, B(e−γ′ , uβ′)−1 = B(−e−γ′ , uβ′) by (20.7.4) which implies

B(e−γ′ , uβ′)−1e−γ = e−γ + e−γ′ uβ′ e

−γ +Q(e−γ′)Q(uβ′)e

−γ .

Since β′ γ, we have 2β′ − γ /∈ Γ by Lemma 28.2 and therefore Q(uβ′)e−γ ∈

V +2β′−γ = 0. This shows B(e−γ′ , uβ′)

−1e−γ = e−γ +yβ by (4), so we obtain, by (11.5.1)for the Peirce grading Pα,(((((((

b(uβ′ , e−γ′), b(xα, e

−γ ))))))))

= b(xα, e−γ + yβ) · b(xα, e

−γ )−1 = b(xα, yβ). (5)

As in (29.9.3), let Z(ε) = % ∈ R : ε+ % /∈ R. We claim that

for any edge α1 α2 contained in (1), α1 − α2 ∈ Z(ε). (6)

By symmetry, it suffices to prove this for α1 = α and α2 = β. Assuming δ :=α− β + ε ∈ R we obtain δ ∈ R1 = Γ by applying the minuscule coweight definingthe 3-grading of R. Then

〈δ, ε∨〉 = 〈α− β + ε, ε∨〉 = 1− 1 + 2 = 2,

and thus δ → ε. In the same way, 〈δ, α∨〉 = 2 − 1 + 1 = 2 implies δ → α. Now itfollows from Lemma 15.6 that ε α, contradiction. Applied to the edges α γand β′ γ′, (6) yields

α− γ ∈ Z(ε) and β′ − γ′ ∈ Z(ε).

As in Lemma 29.9(c), let H = UZ(ε) and U = Uε. Since b(uβ′ , e−γ′) ∈ Uβ′−γ′ ⊂ H

and b(xα, e−γ ) ∈ Uα−γ ⊂ H, we have b(xα, yβ) ∈ D(H) by (5). Let zε ∈ V +

ε .Then x+(zε) ∈ U and therefore E(zε) = p−1(x+(zε)) ⊂ p−1(U) by (29.1.2), whileE(xα, yβ) = p−1

(b(xα, yβ)

)⊂ p−1

(D(H)

). Hence⟨

〈xα, yβ , zε〉⟩

=(((((((E(xα, yβ), E(zε)

)))))))∈(((((((p−1(D(H)

), p−1(U)

)))))))= 1

by (29.9.2). This proves (2).

We come to the proof of (3). Let xα, yβ , zγ in the respective root spaces. Sincethe cog E is defined on ε, we have an idempotent E (ε) = eε ∈ Vε. Consider theelement

u := Q(e+γ′)Q(e−ε )zγ .

Then u ∈ V +2γ′−2ε+γ = V +

γ′ because γ + γ′ = 2ε by (15.3.1), so we write u = uγ′ .By computing in Vγ ⊕ Vε ⊕ Vγ′ , one sees easily that

Q(e+ε )Q(e−γ′)uγ′ = Q(e+

ε )Q(e−γ′)Q(e+γ′)Q(e−ε )zγ = zγ . (7)

Let b ∈ E(e+ε , e−γ′) and c′ ∈ E(uγ′). Then wε := −e+

ε e−γ′ uγ′ ∈ V +

ε , and

§29] Vanishing of the binary symbols 375(((((((b, c′

)))))))=⟨〈e+ε , e−γ′ , uγ′〉

⟩=(((((((E(e+

ε , e−γ′), E(uγ′)

)))))))⊂ E

(− e+

ε e−γ′ uγ′+Q(e+

ε )Q(e−γ′)uγ′)

= E(wε + zγ) = E(wε) · E(zγ), (8)

by (29.3.5), (29.1.3) and (7).Let µ = α− β and let Z(µ) = % ∈ R : µ+ % /∈ R. We claim that

γ′ ∈ Z(µ) and ε− γ′ ∈ Z(µ). (9)

Indeed, the first statement follows from µ+γ′ = α−β+γ′ /∈ R since α, β, γ′ is aclosed triangle. To prove ε−γ′ ∈ Z(µ), we first remark that ε−γ′ ∈ R which followsfrom ε ∼ γ′ and 15.9. Now assume, aiming for a contradiction, that ε− γ′ /∈ Z(µ),so ξ := µ+ ε− γ′ ∈ R. Then

〈ξ, γ′∨〉 = 〈α− β + ε− γ′, γ′∨〉 = 1− 1 + 1− 2 = −1.

This implies sγ′(ξ) = ξ−〈ξ, γ′∨〉γ′ = ξ+ γ′ = µ+ ε = α− β+ ε ∈ R, contradicting(6).

Let U = Uµ and H = UZ(µ). Then(((((((U, H

)))))))= 1 by Lemma 29.9(c) and therefore(((((((

p−1(U), p−1(H))))))))⊂ Z (E) by (29.9.1). Also, b(xα, yβ) ∈ U , and by (9) and

(21.1.2),

b(e+ε , e−γ′) ∈ Uε−γ′ ⊂ H, x+(uγ′) ∈ Uγ′ ⊂ H.

Let a ∈ E(xα, yβ) = p−1(b(xα, yβ)

). Since b(xα, yβ) ∈ Uµ, we have a ∈ p−1(U)

while b ∈ E(e+ε , e−γ′) = p−1

(b(e+

ε , e−γ′))⊂ p−1(H), and c′ ∈ E(uγ′) ⊂ p−1(H).

Hence by (29.9.1),(((((((a, b)))))))∈ Z (E) and

(((((((a, c′

)))))))∈ Z (E) which implies

(((((((a,(((((((b, c′

))))))))))))))=

1 by (29.6.11). Let c ∈ E(zγ) and d ∈ E(wε). Then(((((((b, c′

)))))))≡ dc mod Z (E) by (8).

This shows

1 =(((((((a,(((((((b, c′

))))))))))))))=(((((((a, dc

)))))))=(((((((a, d)))))))· Int(d)

(((((((a, c)))))))

(by (3.6.3))

=⟨〈xα, yβ , wε〉

⟩· Int(d)

⟨〈xα, yβ , zγ〉

⟩(by (29.3.5))

= 1 · Int(d)⟨〈xα, yβ , zγ〉

⟩, (by (2))

and therefore⟨〈xα, yβ , zγ〉

⟩= 1.

29.12. Theorem. Let Γ be a Jordan graph with associated 3-graded root sys-tem (R,R1). Let R = (Vα)α∈Γ be a root grading of a Jordan pair V , compatiblewith a cog E defined on a subset ∆ of Γ containing Γ 0. In addition, we make thefollowing assumptions:

(i) Γ contains no connected component of rank 6 2,

(ii) every connected component of rank 3 is isomorphic to K3,

(iii) if Σ is a connected component of type O3 or T4 then Σ ⊂ ∆,

(iv) Γ contains no connected component of type T4.

Let p: E → G be a central extension of a group G ∈ st(V,R,E ). Then the groupsp−1(Uσ) are abelian.


Proof. By Lemma 29.2(b), we must show that the map F of (29.2.4) vanishes.Assumption (i) guarantees that Γiso = ∅. Hence by Lemma 28.8, we may assumeΓ connected. Then, by the classification of 3-graded root systems in 14.7 and thetable (14.20.12) , the possibilities for Γ and ∆ are as follows:

(a) Γ = KI KJ with 16 |I|6 |J | and |J |> 3,

(b) Γ = TI or TI with |I|> 5,

(c) Γ = OI or OI with |I|> 4,

(d) Γ = ∆ = O3 or Γ = ∆ = T4,

(e) Γ = Cl or Sch.

Since F is EA(V,R)-invariant, the cases (a), (b), (d) for Γ = ∆ = T4 and (e) follow

from 28.13. In the remaining cases (c) and (d) for Γ = ∆ = O3 we show first that

all triangles α, β, γ are closed and satisfy⟨〈V σα , V −σβ , V σγ 〉

⟩= 1. (1)

Indeed, in case (c) it was shown in Example 19.5(c) that every triangle embeds in

a tetrahedron. Hence (1) follows from Lemma 29.10. In Case (d), Γ = ∆ = O3 isthe graph (29.11.1), so (1) holds by Lemma 29.11.

We can now finish both cases simultaneously, assuming Γ = OI with |I| > 3and (1). Let α, α′ be a pair of orthogonal vertices in Γ 0. Since Γ 0 ∼= OI , sucha pair embeds in an octahedron (29.8.1). By (1), Proposition 29.8 is applicableand shows F (V σα , V

σα′) = 0. Now F = 0 follows from Proposition 28.10(b), so the

groups p−1(Uσ) are abelian by Lemma 29.2.

§30. Vanishing of the ternary symbols

30.1. Throughout this section, we assume that Γ is a Jordan graph, V is aJordan pair with a Γ -grading R = (Vα)α∈Γ , E : ∆ → Idp(V ) is a cog compatiblewith R, defined on a subset ∆ of Γ containing Γ 0, and G is a group in st(V,R,E ),unless specified otherwise. The Weyl element corresponding to a root δ ∈ ∆ isdenoted wδ as in (24.1.1). Recall from (24.1.3) and Proposition 24.6 that

G ∈ st(V,R,E ) ⇐⇒G ∈ st(V,R) and wδ is a Weyl elementfor (G, (U%)%∈R

), for all δ ∈ ∆

⇐⇒

G ∈ st(V,R) and G ∈ st(V, eδ),for all δ ∈ ∆

.

Let us remark here that the case of a Jordan pair with an idempotent Peircegrading V = V2(e) ⊕ V1(e) ⊕ V0(e) and a group G ∈ st(V, e) falls under these

assumptions, as the special case Γ = T2 a collision, Γ 0 = ∅ and ∆ = 2ε1, see20.2(b), 12.1 and 24.1(c). Also, the extreme case st(V,R,E ) = st(V ) is included,see 24.1(a).

We consider a central extension

0 // Ai // E

p // G // 1

of G as in (27.9.1). The notations introduced in 29.1, Lemma 29.2 and 29.3 areused throughout.

§30] Vanishing of the ternary symbols 377

30.2. Lemma. Let e be an idempotent of V and G ∈ st(V, e). Then forxk ∈ V σk (e), yl ∈ V −σl (e), zm ∈ V σm(e), for j ∈ 0, 1 and arbitrary z ∈ V σ,⟨

〈x2, y0, z〉⟩

=⟨〈x0, y2, z〉

⟩= 1, (1)⟨

〈xj+1, yj , z〉⟩

=⟨〈eσ, e−σ xj+1 yj, z〉

⟩, (2)⟨

〈xj , yj+1, z〉⟩

=⟨〈xj yj+1 eσ, e−σ, z〉

⟩, (3)

Int(we) · E(xj+1, yj) = E(−e−σ xj+1 yj), (4)

Int(we) · E(xj , yj+1) = E(xj yj+1 eσ), (5)

Int(we) · E(z1) = E(z1,−e−σ) = E(−z1, e−σ) = E(z1, e−σ)−1, (6)⟨〈xj+1, yj , z2〉

⟩=[[Qe−σz2, e−σ xj+1 yj ]

], (7)⟨

〈xj , yj+1, z0〉⟩

=[[ eσ yj+1 xj, z0 ]

], (8)(((((((

E(x2), E(y0))))))))

=[[Qe−σx2, y0 ]

], (9)

Int(we) ·⟨〈xj+1, yj , z1〉

⟩−1=⟨〈e−σ, z1,−e−σ xj+1 yj〉

⟩, (10)

Int(we) ·⟨〈xj , yj+1, z1〉

⟩−1=⟨〈−z1, e−σ, xj yj+1 e−σ〉

⟩. (11)

Proof. In G we have bσ(x2, y0) = 1 by Theorem 11.2(iv), hence E(x2, y0) ⊂Z (E), and therefore

⟨〈x2, y0, z〉

⟩=(((((((E(x2, y0), E(z)

)))))))= 1. The second formula

of (1) is proved in the same way. By Theorem 12.5, G satisfies all shift relationsof 12.3. In particular, from the shift formulas S21 and S10 in case σ = + resp.S12 and S01 in case σ = − we have E(xj+1, yj) = E(eσ, e−σ x2 y1). Taking thecommutator with E(z) yields (2). Formula (3) is proved similarly, and (4) and (5)follow from the corresponding shift formulas S′′j+1,j and S′′j,j+1 and (27.9.4).

By (11.5.2),

bσ(z1, e−σ)−1 = bσ(−z1, e−σ) = bσ(z1,−e−σ) (12),

and by (12.2.2), Int(we) · xσ(z1) = bσ(−z1, e−σ). Hence (6) follows again from(27.9.4).

For (7) we use (4) and the Weyl relations in G which show Int(we) · E(z2) =E(Qe−σz2). This implies, by (27.10.2) and (29.2.1),


⟩=[[−e−σ xj+1 yj, Qe−σz2 ]

]=[[Qe−σz2, e−σ xj+1 yj ]

].

Since the right hand side is in Z (E) by (29.1.5), conjugating with w−1e yields (7).

The proof of (8) is similar, using (5) and Int(we) · E(z0) = E(z0).For (9) we use again

[[Qe−σx2, y0 ]

]∈ Z (E). Hence,[

[Qe−σx2, y0 ]]

= Int(we) ·[[Qe−σx2, y0 ]

]= Int(we) ·

(((((((E(Qe−σx2), E(y0)

)))))))=(((((((E(QeσQe−σx2), E(y0)

)))))))=(((((((E(x2), E(y0)

))))))),

since QeσQe−σ acts like the identity on V σ2 by (6.14.7).Replacing z1 by −z1 in (12) shows E(z1, e−σ) = E(−z1,−e−σ) = E(e−σ, z1)−1

(by (29.3.3)). Now (10) follows from



⟩−1= Int(we) ·

(((((((E(z1), E(xj+1, yj)

)))))))(by (3.6.1))

=(((((((

Int(we) · E(z1), Int(we) · E(xj+1, yj))))))))

=(((((((E(z1, e−σ)−1, E(−e−σ xj+1 yj)

)))))))(by (6) and (4))

=(((((((E(e−σ, z1), E(−e−σ xj+1 yj)

)))))))=⟨〈e−σ, z1,−e−σ xj+1 yj〉

⟩(by (29.3.5)).

The proof of (11) follows the same pattern:

Int(we) ·⟨〈xj , yj+1, z1〉

⟩−1= Int(we) ·

(((((((E(z1), E(xj , yj+1)

)))))))=(((((((

Int(we) · E(z1), Int(we) · E(xj , yj+1))))))))

=(((((((E(−z1, e−σ), E(xj yj+1 e−σ)

)))))))=⟨〈−z1, e−σ, xj yj+1 e−σ〉

⟩.

30.3. Proposition. Assume Γiso = ∅. Then the following conditions are equiv-alent:

(i) for all α, β in Γ with α ⊥ β and all σ ∈ +,−,(((((((E(V −σα ), E(V σβ )

)))))))= 1, (1)

(ii) for all α ∈ ∆ and β ∈ Γ with α ⊥ β and all σ ∈ +,−,[[V σ2 (eα), V σβ ]

]= 1, (2)

(iii) (2) holds for all α ∈ Γ 0 and β ∈ Γ with α ⊥ β and all σ ∈ +,−.

Proof. (i) =⇒ (ii): Let α ∈ Γ . Recall the definition of Γi(α) from (16.7.1), inparticular, that Γ2(α) consists of α and the initial points of all arrows ξ → α. By(C1) and (15.3.1), we then have 2α − ξ → α, whence also 2α − ξ ∈ Γ2(α), and forξ = α this holds trivially, so the map ξ 7→ 2α− ξ is a bijection of Γ2(α) onto itself.From β ⊥ α we see β ∈ Γ0(α) and therefore β ⊥ Γ2(α) by Lemma 16.8(a).

It follows from (23.8.6) and (23.8.7) that V2(eα) =∑ξ∈Γ2(α) Vξ. Since the

binary symbols are multiplicative ((29.2.2)), it suffices to prove[[V σξ , V

σβ ]]

= 1

for all ξ ∈ Γ2(α). By 6.14, Q(eσα): V −σ2 (eα) → V σ2 (eα) is a bijection. SinceQ(eσα)V −σξ ⊂ V σ2α−ξ we get V σξ = Q(eσα)V −σ2α−ξ. Therefore, by (30.2.9) and (1),for all ξ ∈ Γ2(α),[

[V σξ , Vσβ ]]

=[[QeσαV

−σ2α−ξ, V

σβ ]]

=(((((((E(V −σ2α−ξ), E(V σβ )

)))))))= 1.

The implication (ii) =⇒ (iii) being obvious, let us prove (iii) =⇒ (i). Fix α,β ∈ Γ with α ⊥ β and assume (2) for all (γ, δ) ∈ Γ 0×Γ with γ ⊥ δ. We first show:

if α ∈ Γ2(γ) and β ⊥ γ for some γ ∈ Γ 0 then (1) holds for α, β. (3)

Indeed, V −σα ⊂ V −σ2 (eγ) since α ∈ Γ2(γ), whence, by (30.2.9),(((((((E(V −σα ), E(V σβ )

)))))))=[

[QeσγV−σα , V σβ ]

]= 1. In particular, (3) implies (1) in case α ∈ Γ 0 by taking α = γ.


We also get (1) for β ∈ Γ 0 because(((((((E(V −σα ), E(V σβ )

)))))))−1 =(((((((E(V σβ ), E(V −σα )

)))))))= 1

by (3) with α and β = γ exchanged and σ replaced by −σ. Therefore we mayassume α, β ∈ Γ Γ 0 = ∂Γ from now on.

Let Σ be the connected component of Γ containing α. Then Σiso ⊂ Γiso = ∅and ∂Σ 6= ∅ because α ∈ Σ ∩ ∂Γ = ∂Σ. Thus Σ = Σher or Σ = Σorth byProposition 17.7. In the first case, Σ ∼= TI with |I|> 3 by Proposition 17.10. After

identifying Σ with TI we have α = i and β = j for distinct i, j ∈ I. Since|I|> 3, there exists k ∈ I i, j and then γ = i, k ∈ Γ 0 satisfies the assumptionin (3), proving (1) in this case.

We are left with Σ = Σorth. Then, by Proposition 17.12, α is the endpoint ofan orthogonal arrow. By (17.2.3), α embeds in a pyramid

γ

???? δ

α

δ′

??γ′

__????

Hence γ and γ′ belong to Γ 0 ⊂ ∆, so there are corresponding idempotents eγ andeγ′ . Clearly V σα ⊂ V σ1 (eγ) ∩ V σ1 (eγ′). Hence by (30.2.6), (30.2.4) and (23.22.3),

Int(wγ′) Int(wγ) · E(V σα ) = Int(wγ′) · E(V σα ,−e−σγ )

= E(e−σγ′ Vσα e−σγ ) = E(V −σα ).

Since β ⊥ γ and β ⊥ γ′, Int(wγwγ′) · x−(V σβ ) = x−(V σβ ) and therefore alsoInt(wγwγ′) · E(V σβ ) = E(V σβ ). By Lemma 28.6(c), F (V σα , V

σβ ) = 0, equivalently,[

[V σα , Vσβ ]]

=(((((((E(V σα ), E(V σβ )

)))))))= 1. We conjugate this with wγ′wγ and obtain

1 = Int(wγ wγ′) ·(((((((E(V σα ), E(V σβ )

)))))))=(((((((E(V −σα ), E(V σβ )

))))))),

as desired.

30.4. Lemma. Let (α, β, α′, β′) be a square in Γ . Then⟨〈V σα , V −σβ , V σβ′〉

⟩= 1. (1)

Proof. Let xα ∈ V σα , yβ ∈ V −σβ and zβ′ ∈ V σβ′ . Then eσβ , yβ , xα ∈ V σα by

(20.1.1), and xα ∈ V σ1 (eβ), yβ ∈ V −σ2 (eβ) and zβ′ ∈ V σ0 (eβ). Hence (30.2.8) shows⟨〈xα, yβ , zβ′〉

⟩=[[ eσβ , yβ , xα, zβ′ ]

]∈[[V σα , V

σβ′ ]].

By Lemma 28.5(a), F (V σα , Vσβ′) = 0 and therefore

[[V σα , V

σβ′ ]]

= 1, which proves(1).

30.5. Lemma. (a) Let (x, y) ∈ V σ × V −σ be quasi-invertible, and suppose Gsatisfies the relations Bσ(x, y) of (9.10.9). Then the symbol

⟨〈x, y, z〉

⟩behaves as

follows in the second and third variable:

380 CENTRAL CLOSEDNESS [Ch. VI⟨〈x, y, u+ z〉

⟩=⟨〈x, y, u〉

⟩·⟨〈x, y, z〉

⟩·[[ x y z −QxQyz, u ]

], (1)⟨

〈x, y + v, z〉⟩

=⟨〈x, y,−xy, v, z+Q(xy)Qvz〉

⟩·⟨〈xy, v, z〉

⟩·⟨〈x, y, z〉

⟩. (2)

for all u, z ∈ V σ and v ∈ V −σ, provided (x, y + v) is quasi-invertible.

(b) Let e be an idempotent of V , let G ∈ st(V, e), let x1, u1 ∈ V σ1 (e), y2 ∈V −σ2 (e) and z ∈ V σ. Then⟨〈x1 + u1, y2, z〉

⟩=⟨〈x1, y2,−u1 y2 z+Qu1

Qy2z〉⟩·⟨〈u1, y2, z〉

⟩·⟨〈x1, y2, z〉

⟩. (3)

In particular, for z1 ∈ V σ1 and z2 ∈ V σ2 ,⟨〈x1 + u1, y2, z1〉

⟩=[[ u1 y2 z1, eσ y2 x1 ]

]·⟨〈u1, y2, z1〉

⟩·⟨〈x1, y2, z1〉

⟩, (4)⟨

〈x1 + u1, y2, z2〉⟩

=⟨〈x1, y2,−u1 y2 z2〉

⟩·⟨〈u1, y2, z2〉

⟩·⟨〈x1, y2, z2〉

⟩·[[ eσ, y2, x1, Qu1Qy2z2 ]

]. (5)

Proof. It suffices to prove this for σ = +, the case σ = − then follows by passingto V op.

(a) We use the commutator formula (3.6.3),(((((((a, bc

)))))))=(((((((a, b)))))))·(((((((a, c)))))))·((((((((((((((c, a))))))), b))))))),

where a ∈ E(x, y), b ∈ E(u) and c ∈ E(z). Then(((((((a, bc

)))))))is the left hand side of

(1), while(((((((a, b)))))))

=⟨〈x, y, u〉

⟩,(((((((a, c)))))))

=⟨〈x, y, z〉

⟩, and(((((((

c, a)))))))

=(((((((a, c)))))))−1 =

⟨〈x, y, z〉

⟩−1 ∈ E(x y z −QxQyz

)by (29.3.5). Hence((((((((((((((

c, a))))))), b)))))))

=(((((((E(x y z −QxQyz

), E(u)

)))))))=[[ x y z −QxQyz, u ]

],

so (1) follows.

For (2), we have b(x, y + v) = b(x, y)b(xy, v) by (9.9.2), hence⟨〈x, y + v, z〉

⟩=(((((((E(x, y + v), E(z)

)))))))=(((((((E(x, y)E(xy, v), E(z)

))))))).

Let a ∈ E(x, y), b ∈ E(xy, v) and c ∈ E(z). The commutator formula (3.6.2) says⟨〈x, y + v, z〉

⟩=(((((((ab, c

)))))))=(((((((a,(((((((b, c))))))))))))))·(((((((b, c)))))))·(((((((a, c))))))).

By (29.3.5), (((((((b, c)))))))

=⟨〈xy, v, z〉

⟩∈ E(−xy, v, z+Q(xy)Qvz).

Hence (((((((a,(((((((b, c))))))))))))))

=⟨〈x, y,−xy, v, z+Q(xy)Qvz〉

⟩,

while(((((((a, c)))))))

=⟨〈x, y, z〉

⟩. This proves (2).

(b) Let a ∈ E(x1, y2), b ∈ E(u1, y2), c ∈ E(z), and compute the terms in thecommutator formula (3.6.2). By (9.9.1) and since yu1

2 = y2 by Corollary 10.7(a), we


have b(x1+u1, y2) = b(x1, y2)·b(u1, y2), so ab ∈ E(x1, y2)E(u1, y2) = E(x1+u1, y2).This implies

(((((((ab, c

)))))))=⟨〈x1 + u1, y2, z〉

⟩. Furthermore, since the relations B(u1, y2)

hold, (((((((b, c)))))))

=⟨〈u1, y2, z〉

⟩∈ E(−u1 y2 z+Qu1Qy2z),(((((((

a,(((((((b, c))))))))))))))

=⟨〈x1, y2,−u1 y2 z+Qu1

Qy2z〉⟩,

which implies (3). For z = z1 ∈ V σ1 we have u1 y2 z1 ∈ V σ0 andQy2z1 ∈ V σ3 = 0, so(4) follows from (3), (30.2.8) and (29.2.1). If z = z2 ∈ V σ2 then u′1 := −u1 y2 z2 ∈V σ1 and z0 := Qu1Qy2z2 ∈ V σ0 . Hence (3) yields⟨

〈x1 + u1, y2, z2〉⟩

=⟨〈x1, y2, u

′1 + z0〉

⟩·⟨〈u1, y2, z2〉

⟩·⟨〈x1, y2, z2〉

⟩. (6)

We use (1), (30.2.8) and the Peirce relations to simplify the first factor on the right:⟨〈x1, y2, u

′1 + z0〉

⟩=⟨〈x1, y2, u

′1〉⟩·⟨〈x1, y2, z0〉

⟩·[[ x1 y2 z0 −Qx1Qy2z0, u

′1 ]]

=⟨〈x1, y2, u

′1〉⟩·[[ eσ, y2, x1, z0 ]

]=⟨〈x1, y2,−u1 y2 z2〉

⟩·[[ eσ, y2, x1, Qu1

Qy2z2 ]]. (7)

Now (5) follows from (6) and (7) and the fact that the binary symbols are central.

30.6. Lemma. Let α, β, γ be a closed triangle in Γ , and let xα ∈ V σα , yβ ∈V −σβ , zγ ∈ V σγ . Let u ∈ V σ and v1 ∈ V −σ1 (eα). Then⟨

〈xα, yβ , zγ + u〉⟩

=⟨〈xα, yβ , zγ〉

⟩·⟨〈xα, yβ , u〉

⟩, (1)⟨

〈xα, yβ + v1, zγ〉⟩


⟩·⟨〈xα, v1, zγ〉

⟩, (2)

and the order of the factors on the right in both formulas is immaterial since⟨〈xα, yβ , zγ〉

⟩∈ Z (E) by (29.5.2).

Proof. (1) follows from (30.5.1) by putting x = xα, y = yβ and z = zγ ,and observing xα yβ zγ ∈ V σα−β+γ = 0 and Q(xα)Q(yβ)zγ ∈ V σ2(α−β)+γ = 0

because α − β + γ /∈ Γ and 2(α − β) + γ 6∈ Γ by Lemma 19.9(a). Similarly,put x = xα, v = yβ , y = v1 and z = zγ in (30.5.2). Then xα = x2 ∈ V σ2 (eα),so xy = xv12 = x2 + Qx2y1 = x2 by (10.7.1) since Qx2y1 ∈ V σ3 (eα) = 0. Also,xy, v, z = xv1α , yβ , zγ = xα, yβ , zγ = 0, Qvz = Qyβzγ = 0, since β γand therefore 2β − γ /∈ Γ by Lemma 28.2. Now (2) follows from (30.5.2).

30.7. Lemma. Let G ∈ st(V, e). In the following formulas, the index j takesvalues in 0, 1, and subscripts at x, y, z, u indicate membership in the correspondingPeirce space of e. Then the ternary symbols

⟨〈xj+1, yj , z1〉

⟩and

⟨〈xj , yj+1, z1〉

⟩satisfy⟨

〈xj+1, yj , z1〉⟩

=⟨〈eσ, e−σ xj+1 yj, z1〉

⟩∈ E(−xj+1 yj z1), (1)⟨

〈xj , yj+1, z1〉⟩

=⟨〈xj yj+1 eσ, e−σ, z1〉

⟩∈ E(−xj yj+1 z1), (2)⟨

〈xj+1, yj , u1 + z1〉⟩

=⟨〈xj+1, yj , u1〉

⟩·⟨〈xj+1, yj , z1〉

⟩·[[ xj+1 yj z1, u1 ]

], (3)⟨

〈xj , yj+1, u1 + z1〉⟩

=⟨〈xj , yj+1, u1〉

⟩·⟨〈xj , yj+1, z1〉

⟩·[[ xj yj+1 z1, u1 ]

]. (4)

Proof. By (29.3.5), we have⟨〈xj+1, yj , z1〉

⟩∈ E

(− xj+1 yj z1+Qxj+1

Qyjz1

).

Here Qxj+1Qyjz1 ∈ V +

3 = 0 by the Peirce rules, so (1) follows from (30.2.2), andthe proof of (2) is similar. Formulas (3) and (4) follow from (30.5.1).


30.8. Lemma. Let (α, β, γ, δ) be a square in Γ and assume[[V σα , V

σγ ]]

= 1. Let

xα ∈ V σα , yβ ∈ V −σβ and zβ ∈ V σβ . Then the symbol⟨〈xα, yβ , zβ〉

⟩is multiplicative

in each variable, and satisfies⟨〈xα, yβ , zβ〉

⟩=⟨〈zβ e−σγ eσδ , e−σδ , xα yβ eσγ〉

⟩. (1)

Proof. We prove the lemma for σ = +, the case σ = − then follows by passingto V op.

We show first that⟨〈xα, yβ , zβ〉

⟩is multiplicative in the last variable zβ ∈ V +

β .Indeed, replace zβ by zβ+uβ . Then (30.5.1) shows that multiplicativity is equivalentto [

[ xα yβ zβ −QxαQyβzβ , uβ ]]

= 1.

Here xα yβ zβ ∈ V +α and QxαQyβzβ ∈ V +

2α−β = 0 by Lemma 28.2. Now

Lemma 28.5(a) shows[[V +α , V

+β ]]

= i(F (V +

α , V+β ))

= 1, as desired.

Put a :=⟨〈xα, yβ , zβ〉

⟩. Then a ∈ E(−xα yβ zβ) ⊂ E(V +

α ) by (29.3.5). Letwγ = x−(e−γ )x+(e+

γ )x−(e−γ ) ∈ G be the Weyl element associated with the idempo-tent eγ ∈ Vγ , and let cσ ∈ E(eσγ ) be lifts of xσ(eσγ ) ∈ G to E. By the assumption[[V +α , V

+γ ]]

= 1, (((((((c+, a

)))))))=[[ e+γ , −xα, yβ , zβ ]

]∈[[V +γ , V

+α ]]

= 1.

Also, by the Weyl relation W(eγ),

Int(wγ) · c− ∈ E(Int(wγ) · x−(e−γ )) = E(x+(e+γ )),

which implies Int(wγ) · c− ≡ c+ mod Z (E). For uα ∈ V +α we have Int(wγ) ·x+(uα)

= x+(uα), since α ⊥ γ and since wγ is a Weyl element for G. Hence Int(wγ) · a ≡a mod Z (E) and

Int(wγ) ·(((((((c−, a

)))))))=(((((((

Int(wγ) · c−, Int(wγ) · a)))))))

=(((((((c+, a

)))))))= 1,

It follows that(((((((c−, a

)))))))= 1, and therefore

Int(wγ) · a = Int(c− c+ c−) · a = a.

On the other hand, xα ∈ V0(eγ) and yβ , zβ ∈ V ±1 (eγ). Hence by (30.2.11), appliedto e = eγ and for j = 0, and by multiplicativity in zβ ,⟨

〈xα, yβ ,−zβ〉⟩

= a−1 = Int(wγ) ·⟨〈xα, yβ , zβ〉

⟩−1=⟨〈−zβ , e−γ , xα yβ e+

γ 〉⟩.

Let us put uδ := xα yβ e+γ ∈ V +

α−β+γ = V +δ (by (15.3.2)). Then zβ ∈ V +

0 (eδ) and

e−γ ∈ V −1 (eδ) while uδ ∈ V +2 (eδ). Hence (30.2.3) applied to e = eδ shows⟨

〈xα, yβ ,−zβ〉⟩

=⟨〈−zβ , e−γ , uδ〉

⟩=⟨〈−zβ e−γ e+

δ , e−δ , uδ〉

⟩,

which is (1) after replacing zβ with −zβ .Finally, x′α := zβ e−γ e+

δ ∈ V +α and xα yβ e+

γ ∈ V +δ , since (α, β, γ, δ) is a

square. The situation is symmetric in β and δ, so⟨〈x′α, e−δ , uδ〉

⟩is multiplicative in

uδ ∈ V σδ . Hence the multiplicativity of⟨〈xα, yβ , zβ〉

⟩in xα and yβ follows from (1)

and the fact that the Jordan triple product xα yβ e+γ is trilinear.


30.9. Corollary. Let (α, β, γ, δ) be a square in Γ satisfying[[V σα , V

σγ ]]

= 1and assume that (eα, eβ , eγ , eδ) ∈ Vα×Vβ×Vγ ×Vδ is a quadrangle of idempotents.Then ⟨

〈xα, e−σβ , eσβ〉⟩

=⟨〈eσα, e−σδ , xα e−σβ eσγ〉

⟩for all xα ∈ V σα .

Proof. This is the special case yβ = e−σβ , zβ = eσβ of Lemma 30.8, since

eσβ e−σγ eσδ = eσα by (25.5.1).

30.10. Lemma. Consider an induced subgraph

Ω =

α

MMMMMMMMMMMMMM β

pppppppppppppp

4444444

γ′

555555

MMMMMMMMMMMMMM γ

β′

qqqqqqqqqqqqqqα′

(1)

of Γ and assume that all triangles in Ω are closed. Then⟨〈V σα , V −σγ′ , V

σβ′〉⟩

=⟨〈eσα, e−σβ , V σγ 〉

⟩. (2)

Proof. We prove this as usual only for σ = +. Let e = eα. Then α γ′

implies Vγ′ ⊂ V1(e), so by (30.2.2),⟨〈V +α , V

−γ′ , V

+β′ 〉⟩

=⟨〈e+α , e−α V +

α V −γ′ , V+β′ 〉⟩.

From (23.22.1) we see e+α V

−α V +

γ′ = V +γ′ , whence⟨

〈V +α , V

−γ′ , V

+β′ 〉⟩

=⟨〈e+α , V

−γ′ , V

+β′ 〉⟩. (3)

Let zβ′ ∈ V +β′ . From (1) it is evident that (α, β, α′, β′) is a square. Hence

Lemma 30.4 yields ⟨〈e+α , e−β , zβ′〉

⟩= 1. (4)

Let vβ′ ∈ V −β′ and conjugate (4) with h := b(e+γ , vβ′). By (29.3.9), we obtain

Int(h) ·⟨〈e+α , e−β , zβ′〉

⟩=⟨〈h · e+

α , h · e−β , h · zβ′〉⟩

= 1. (5)

Now we compute the entries of the ternary symbol as follows. First,

h · e+α = B(e+

γ , vβ′)e+α = e+

α − e+γ , vβ′ , e

+α+Qe+γ Qvβ′ e

+α = e+

α ,

since e+γ vβ′ e

+α ∈ V +

γ−β′+α = 0 and also Qe+γ Qvβ′ e+α ∈ V +

2γ−2β′+α = 0, because

α, β′, γ is a closed triangle by assumption and therefore γ − β′ + α /∈ Γ and2(γ − β′) + α 6∈ Γ , by Lemma 19.9. Next,

h · e−β = B(−vβ′ , e+γ ) · e−β = e−β + vβ′ e+

γ e−β +Qvβ′Qe+γ e

−β = e−β + vβ′ e+

γ e−β .

Indeed, γ β implies 2γ−β /∈ Γ by Lemma 28.2, so Qe+γ e−β ∈ V

−2γ−β = 0. Finally,


h · zβ′ = B(e+γ , vβ′)zβ′ = zβ′ − e+

γ vβ′ zβ′,

because Qe+γ Qvβ′ zβ′ ∈ V+2γ−β′ = 0, which follows as before from γ β′.

Put sγ := e+γ vβ′ zβ′ ∈ V +

γ and tγ′ := vβ′ e+γ e−β ∈ V

−β′−γ+β = V −γ′ , because

(β′, γ, β, γ′) is a square. Then (5) says⟨〈e+α , e

−β + tγ′ , zβ′ − sγ〉

⟩= 1. (6)

With respect to e := eα, it follows from (1) that y1 := e−β + tγ′ ∈ V −1 (e) and

z1 := zβ′ and u1 = sγ are in V +1 (e). Also,

e+, y1, z1 = e+α , e−β , zβ′+ e+

α , tγ′ , zβ′ = 0, (7)

e+, y1, u1 = e+α , e−β , sγ+ e+

α , tγ′ , sγ = 0. (8)

Indeed, (7) follows from β ⊥ β′ and the fact that α, γ′, β′ is a closed triangle,while for (8) one uses γ′ ⊥ γ and that α, β, γ is a closed triangle.

Now (6), (30.7.3) and (7) show that

1 =⟨〈e+, y1, z1〉

⟩·⟨〈e+, y1,−u1〉

⟩·[[ e+, y1, z1,−u1 ]

]=⟨〈e+, y1, z1〉

⟩·⟨〈e+, y1,−u1〉

⟩.

Moreover,⟨〈e+, y1,−u1〉

⟩=⟨〈e+, y1, u1〉

⟩−1, which follows again from (30.7.3) and

(8):

1 =⟨〈e+, y1, u1 − u1〉

⟩=⟨〈e+, y1, u1〉

⟩⟨〈e+, y1,−u1〉

⟩·[[ e+, y1, u1,−u1 ]

]=⟨〈e+, y1, u1〉

⟩⟨〈e+, y1,−u1〉

⟩.

Altogether, (6) implies

l :=⟨〈e+α , e

−β + tγ′ , zβ′〉

⟩=⟨〈e+α , e

−β + tγ′ , sγ〉

⟩=: r. (9)

By assumption, α, γ′, β′ is a closed triangle, and we have eβ ∈ V1(eα). Hence(30.6.2) shows

l =⟨〈e+α , e−β , zβ′〉

⟩·⟨〈e+α , tγ′ , zβ′〉

⟩,

Similarly, using the closed triangle α, β, γ′ and again (30.6.2), we get

r =⟨〈e+α , tγ′ , sγ〉

⟩·⟨〈e+α , e−β , sγ〉

⟩.

Since (α, β, α′, β′) and (α, γ, α′, γ′) are squares, (30.4.1) shows that the first factorsin l and r are 1, so we obtain⟨

〈e+α , tγ′ , zβ′〉

⟩=⟨〈e+α , e−β , sγ〉

⟩.

Explicitly, this says,⟨〈e+α , e−β e

+γ vβ′, zβ′〉

⟩=⟨〈e+α , e

−β , e

+γ vβ′ zβ′〉

⟩. (10)

Since (γ′, β′, γ, β) is a square, (23.22.4) shows Vγ′ = eβ eγ Vβ′. Hence (10) implies⟨〈eα, Vγ′ , Vβ′〉

⟩⊂⟨〈eα, eβ , Vγ〉

⟩, because eγ vβ′ zβ′ ∈ Vγ . The reverse inclusion

follows from (23.22.4) as well. Thus⟨〈eα, Vγ′ , Vβ′〉

⟩=⟨〈eα, eβ , Vγ〉

⟩. Now (2) is a

consequence of (3).


30.11. Proposition. Let T = α1, α2, α3 be a closed triangle of type 2 asdefined in 19.4, and assume that T is contained in a connected component of rank> 5 of Γ . Then ⟨

〈V σα1, V −σα2

, V σα3〉⟩

= 1. (1)

Proof. Let Σ be the connected component of Γ containing T . Then T ⊂ Σ0,and by Propositions 19.7 and 19.5(b), Σ0 ∼= TI where I > 5, under an isomorphismmapping T to i, j, j, k, k, i for distinct indices i, j, k. If Σ 6= Σ0 then

Σ ∼= TI by Propositions 17.10 and 17.12, and then T is not closed by 19.10. ThusΣ = Σ0 and therefore all triangles in Σ are closed by Lemma 19.9(b).

Without loss of generality, we may assume Σ contains T5 and that α1 = 2, 3,α2 = 3, 4, α3 = 2, 4. Then T is embedded in the following octahedron wherewe abbreviated i, j to ij:

Ω =

23

OOOOOOOOOOOOO 13

ooooooooooooo

8888888

34

8888888

OOOOOOOOOOOOO 12

24

ooooooooooooo14

=

α

MMMMMMMMMMMMMM β

pppppppppppppp

4444444

γ′

555555

MMMMMMMMMMMMMM γ

β′

qqqqqqqqqqqqqqα′

(2)

as T = α, γ′, β′.As usual, we prove (1) for σ = +. By Lemma 30.10, (1) is equivalent to⟨

〈e+α , e−β , V

+γ 〉⟩

= 1. (3)

Consider the elementh = Int(wα) · wβ ∈ G.

By Lemma 13.10, h = weβ ,eα , so h ∈ N ∩ G0 and therefore (29.3.9) is applicable.Let uγ ∈ V +

γ and put

aγ =⟨〈uγ , e−β , e

+β 〉⟩∈ E.

By (30.7.1) applied to e = eγ , we have aγ ∈ E(−uγ , e−β , e+β ). Also, γ β shows

uγ ∈ V σ1 (eβ), so uγ = uγ , e−β , e+β and therefore

aγ ∈ E(−uγ). (4)

For the proof of (3), we will compute Int(h) · aγ in two different ways. Comparingthe results will yield, after more computation, the desired result. This follows thestrategy used in the proof of Proposition 29.8. For better readability we will omitthe superscripts ± in

⟨〈uγ , e−β , e

+β 〉⟩

and similar expressions. This should not cause

any confusion since the parity of the arguments of the symbols⟨〈x, y, z〉

⟩alternates.

At this point, we remark that Lemma 30.8 is applicable to every square con-tained in Ω. Indeed, since Σ ∼= TI and |I| > 5, we have c(Σ) =

(|I|−22

)> 3 by

(17.14.4). Taking for example the square (α, β, α′, β′), it follows from Proposi-tion 28.12 that F (V σα , V

σα′) = 0 and therefore

[[V σα , V

σα′ ]]

= 1, so that Lemma 30.8


and its corollary 30.9 apply. In particular, the symbol⟨〈xα, yβ , zβ〉

⟩is multiplicative

in each variable. By symmetry, the same holds for any other edge instead of α βcontained in Ω, since by (2), any edge in Ω is contained in a square. During theproof, we also make use several times of Lemma 30.6 applied to triangles containedin Σ resp. Ω.

(a) First computation of Int(h) · aγ . By (29.3.9),

Int(h) · aγ =⟨〈h · uγ , h · eβ , h · eβ〉

⟩.

By Proposition 10.12, h acts on V σ via ωf,e where f = eβ and e = eα. Wehave uγ ∈ V +

(11) = V +1 (f) ∩ V +

1 (e) = V +1 (eβ) ∩ V +

1 (eα) because α γ β.

As remarked above, all triangles in Σ are closed. In particular, this holds for thetriangle α, β, γ, see (2), so α−β+γ /∈ Γ , which implies Vα Vβ Vγ ⊂ Vα−β+γ = 0.Hence ωf,e ·uγ = uγ−f e e f uγ = uγ , because e f uγ = eα eβ uγ = 0. Also,eβ ∈ V(21) = V2(eβ) ∩ V1(eα) implies ωf,e · eβ = −e f eβ = −eα eβ eβ = −eαbecause eα and eβ are collinear idempotents. This shows

Int(h) · aγ =⟨〈uγ , −eα, −eα〉

⟩=⟨〈uγ , eα, eα〉

⟩,

since, as noted before,⟨〈uγ , eα, eα〉

⟩is multiplicative in each variable.

By (2), γ, β′, α = 12, 24, 23 is a triangle, contained in the tetrahedron12, 24, 23, 25. Hence by Lemma 29.10,

⟨〈Vγ , Vβ′ , Vα〉

⟩= 1, so (29.6.5) shows

Int(h) · aγ =⟨〈uγ , eα, eα〉

⟩=⟨〈uγ , eβ′ , eβ′〉

⟩.

By (2), (β, γ, β′, γ′) is a square, so β − γ + β′ = γ′ by (15.3.2). Define e′γ′ =eβ eγ eβ′ ∈ Vβ−γ+β′ = Vγ′ where the right hand side is understood component-wise. By Lemma 25.6(b), eγ′ and e′γ′ are associated idempotents. By Lemma 23.16,we may replace eγ′ by e′γ′ and thus assume that (eβ , eγ , eβ′ , eγ′) is a quadrangle ofidempotents. But also (γ, β′, γ′, β) is a square. Put zβ := uγ eβ′ eγ′ ∈ Vγ−β′+γ′ =Vβ . As noted before, Corollary 30.9 is applicable to the square (γ, β′, γ′, β). Thisshows

Int(h) · aγ =⟨〈uγ , eβ′ , eβ′〉

⟩=⟨〈eγ , eβ , uγ eβ′ eγ′〉

⟩=⟨〈eγ , eβ , zβ〉

⟩.

Since γ, α′, β = 12, 14, 13 is a triangle contained in the tetrahedron 12, 13,14, 15, Lemma 29.10 shows that

⟨〈Vγ , Vα′ , Vβ〉

⟩= 1. Hence formula (29.6.4) (with

α, β replaced by γ, β) can be applied and yields⟨〈eγ , eβ , zβ〉

⟩=⟨〈eγ eβ zβ, eβ , eβ〉

⟩=⟨〈uγ , eβ , eβ〉

⟩= aγ ,

since eγ eβ zβ = eγ eβ uγ eβ′ eγ′ = uγ by (25.5.5). This proves

Int(h) · aγ = aγ . (5)

In the course of the proof we have obtained

aγ =⟨〈uγ , eβ , eβ〉

⟩=⟨〈uγ , eα, eα〉

⟩. (6)


(b) Second computation of Int(h) · aγ . By Lemma 13.10 we have h = b1b2b1where b1 = b(eα, eβ) and b2 = b(−eβ , eα). Then by (4) and (29.3.5),(((((((

b1, aγ)))))))

=(((((((E(eα, eβ), E(−uγ)

)))))))=⟨〈eα, eβ ,−uγ〉

⟩=⟨〈eα, eβ , uγ〉

⟩−1 ∈ Z (E), (7)

where (7) follows from Lemma 30.6. We also have, by (29.3.9),

Int(b1) · aγ = Int(b1) ·⟨〈uγ , eβ , eβ〉

⟩=⟨〈B(eα, eβ)uγ , B(−eβ , eα)eβ , B(eα, eβ)eβ〉

⟩=⟨〈uγ , eβ , eβ − eα〉

⟩=⟨〈uγ , eβ , eα〉

⟩−1 · aγ ,using (30.6.1) in the last line, so that(((((((

b1, aγ)))))))

=⟨〈uγ , eβ , eα〉

⟩−1. (8)

Next, we use (6) and compute similarly, this time using (30.6.2) for γ, β, α:

Int(b1) · aγ = Int(b1) ·⟨〈uγ , eα, eα〉

⟩=⟨〈uγ , B(−eβ , eα)eα, B(eα, eβ)eα〉

⟩=⟨〈uγ , eα + eβ , eα〉

⟩=⟨〈uγ , eβ , eα〉

⟩· aγ ,

so that (((((((b1, aγ

)))))))=⟨〈uγ , eβ , eα〉

⟩. (9)

Comparing (7), (8) and (9) shows

z1 :=⟨〈eα, eβ , uγ〉

⟩−1=⟨〈uγ , eβ , eα〉

⟩−1=⟨〈uγ , eβ , eα〉

⟩∈ Z (E),

whenceInt(b1) · aγ = z1aγ and z2

1 = 1. (10)

The computation for b2 follows the same lines:(((((((b2, aγ

)))))))=(((((((E(−eβ , eα), E(uγ)

)))))))=⟨〈−eβ , eα, uγ〉

⟩and, again by Lemma 30.6,

Int(b2) · aγ =⟨〈B(−eβ , eα)uγ , B(eα, eβ)eβ , B(−eβ , eα)eβ〉

⟩=⟨〈uγ , eβ − eα, eβ〉

⟩=⟨〈uγ ,−eα, eβ〉

⟩· aγ (by (30.6.2))

= Int(b2) ·⟨〈uγ , eα, eα〉

⟩=⟨〈B(−eβ , eα)uγ , B(eα, eβ)eα, B(−eβ , eα)eα〉

⟩=⟨〈uγ , eα, eα + eβ〉

⟩=⟨〈uγ , eα, eβ〉

⟩· aγ , (by (30.6.1))

which yields

z2 :=⟨〈−eβ , eα, uγ〉

⟩=⟨〈uγ ,−eα, eβ〉

⟩=⟨〈uγ , eα, eβ〉

⟩∈ Z (E),

thusInt(b2) · aγ = z2aγ and z2

2 = 1. (11)

The group G satisfies the Weyl relation W(f), so wf = wfop is symmetric by9.17. Hence Lemma 13.10 shows that also h = b2b1b2. Now (5), (10) and (11) yield

aγ = Int(h) · aγ =(

Int(b1) Int(b2) Int(b1))· aγ = z2

1z2aγ = z2aγ

=(

Int(b2) Int(b1) Int(b2))· aγ = z1z

22aγ = z1aγ .

This implies z1 = z2 = 1, so we have (3), as desired.


30.12. Lemma. Let

Ψ =

α

3333333

NNNNNNNNNNNNNNN β

ooooooooooooooo

4444444

γ′ //

555555

OOOOOOOOOOOOOO ε γoo

β′

EE

pppppppppppppppα′

YY3333333

(1)

be an induced subgraph of type O3 of Γ as in Lemma 29.11. Then⟨〈V σξ , V −ση , V σζ 〉

⟩= 1 (2)

for any permutation ξ, η, ζ of α, β, ε, and⟨〈xα, yε, zε〉

⟩=⟨〈xα yε zε, e−σβ , eσβ〉

⟩. (3)

for xα, yε, zε in the respective root spaces.

Proof. We proved⟨〈Vα, Vβ , Vε〉

⟩= 1 in (29.11.2), whence by symmetry in α and

β also⟨〈Vβ , Vα, Vε〉

⟩= 1. Since 〈ε, α∨〉 = 〈β, α∨〉 = 1, we have yε and zβ in V1(eα),

so by (30.2.10) for j = 1,

Int(wα) ·⟨〈xα, yε, zβ〉

⟩−1=⟨〈eσα, zβ ,−eσα xα yε〉

⟩∈⟨〈Vα, Vβ , Vε〉

⟩= 1,

which proves⟨〈Vα, Vε, Vβ〉

⟩= 1, and

⟨〈Vβ , Vε, Vα〉

⟩= 1 follows by symmetry in α and

β. Finally, let wβ be the Weyl element defined by eβ . Then zε, xα ∈ V1(eβ), so by(30.2.11) for j = 1,

Int(wβ) ·⟨〈zε, yβ , xα〉

⟩−1=⟨〈−xα, e−σβ , zε yβ eσβ〉

⟩∈⟨〈Vα, Vβ , Vε〉

⟩= 1,

because zε yβ eβ ∈ Vε. Hence⟨〈Vε, Vβ , Vα〉

⟩= 1, and

⟨〈Vε, Vα, Vβ〉

⟩= 1 holds again

by symmetry in α and β. This proves (2).

To prove (3), we use Lemma 29.6(b), after replacing γ and β there by β and ε,respectively. Then the assumptions (29.6.1) and (29.6.7) become

⟨〈Vα, Vβ , Vε〉

⟩= 1,⟨

〈Vα, Vε, Vβ′〉⟩

= 1, and⟨〈Vα, Vβ , Vβ′〉

⟩= 1. The first two conditions hold by (2). The

third condition follows from Lemma 30.4. Now (29.6.3) shows⟨〈xα, yε, zε vβ uβ〉

⟩=⟨〈xα yε zε, vβ , uβ〉

⟩.

Taking here (uβ , vβ) = (eσβ , e−σβ ) proves (3).

30.13. Lemma. Let α, β, γ be a closed triangle in Γ which embeds in an

induced subgraph Υ ∼= T4 and assume Υ ⊂ ∆. Then⟨〈V σα , V −σβ , V σγ 〉

⟩= 1. (1)


Proof. We picture Υ as in (14.18.4), where ij stands for i, j. After a renum-bering we may assume (α, β, γ) = (14, 24, 34). Put δ = 11, ζ = 33 and η = 44:

δ

444444444444444

α

ooooooooOOOOOOOO

444444444444444

12

444444444444444 13

η

OO

wwooooooooo

''OOOOOOOOO

β γ

22 //

77oooooooo

DD23

OOOOOOOOoooooooo

ζoo

ggOOOOOOOOO

ZZ444444444444444

First, we show ⟨〈V σα , V −σβ , V ση 〉

⟩= 1. (2)

Indeed, let e = eα. Then η → α and β α, whence Vη ⊂ V2(e) and Vβ ⊂ V1(e).By (30.2.7),⟨

〈V σα , V −σβ , V ση 〉⟩

=[[Q(e−σα )V ση , e−σα V σα V

−σβ ]

]⊂[[V −σδ , V −σβ ]

]= 1,

since F (V −σδ , V −σβ ) = 0 by Lemma 28.9(a).

By our assumption Υ ⊂ ∆ we have an idempotent eζ = E (ζ) ∈ Vζ , and ζ → γimplies Vγ ⊂ V1(eζ). Hence zγ e−σζ eσζ = zγ , so (9.10.10) yields(((((((

bσ(−zγ , e−σζ ), xσ(eσζ ))))))))

= xσ(zγ) · xσ(uη), (3)

where uη = xσ(Q(zγ)e−σζ

)∈ V σ2γ−ζ = V ση . Let µ = α − β = ε1 − ε2 and Z(µ) =

% ∈ R : µ + % /∈ R. Then ζ and γ − ζ = ε4 − ε3 belong to Z(µ). As inLemma 29.9 let U = Uµ and H = UZ(µ). Then (3) shows xσ(zγ) · xσ(uη) ∈ D(H)while bσ(xα, yβ) ∈ Uµ, so by (29.9.2),(((((((

E(xα, yβ), E(zγ)E(uη))))))))

= 1.

Now use the commutator formula (3.6.3), namely(((((((a, bc

)))))))=(((((((a, b)))))))·Int(b)

(((((((a, c)))))))

, with

a ∈ E(xα, yβ), b ∈ E(zγ), and c ∈ E(uη) ⊂ E(V ση ). Then(((((((a, c)))))))

=⟨〈xα, yβ , uη〉

⟩= 1

by (2), so 1 =(((((((a, bc

)))))))=(((((((a, b)))))))


⟩.

30.14. Theorem. Let V, Γ,R, G,∆,E , E be as in 30.1. In addition, we makethe following assumptions:

(i) every connected component of Γ has rank > 4,

(ii) if Σ is a connected component of type O3 or T4 then Σ ⊂ ∆,

(iii) Γ contains no connected components of type T4.


Then the pre-images p−1(Uσ) are abelian, and if α, β, γ ∈ Γ satisfy

α− β + γ /∈ Γ (1)

then ⟨〈V σα , V −σβ , V σγ 〉

⟩= 1. (2)

Proof. By Theorem 29.12, the groups p−1(Uσ) are abelian, whence[[V σ, V σ ]

]= 1

by Lemma 29.2. If α ⊥ β then bσ(xα, yβ) = 1 in G by Theorem 21.7(iv), soE(xα, yβ) ⊂ Z (E) and therefore (2) holds by (29.3.5). Observe that α 6= β 6= γfollows from (1), but α = γ is allowed. It remains to consider the following twocases:

Case (a): α β, Case (b): α→ β or α← β.

Observe that α and β are in Γ 0 in Case (a), so we have corresponding idempotentseα and eβ . Case (a) splits into the following subcases.

Case (a1): α β ⊥ γ. Relative to eβ we have V σα ×V −σβ ×V σγ ⊂ V σ1 ×V −σ2 ×V σ0 ,

so by (30.2.8),⟨〈V σα , V −σβ , V σγ 〉

⟩=[[ V σα V −σβ eσβ, V σγ ]

]= 1.

Case (a2): α β γ. If γ ⊥ α then α− β + γ ∈ Γ by the closure condition(C2) of a Jordan graph, contradiction. Thus either γ = α or γ α or γ → αor γ ← α. But the last two cases lead to a subgraph of type (15.6.4) which isimpossible by Lemma 15.6. If γ = α then by (30.2.7) with respect to e = eα:⟨

〈V σα , V −σβ , V σα 〉⟩

=[[Qe−σα V σα , e−σα V σα V

−σβ ]

]= 1.

Finally, if γ α, then (1) says that T = α, β, γ is a closed triangle. IfT is contained in a tetrahedron or a prism then (2) follows from Lemma 29.10.Otherwise, T is of type 2, see 19.4. Let Σ be the connected component of Γcontaining T . By Proposition 19.7, Σ0 ∼= TI , |I| > 4. If |I| = 4 then by (ii) and

(iii), Σ ∼= O3 or Σ ∼= T4. In the first case, (2) follows from Lemma 29.11 and in thesecond case, it follows from Lemma 30.13. Finally, if |I| > 5 then (2) follows fromProposition 30.11.

Case (a3): α β ← γ. then γ → α or α ⊥ γ by Lemma 15.6. In the first casewe have V σα × V −σβ × V σγ ⊂ V σ2 × V −σ1 × V σ2 relative to eα, so⟨

〈V σα , V −σβ , V σγ 〉⟩

=[[Qe−σα V σγ , e−σα , V σα , V

−σβ ]

]= 1

by (30.2.7). In the second case, α, β, γ generate a kite (γ;β, α, δ) by (C3) whereδ = α− β + γ ∈ Γ by (15.3.2), contradicting (1).

Case (a4): α β → γ. Let us write ε instead of γ. Then it follows fromLemma 15.6 that α → ε, so β → ε is an arrow of orthogonal type by (17.2.3). By

Proposition 17.12, the connected component Σ containing α, β, ε is of type OI , andby assumption (i), |I|> 3. Hence (2) follows from (30.12.2).


Case (b): β → α or α → β. By (i), Γ contains no isolated vertices or arrows.Hence an arrow in Γ is either of orthogonal or of hermitian type. Accordingly, wedistinguish the following subcases.

Case (b1): β → α is an arrow of orthogonal type. In order to conform to theusual notation, we write α = ε, and put β′ = 2ε − β. Since β is in Γ 0 we have acorresponding idempotent eβ . There are two possibilities.

Case (b1.1): β ⊥ γ. With respect to eβ , Vε × Vβ × Vγ ⊂ V1 × V2 × V0, so by(30.2.8), ⟨

〈V σε , V −σβ , V σγ 〉⟩

=[[ eσβ V −σβ V σε , V σγ ]

]= 1.

Case (b1.2): β ∼ γ. First, γ = ε is impossible because it would imply ε−β+γ =2ε − β = β′ ∈ Γ , contradicting (1). Likewise, β = γ is impossible. Now it followsfrom Lemma 15.6 and the fact that β → ε is of orthogonal type that

β

???? γ

ε

Since the connected component of Γ containing ε, β, γ is of type OI with |I|> 3 byassumption (i), (2) holds by (30.12.2).

Case (b2): α → β is an arrow of orthogonal type. Again, we write β = ε, putα′ = 2ε− α, and distinguish the following subcases.

Case (b2.1): γ ⊥ α′. Then Vα × Vε × Vγ ⊂ V0 × V1 × V0 with respect to eα′ , so⟨〈V σα , V −σε , V σγ 〉

⟩=[[ eσα′ , V −σε , V σα , V σγ ]

]= 1

by (30.2.8).

Case (b2.2): γ ∼ α′. Then γ belongs to the connected component Σ of Γ

containing α, ε and α′ which, by assumption (i), is of type OI with |I| > 3. Weclaim that γ /∈ ε, α, α′. Indeed, γ = ε implies α − ε + γ = α ∈ Γ , contradicting(1), and γ = α implies γ ⊥ α′ which contradicts γ ∼ α′. Finally, γ = α′ impliesα − ε + γ = α + α′ − ε = 2ε − ε = ε ∈ Γ by (15.3.1), which is impossible by(1). Now it follows from Propositions 17.12 and 17.11 that α γ → ε. HenceLemma (30.12.2) (applied to α, γ, ε) yields (2).

Case (b3): β → α is a hermitian arrow. By 17.2, β → α embeds in a kite(β, α, δ, ε):

β

????

α

???? ε

δ

Then α− β = δ − ε by (15.3.2), hence δ − ε+ γ /∈ Γ and δ ε. With respect tothe idempotent eδ we have Vα ⊂ V1 and Vβ ⊂ V0. By (30.2.2),⟨

〈V σα , V −σβ , V σγ 〉⟩

=⟨〈eσδ , e−σδ V σα V

−σβ , V σγ 〉

⟩,

and e−σδ V σα V−σβ ∈ V −σδ−α+β = V −σε . Hence

⟨〈V σα , V −σβ , V σγ 〉

⟩= 1 by Case (a)

applied to the triple (δ, ε, γ).

Case (b4): α→ β is a hermitian arrow. This case can be dealt with in the sameway as (b3), using a kite (α, β, δ, ε) and (30.2.3).


30.15. Corollary. Under the hypotheses of Theorem 30.14, let α β be anedge contained in a closed triangle. Then⟨

〈xα, yβ , zβ〉⟩

=⟨〈xα yβ zβ, e−σβ , eσβ〉

⟩. (1)

Proof. This is (29.6.4), which can be applied because we have just shown in(30.14.2) that the assumption (29.6.1) holds in our setting.

§31. Definition of the partial sections

31.1. Preliminaries. In this section we carry out Step 2 of the proof of The-orem 27.4 as outlined in 27.10. Thus, assuming the hypotheses of Theorem 27.4,we let G be the Steinberg group, p: E → G a central extension, and define sectionssσ: Uσ → E of p: p−1(Uσ)→ Uσ. It turns out that the proofs work just as well forany group G in st(V,R); the fact that G is the Steinberg group, thus defined bygenerators and relations as in 22.1, does not become relevant until the next section§32.

The hypotheses of Theorem 27.4 are identical with those of Theorem 30.14. Forthe convenience of the reader, we repeat them here:

(i) Γ is a Jordan graph,

(ii) V is a Jordan pair with a Γ -grading R which is idempotent with respectto a cog E defined on ∆ with Γ 0 ⊂ ∆ ⊂ Γ ,

(iii) p: E → G is a central extension of a group G ∈ st(V,R,E ),

(iv) every connected component of Γ has rank > 4,

(v) if Σ is a connected component of type T4 or O3 then Σ ⊂ ∆,

(vi) there are no components of type T4.

Hence the conclusions of Theorem 30.14 apply, so the groups p−1(Uσ) ⊂ E areabelian, and α−β+γ /∈ Γ implies

⟨〈Vα, Vβ , Vγ〉

⟩= 1. The notations and conventions

of 29.1 and 29.2 will be used without further comment. We also remark that thehypotheses of Theorem 24.2 are fullfilled, so that

st(V,R,E ) = st(V,E ) = st(V,E 0) = st(V,R). (1)

In view of the isomorphisms xσ: V σ → Uσ, finding the sections sσ amounts tofinding homomorphisms ϕσ: V σ → E satisfying p(ϕσ(x)) = xσ(x) for all x ∈ V σ,σ ∈ +,−. Since p−1(Uσ) is abelian, it suffices to define, for each α ∈ Γ ,homomorphisms ϕσα: V σα → E with p(ϕσα(xα)) = xσ(xα) for all xα ∈ V σα , andthen multiply these maps together (31.10). The definition of ϕσα for α ∈ Γ 0 is givenin 31.4, while for α ∈ ∂Γ it is done in Proposition 31.6 for the orthogonal case andin Proposition 31.9 for the hermitian case.

31.2. Lemma. If α β and α δ then⟨〈xα, e−σβ , eσβ〉

⟩=⟨〈xα, e−σδ , eσδ 〉

⟩. (1)

Proof. For better readability, we suppress the superscripts ±σ in this proof,trusting the reader to recall that the parity of the arguments alternates in the

§31] Definition of the partial sections 393

ternary symbols⟨〈−,−,−〉

⟩and in the Jordan triple products −,−,−. By

Lemma 15.6, there cannot be an arrow between β and δ, so either β ⊥ δ or β δ.Accordingly, we distinguish the following cases.

Case 1: β ⊥ δ. Then by the axiom (C2) of a Jordan graph, α, β, δ generate asquare Q = (α, β, γ, δ). By Lemma 19.11(b), applied to the connected componentΣ of Γ containing Q, there exists ε such that either (α, δ, ε) or (α, β, ε) is a closedtriangle. By symmetry, it suffices to do the first case.

By Lemma 25.6 and Corollary 23.16 we may replace eα by the idempotente′α = eβ eγ eδ ∈ Vα, and thus assume that (eα, eβ , eγ , eδ) is a quadrangle ofidempotents. Put uδ := xα eβ eγ ∈ Vα−β+γ = Vδ. Then⟨

〈xα, eβ , eβ〉⟩

=⟨〈eα, eδ, xα eβ eγ〉

⟩=⟨〈eα, eδ, uδ〉

⟩by Corollary 30.9. Since α δ is contained in the closed triangle (α, δ, ε), itfollows from Corollary 30.15 that⟨

〈eα, eδ, uδ〉⟩

=⟨〈eα eδ uδ, eδ, eδ〉

⟩,

and by (25.5.5), we have eα eδ uδ = eα eδ eγ eβ xα = xα, which shows (1).

Case 2: β δ, so T = α, β, δ is a triangle. By Lemma 19.9(b), T is eitherclosed or hermitian.

Case 2.1: T is closed. Then (1) follows from Lemma 29.7(b).

Case 2.2: T is hermitian. Then by Lemma 19.9, the connected component Σ ofΓ containing T is isomorphic to TI , and by our general assumptions, |I| > 4. By

Example 19.10, the hermitian triangles in TI are of type ij, jk, ki for i, j, k 6=,whereas the closed triangles are of type ij, ik, il for i, j, k, l 6=.

Without loss of generality, we may assume that I contains the set 1, 2, 3, 4.Hence we may identify Σ with TI = i, j : i, j ∈ I in such a way that α = 1, 2,β = 1, 3 and δ = 2, 3. Let α′ = 3, 4 and β′ = 2, 4. Then Σ contains thesubgraph

11

444444444444444

14

ooooooooOOOOOOOO

444444444444444

α=12

444444444444444 β=13

44

OO

wwoooooooo

''OOOOOOOO

β′=24 α=34

22 //

77oooooo

DDδ=23

OOOOOOooooooo

33oo

ggOOOOOOO

ZZ44444444444444

see (14.18.4). Here (α, β, α′, β′) is a square, and (α, β′, δ) is a closed triangle. Hence⟨〈xα, eβ , eβ〉

⟩=⟨〈xα, eβ′ , eβ′〉

⟩(by Case 1) =

⟨〈xα, eδ, eδ〉

⟩(by Case 2.1),

as desired.


31.3. Lemma. For any edge α β, we have⟨〈xα, yβ , zβ〉

⟩=⟨〈xα yβ zβ, e−σβ , eσβ〉

⟩.

Proof. As before, we suppress superscripts ±σ. If α β is contained ina closed triangle then the assertion follows from Corollary 30.15. Otherwise, byLemma 19.11(a) applied to the connected component of Γ containing α and β, wehave the configuration (19.11.1):

α

???? β

ε

ζ

????

δ γ

Hence by Lemma 30.8 for the square (α, β, γ, δ),⟨〈xα, yβ , zβ〉

⟩=⟨〈zβ eγ eδ, eδ, xα yβ , eγ〉

⟩=: A.

Put z′α = zβ eγ eδ and uδ = xα yβ eγ. Then

A =⟨〈z′α, eδ, uδ〉

⟩=⟨〈z′α eδ uδ, eδ, eδ〉

⟩=: B,

by Corollary 30.15 applied to the closed triangle (α, δ, ε). We claim that

z′α eδ uδ = xα yβ zβ.

Indeed, we apply the shift formula (10.8.2) twice, first with respect to eδ and thenwith respect to eγ :

z′α eδ uδ =zβ eγ eδ eδ uδ

=zβ eγ eδ eδuδ

= zβ eγ uδ

=zβ eγ xα yβ eγ

=xα yβ eγ eγ zβ

=xα yβ eγ eγ zβ

= xα yβ zβ.

Hence B =⟨〈xα yβ zβ, eδ, eδ〉

⟩=⟨〈xα yβ zβ, eβ , eβ〉

⟩by (31.2.1).

31.4. Proposition Let α ∈ Γ 0, choose β ∈ Γ 0 satisfying α β and defineϕσα: V σα → E by

ϕσα(xα) :=⟨〈−xα, e−σβ , eσβ〉

⟩. (1)

Then ϕσα is independent of the choice of β and does not change if eβ is replaced byan associated idempotent in Vβ. Moreover, ϕσα: V σα → E is a group homomorphismsatisfying ϕσα(xα) ∈ E(xα).

Proof. By Lemma 31.2 this definition is independent of the choice of β. If eβ ∈Vβ is associated with eβ then xα = xα e−σβ eσβ (since Vα ⊂ V1(eβ)) = xα e−σβ eσβ(by (6.17.1)), so Lemma 31.3 shows⟨

〈−xα, e−σβ , eσβ〉⟩

=⟨〈−xα e−σβ eσβ, e−σβ , eσβ〉

⟩=⟨〈−xα, e−σβ , eσβ〉

⟩.

Since p−1(Uσ) is abelian and therefore the binary symbols are 1, it follows from(30.5.5), applied to the idempotent e = eβ , that ϕσα is a group homomorphism sincethe last factor on the right hand side of (30.5.5) is 1 because of

[[V σ, V σ ]

]= 1.

Moreover,ϕσα(xα) ∈ E(xα e−σβ eσβ) = E(xα) = p−1(xσ(xα)),

by (30.7.1).


31.5. Lemma. Let α β and put h = weβ ,eα = Int(weα) · weβ ∈ G.

(a) Then h ∈ N ∩G0 and

Int(h) · ϕσα(xσα) = ϕσβ(h · xα) = ϕσβ(eσβ e−σα xα

). (1)

If (α, β, α′, β′) is a square then

Int(h) · ϕσα′(zα′) = ϕσβ′(h · zα′) = ϕσβ′(− eσα e−σβ zα′

). (2)

(b) Let

α

???? β

ε

β′

??α′

__????

be a pyramid in Γ . Then

Int(h) · c = c (3)

for all c ∈ E(V σε ).

Proof. (a) Let e = eα and f = eβ . We have h ∈ N ∩G0, and h acts on V viaωf,e, by Lemma 13.10. By (31.4.1) and (29.3.9),

Int(h) · ϕσα(xα) = Int(h) ·⟨〈−xα, e−σβ , eσβ〉

⟩=⟨〈−h · xα, h · e−σβ , h · eσβ〉

⟩.

Recall the notation V(ij) = Vi(f) ∩ Vj(e). Then xα ∈ V(12) and eβ ∈ V(21), soby Proposition 10.12, h · xα = fσ, e−σ, xα = eσβ e−σα xα ∈ V σβ and h · eσβ =

−eσα e−σβ eσβ = −eσα. Hence Int(h) ·ϕσα(xα) =⟨〈h · xα,−e−σα ,−eσα〉

⟩= ϕσβ(h · xα) by

Proposition 31.4, since −eα is associated with eα.

By (31.4.1), ϕσα′(zα′) =⟨〈−zα′ , e−σβ , eσβ〉

⟩, and zα′ ∈ V σ(10), hence h · zα′ =

−eσα, e−σβ , zα′ ∈ V σα−β+α′ = V σβ′ , by Proposition 10.12. It follows that

Int(h) · ϕσα′(zα′) =⟨〈−h · zα′ ,−e−σα ,−eσα〉

⟩= ϕσβ′(h · zα′)

again by Proposition 31.4, since β′ α and −eα is associated with eα.

(b) Lemma 13.10 shows h = b1b2b1 where b1 = b(e+α , e−β ) and b2 = b(−e+

β , e−α ).

Let xε ∈ V σε and c ∈ E(xε). Then, for σ = +,(((((((b1, c

)))))))=(((((((

b(e+α , e−β ), c

)))))))=⟨〈e+α , e−β , xε〉

⟩= 1,(((((((

b2, c)))))))

=(((((((

b(−e+β , e−α ), c

)))))))=⟨〈−e+

β , e−α , xε〉

⟩= 1,

by (30.12.2). A similar argument, using b1 = b−(−e−β , e+α ) and b2 = b−(e−α , e

+β ),

shows(((((((b1, c

)))))))= 1 =

(((((((b2, c

)))))))in case σ = −. This proves (3).


31.6. Proposition. Let α → ε be an arrow of orthogonal type. Defineϕσε,α: V σε → E by

ϕσε,α(xε) =⟨〈−xε, e−σα , eσα〉

⟩· ϕσα′

(−Qxεe−σα

)(1)

where α′ = 2ε − α. If β → ε is a second arrow of orthogonal type with thesame endpoint ε then ϕσε,α = ϕσε,β =: ϕσε . The map ϕσε : V σε → E is a grouphomomorphism satisfying p(ϕσε (x)) = xσ(x), and ϕσε does not change if E is replacedby an associated cog.

Proof. First, an arrow of orthogonal type embeds, by 17.2, in a pyramid. Thisshows that α′ ∈ Γ 0, so the map ϕσα′ occurring in (1) is well defined. Next,c := ϕσε,α(xε) ∈ E(xε). This follows (29.3.6) and Proposition 31.4:

p(c) = xσ(xε e−σα eσα+QxεQe−σα eσα

)· xσ(−Qxεe−σα

)= xσ

(xε +Qxεe

−σα −Qxεe−σα

)= xσ(xε).

Now we show that ϕσε,α is independent of the choice of α. If β → ε and α 6=β ∼ α then α β by Lemma 15.6. Hence α, β, ε generate a pyramid as inLemma 31.5(b). Let h = Int(wα) ·wβ . Then Int(h) ·c = c by (31.5.3). On the otherhand, (29.3.9) and (31.5.2) show

Int(h) · c =⟨〈−h · xε, h · e−σα , h · eσα〉

⟩· ϕβ′(−h ·Qxεe−σα ).

We have xε ∈ V σ1 (eβ) ∩ V σ1 (eα). Hence by Proposition 10.12, h · xε = xε −eσβ e−σα eσα e

−σβ xε = xε, because eσα e−σβ xε ∈ V σα−β+ε = 0 which follows from

α−β+ ε /∈ Γ by Proposition 16.1(b). We also know h · eα = eβ by (10.12.2). Sinceh acts on V as a Jordan pair automorphism,

h ·Qxεe−σα = Qh·xεh · e−σα = Qxεe−σβ ,

so we get

c = Int(h) · c =⟨〈−xε, e−σβ , eσβ〉

⟩· ϕσβ′(−Qxεe−σβ ) = ϕσε,β(xε).

This proves ϕε,α = ϕε,β in case β α. If β ⊥ α then β = 2ε− α = α′. But thenagain α, ε, α′ embed in a pyramid, so there exists γ ∈ Γ with α γ α′ whichreduces us to the previous case.

Now let F be a cog associated with E . To indicate the dependence of ϕσε,α on

E resp. F , we write ϕE ,σε,α resp. ϕF ,σ

ε,α . By (1) and (31.4.1),

ϕE ,σε,α (x) =

⟨〈−xε, e−σα , eσα〉

⟩·⟨〈Qxe−σα , e−σβ , eσβ〉

⟩(2)

depends only the idempotents eα = E (α) and eβ = E (β). By 23.4 the Peirce spacesof E and F coincide, in particular F (δ) ∈ Vδ for all δ ∈ ∆. Moreover, by 23.15,(V,R) is also idempotent with respect to F so that the assumptions in 31.1 alsohold for F replaced by E . Let E ′ be the family defined by

E ′(δ) =

E (δ) if δ ∈ α, βF (δ) otherwise

.


By Corollary 23.16, E ′ is a cog associated with E , and everything we said about Falso holds for E ′. We have ϕE ,σ

ε,α = ϕE ′,σε,α because E (α) = E ′(α) and E (β) = E ′(β).

Next, ϕE ′,σε,α = ϕE ′σ

ε,α′ by what we proved before, but now applied to E ′ instead of E .

Further, ϕE ′,σε,α′ = ϕF ,σ

ε,α′ by (2), since F (α′) = E ′(α′) and F (β′) = E ′(β′). Finally,

ϕF ,σε,α′ = ϕF ,σ

ε,α by the independence of ϕσε,α of the choice of α.

It remains to verify that ϕσε is a homomorphism. Let xε, zε ∈ V σε . Then by(30.5.5) and since xε ∈ V1(eα),

⟨〈−xε − zε, e−σα , eσα〉

⟩=⟨〈−xε, e−σα , zε e−σα eσα〉

⟩·⟨〈−zε, e−σα , eσα〉

⟩·⟨〈−xε, e−σα , eσα〉

⟩=⟨〈−xε, e−σα , zε〉


⟩·⟨〈−xε, e−σα , eσα〉

⟩.

We claim that ⟨〈−xε, e−σα , zε〉

⟩= ϕσα′(xε e−σα zε).

Indeed,⟨〈−xε, e−σα , zε〉

⟩is of type

⟨〈V σ1 , V −σ0 , V σ1 〉

⟩with respect to eα′ . Hence

⟨〈−xε, e−σα , zε〉

⟩=⟨〈eσα′ , e−σα′ ,−xε, e

−σα , zε〉

⟩(by (30.2.2))

=⟨〈eσα′ e−σα′ ,−xε, e

−σα zε

, e−σβ , eσβ〉

⟩(by (30.12.3))

=⟨〈−xε e−σα zε, e−σβ , eσβ〉

⟩(by (10.8.2))

= ϕσα′(xε e−σα zε

)(by (31.4.1)).

Now it follows from (1) and the fact that ϕσα′ is a homomorphism by Proposition 31.4that

ϕσε (xε + zε) =⟨〈−xε − zε, e−σα , eσα〉

⟩· ϕσα′(−Q(xε + zε)e

−σα )

= ϕσα′(zε e−σα xε) ·⟨〈−xε, e−σα , eσα〉


⟩· ϕσα′(−Qxεe−σα ) · ϕσα′(−Qzεe−σα ) · ϕσα′(−xε e−σα zε)

= ϕσε (xε) · ϕσε (zε),

since all terms involved commute.

31.7. Lemma. Let Γ be a Jordan graph and let α, β, γ, δ in Γ satisfy αβ ⊥ δ ⊥ γ α, δ → α and β 6= γ:

δ

α

????

β γ

(1)

Then T = α, β, γ is a closed triangle, and the Jordan subgraph generated by

α, β, γ, δ is isomorphic to T4:


δ

44444444444444

α

ooooooooOOOOOOOO

444444444444444

γ′

44444444444444 β′

η

OO

wwooooooooo

''OOOOOOOOO

β γ

ε //

77ooooooooo

DDα′

OOOOOOOOoooooooo

ζoo

ggOOOOOOOOO

ZZ44444444444444

(2)

Proof. By Lemma 16.4, β ∼ γ, and by Lemma 15.6, β γ, so T is a triangle.In the course of the proof, we will see that it is closed.

After replacing Γ by the connected component containing (1) we may assumeΓ connected. Since Γ contains an arrow, Proposition 17.5 implies that Γ = Γiso

or Γ = Γorth or Γ = Γher. The first case is impossible since |Γiso| 6 3. In thesecond case, δ → α would be an arrow of orthogonal type, whence α ∈ ∂Γ byProposition 17.7. But α β shows α ∈ Γ 0, contradiction. Hence Γ = Γher

∼= TI ,|I|> 3, by Proposition 17.10.

We identify (1) with a subgraph of TI , defined as in 14.18. From the relations(1) we get δ = i, α = i, l, β = j, l for distinct i, j, l ∈ I, and similarlyγ = k, l for distinct i, k, l ∈ I. Since β 6= γ, we must have j 6= k. Thus the i, j, k, lare distinct. Without loss of generality, we may assume i, j, k, l = 1, 2, 3, 4. Thenthe Jordan subgraph of Γ generated by α, β, γ, δ is T4, pictured in (14.18.4). Inparticular α, β, γ = 1, 4, 2, 4, 3, 4 is a closed triangle by 19.10.

31.8. Lemma. With the notation of Lemma 31.7, let xδ ∈ V σδ . Then

eσα, e−σβ , QeσβQe−σα xδ = eσβ , e−σα , xδ ∈ V σγ′ , (1)

QeσαQe−σβQeσ

βQe−σα xδ = xδ. (2)

Proof. By (10.8.1),

eσα, e−σβ , QeσβQe−σα xδ = eσα, Qe−σα xδ, e

σβ = xδ e−σα eσβ ∈ V σδ−α+β .

By (31.7.2), (δ, α, β, γ′) is a kite. Hence δ − α + β = γ′ by (15.3.2). For (2),observe that xδ ∈ V σ2 (eα) and Qe−σα xδ ∈ V −σ2α−δ = V −ση ⊂ V −σ2 (eβ). By (6.14.7),

Qe−σβQeσ

βacts like the identity on V −σ2 (eβ), and so does QeσαQe−σα on V σ2 (eα). Hence

Qeσα

((Qe−σ

βQeσ

β

)Qe−σα xδ

)= Qeσα

(Qe−σα xδ

)= xδ.

31.9. Proposition. Let δ be the initial point of a hermitian arrow. Choose anembedding of δ in a kite K = (δ, α, β, γ′ := δ − α+ β) and define, for xδ ∈ V σδ ,

Φσ1 (xδ; eα, eβ) =⟨〈eσα, e−σβ , Qeσ

βQe−σα xδ〉

⟩, (1)

Φσ2 (xδ; eα, eβ) = ϕσγ′(eσβ e−σα xδ) =⟨〈−eσβ , e−σα , xδ, e−σα , eσα〉

⟩, (2)

ϕσδ (xδ) = Φσ1 (xδ; eα, eβ) · Φσ2 (xδ; eα, eβ). (3)


Then ϕσδ : V σδ → E is a homomorphism satisfying ϕσδ (xδ) ∈ E(xδ). It is independentof the embedding of δ in K and does not change when E is replaced by an associatedcog.

Proof. To improve readability, we will suppress superscripts ±σ in the proof. By17.2, δ does embed in a kite so the definition makes sense. Also, since eβ eα xδ ∈Vβ−α+δ = Vγ′ , and γ′ α by (31.7.2), (2) makes sense.

We show first that ϕδ(xδ) ∈ E(xδ). By (29.3.5), Lemma 31.8 and Proposi-tion 31.4,

Φ1(xδ; eα, eβ) ∈ E(− eα, eβ , QeβQeαxδ+QeαQeβQeβQeαxδ

)= E

(− eβ eα xδ+ xδ

),

Φ2(xδ; eα, eβ) = ϕγ′(eβ eα xδ

)∈ E(eβ eα xδ).

Hence ϕδ(xδ) ∈ E(xδ) by (29.1.3).From (30.5.1) it follows that Φ1 is multiplicative in xδ, and Φ2 is so by Propo-

sition 31.4. Since both factors belong to the abelian group p−1(Uσ), ϕδ is a homo-morphism. The independence of ϕδ of the various choices will be done in severalsteps. We write

Φ = Φ1 · Φ2

for the right hand side of (3).

(a) Let K = K1, and let K2 be a kite of the form K2 = (δ, α, γ, β′ := δ−α+γ)with β 6= γ, thus having the same first two entries as K1. Then the vertices α, β, γ, δsatisfy the assumptions of Lemma 31.7. We claim

Φ(xδ; eα, eβ) = Φ(xδ; eα, eγ). (4)

This will be done by conjugating Φ(xδ; eα, eβ) with

h = Int(weβ ) · weγ ∈ G

and computing the result in two different ways. First, by Lemma 13.10, h = b1b2b1where b1 = b(eβ , eγ) and b2 = b(−eγ , eβ). Let d ∈ E(xδ). Then(((((((

b1, d)))))))

=⟨〈eβ , eγ , xδ〉

⟩= 1

by (30.14.2) and β−γ+ δ /∈ Γ since, by the relations between the vertices picturedin (31.7.2), 〈β − γ + δ, γ∨〉 = 1 − 2 + 0 = −1. In the same way, 〈γ − β + δ, β∨〉 =1− 2 + 0 = −1 shows γ − β + δ /∈ Γ and therefore

(((((((b2, d

)))))))=⟨〈−eγ , eβ , xδ〉

⟩= 1, so

we have Int(h) · d = d, and for d = Φ(xδ; eα, eβ) we obtain

Int(h) · Φ(xδ; eα, eβ) = Φ(xδ; eα, eβ). (5)

On the other hand, h ∈ N ∩ G0 by Lemma 13.10. Let e = eβ and f = eγ . ByLemma 13.10, h acts on V by the automorphism ωf,e described in Proposition 10.12.Let V(ij) = Vi(f) ∩ Vj(e). We claim that

h · eα = eα, h · eβ = eγ , h · xδ = xδ. (6)


Indeed, eα ∈ V(11), eβ ∈ V(12), and xδ ∈ V(00). Hence 10.12 shows

h · eα = eα − eγ eβ , eβ , eγ , eα.

Since α, β, γ is a closed triangle by Lemma 31.7, eβ , eγ , eα ∈ Vβ−γ+α = 0 andtherefore h · eα = eα. Furthermore, h · eβ = f, e, eβ = eγ , eβ , eβ = eγ becauseeβ is collinear to eγ , and finally h · xδ = xδ follows from xδ ∈ V(00). Since h actsas an automorphism of V , (6) implies h · QeβQeαxδ = QeγQeαxδ. Now we obtainfrom (29.3.9)

Int(h) · Φ1(xδ; eα, eβ) =⟨〈h · eα, h · eβ , h ·QeαQeβxδ〉

⟩=⟨〈eα, eγ , QeγQeαxδ〉

⟩= Φ1(xδ; eα, eγ),

Int(h) · Φ2(xδ; eα, eβ) =⟨〈−h · eβ , h · eα, h · xδ, h · eα, h · eα〉

⟩=⟨〈−eγ , eα, xδ, eα, eα〉

⟩= Φ2(xδ; eα, eγ),

whenceInt(h) · Φ(xδ; eα, eβ) = Φ(xδ; eα, eγ). (7)

From (5) and (7) we have (4). This not only shows that Φ(xδ; eα, eβ) is independentof the embedding of δ → α into a kite (δ, α, β, γ′) but also independent of theidempotent eβ in the sense that

Φ(xδ; eα, eβ) = Φ(xδ; eα, e′β) (8)

where e′β ∈ Vβ is an idempotent associated with eβ . Indeed, by Corollary 23.16,replacing eβ by e′β yields a cog E ′ associated with E and G ∈ st(V,R,E ′). We arethus entitled to write

Φ(xδ; eα) := Φ(xδ; eα, eβ). (9)

(b) It remains to show the independence of α and E . Let δ → ω be anotherhermitian arrow starting from δ. Then it follows from Lemma 15.6 that α ω, soby (C4) of 15.3, there exists γ ∈ Γ such that (δ, α, γ, ω) is a kite. By our standingassumptions in 31.1, the connected component of Γ containing δ is isomorphic toTI with |I|>4. Treating the isomorphism as an identification, it is no restriction ofgenerality to assume δ = 1, α = 1, 4, γ = 3, 4 and ω = 1, 3. Then we havethe configuration (14.18.4). It will be convenient to use the notation of (31.7.2) andwrite ω = β′. From (31.7.2) it is evident that the arrow δ → β′ is contained in thekites K3 = (δ, β′, α′, γ′) and K4 = (δ, β′, γ, α). From (4) applied to these kites itfollows

Φ(xδ; eβ′ , eα′) = Φ(xδ; eβ′ , eγ). (10)

Arguing as for (8) and (9), we have, for an idempotent e′α′ ∈ Vα′ associated witheα′ , that

Φ(xδ; eβ′) := Φ(xδ; eβ′ , eα′) = Φ(xδ; eβ′ , e′α′). (11)

(c) Consider the square (β′, α, β, α′) in (31.7.2). By Lemma 25.6,

e′α′ := eβ′ , eα, eβ ∈ Vβ′−α+β = Vα′


is an idempotent associated with eα′ . We claim that

Φ(xδ; eα, eβ) = Φ(xδ; eβ′ , e′α′). (12)

The proof is similar to that of (4), using conjugation by the element

h′ = Int(weα) · weβ′ ∈ G.

As before, Lemma 13.10 shows h′ = b′1b′2b′1 with b′1 = b(eα, eβ′) and b′2 =

b(−eβ′ , eα). From (31.7.2) we see 〈α−β′+δ, α∨〉 = 2−1+2 = 3, so α−β′+δ /∈ Γ .Hence, again for d ∈ E(xδ),

(((((((b′1, d

)))))))=⟨〈eα, eβ′ , xδ〉

⟩= 1 by (30.14.2). Similarly,

β′ − α+ δ /∈ Γ which implies(((((((b′2, d

)))))))=⟨〈−eβ′ , eα, xδ〉

⟩= 1, so for d = Φ(xδ; eα, eβ)

we haveInt(h′) · Φ(xδ; eα, eβ) = Φ(xδ; eα, eβ). (13)

On the other hand, let e = eα and f = eβ′ . By Lemma 13.10, h′ ∈ N ∩G0 and hacts on V by ωf,e. We claim that

h′ · eα = eβ′ , h′ · eβ = e′α′ , h′ · xδ = xδ. (14)

Indeed, let V(ij) = Vi(f)∩Vj(e). Then eα ∈ V(12), eβ ∈ V(01) and xδ ∈ V(22). HenceProposition 10.12 shows

h′ · eα = f, e, eα = eβ′ , eα, eα = eβ′

since eα is collinear to eβ′ . Moreover,

h′ · eβ = f, e, eβ = eβ′ , eα, eβ = e′α′ ,

and h′ · xδ = xδ. This implies, again by (29.3.9),

Int(h′) · Φ(xδ; eα, eβ) = Φ(h′ · xδ; h′ · eα, h′ · eβ) = Φ(xδ; eβ′ , e′α′). (15)

Together with (13), this proves (12). Now

ϕδ(xδ) = Φ(xδ; eα) = Φ(xδ; eβ′) (16)

follows from

Φ(xδ; eα) = Φ(xδ; eα, eβ) (by (9)) = Φ(xδ; eβ′ , e′α′) (by (12))

= Φ(xδ; eβ′ , eα′) = Φ(xδ; eβ′) (by (11)).

This proves that ϕδ is indeed independent of the choice of α and the idempotenteα.

(d) Finally, let E be a cog associated with E . To see that ϕδ does not changeupon replacing E by E , define a cog F by

F (δ) =

eδ if δ = αeδ otherwise

.

By 23.4, eδ ∈ Vδ for all δ ∈ ∆, so that F is indeed a cog by Corollary 23.16.We indicate the dependence on the cog used by a superscript E , E or F . ThenϕEδ (xδ) = Φ(xδ; eα) = ϕF

δ (xδ) by (16) because E and F agree at α. Further,

ϕFδ (xδ) = Φ(xδ; eβ′) = ϕE

δ (xδ) by (16) and since F and E agree at β′.


31.10. Definition. Let V, Γ,R,E , G,E satisfy our standing assumptions de-scribed in 31.1. We now combine Propositions 31.4, 31.6 and 31.9 and defineϕσ: V σ → E by

ϕσ(∑α∈Γ

xα

)=∏α∈Γ

ϕσα(xα).

Since p−1(Uσ) is abelian and the sum involves only finitely many non-zero terms,this is a well-defined homomorphism ϕσ: V σ → E, it does not change when E isreplaced by an associated cog, and satisfies

p(ϕσ(x)) = xσ(x),

for all x ∈ V σ. In view of the isomorphism xσ: V σ → Uσ, this yields sectionssσ: Uσ → p−1(Uσ) of p: p−1(Uσ) → Uσ by sσ(xσ(x)) = ϕσ(x). In particular, ifG = St(V,R) is the Steinberg group of (V,R) this completes the proof of Step 2 of27.10.

31.11. Proposition. Recall the group M0(G) ⊂ G defined in 25.11. With thenotation introduced in 31.10, the homomorphisms ϕσ: V σ → E are equivariant withrespect to the action of M0(G) on V σ given by the canonical projection π: M0(G)→PE0(V ) ⊂ Aut(V ) and the action on E defined in (27.9.3):

Int(g) · ϕσ(x) = ϕσ(ψ(g) · x), (1)

for all g ∈ M0(G,R) and x ∈ V σ.

Proof. It suffices to prove this for x = xγ ∈ V σγ and all γ ∈ Γ . We distinguishthe following cases. To simplify notation, we drop the superscripts σ at ϕ andthe idempotents in E , and write h = ψ(g). Let θ(g) = t ∈ W (S0) ⊂ Aut(Γ ) asin Proposition 25.12. Then (t, h) ∈ Aut(Γ, (V,R)) by Corollary 25.13, so we haveh · Vγ = Vt(γ) for all γ ∈ Γ .

(a) γ ∈ Γ 0. Then by Proposition 31.4, ϕ(xγ) =⟨〈−xγ , eδ, eδ〉

⟩where δ γ,

and therefore, by (29.3.9),

Int(g) · ϕ(xγ) =⟨〈−h · xγ , h · eδ, h · eδ〉

⟩.

From h · xγ ∈ Vt(γ) it follows that ϕ(h · xγ) =⟨〈−h · xγ , eζ , eζ〉

⟩for any ζ with

ζ t(γ). Since t is an automorphism of Γ , γ δ implies t(γ) t(δ).Thus we may choose ζ = t(δ) and the idempotent h(eδ) ≈ eζ . Then ϕ(h · xγ) =⟨〈−h · xγ , h · eδ, h · eδ〉

⟩(by (31.4.1)) = Int(g) · ϕ(xγ), as desired.

(b) γ = ε is the end point of an arrow δ → ε of orthogonal type. By (31.6.1),

ϕ(xε) =⟨〈−xε, eδ, eδ〉

⟩· ϕ2ε−δ(−Qxεeδ),

and therefore, by (29.3.9), by what we proved in (a), and since h acts as anautomorphism of V ,

Int(g) · ϕ(xε) =⟨〈−h · xε, h · eδ, h · eδ〉

⟩· ϕ(−h ·Qxεeδ)

=⟨〈−h · xε, h · eδ, h · eδ〉

⟩· ϕ(−Qh·xεh · eδ).

§32] Proof of the relations 403

Since t is an automorphism of Γ , t(δ)→ t(ε) is an arrow of orthogonal type. Hencewe may use t(δ) → t(ε) and the idempotent h · eδ ≈ et(δ) to compute ϕ(h · xε), soby (31.6.1),

ϕ(h · xε) =⟨〈−h · xε, h · eδ, h · eδ〉

⟩· ϕ(−Qh·xεh · eδ) = Int(g) · ϕ(xε).

(c) Let γ = δ be the initial point of a hermitian arrow δ → ξ, embedded in akite (δ, ξ, %, ω). Then by Proposition 31.9, ϕ(xδ) = Φ1 · Φ2 where

Φ1 =⟨〈eξ, e%, Qe%Qeξxδ〉

⟩, Φ2 = ϕω(e%, eξ, xδ).

Hence by (29.3.9), (a), and since h acts as a Jordan pair automorphism,

Int(g) · Φ1 =⟨〈h · eξ, h · e%, Qh·e%Qh·eξh · xδ〉

⟩,

Int(g) · Φ2 = ϕt(ω)(h · e%, h · eξ, h · xδ).

Since t is an automorphism of Γ , (t(δ), t(ξ), t(%), t(ω)) is a kite as well. Hence,again by Proposition 31.9, and of course since h · eξ ≈ et(ξ, h · e% ≈ et(%), we haveϕ(h · xδ) = Φ′1 · Φ′2 where

Φ′1 =⟨〈h · eξ, h · e%, Qh·e%Qh·eξh · xδ〉

⟩,

Φ′2 = ϕt(ω)(h · e%, h · eξ, h · xδ).


§32. Proof of the relations

32.1. Generalities. We now come to Step 3 of the program outlined in 27.10and will work in the setup of 31.1, except that the group G, which in §§30 and 31was any group in st(V,R,E ), is now assumed to be the Steinberg group St(V,R).Again, p: E → G is a central extension. As usual, we denote the abelian subgroupsof G by Uσ. By Theorem 29.12, the groups p−1(Uσ) are abelian, so the binarysymbols of (27.10.2) are 1.

By definition in 22.1, G = St(V,R) is presented by generators and relations asfollows. The generators are xσ(u), u ∈ V σ, σ ∈ +,−, and the first group ofrelations is

xσ(u+ x) = xσ(u) · xσ(x),

for all u, x ∈ V σ. For (x, y) ∈ V quasi-invertible, b(x, y) ∈ G was defined in (9.7.1)by

x+(x) · x−(y) = x−(yx) · b(x, y) · x+(xy). (1)

The next group of relations are the relations B(xα, yβ) (see 9.8) for all α ∼ β 6= αin Γ and all xα ∈ V +

α , yβ ∈ V −β . Explicitly,

B(xα, yβ) :

(((((((b(xα, yβ), x+(z)

)))))))= x+

(− xα yβ z+QxαQyβz

)(((((((b(xα, yβ)−1, x−(v)

)))))))= x−

(− yβ xα v+QyβQxαv

) , (2)

for all z ∈ V +, v ∈ V −. Finally, we require


b(V +α , V

−β ) = 1, (3)

for all α ⊥ β in Γ . Of course, (2) corresponds to (StR1) and (3) to (StR2) ofTheorem 21.7. Since x

yβα = xα and yxαβ = yβ for α ⊥ β by (20.7.3), (3) is (by (1))

equivalent to (((((((x+(V +

α ), x−(V −β ))))))))

= 1 for α ⊥ β. (4)

It will be convenient to write (2) in a more compact form. For σ ∈ +,− andquasi-invertible (x, y) ∈ V σ × V −σ let, as in 9.10, bσ(x, y) be defined by

xσ(x) · x−σ(y) = x−σ(yx) · bσ(x, y) · xσ(xy).

Then one checks easily, using (9.10.8), the fact that the V σα are subgroups of theadditive group V σ and that the Bergmann operators satisfy B(−x,−y) = B(x, y),that (2) is equivalent to the relations(((((((

bσ(xα, yβ), xσ(z))))))))

= xσ(− xα yβ z+QxαQyβz

), (5)

for all α ∼ β 6= α in Γ , xα ∈ V σα , yβ ∈ V −σβ , z ∈ V σ and σ ∈ +,−.In 31.10, we constructed homomorphisms ϕσ: V σ → E satisfying p(ϕσ(x)) =

xσ(x), and therefore partial sections sσ: Uσ → E of the projection p: p−1(Uσ) →Uσ given by sσ(xσ(x)) = ϕσ(x). Our task is to show that the sσ induce a sections: G→ E of p: E → G.

For (x, y) ∈ V σ×V −σ quasi-invertible, define bϕσ(x, y) ∈ E in analogy to bσ(x, y)by

ϕσ(x)ϕ−σ(y) = ϕ−σ(yx) · bϕσ(x, y) · ϕσ(xy). (6)

In view of the presentation of G described above, the sσ will induce a section s ifand only if (((((((

ϕ+(V +α ), ϕ−(V −β )

)))))))= 1 (7)

for α ⊥ β, and the relations

B(α, β) :(((((((

bϕσ(xα, yβ), ϕσ(z))))))))

= ϕσ(− xα yβ z+QxαQyβz

)(8)

hold for all α ∼ β 6= α, σ ∈ +,−, z ∈ V σ and (xα, yβ) ∈ V σα × V −σβ .We rewrite these relations as follows. Recall the notation Eσ(z) and Eσ(x, y)

of (29.1.2) and (29.3.2) and the ternary symbols⟨〈x, y, z〉

⟩, introduced in (29.3.5).

From p(ϕσ(u)) = xσ(u) we have ϕσ(x) ∈ Eσ(x), and applying p to (6) shows thatp(bϕσ(x, y)) = bσ(x, y) and therefore bϕσ(x, y) ∈ Eσ(x, y). Hence (7) is equivalent to(((((((

E(V +α ), E(V −β )

)))))))= 1, (9)

which holds by Proposition 30.3, and (5) is equivalent to

B(α, β) :⟨〈xα, yβ , z〉

⟩= ϕσ

(− xα yβ z+QxαQyβz

), (10)

for all α ∼ β 6= α, xα ∈ V σα , yβ ∈ V −σβ , z ∈ V σ, σ ∈ +,−.Let also γ ∈ Γ and define relations B(α, β, γ) by

B(α, β, γ) :⟨〈xα, yβ , zγ〉

⟩= ϕσ

(− xα yβ z+QxαQyβzγ

)(11)


for all xα, yβ , zγ in the appropriate root spaces. By (30.5.1) and the vanishing ofthe binary symbols the left hand side of (10) is multiplicative in z, and so is theright hand side because ϕσ is a homomorphism. Since V σ is the direct sum of theV σγ it is clear that

B(α, β) ⇐⇒ B(α, β, γ) for all γ ∈ Γ . (12)

Let us put α1 = α, α2 = β, and for α3 ∈ Γ define

α4 := α1 − α2 + α3, α5 := 2(α1 − α2) + α3.

By (20.1.1), xα1yα2

zα3 ∈ V σα4

and Q(xα1)Q(yα2

)zα3∈ V σα5

. By Lemma 19.9,α4 /∈ Γ implies α5 /∈ Γ . Therefore, the relations B(α1, α2, α3) split into threecases:⟨〈xα1 , yα2 , zα3〉

⟩= 1 if α4 6∈ Γ , (13)⟨

〈xα1, yα2

, zα3〉⟩

= ϕσ(− xα1

yα2zα3)

if α4 ∈ Γ , α5 6∈ Γ , (14)⟨〈xα1

, yα2, zα3〉⟩

= ϕσ(− xα1

yα2zα3+Qxα1

Qyα2zα3

)if α5 ∈ Γ . (15)

We have already established (13) in Theorem 30.14. The remaining relations willbe proved in 32.5, 32.7, 32.9, 32.10 and 32.11.

32.2. Lemma. Let Γ be a Jordan graph and let α1, α2, α3, α4 ∈ Γ with α1 ∼α2 6= α1 and

∑4i=1(−1)iαi = 0. Put α5 := 2(α1 − α2) + α3 = 2α4 − α3. Then

exactly the cases in the following table are possible. A subscript “orth” resp. “her”in the cases (ii) – (v) indicates that the arrows occurring in the third column of thetable are of orthogonal resp. hermitian type.

Proof. LetΣ = α1, . . . , α4. We distinguish cases according to CardΣ. For thefourth column we use the criterion 16.11: α5 = 2α4−α3 ∈ Γ if and only if α4 ← α3.By Proposition 16.1 and since α1 6= α2, there are the following possibilities:

CardΣ = 2: then α1, α2 = α3, α4. Here α1 = α3 and α2 = α4 is impossiblebecause the alternating sum of the αi is zero. Hence α1 = α4 and α2 = α3, whichyields the cases (i), (ii), (iii).

CardΣ = 3: then either α1 = α3 and α2 → α3 ← α4 which is Case (iv), orα2 = α4 and α1 → α1 ← α3 which is Case (v).

CardΣ = 4: then either (α1, . . . , α4) is a square (Case (vi)) or a cyclic permu-tation is a kite (Cases (vii) – (x)).

32.3. Lemma. (a) Let α→ ε be an arrow of orthogonal type, completed to acollision α→ ε← α′ = 2ε−α. Then the relation B(ε, α) of (32.1.10) holds if andonly if B(α′, ε) holds.

(b) Let δ → α be an arrow of hermitian type, embedded in a kite (δ, α, β, γ),cf. 17.2, and let % ∈ Γ . Then B(γ, β, %) implies B(δ, α, %), and B(β, γ, %) impliesB(α, δ, %).

Proof. For better readability we suppress the superscripts ±σ in this and manyof the following proofs.


Case descriptioninduced subgraph

on α1, α2, α3α5? Proof

(i) α1 = α4, α2 = α3 α1 α2 = α3 α5 /∈ Γ 32.5

(ii)orth α1 = α4, α2 = α3 α1 → α2 = α3 α5 /∈ Γ 32.5

(ii)her α1 = α4, α2 = α3 α1 → α2 = α3 α5 /∈ Γ 32.11

(iii)orth α1 = α4, α2 = α3 α1 ← α2 = α3 α5 ∈ Γ 32.7

(iii)her α1 = α4, α2 = α3 α1 ← α2 = α3 α5 ∈ Γ 32.11

(iv)orth α1 = α3 α2 → α1 = α3 α5 /∈ Γ 32.5

(iv)her α1 = α3 α2 → α1 = α3 α5 /∈ Γ 32.11

(v)orth α2 = α4 α1 → α2 ← α3 α5 = α1 ∈ Γ 32.7

(v)her α2 = α4 α1 → α2 ← α3 α5 = α1 ∈ Γ 32.11

(vi) (α1, . . . , α4) square α1 α2 α3 α5 /∈ Γ 32.5

(vii) (α1, . . . , α4) kite α1 → α2 α3 α5 /∈ Γ 32.5

(viii) (α2, α3, α4, α1) kite

α2

????

α3 α1

α5 /∈ Γ 32.5

(ix) (α3, α4, α1, α2) kite α1 α2 ← α3 α5 ∈ ∂Γ 32.9

(x) (α4, α1, α2, α3) kite

α3

????

α1 α2

α5 /∈ Γ 32.10

(a) Suppose B(ε, α) holds, so we have⟨〈xε, yα, z〉

⟩= ϕ

(− xε yα z+QxεQyαz

)(1)

for all xε, yα in the respective root spaces, and all z. We show B(α′, ε), that is,

a :=⟨〈uα′ , vε, z〉

⟩= ϕ

(− uα′ vε z+Quα′Qvεz

). (2)

Since α → ε is of orthogonal type there exists β such that (17.2.4) holds; inparticular, α is in Γ 0. Hence we have an idempotent e = eα, and with respectto eα,

⟨〈uα′ , vε, z〉

⟩is of type

⟨〈V0, V1, V 〉

⟩. By (30.2.3) for j = 0, it follows that

a =⟨〈uα′ vε eα, eα, z〉

⟩.

Now uα′ vε eα ∈ Vα′−ε+α = Vε, so from B(ε, α) we obtain

a = ϕ(− uα′ vε eα eα z+Q(uα′ vε eα)Q(eα)z

). (3)


By (10.8.2) and (10.8.3) we have

uα′ vε z = uα′ vε eα eα z and Q(uα′)Q(vε) = Q(uα′ vε eα)Q(eα).

Substituting this in (3) shows (2).

Now assume that B(α′, ε) holds. Then⟨〈xε, yα, z〉

⟩is of type

⟨〈V1, V0, V 〉

⟩with

respect to eα′ . Hence by (30.2.2) for j = 0,

a :=⟨〈xε, yα, z〉

⟩=⟨〈eα′ , eα′ xε yα, z〉

⟩,

and tε := eα′ xε yα ∈ Vα−ε+α′ = Vε. Hence by B(α′, ε),

a = ϕ(− eα′ tε z+Qeα′Qtεz

).

Again by (10.8.2) and (10.8.3),

eα′ tε z = xε yα z and Q(eα′)Q(tε) = QxεQyα ,

so the assertion follows.

(b) The proof is similar to that of (a). We do the first case, and leave thesecond to the reader. If B(γ, β, %) holds then⟨

〈xγ , yβ , z〉⟩

= ϕ(− xγ yβ z+QxγQyβz

)for all xβ , yγ , z = z% in the respective root spaces. To prove B(δ, α, %) we mustshow

a :=⟨〈uδ, vα, z〉

⟩= ϕ

(− uδ vα z+QuδQvαz

). (4)

Since β ∈ Γ 0, we have the idempotent e = eβ , and the left hand side of (4) is oftype

⟨〈V0, V1, V 〉

⟩with respect to e. Hence by (30.2.3),

a =⟨〈uδ vα eβ, eβ , z〉

⟩and tγ := uδ vα eβ ∈ Vδ−α+β = Vγ . Hence by B(γ, β, %),

a = ϕ(− tγ eβ z+QtγQeβz

).

Again by (10.8.2) and (10.8.3), it follows that

a = ϕ(− uδ vα z+QuδQvαz

),

as desired.

32.4. Lemma. Let (α, β, γ, δ) be a square in Γ . Then for all elements in theappropriate root spaces, ⟨

〈xα, yβ , zγ〉⟩

= ϕσδ(− xα yβ zγ

).

Proof. The left hand side is of type⟨〈V σ0 , V −σ1 , V σ2 〉

⟩with respect to eγ . Since

xα yβ eσγ ∈ V σα−β+γ = V σδ , we have⟨〈xα, yβ , zγ〉

⟩=⟨〈xα yβ eσγ, e−σγ , zγ〉

⟩(by (30.2.3))

=⟨〈xα yβ eσγ e−σγ zγ, e−σγ , eσγ 〉

⟩(by 31.3)

=⟨〈xα yβ zγ, e−σγ , eσγ 〉

⟩(by (10.8.2))

= ϕσδ(− xα yβ zγ

)(by (31.4.1)),

as desired.


32.5. Proposition. The relations (32.1.14) hold in the cases (i), (ii)orth,(iv)orth, (vi), (vii) and (viii) of Lemma 32.2.

Proof. To simplify notation, write α1 = α, α2 = β, α3 = γ and α4 = δ. Letxα ∈ V σα , yβ ∈ V −σβ , zγ ∈ V σγ and σ = ±. We suppress again the superscripts ±σ.By (32.1.14) we have to show⟨

〈xα, yβ , zγ〉⟩

= ϕ(− xα yβ zγ

). (1)

Case (i), δ = α β = γ: here (1) follows from Lemma 31.3:⟨〈xα, yβ , zβ〉

⟩=⟨〈xα yβ zβ, eβ , eβ〉

⟩= ϕ(−xα yβ zβ),

by (31.4.1).

Case (ii)orth, δ = α → β = γ: since α → β is an arrow of orthogonal type,

the connected component of Γ containing α, β is isomorphic to BqfI with |I| > 4.

Changing notation so as to conform with that of Lemma 30.12, we have to show⟨〈xα, yε, zε〉

⟩= ϕ

(− xα yε zε

). By (31.4.1),

ϕ(− xα yε zε

)=⟨〈xα yε zε, eβ , eβ〉

⟩and this equals

⟨〈xα, yε, zε〉

⟩by (30.12.3).

Case (iv)orth, β → α ← δ is a collision, γ = α: by Lemma 32.3(a) this case isequivalent to Case (ii)orth.

Case (vi), (α, β, γ, δ) a square: this follows from Lemma 32.4.

Case (vii), (α, β, γ, δ) a kite, and (viii) (β, γ, δ, α) a kite: this case follows fromCase (i), by Lemma 32.3(b).

32.6. Lemma. Letα

???? β

ε

β′

??α′

__????(1)

be a pyramid in Γ . For xε ∈ V σε , zα ∈ V σα , yα ∈ V −σα , put uε = xε yα zα ∈ V σεand tα′ = QxεQyαzα ∈ V σ2ε−α = V σα′ . Then⟨

〈xε, yα, zα〉⟩

= ϕσβ′(tα′ e−σβ zα

)·⟨〈uε, e−σβ , eσβ〉

⟩· ϕσα′

(tα′). (2)

Proof. We prove this for σ = +, the case σ = − follows by passing to V op. Theproof of (2) is another instance of the formula (3.6.9),(((((((

a,(((((((b, c))))))))))))))

=((((((( (((((((

a, b))))))),(((((((b, c))))))) )))))))·((((((( (((((((

b, c))))))),((((((((((((((a, b))))))), c))))))) )))))))·((((((((((((((a, b))))))), c))))))), (3)

this time for a ∈ E(xε, yα), b ∈ E(−zα, e−β ), and c ∈ E(e+β ). We have

(((((((a, c)))))))

=⟨〈xε, yα, e+

β 〉⟩

= 1 by (30.12.2), so (3.6.9) is applicable. Now compute as follows:

§32] Proof of the relations 409(((((((b, c)))))))

=⟨〈−zα, e−β , e

+β 〉⟩∈ E(zα),

(((((((a,(((((((b, c))))))))))))))

=⟨〈xε, yα, zα〉

⟩.

We compute p((((((((a, b)))))))

) ∈ G, using (9.9.5):

p((((((((a, b)))))))

) =(((((((

b(xε, yα), b(−zα, e−β ))))))))

= b(B(xε, yα) · (−zα), B(yα, xε)

−1 · e−β)· b(−zα, e−β )−1.

Here B(xε, yα)·(−zα) = −zα+uε−tα′ by definition of uε and tα′ . Also, ξ := α−ε+β /∈ Γ by Proposition 16.1(b), and therefore 2α−2ε+β /∈ Γ by Lemma 19.9(a). Thisimplies B(yα, xε)

−1 · e−β = e−β . With respect to eβ we have Vα⊕ Vε⊕ Vα′ ⊂ V1(eβ).Hence (11.5.1) yields

p((((((((a, b)))))))

) = b(−zα + uε − tα′ , e−β ) · b(−zα, e−β )−1 = b(uε − tα′ , e−β ).

By (29.3.5), this together with the previous computations implies((((((((((((((a, b))))))),(((((((b, c))))))))))))))

=⟨〈uε − tα′ , e−β , zα〉

⟩,

((((((((((((((a, b))))))), c)))))))

=⟨〈uε − tα′ , e−β , e

+β 〉⟩.

Since p−1(U+) is abelian, the middle term on the right of (3) is 1, so we obtain⟨〈xε, yα, zα〉

⟩=⟨〈uε − tα′ , e−β , zα〉

⟩·⟨〈uε − tα′ , e−β , e

+β 〉⟩. (4)

The first factor on the right is of type⟨〈V +

1 , V −2 , V +1 〉⟩

with respect to eβ . Hence by(30.5.4) and since the binary symbols are 1, it is multiplicative in the first variable.

The connected component of Γ containing the pyramid (1) is of type OI where|I|> 3. Hence Lemma 30.12 is applicable, so (30.12.2) yields

⟨〈uε, e−β , zα〉

⟩= 1.

The second factor is of type⟨〈V +

1 , V −2 , V +2 〉⟩

with respect to eβ , so by (30.5.5)and since Vα′ ⊂ V1(eβ),⟨

〈uε − tα′ , e−β , e+β 〉⟩

=⟨〈uε, e−β , tα′〉

⟩·⟨〈uε, e−β , e

+β 〉⟩·⟨〈−tα′ , e−β , e

+β 〉⟩.

Here the first factor is in⟨〈V +ε , V

−β , V

+α′ 〉⟩

and therefore equals 1 by Lemma 30.12.Now (4) becomes⟨

〈xε, yα, zα〉⟩

=⟨〈−tα′ , e−β , zα〉

⟩·⟨〈uε, e−β , e

+β 〉⟩·⟨〈−tα′ , e−β , e

+β 〉⟩

= ϕ+β′

(tα′ e−β zα

)·⟨〈uε, e−β , e

+β 〉⟩· ϕ+

α′(tα′),

where we used Lemma 32.4 for the square (α, β, α′, β′) on the first factor, and(31.4.1) on the third factor. This proves (2).

32.7. Proposition. The relations (32.1.15) hold in the cases (iii)orth and(v)orth of Lemma 32.2.

Proof. In Case (iii)orth we have α3 = α2 → α1 = α4 ← α5, since α5 = 2α4−α3 =2α1 − α2. Putting here ε = α1 and α = α2, we must show B(ε, α). In Case (v)orth

we have α3 → α2 = α4 ← α1 = α5. Putting α = α3, ε = α2 and α′ = α1, wehave the collision α→ ε← α′, and have to prove the relations B(α′, ε). Since thearrows are of orthogonal type, Lemma 32.3(a) shows B(ε, α) ⇐⇒ B(α′, ε), so itsuffices to deal with Case (iii)orth.


With the notation just introduced, (32.1.15) reads

A :=⟨〈xε, yα, zα〉

⟩= ϕ

(− xε yα zα+QxεQyαzα

)=: B. (1)

Here xε yα zα ∈ Vε and tα′ := QxεQyαzα ∈ V2ε−α = Vα′ . Hence the right handside of (1) becomes

B = ϕε(− xε yα zα

)· ϕα′(tα′).

To compute ϕε, choose a vertex β α (which is possible by (17.2.3) since α→ εis of orthogonal type). We use the definition of ϕε in (31.6.1), but replace the vertexα by β which is allowed by Proposition 31.6. Then

B =⟨〈xε yα zα, eβ , eβ〉

⟩· ϕβ′

(−Q(xε yα zα)eβ

)· ϕα′(tα′).

On the other hand, by (32.6.2),

A =⟨〈xε yα zα, eβ , eβ〉

⟩· ϕβ′

(tα′ , eβ , zα

)· ϕα′(tα′).

Hence it remains to show that

−Q(xε yα zα)eβ = QxεQyαzα, eβ , zα.

This follows from the Jordan identity (JP21):

Q(QxεQyαzα, zα) +Q(xε yα zα) = QxεQyαQzα +QzαQyαQxε

+D(xε, yα)QzαD(yα, xε),

applied to eβ , since Qzαeβ ∈ V2α−β = 0, QyαQxεeβ ∈ V2α−2ε+β = 0 andD(yα, xε)eβ = 0.

32.8. Lemma. Consider a Jordan subgraph of Γ isomorphic to T4 as in(31.7.2):

δ

44444444444444

α

ooooooooOOOOOOOO

444444444444444

γ′

44444444444444 β′

η

OO

wwooooooooo

''OOOOOOOOO

β γ

ε //

77ooooooooo

DDα′

OOOOOOOOoooooooo

ζoo

ggOOOOOOOOO

ZZ44444444444444

(1)

Let yγ ∈ V −σγ and zζ ∈ V σζ and define

uε := QeσβQyγzζ ∈ V σε .


Then⟨〈eσα, e−σβ , uε〉

⟩· ϕσγ′

(QeσαQyγzζ , e

−σα , eσβ

)=⟨〈eσα, yγ , zζ〉

⟩· ϕσβ′

(eσα yγ zζ

). (2)

Proof. As usual, we may assume σ = +. From (1) it is evident that (ζ, γ, β, α′)and (ε, α′, γ, β) are kites. Hence 2β−2γ+ζ = (ζ−γ+β)+(β−γ) = α′+(ε−α′) = ε(by (15.3.2)) which shows that uε lies indeed in V +

ε . One proves in the same waythat Qe+αQyγzζ , e

−α , e

+β ∈ V

+γ′ and e+

α yγ zζ ∈ V +β′ . Define

uα′ := e+β yγ zζ ∈ V

+α′ , u := B(−e+

β , yγ)zζ − zζ = uα′ + uε.

We first prove the formula⟨〈e+α , e−β , uε〉

⟩· ϕσγ′

(e+α yγ uα′

)=⟨〈e+α , yγ , zζ〉

⟩· ϕσβ′

(e+α e−β uα′

). (3)

The proof uses again (3.6.9),(((((((a,(((((((b, c))))))))))))))

=((((((( (((((((

a, b))))))),(((((((b, c))))))) )))))))·((((((( (((((((

b, c))))))),((((((((((((((a, b))))))), c))))))) )))))))·((((((((((((((a, b))))))), c))))))), (4)

this time fora ∈ E(e+

α , e−β ), b ∈ E(−e+

β , yγ), c ∈ E(zζ).

We must check(((((((a, c)))))))∈ Z (E) which follows from

(((((((a, c)))))))

=⟨〈e+α , e−β , zζ〉

⟩= 1 by

(30.2.8), since α β ⊥ ζ and the binary symbols are 1. Now we compute theterms occurring in (4), similarly to the calculation in the proof of 32.6 and usingthat α, β, γ is a closed triangle:(((((((

b, c)))))))

=⟨〈−e+

β , yγ , zζ〉⟩∈ E(e+

β yγ zζ+Qe+βQyγzζ) = E(u),(((((((

a,(((((((b, c))))))))))))))

=⟨〈e+α , e−β , u〉

⟩,

p((((((((a, b))))))))

=(((((((

b(e+α , e−β ),b(−e+

β , yγ))))))))

= b(−B(e+

α , e−β )e+

β , B(e−β , e+α )−1yγ

)· b(−e+

β , yγ)−1

= b(e+α − e+

β , yγ) · b(−e+β , yγ)−1 = b(e+

α , yγ),((((((((((((((a, b))))))),(((((((b, c))))))))))))))

=⟨〈e+α , yγ , u〉

⟩,((((((((((((((

a, b))))))), c)))))))

=⟨〈e+α , yγ , zζ〉

⟩∈ E(V +).

Since E(V +) = p−1(U+) is abelian, the middle term on the right of (4) is 1 and weobtain ⟨

〈e+α , e−β , u〉

⟩=⟨〈e+α , yγ , u〉

⟩·⟨〈e+α , yγ , zζ〉

⟩. (5)

Evaluate the left hand side of (5) using (30.5.1):⟨〈e+α , e−β , u〉

⟩=⟨〈e+α , e−β , uα′〉

⟩·⟨〈e+α , e−β , uε〉

⟩= ϕ+

β′

(− e+

α e−β uα′

)·⟨〈e+α , e−β , uε〉

⟩,

where in the last equality we used Lemma 32.4 and the fact that (α, β, α′, β′) is asquare. We proceed in the same way with the first factor on the right hand side of(5):

412 CENTRAL CLOSEDNESS [Ch. VI⟨〈e+α , yγ , u〉

⟩=⟨〈e+α , yγ , uα′〉

⟩·⟨〈e+α , yγ , uε〉

⟩.

Since (α, γ, α′, γ′) is a square, the first factor equals ϕ+γ′(−e+

α , yγ , uα′), again byLemma 32.4. The second factor on the right hand side of (5) is 1 by (30.14.2)because α− γ + ε /∈ Γ which follows from γ ⊥ ε and Proposition 16.1(b). Puttingeverything together we have now shown

ϕ+β′

(− e+

α e−β uα′

)·⟨〈e+α , e−β , uε〉

⟩= ϕγ′

(− e+

α yγ uα′)·⟨〈e+α , yγ , zζ〉

⟩.

Since all terms belong to the abelian group p−1(U+) and ϕ+β′ and ϕ+

γ′ are grouphomomorphisms, this is equivalent with (3).

A comparison of (3) with (2) shows that it remains to prove:

e+α , e−β , e

+β , yγ , zζ = e+

α , yγ , zζ, (6)

Qe+αQyγzζ , e−α , e

+β = e+

α , yγ , e+β , yγ , zζ. (7)

For the first formula we use (JP15):

e+α , e−β , e

+β , yγ , zζ = e+

α , e−β , e

+β , yγ , zζ

− e+β , e

−β e

+α yγ, zζ+ e+

β , yγ , e+α , e−β , zζ. (8)

Since eα and eβ are collinear we have e+α , e−β , e

+β = e+

α . By Lemma 31.7, α, β, γis a closed triangle. Thus β − α+ γ /∈ Γ so the second term in (8) is zero, and thethird term vanishes because β ⊥ ζ.

By (1), we have 2γ − ζ = η. Put vη = Qyγzζ and compute the right hand sideof (7) as follows:

e+α , yγ , e−β , yγ , zζ = D(e+

α , yγ)D(e+β , yγ)zζ

= e+α , Qyγzζ , e

+β + e+

α , Qyγe−β , zζ (by (JP9))

= e+α , vη, e

+β ,

since Qyγe+β ∈ V

−2γ−β = 0. On the left of (7) we use (JP12) and obtain

Qe+α vη, e−α , e

+β = D(e+

β , e−α )Qe+α vη

= e+α , vη, e

+β −Qe+α e

−α , e

+β , vη.

By Proposition 16.1(b), α − β + η /∈ Γ , so e−α , e+β , vη ∈ V −α−β+η = 0. This

completes the proof of (7).

32.9. Lemma. The relation (32.1.15) holds in case (ix) of Lemma 32.2.

Proof. Here K := (α3, α4, α1, α2) is a kite. Since the connected component of

Γ containing K is isomorphic to TI with |I| > 4, we can embed K in a subgraph

isomorphic to T4 as in (32.8.1) by putting

α1 = α, α2 = γ, α3 = ζ, α4 = β′,


Let xα ∈ Vα, yγ ∈ Vγ and zζ ∈ Vζ . Then xα yγ zζ ∈ Vβ′ since (ζ, γ, α, β′)is a kite and thus α − γ + ζ = β′. Also, because (δ, α, γ, β′) is a kite we get2(α − γ) + ζ = (α − γ) + (α − γ + ζ) = α − γ + β′ = δ, whence QxαQyβzζ ∈ Vδ.Therefore the relation (32.1.15) becomes⟨

〈xα, yγ , zζ〉⟩

= ϕβ′(− xα yγ zζ

)· ϕδ

(QxαQyγzζ

). (1)

We first prove (1) in the special case xα = eα and put uδ = QeαQyγzζ . Afterbringing the first factor of the right hand side of (1) to the left, we have to show⟨

〈eα, yγ , zζ〉⟩· ϕβ′

(eα yγ zζ

)= ϕδ(uδ). (2)

Comparing (32.8.2) with (2) we see that it remains to prove

ϕδ(uδ) =⟨〈eα, eβ , QeβQyγzζ〉

⟩· ϕγ′

(uδ, eα, eβ

).

By Proposition 31.9,

ϕδ(uδ) =⟨〈eα, eβ , QeβQeαuδ〉

⟩· ϕγ′(QeβQeαuδ, eβ , eα),

so we are reduced to showing

QeβQyγzζ = QeβQeαuδ and uδ, eα, eβ = QeβQeαuδ, eβ , eα.

The first equation follows from Qyγzζ ∈ V2γ−ζ = Vη ⊂ V2(eα) and QeβQeαuδ =QeβQeαQeαQyγzζ = QeβQyγzζ since QeαQeα acts like the identity on V2(eα), by(6.14.7). For the proof of the second equation we put wη = Qyγzζ ∈ Vη and use(10.8.1) twice, first for eα and then for eβ ,

uδ, eα, eβ = QeαQyγzζ , eα, eβ = Qeαwη, eα, eβ= eα, wη, eβ = eβ , wη, eα= Qeβwη, eβ , eα = QeβQeαuδ, eβ , eα.

This proves (2). For the general case, let y′γ = eα xα yγ. Then

eα, y′γ , zζ = xα yγ zζ and QeαQeα xα yγ = QxαQyγ

by (10.8.2), and (10.8.4). Hence⟨〈xα, yγ , zζ〉

⟩=⟨〈eα, eα xα yγ, zζ〉

⟩=⟨〈eα, y′γ , zζ〉

⟩(by (30.2.2))

= ϕβ′(− eα y′γ zζ

)· ϕδ

(QeαQ(y′γ)zζ

)(by (2))

= ϕβ′(− xα yγ zζ

)· ϕδ

(QxαQyγzζ

).

32.10. Lemma. The relation (32.1.14) holds in case (x) of Lemma 32.2.

Proof. Let (α4, α1, α2, α3) = (δ, α, β, γ) be a kite. Then we have to show

ϕδ(− xα, yβ , zγ

)=⟨〈xα, yβ , zγ〉

⟩,


for all xα, yβ , zγ in the appropriate root spaces. We first reduce this to the specialcase yβ = eβ :

ϕδ(− xα eβ zγ

)=⟨〈xα, eβ , zγ〉

⟩. (1)

Indeed, for arbitrary yβ ∈ Vβ we have x′α := xα yβ eβ ∈ Vα,⟨〈xα, yβ , zγ〉

⟩=⟨〈xα yβ eβ, eβ , zγ〉

⟩=⟨〈x′α, eβ , zγ〉

⟩by (30.2.3) (with respect to the idempotent eβ), and

x′α, eβ , zγ = xα, yβ , eβ, eβ , zγ = xα, yβ , zγ

by (10.8.2), so assuming (1),⟨〈xα, yβ , zγ〉

⟩=⟨〈x′α, eβ , zγ〉

⟩= ϕδ

(− x′α eβ zγ

)= ϕδ

(− xα yβ zγ

).

To prove (1), put wδ = xα eβ zγ ∈ Vδ. We use the kite (δ, γ, β, α) andProposition 31.9 to compute ϕδ(−wδ). Define

X :=⟨〈eγ , eβ , QeβQeγ (−wδ)〉

⟩, Y := ϕα

(eβ , eγ ,−wδ

). (2)

Then we haveϕδ(−wδ) = X · Y. (3)

Let e = eγ , f = eβ and let h = ωf,e ∈ Aut(V ) as in Proposition 10.12. PutV(ij) = Vi(f) ∩ Vj(e). Then

xα ∈ V(11), zγ ∈ V(12), wδ ∈ V(02).

Hence by 10.12,

xα : = h · xα = xα − f e e f xα ∈ Vα, h · eβ = −eγ ,zβ : = h · zγ = eβ eγ zγ ∈ Vβ , h · wδ = QeβQeγwδ.

Since h is an automorphism of V , we also have

h · wδ = h · xα, h · eβ , h · zγ = xα, −eγ , zβ.

Then the above formulas show QeβQeγwδ = −xα, −eγ , zβ, so

X =⟨〈eγ , eβ , xα, eγ , zβ〉

⟩.

By definition of the ternary symbols in (29.3.5),

X =(((((((E(eγ , eβ), c

)))))))for any c ∈ p−1

(xσ(xα, eγ , zβ)

). We may choose

c =⟨〈xα, eγ ,−zβ〉

⟩


because p(⟨〈xα, eγ ,−zβ〉

⟩) = xσ(−xα, eγ ,−zβ + QxαQeγ zβ) = xσ(xα, eγ , zβ),

since γ β implies Qeγ zβ = 0. On the other hand, consider the elementb = b(eγ , eβ) ∈ G0 ∩N . Then we also have

X =(((((((b, c)))))))

=(

Int(b) · c)· c−1. (4)

We now put a := Int(b) · c and work this out. By (29.3.9) and B(eβ , eγ)−1 · eγ =B(eβ ,−eγ) · eγ = eγ + eβ eγ eγ+QeβQeγeγ = eγ + eβ we get

a =⟨〈B(eγ , eβ)xα, B(eβ , eγ)−1 · eγ ,−B(eγ , eβ) · zβ〉

⟩=⟨〈xα − eγ eβ xα, eγ + eβ , −zβ + eγ eβ zβ〉

⟩.

The third entry can be simplified as follows. Since zβ ∈ Vβ ⊂ V(21) and zγ ∈ V(12),

−eγ , eβ , zβ = h · zβ = h2 · zγ = (−1)1+2zγ = −zγ

by Proposition 10.12. Hence

a =⟨〈xα − eγ eβ xα, eγ + eβ , zγ − zβ〉

⟩.

We introduce the abbreviations

qδ = eγ , eβ , xα and u = zγ − zβ .

Thena =

⟨〈xα − qδ, eγ + eβ , u〉

⟩.

From δ → α and γ α β it follows that a is of type⟨〈V2, V1, V1〉

⟩with respect

to eα, so by (30.7.3) and since the binary symbols vanish, it is multiplicative in thethird variable. From (30.2.10) we then see that it is multiplicative in the first andsecond variable as well. Hence

a =⟨〈xα − qδ, eγ + eβ , u〉

⟩=⟨〈xα, eγ , u〉

⟩·⟨〈xα, eβ , u〉

⟩·⟨〈−qδ, eγ , u〉

⟩·⟨〈−qδ, eβ , u〉

⟩,

where the order of the factors is irrelevant because they all belong to the abeliangroup p−1(Uσ). Since δ ⊥ β, the last factor is of type

⟨〈V0, V2, V 〉

⟩with respect to

eβ and therefore equals 1 by (30.2.1). The third factor is of type⟨〈V0, V1, V 〉

⟩with

respect to eβ . Hence by (30.2.3), it equals⟨〈−qδ, eγ , eβ, eβ , u〉

⟩. By additivity in

the first variable, we collect the second and third term and obtain

a =⟨〈xα, eγ , u〉

⟩·⟨〈xα − eβ eγ qδ, eβ , u〉

⟩.

To simplify the first entry of the second factor, observe that xα and xα are in V(11).Hence by Proposition 10.12,

xα − eβ eγ qδ = xα − eβ eγ eγ eβ xα = h · xα= h2 · xα = (−1)1+1xα = xα,

so we obtain


a =⟨〈xα, eγ , u〉

⟩·⟨〈xα, eβ , u〉

⟩.

Next, recall u = zγ − zβ , use multiplicativity in the third variable, and regroup theresulting factors:

a =⟨〈xα, eβ , zγ〉

⟩·⟨〈xα, eγ , zγ〉

⟩·⟨〈xα, eβ ,−zβ〉

⟩·⟨〈xα, eγ ,−zβ〉

⟩. (5)

The last factor is c. By 31.3 for the edge α γ and Proposition 31.4 we have⟨〈xα, eγ , zγ〉

⟩=⟨〈xα, eγ , zγ, eγ , eγ〉

⟩= ϕα

(− xα, eγ , zγ

),

and analogously⟨〈xα, eβ ,−zβ〉

⟩= ϕα

(xα, eβ , zβ

). Hence the product of the sec-

ond and third factor in (5) is⟨〈xα, eγ , zγ〉

⟩·⟨〈xα, eβ ,−zβ〉

⟩= ϕα

(− xα, eγ , zγ

)· ϕα

(xα, eβ , zβ

)= ϕα

(− xα, eγ , zγ+ xα, eβ , zβ

). (6)

We claim that−xα, eγ , zγ+ xα, eβ , zβ = eβ eγ wδ. (7)

Indeed, put t = eβ eγ xα. Then the right hand side of (7) is, by (JP15) and sincewδ = xα eβ zγ,

R = eβ eγ xα eβ zγ = eβ eγ xα eβ zγ − xα eγ eβ eβ zγ+ xα eβ zβ= t eβ zγ − xα eγ zγ+ xα eβ zβ,

while the left hand side is, by definition of xα,

L = xα eβ zβ − zγ eγ xα= xα eβ zβ − zγ eγ xα+ zγ eγ eβ eγ eγ eβ xα.

By (JP15) the double Jordan triple product in the third summand is

eβ eγ eγ eβ xα = eβ eγ eγ eβ xα− eγ eγ eβ eβxα+ eγ eβ eβ eγ xα

= eβ eβ xα − eγ eγ xα+ eγ eβ t= xα − xα + eγ eβ t = eγ eβ t.

HenceL = xα eβ zβ − zγ eγ xα+ zγ eγ eγ eβ t.

This shows R − L = t eβ zγ − t eβ eγ eγ zγ. Embed the kite (δ, α, β, γ) in ahexagram as in (15.4.3). Then t = eβ eγ xα ∈ Vδ′ and δ′ ⊥ γ. Hence R = Lfollows from (10.8.2) for the idempotent eγ , since t ∈ V0(eγ) and eβ ∈ V1(eγ).

From (6) and (7) we now obtain⟨〈xα, eγ , zγ〉

⟩·⟨〈xα, eβ ,−zβ〉

⟩= ϕα

(eβ eγ wδ

)= Y −1,

by definition of Y in (2) and because ϕα is a group homomorphism. We have shown

a =⟨〈xα, eβ , zγ〉

⟩· Y −1 · c,

and thereforeX = a · c−1 =

⟨〈xα, eβ , zγ〉

⟩· Y −1.

Hence⟨〈xα, eβ , zγ〉

⟩= X · Y = ϕδ(−wδ) (by (3)) which is (1).


32.11. Proposition. The relations (32.1.14) and (32.1.15) hold in the cases(ii)her, (iii)her, (iv)her and (v)her of Lemma 32.2.

Proof. Case (ii)her: Here α1 → α2 = α3 and α4 = α1. Since the arrow α1 → α2

is hermitian, it is embeddable in a kite. We write δ = α1, α = α2, embedded in thekite (δ, α, β, γ) which in turn embeds in a hexagram:

δ

????

α γ

δ′

??// β

???? δ′′

__???oo

(1)

Then B(α1, α2, α3) = B(δ, α, α) which by Lemma 32.3(b) follows from B(γ, β, α).But B(γ, β, α) holds by Lemma 32.10 for the kite (δ, γ, β, α).

Case (iv)her: Here we put α1 = α, α2 = δ, embed the hermitian arrow δ →α in a kite (δ, α, β, γ) and then in the hexagram (1). Then we have to showB(α, δ, α) which by Lemma 32.3(b) follows from B(β, γ, α). But B(β, γ, α) holdsby Lemma 32.10 for the kite (δ′, β, γ, α).

Case (iii)her: Put δ = α2, α = α1. Then δ → α is a hermitian arrow, embeddablein a kite (δ, α, β, γ), and B(α1, α2, α3) = B(α, δ, δ). By Lemma 32.3(b), it sufficesto prove B(β, γ, δ). But this is precisely Lemma 32.9 in the present notation.

Case (v)her: Put δ = α1, α = α2, δ′ = α3 and embed δ → α in a kite (δ, α, β, γ)and further in the hexagram (1). Then we have to show B(δ, α, δ′) which byLemma 32.3(b) follows from B(γ, β, δ′). The latter follows from Lemma 32.9 forthe kite (δ′, α, γ, β).

32.12. Summary. If Γ, V,R satisfy the assumptions of Theorem 27.4 then theSteinberg group St(V,R) is centrally closed.

Proof. Let p: E → G = St(V,R) be a central extension. By Lemma 32.2 andwhat we proved in 30.14, 32.5, 32.7, 32.9, 32.10 and 32.11, the partial sectionssσ: Uσ → E constructed in 31.10 satisfy the defining relations of G and thereforeyield a section s: G→ E of p.

Notes

§27. The theory of central extensions of groups is standard and available in many references,besides the ones given in 27.2. Regarding the background of Theorem 27.4 and Corollary 27.6,refer to the detailed review 27.11.

§29. The proof of Proposition 29.8 is inspired by an argument of R. Steinberg for the caseof Chevalley groups of type Dn over the field F2 [93, 2.3.(6)]. The subsets Z(α) of (29.9.3) are

studied in [91, (5.7)–(5.12)]. The proof of (29.11.3) is a generalization of an argument used in the

proof of [91, Lemma 6.2].

Bibliography

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Subject Index

Additively closed . . . . . . . . . . . . . . . . . . . . . . . . 4

alternating— map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

— matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

annihilator— of subset of Jordan pair . . . . . . . . . . 135

arrow

—, embedded . . . . . . . . . . . . . . . . . . . . . . . 220—, hermitian, of hermitian type . . . . . 221

—, isolated . . . . . . . . . . . . . . . . . . . . . . . . . . 222

—, of orthogonal type . . . . . . . . . . . . . . . 221

Balanced . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

basis—, grid basis . . . . . . . . . . . . . . . . . . . . . . . . 236

— in free abelian group . . . . . . . . . . . . . 233Bergmann operator . . . . . . . . . . . . . . . . . . . . . 81

—, generalized . . . . . . . . . . . . . . . . . . . . . . 110

bi-multiplicative . . . . . . . . . . . . . . . . . . . . . . . 149big cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

binary symbols . . . . . . . . . . . . . . . . . . . . . . . . 350

Category

— of coverings . . . . . . . . . . . . . . . . . . . . 40, 55

— of groups with commutator relations32, 44

— of root graded Jordan pairs . . . . . . . 256

central chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26central series

—, lower . . . . . . . . . . . . . . . . . . . . . . . . . . . 5, 26—, upper . . . . . . . . . . . . . . . . . . . . . . . . . . 6, 26

central trick . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

centrally closed . . . . . . . . . . . . . . . . . . . . . . . . 345cleavage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

cogs

—, associated . . . . . . . . . . . . . . . . . . . . . . . 291—, definition . . . . . . . . . . . . . . . . . . . . . . . . 289

—, maximal . . . . . . . . . . . . . . . . . . . . . . . . . 303—, partial order . . . . . . . . . . . . . . . . . . . . . 293— for hermitian matrices . . . . . . . . . . . . 308

— for Jordan pairs of quadratic forms

309— for rectangular matrices . . . . . . . . . . 307

collision . . . . . . . . . . . . . . . . . . . . . . . . . . 189, 194commutator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

— formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

— relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 22— set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

coreflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

coroot system . . . . . . . . . . . . . . . . . . . . . . . . . . . 16covering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

coweight

—, minuscule . . . . . . . . . . . . . . . . . . . . . . . . 178Coxeter

— group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

— of Jordan pair . . . . . . . . . . . . . . . . . . . . . 80diagonal subgroup . . . . . . . . . . . . . . . . . . . . . . 96

diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

diamond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

division of roots . . . . . . . . . . . . . . . . . . . . . . . . 21

—, indivisibility . . . . . . . . . . . . . . . . . . . . . . 27

Elementary automorphism . . . . . . . . . . . . 141

—, with respect to root grading . . . . . 258elementary group

—, example . . . . . . . . . . . . . . . . . . . . . . . . . 104

—, generalized . . . . . . . . . . . . . . . . . . . . . . . 72—, linear . . . . . . . . . . . . . . . . . . . . . . 30, 43, 50

—, linear projective . . . . . . . . . . . . . . . . . . 61—, orthogonal . . . . . . . . . . . . . . . . . . . . . . . . 72

—, projective . . . . . . . . . . . . . . . . . . . . . . . . . 96

—, symplectic . . . . . . . . . . . . . . . . . . . . . . . . 72—, unitary . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

— of Morita context . . . . . . . . . . . . . . . . . . 71

— of special Jordan pair . . . . . . . . . 96, 119—, Weyl elements . . . . . . . . . . . . . . . . . 127

exponential map

— in TKK-algebra . . . . . . . . . . . . . . . . . . . 94extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

—, central . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

—, split . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345—, universal central . . . . . . . . . . . . . . . . . 345

extreme radical . . . . . . . . . . . . . . . . . . . . . . . . 112

Fibre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

form ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76free product . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

fundamental formula . . . . . . . . . . . . . . . . . . . . 75

Grading

—, Peirce . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

—, alternating . . . . . . . . . . . . . . . . . . . . . . 177—, even quadratic form . . . . . . . . . . . . . 177

—, hermitian . . . . . . . . . . . . . . . . . . . 177, 254

—, associated with root grading . . . 258—, induced . . . . . . . . . . . . . . . . . . . . . . . . . . 256

—, odd quadratic form . . . . . . . . . . . . . . 177—, rectangular . . . . . . . . . . . . . . . . . 176, 253

— of Jordan pair . . . . . . . . . . . . . . . . . . . . 133

— by abelian group . . . . . . . . . . . . . . . 253— by Jordan graph . . . . . . . . . . . . . . . 253

grading derivation . . . . . . . . . . . . . . . . . . . . . . 92

graph—, Clebsch . . . . . . . . . . . . . . . . . . . . . . . . . . 191

—, Jordan . . . . . . . . . . . . . . . . . . . . . . . . . . 195

—, Schlafli . . . . . . . . . . . . . . . . . . . . . . . . . . 192—, complete . . . . . . . . . . . . . . . . . . . . . . . . 181

—, coproduct . . . . . . . . . . . . . . . . . . . . . . . 184

425

426 SUBJECT INDEX

—, extended octahedral . . . . . . . . . . . . . 190—, extended triangular . . . . . . . . . . . . . . 188

—, mixed, partially directed . . . . . . . . . 181

—, morphism . . . . . . . . . . . . . . . . . . . . . . . 183—, octahedral . . . . . . . . . . . . . . . . . . . . . . . 189

—, rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183—, rectangular . . . . . . . . . . . . . . . . . . . . . . 187

—, simply laced . . . . . . . . . . . . . . . . . . . . . 181

—, triangular . . . . . . . . . . . . . . . . . . . . . . . 188— associated with 3-graded root system

185

grid basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236Grothendieck construction . . . . . . . . . 47, 281

group

—, algebraic . . . . . . . . . . . . . . . . . . . . . . 22, 61—, monomial . . . . . . . . . . . . . . . . . . . . . . . . 329

— of rank one . . . . . . . . . . . . . . . . . . . . . . . . 64

— over V . . . . . . . . . . . . . . . . . . . . . . . . . . . 117—, induced by subpair . . . . . . . . . . . . 121

— over V op . . . . . . . . . . . . . . . . . . . . . . . . . 124— scheme . . . . . . . . . . . . . . . . . . . . . . . . 23, 62

Height . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234hexagram . . . . . . . . . . . . . . . . . . . . . . . . . 189, 197

Idempotent— of Jordan pair

—, definition . . . . . . . . . . . . . . . . . . . . . . . 87

—, invertible . . . . . . . . . . . . . . . . . . . . . . 301idempotent root grading . . . . . . . . . . . . . . . 293

—, examples . . . . . . . . . . . . . . . . . . . . . . . . 295

idempotents—, associated . . . . . . . . . . . . . . . . . . . . . 89, 91

—, collinear . . . . . . . . . . . . . . . . . . . . . . . . . . 89

—, compatibility . . . . . . . . . . . . . . . . . . . . . 88—, governing . . . . . . . . . . . . . . . . . . . . . . . . . 89

—, orthogonal . . . . . . . . . . . . . . . . . . . . . . . . 89—, orthogonal system . . . . . . . . . . . . . . . . 90

identities— in Jordan pairs . . . . . . . . . . . . . . . . . . . . 78

inner automorphism . . . . . . . . . . . . . . . . . . . . 21— of Jordan pair . . . . . . . . . . . . . . . . . . . . . 84

inner ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76inner product

—, normalized . . . . . . . . . . . . . . . . . . . . . . . . 16invertible

— element of Jordan pair . . . . . . . . . . . . . 85

Jordan algebra . . . . . . . . . . . . . . . . . . . . . . . . . 78—, exceptional . . . . . . . . . . . . . . . . . . . . . . . 78

—, unital quadratic . . . . . . . . . . . . . . . . . . . 86Jordan graph . . . . . . . . . . . . . . . . . . . . . . . . . . 195

—, Jordan subgraph . . . . . . . . . . . . . . . . . 205—, automorphisms . . . . . . . . . . . . . . . . . . 205—, claw-free . . . . . . . . . . . . . . . . . . . . . . . . . 212

—, forbidden subgraphs . . . . . . . . . . . . . 198Jordan pair

—, abstract . . . . . . . . . . . . . . . . . . . . . . . . . . 75

—, concrete . . . . . . . . . . . . . . . . . . . . . . . . . . 74—, division . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

—, exceptional . . . . . . . . . . . . . . . . . . . . 76, 77

—, identities . . . . . . . . . . . . . . . . . . . . . . . . . 78—, opposite . . . . . . . . . . . . . . . . . . . . . . . . . . 76

—, special . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76—, subpair . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

— homomorphism . . . . . . . . . . . . . . . . . . . . 75

— of alternating matrices . . . . . . . . . . . . . 76— of hermitian matrices over a form ring

76

— of quadratic form . . . . . . . . . . . . . . . . . . 78— of rectangular matrices . . . . . . . . . . . . 76

— of symmetric matrices . . . . . . . . . . . . . 76

Jordan triple— product . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

— system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Kernel

— of subset of Jordan pair . . . . . . . . . . 135kite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120lift

— of Jordan pair homomorphism . . . . 120

locally— finite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

— nilpotent . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Monomial

— group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

— matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 329Morita context . . . . . . . . . . . . . . . . . . . . . . . . . . 71

—, elementary group . . . . . . . . . . . . . . . . . 71morphism

—, bijective on root groups . . . . . . . . . . . 33

—, injective on root groups . . . . . . . . . . . 33—, surjective on root groups . . . . . . . . . . 33

Moufang building . . . . . . . . . . . . . . . . . . . 23, 52

Nilpotent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

— group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

— pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21— set of roots . . . . . . . . . . . . . . . . . . . . . . . . . 2

Opcartesian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45opfibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

opposite— of Jordan pair . . . . . . . . . . . . . . . . . . . . . 76

Peirce— decomposition . . . . . . . . . . . . . . . . . . . . . 87

—, joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

—, with respect to cog . . . . . . . . . . . . 291— grading . . . . . . . . . . . . . . . . . . . . . . . . . . 133

—, examples . . . . . . . . . . . . . . . . . . . . . . 134

— associated with root grading . . . . 258— reflection . . . . . . . . . . . . . . . . . . . . . . . . . . 88

SUBJECT INDEX 427

— spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Peirce grading

—, reverse . . . . . . . . . . . . . . . . . . . . . . . . . . 134

perfect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345positive functional . . . . . . . . . . . . . . . . . . . . . . . 3

prenilpotent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6— pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

prism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

projective elementary group . . . . . . . . . . . . . 96—, Faulkner’s . . . . . . . . . . . . . . . . . . . . . . . 114

— example . . . . . . . . . . . . . . . . . . . . . . . . 116

—, centre . . . . . . . . . . . . . . . . . . . . . . . . . . . 112—, example . . . . . . . . . . . . . . . . . . . . . . . . . 104

—, normalizer of U± . . . . . . . . . . . . . . . . 112

— of a direct sum . . . . . . . . . . . . . . . . . . . 108— of a subpair . . . . . . . . . . . . . . . . . . . . . . 107

— of special Jordan pair . . . . . . . . . . . . . . 97

pyramid . . . . . . . . . . . . . . . . . . . . . . . . . . 190, 196

Quadrangle . . . . . . . . . . . . . . . . . . . . . . . . . . . 331quasi-inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

—, generalized . . . . . . . . . . . . . . . . . . . . . . 111

quasi-invertible pair . . . . . . . . . . . . . . . . . . . . . 81

Rank

— of root system . . . . . . . . . . . . . . . . . . . . . 14reduced . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11, 32

reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

reflection system . . . . . . . . . . . . . . . . . . . . . . . . 10—, direct sum . . . . . . . . . . . . . . . . . . . . . . . . 12

—, morphism . . . . . . . . . . . . . . . . . . . . . . . . . 11

—, reduced . . . . . . . . . . . . . . . . . . . . . . . . . . . 11—, subsystem . . . . . . . . . . . . . . . . . . . . . . . . 12

relation

—, shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156relations

— B(x, y) . . . . . . . . . . . . . . . . . . . . . . . . . . . 123— W(e) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

— (StP1), (StP2) . . . . . . . . . . . . . . . . . . . 147— (StR1), (StR2) . . . . . . . . . . . . . . . . . . . 263

root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4—, long . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

—, short . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16— grading . . . . . . . . . . . . . . . . . . . . . . . . . . 253

—, idempotent . . . . . . . . . . . . . . . . . . . . 293root basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15root group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22root interval

—, closed . . . . . . . . . . . . . . . . . . . . . . . . . . 4, 21—, open . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

root system—, 3-graded

—, classical . . . . . . . . . . . . . . . . . . . . . . . 176—, classification . . . . . . . . . . . . . . . . . . . 179—, embedding . . . . . . . . . . . . . . . . . . . . 175

—, examples . . . . . . . . . . . . . . . . . . . . . . 176—, 3-grading . . . . . . . . . . . . . . . . . . . . . . . . 175—, classical . . . . . . . . . . . . . . . . . . . . . . . . . . 19

—, classification . . . . . . . . . . . . . . . . . . . . . . 18—, embedding . . . . . . . . . . . . . . . . . . . . . . . . 18

—, exceptional . . . . . . . . . . . . . . . . . . . . . . . 19

—, irreducible . . . . . . . . . . . . . . . . . . . . . . . . 15—, irreducible component . . . . . . . . . . . . 16

—, locally finite, finite . . . . . . . . . . . . . . . . 14—, reduced . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

roots

—, connected . . . . . . . . . . . . . . . . . . . . . . . . . 15—, orthogonal . . . . . . . . . . . . . . . . . . . . . . . . 15

Set— in free abelian group . . . . . . . . . . . . . . . . 4

shifting principle . . . . . . . . . . . . . . . . . . . . . . . . 85

simply connected . . . . . . . . . . . . . . . . . . . . . . . 35split

— opfibration . . . . . . . . . . . . . . . . . . . . . . . . 45

square . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187, 194Steinberg

— central trick . . . . . . . . . . . . . . . . . . . . . . 349Steinberg group

— St2(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

— Stn(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . 131— of (V,P) . . . . . . . . . . . . . . . . . . . . . . . . . 152

— of (V,R) . . . . . . . . . . . . . . . . . . . . . . . . . 277

— of a set of idempotents . . . . . . . . . . . 159— of group with commutator relations 40

strictly positive . . . . . . . . . . . . . . . . . . . . . . . . . . 4

structural transformation . . . . . . . . . . . . . . . 84subcategory

—, coreflective . . . . . . . . . . . . . . . . . . . . . . . 35

—, reflective . . . . . . . . . . . . . . . . . . . . . . . . . . 56subset

—, N-free . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2—, additively closed . . . . . . . . . . . . . . . . . . . 4

—, normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233symmetry principle . . . . . . . . . . . . . . . . . . . . . 83

Ternary symbols . . . . . . . . . . . . . . . . . . . . . . 350tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187Tits-Kantor-Koecher algebra . . . . . . . . . . . . 92

TKK-algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 92—, Faulkner’s . . . . . . . . . . . . . . . . . . . . . . . 114

— of simple Jordan pairs in characteristic

zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94— of special Jordan pair . . . . . . . . . . . . . . 93

trace form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

transvection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

—, closed, hermitian . . . . . . . . . . . . . . . . 247— of type i . . . . . . . . . . . . . . . . . . . . . . . . . 243

Unique factorization . . . . . . . . . . . . . . . . . . . . 29

Vertex

—, internal, external . . . . . . . . . . . . . . . . 224—, type of a vertex . . . . . . . . . . . . . . . . . . 223

428 SUBJECT INDEX

Weyl— element . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

— defined by idempotent . . . . . . . . . . 100

— group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12— relations . . . . . . . . . . . . . . . . . . . . . . . . . 130

— triple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Notation Index

We first list some general notations.

Set theory

X ⊂ Y (X & Y ) X is a (proper) subset of YCard(X), |X| cardinality of set XX Y difference of X and YX 4 Y symmetric difference of sets X and YXY maps from Y to XP(X) power set of a set XSX finitary symmetric group of a set XS(X) full symmetric group of a set X⋃i∈I Xi disjoint union of (Xi)i∈I

Algebra

Z rational integersN = Z+ natural numbers including 0N+ = Z++ positive natural numbersQ, R rational (real) numbersR+, R++ non-negative (positive) real numbersFq field with q elementsA× units of a ring AMatpq(A) p× q-matrices over a ring AGLn(A) general linear group over a ring ASLn(k) special linear group over a commutative ring kxT transpose of a matrix x

Category theory

MorC(X,Y ) morphisms from X to Y in the category CCat category of categoriesgrp category of groupsab category of abelian groupsk-alg category of associative commutative k-algebras∐i∈I Xi coproduct of objects (Xi)i∈I

Specific notations ocurring in the text are listed in the following table.

Symbol Explanation Page(((((((A,B

)))))))commutator set of A and B 4(((((((

α, β)))))))

open root interval from α to β 4[[[[α, β

]]]]closed root interval 4

〈x, f〉 evaluation of linear form f on x 10

(x | y) normalized inner product 16

429

430 NOTATION INDEX(((((((a, b)))))))

aba−1b−1, group commutator of a and b 21⟨Si : i ∈ I

⟩subgroup generated by subsets (Si)i∈I 21(((((((

S, T)))))))

subgroup generated by all(((((((a, b)))))))

, a ∈ S, b ∈ T 21

x, y, z Jordan triple product 74[[x, z ]

]binary symbol 350⟨

〈x, y, z〉⟩

ternary symbol 350

structural transformation 84∫Φ (op-)fibration obtained from Φ 47

α∣∣β α divides β, β ∈ Nα 21

α∨ coroot defined by α 10, 194

An(k) Jordan pair of alternating matrices 76

A− Lie algebra defined by associative algebra A 72

Ac additive closure of A 4

Aind indivisible roots of A 27

A× nonzero elements of A, units of A 4, 50

AI , An root systems of type A, A 19

AIK rectangular grading of AK 176

Alt(V,A)H H-invariant alternating maps 352

AnnX annihilator of a subset S of V σ 135

Apn, Ap

n−1 rectangular grading of An = An−1 177

AcollK , A1

K collinear grading 177

β(x, y) inner automorphism defined by (x, y) 84

β(x) automorphism defined by x 111

B(α, β) relations in central extension of St(V,R) 405

B(α, β, γ) relations in central extension of St(V,R) 405

B(x, y), Bσ(x, y) relations in a group over V 123, 125

B+, B− subsets of basis B 236

B(x) generalized Bergmann operator 110

B(x, y), Bx,y Bergmann operator 81

B(x, y, z, v) four-fold Bergmann operator 110

BI , Bn root systems of type B 19

BCI , BCn root systems of type BC 19

BqfI odd quadratic form grading of BI 177

b(x, y) lift of β(x, y) to a group over V 122

Cl Clebsch graph 191

cog(R) cogs compatible with R 293

c(Γ ) numerical invarant of Γ 230

CI , Cn root systems of type C 19

NOTATION INDEX 431

(C1) – (C4) closure conditions 196, 200

CherI hermitian grading of CI 177

C n(H) upper central series of group H 26

∆max maximal domain of definition of cogs 303

δ(x, y) inner derivation 80

D(x, y), Dx,y Dx,yz = x y z 79

DI , Dn root systems of type D 19

DaltI alternating grading of DI 177

Der derivation algebra 80

dom(E ) domain of definition of E 293

DqfI even quadratic form grading of DI 177

D(G) commutator group of G 345

e(M, V ) elementary Lie algebra of (M, V ) 72

e ≈ f e associated with f 89

e > f e collinear to f 89

e ` f e governs f 89

e ⊥ f e orthogonal to f 89

Ebi6 bi-Cayley grading of E6 178

Ealb7 Albert grading of E7 178

E6,E7,E8,F4,G2 exceptional root systems 19

EA(V,P) P-elementary automorphisms 141

EA(V,R) R-elementary automorphisms 258

EI(A) elementary linear group of a ring A 30

E(M) elementary group of Morita context M 72

E(M, V ) elementary group of V ⊂M 72

EO2n(k) elementary orthogonal group 72

ESp2n(k) elementary symplectic group 72

EU2n(A, J, ε, Λ) elementary unitary group of a form ring 44

expσ exponential map in TKK-algebra 94

exp(x) generalized exponential 110

Extr(V ) extreme radical 112

Eσi Peirce projections 87

f• Φ(f), Φ a functor with values in Cat 47

f•(R) induced root grading 256

FPE(V ) Faulkner’s projective elementary group 114

Fr(V ) free product of V + and V − 120

Γi(ω) ith neighbourhood of ω in Γ 213

432 NOTATION INDEX

Γ 0, ∂Γ internal and external vertices of Γ 224

Γher, Γorth, Γiso, Γlin subgraphs of various types of Γ 223

Γ×(R) subset of Γ defined by R 260

gcjΓ subcategory of gcR defined by Γ 261

gcj category comprising all gcjΓ 279

gcR groups with R-commutator relations 32

gc groups with commutator relations 44

gradjpΓ category of Γ -graded Jordan pairs 256

gradjp category of root graded Jordan pairs 256

gradjp∗, gcj∗ fibrations over jgraphop 281

G (R,R1) graph associated with (R,R1) 185

Hn(A, J, ε, Λ) hermitian matrices over a form ring 77

Hn(k) Jordan pair of symmetric matrices 76

idgradjp category of idempotent root gradings 293

IR embedding of st(V,R) 271

Idp(V ) idempotents of V 289

Inder(V ) inner derivation algebra 80

Inn(V ) inner automorphism group 85

Int(a) · b aba−1, inner automorphism definded by a 21

J(M, q) Jordan pair of a quadratic form q 78

jgraph category of Jordan graphs 196

KI complete graph on a set I 181

KI KJ rectangular graph on I × J 187

KerS kernel of a subset S of V σ 135

L(V ) Tits-Kantor-Koecher algebra of V 92

LFau(V ) Faukner’s TKK-algebra 114

LR functor from st(V,R) to gcjΓ 261

Mpq(A) Jordan pair of rectangular matrices 76

mgraph category of mixed graphs 183

M(G) monomial group of G 329

N[A] submonoid generated by A 2

N(A) free abelian monoid generated by A 2

ωe Weyl element in PE(V ) defined by e 100

Ω big cell 96, 118

NOTATION INDEX 433

ωf,e Int(ωe) · ωf 141

OI octahedral graph on I 189

OI extended octahedral graph on I 190

P(I) power set of I 188

P Peirce grading 133

P(e) Peirce grading defined by e 134

P = (p(α, β)) matrix associated with mixed graph 181

PE(V ) projective elementary group of V 96

PE0(V ) diagonal subgroup of PE(V ) 96

Qxy, Q(x)y quadratic operators 74

ReS category of reflection systems 11

RS3 category of 3-graded root systems 175

R root grading 253

R∨ coroot system 16

Rre reflective roots 10

Rim non-reflective roots 10

(R,R1) 3-graded root system 175

rankΓ rank of graph Γ 183

Re(R) subsystem of reflection system 12

R(Γ ) 3-graded root system defined by Γ 202

Sch Schlafli graph 192

sgcR simply connected groups in gcR 35

SF category of sets in free abelian groups 4

st(G) coverings of G ∈ gcR 40

st(G, X) category defined by a set of Weyl triples 55

st(V ) category of groups over V 117

st(V,E ) subcategory of st(V ) defined by a cog E 292

st(V, e) subcategory of st(V ) defined by e 153

st(V,P) subcategory of st(V ) defined by P 146

st(V,R) subcategory of st(V ) defined by R 259

stbij(V,R) “bijective” subcategory of st(V,R) 274

st(V,R,E ) subcategory of st(V,R) defined by a cog E 310

st(V,S ) subcategory of st(V ) defined by S 159

stw(V,R) subcategory of st(V,R) 328

Sij , S′ij shift relations 156

sα reflection in the root α 10

SΓ functor from gradjpΓ to gcjΓ 278

434 NOTATION INDEX

St2(A) Steinberg group of a ring A 59

Stn(A) Steinberg group of a ring A, n> 3 131

St(G) Steinberg group of G ∈ gcR 40

(StP1), (StP2) relations defining st(V,P) 147

(StR1), (StR2) relations defining st(V,R) 263

S∗ functor from gradjp∗ to gcj∗ 288

St(V,P) Steinberg group of (V,P) 152

St(V,R) Steinberg group of (V,R) 277

St(V,R,E ) Steinberg group of (V,R,E ) 310

St(V,S ) Steinberg group of (V,S ) 160

supp(x) support of x with respect to a basis 233

TI triangular graph on I 188

TI extended triangular graph on I 188

T (x, y, x) ternary operator 110

U family of root subgroups 32

Uσ Uσ 1 64∨x, x∨ maps defined on Uσ for rank 1-groups 65

V σi Peirce spaces 87

VI(E ) intersection of I-Peirce spaces of a cog E 291

V op opposite of Jordan pair V 76

V × invertible idempotents of V 301

W(e) Weyl relations for an idempotent e 130

Wα, Tα Weyl elements (triples) for the root α 49

we Weyl element candidate defined by e 125

wf,e Int(we) · wf 168

W(R) Weyl group 12

x−1, x], x[, x • y algebraic operations on Weyl triples 53

xy quasi-inverse of (x, y) 81

X∨ coroot lattice 179

X∗ dual of abelian group X 9

x± exponential maps in groups over V 118

X•(Γ ), X •(Γ ) span of Pα, Pα, α ∈ Γ 182

X• functor from mgraph to ab 183

ζV central derivation 80

Z[A] abelian group generated by A 2

Z (G) centre of a group G 25

Zn(H) lower central series of group H 26

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