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Classification of complex and realsemisimple Lie Algebras

Diplomarbeit zur Erlangung

des akademischen Grades

Magister der Naturwissenschaften

an der Fakultat fur Naturwissenschaften und Mathematik

der Universitat Wien

Eingereicht von Florian Wisser

Betreut von ao. Prof. Dr. Andreas Cap

Wien, Juni 2001

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Contents

Vorwort 3

Preface 5

Chapter 1. Classification of complex semisimple Lie Algebras 7

Chapter 2. Classification of real semisimple Lie algebras 25

1

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Vorwort

Der vorliegende Text behandelt die Klassifikation von komplexenund reellen halbeinfachen Lie Algebren. Das erste Kapitel behandeltden Fall der komplexen Lie Algebren und beginnt mit elementarenDefinitionen. Wir geben eine Liste aller im Text vorkommenden LieAlgebren von Matrizen. Die Cartan Teilalgebra einer komplexen halbe-infachen Lie Algebra wird eingefuhrt. Diese ist eine maximale abelscheTeilalgebra und fuhrt zur Wurzelraumzerlegung einer halbeinfachen LieAlgebra. Die Wurzeln sind lineare Funktionale auf der Cartan Teilalge-bra, die die Wirkung der adjungierten Darstellung auf die Lie Algebrabeschreiben. Die Wurzeln solch einer Zerlegung bilden eine endlicheTeilmenge eines endlichdimensionalen euklidischen Vektorraumes mitbesonderen Eigenschaften, ein reduziertes Wurzelsystem. Aus einemWurzelsystem lassen sich die Cartan Matrix und das Dynkin Dia-gramm einer Lie Algebra bilden, welche die Eigenschaften der Lie Al-gebra beschreiben. Wir definieren ein abstraktes Dynkin Diagram undumreien die Klassifikation derselben. Wir geben eine Liste aller ab-strakten Dynkin Diagramme. Die Klassifikation der komplexen hal-beinfachen Lie Algebren basiert auf dem Existenzsatz, welcher sagt,da jedes abstrakte Dynkin Diagramm Dynkin Diagramm einer kom-

plexen halbeinfachen Lie Algebra ist, und aus dem Isomorphismus The-orem, welches garantiert, da nichtisomorphe einfache Lie Algebrenverschiedene Dynkin Diagramme besitzen. Am Ende des ersten Kapi-tels steht ein Beispiel.

Das zweite Kapitel wendet sich den reellen halbeinfachen Lie Al-gebren zu. Wir beginnen mit der Definition reeller Formen von kom-plexen Lie Algebren. Zu jeder komplexen Lie Algebra existiert eineSplitform und eine kompakte reelle Form. Die kompakte reelle Formder Komplexifizierung einer reellen halbeinfachen Lie Algebra fuhrt zuden aquivalenten Begriffen der Cartan Involution und der Cartan Zer-legung, welche eine reelle halbeinfache Lie Algebra in eine maximale

kompakte Teilalgebra und einen Vektorteil zerlegt. Die Iwasawa Zer-legung von Lie Gruppen verallgemeinert den Gram-Schidt Orthogonal-isierungsproze und fuhrt zum Begriff der eingeschrankten Wurzeln.Diese sind lineare Funktionale auf einem maximalen abelschen Teil-raum des Vektorteils und bilden ein abstraktes Wurzelsystem. EineCartan Teilalgebra einer reellen halbeinfachen Lie Algebra ist eine

3

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4 VORWORT

Teilalgebra, deren Komplexifizierung eine Cartan Teilalgebra der Kom-plexifizierung der Lie Albegra ist. Im Gegensatz zum komplexen Fallsind nicht alle Cartan Teilalgebren konjugiert. Beim Studium der Car-tan Teilalgebren konnen wir uns jedoch auf solche einschranken, diestabil unter einer Cartan Involution sind. Wenn wir eine maximalkompakte Cartan Teilalgebra wahlen konnen wir ein positives Systemdes Wurzelsystems wahlen, das aus rein imaginaren und komplexenWurzeln besteht. Die Cartan Involution fixiert die rein imaginarenWurzeln und permutiert die komplexen Wurzeln in Orbits aus 2 Ele-menten. Das Vogan Diagramm einer reellen halbeinfachen Lie Algebrabesteht aus dem Dynkin Diagramm der Komplexifizierung und folgen-der zusatzlicher Information. Die komplexen Wurzeln in 2-elementigenOrbits werden durch Pfeile verbunden und die imaginaren Wurzeln wer-den ausgemalt so sie nicht kompakt sind. Zur Klassifizierung dienenwieder Satze, die den Zusammenhang zwischen Vogan Diagrammenund reellen halbeinfachen Lie Algebren herstellen. Jedes Diagramm,das formal wie ein Vogan Diagramm aussieht ist Vogan Diagrammeiner reellen halbeinfachen Lie Algebra. Haben zwei reelle halbein-fache Lie Algebren das selbe Vogan Diagramm, so sind sie isomorph,aber reelle halbeinfache Lie Algebren mit unterschiedlichen Vogan Di-agrammen konnen isomorph sein. Das Problem dieser Redundenz lostdas Theorem von Borel und de Siebenthal. Wir geben eine Liste allerVogan Diagramme, die den Redundenztest dieses Theorems uberstehenund wenden uns schlielich der Realisierung einiger Diagramme als LieAlgebren von Matrizen zu.

Zum Schlu besprechen wir kurz einen alternativen Weg der Klas-sifizierung reeller halbeinfacher Lie Algebren. Nimmt man anstatt dermaximal kompakten eine maximal nichtkompakte Cartan Teilalgebraund legt man eine andere Ordnung der Wurzeln zugrunde erhalt manden Begriff des Satake Diagramms. Eine Auflistung aller auftretendenSatake Diagramme bildet den Schlu.

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Preface

This text deals with the classification of complex and real semisim-ple Lie algebras. In the first chapter we deal the complex case andstart with elementary definitions. We list all matrix Lie algebras whichwe will deal with throughout the text. We introduce Cartan subalge-bras of a semisimple complex Lie algebra, which are maximal abeliansubalgebras and discuss the root space decomposition. The roots arelinear functionals on the Cartan subalgebra describing the action ofthe adjoint representation on the Lie algebra. These roots form re-duced root systems which we describe as a specific finite subset of afinite dimensional vector space with inner product. From the root sys-tems we deduce Cartan matrices and Dynkin diagrams. We introducethe notion of abstract Dynkin diagrams and outline the classificationof these. We give a complete list of abstract Dynkin diagrams. Wemention the Existence theorem, stating that every abstract Dynkin di-agram comes from a complex simple Lie algebra, and the IsomorphismTheorem, which says that nonisomorphic simple Lie algebras have dif-ferent Dynkin diagrams, to obtain the classification. At the end of thefirst chapter we look at an example.

The second chapter deals with real semisimple Lie algebras. We

start with real forms of complex Lie algebras, observing that for everycomplex semisimple Lie algebra there exists a split real form and acompact real form. The compact real form of the complexification of areal semisimple Lie algebra yields to the notion of Cartan involutionsand the equivalent notion of Cartan decompositions, which decomposesreal semisimple Lie algebras in a maximally compact subalgebra anda vector part. The Iwasawa decomposition on group level generalizesthe Gram-Schmidt orthogonalization process and leads to the notionof restricted roots. These are linear functionals on a maximal abeliansubspace of the vector part, which form an abstract root system. ACartan subalgebra of a real semisimple Lie algebra is a subalgebra

whose complexification is a Cartan subalgebra in the complexified Liealgebra. In contrast to the complex case not all Cartan subalgebras areconjugate, but we may restrict to the study of Cartan subalgebras thatare stable under a Cartan involution. When we choose a maximallycompact Cartan subalgebra we can fix a positive system such thatthe simple roots are either purely imaginary or complex. The Cartaninvolution permutes the complex ones in 2-cycles. A Vogan diagram

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6 PREFACE

of a real semisimple Lie algebra consists of the Dynkin diagram of itscomplexification plus additional information. The 2-cycles of complexsimple roots are labeled and the imaginary simple roots are painted ifthey are noncompact. Every diagram that looks formally like a Vogandiagram comes from a real semisimple Lie algebra. Two real semisimpleLie algebras with the same Vogan diagram are isomorphic, but realsemisimple Lie algebras with different Vogan diagrams might also beisomorphic. This redundancy is resolved by the Borel and de SiebenthalTheorem. We give a complete list of Vogan diagrams surviving thisredundancy test an take a look at the matrix realizations of some ofthese.

An alternative way of classifying real semisimple arises from choos-ing a maximally noncompact Cartan subalgebra and another positivesystem which leads to the notion of Satake diagrams.

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CHAPTER 1

Classification of complex semisimple Lie Algebras

We will start with some elementary definitions and notions.A vector space g over the field K together with a bilinear mapping

[, ] : g g g is a Lie algebra if the following two conditions aresatisfied:

1: [X, Y] = [Y, X]2a: [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 02b: [X, [Y, Z]] = [[X, Y], Z] + [Y, [X, Z]]

2a and 2b are equivalent. We will call [, ] bracket. The second condition

is called Jacobi identity. Let g and h be Lie algebras. A homomorphismof Lie algebras is a linear map : g h such that

([X, Y]) = [(X), (Y)]

X, Y g. An isomorphism is a homomorphism that is one-one andonto. An isomorphism : g g is called automorphism ofg. The setof automorphisms of a Lie algebra g over K is denoted by AutK g. Ifaan b are subsets ofg, we write

[a, b] = span{[X, Y]|X a, Y b}A Lie subalgebra (or subalgebra for short) h ofg is a linear subspacesatisfying [h, h] h. Then h itself is a Lie algebra. An ideal h ofg isa linear subspace satisfying [h, g] h. Every ideal ofg is a subalgebraof g. A Lie algebra g is said to be abelian if [g, g] = 0. Let s be anarbitrary subset ofg. We call

Zg(s) = {X g|[X, Y] = 0 Y s}the centralizer ofs in g. The center ofg is Zg(g) denoted Zg. Ifs is aLie subalgebra we call

Ng(s) = {X g|[X, Y] s Y s}

the normalizer ofs in g. Centralizer and normalizer are Lie subalgebrasofg and s Ng(s) always holds. Ifa and b are ideals in a Lie algebrag, then so are a + b, a b and [a, b]. If a is an ideal in g we definethe quotient algebra g/a as the quotient of the vector spaces g and aequipped with the bracket law [X+a, Y+a] = [X, Y]+a. Furthermorelet : g h be a map satisfying a ker . Then factors throughthe quotient map g g/a defining a homomorphism g/a h.

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8 1. CLASSIFICATION OF COMPLEX SEMISIMPLE LIE ALGEBRAS

Let g be a finite-dimensional Lie algebras. We define recursively

g0 = g, g1 = [g, g], gj+1 = [gj, gj ]

The decreasing sequence

g0 g1 g2 . . .is called commutator series for g. Each gj is an ideal in g and g iscalled solvable ifgj = 0 for some j. We define recursively

g0 = g, g1 = [g, g], gj+1 = [g, gj]

The decreasing sequence

g0 g1 g2 . . .is called lower central series for g. Each gj is an ideal in g and g iscalled nilpotent ifgj = 0 for some j. Since gj gj for each j nilpotencyimplies solvability. The sum of two solvable ideals is a solvable ideal.Hence there exists a unique maximal solvable ideal, which we call the

radical rad g ofg.A Lie algebra g is simple if g is nonabelian and has no proper

nonzero ideals. A Lie algebra g is semisimple if it has no nonzero solv-able ideals (i.e.: radg = 0). Every simple Lie algebra is semisimple andevery semisimple Lie algebra has 0 center. Ifg is any finite-dimensionalLie algebra, then g/ rad g is semisimple.

Let V be a vector space over K and let M : V V and N : V Vbe vector space endomorphisms. Let EndK V denote the vector spaceof endomorphisms of V. This is a Lie algebra with bracket defined by

[M, N] := M N N MA derivation of a Lie algebra g is an Endomorphism D EndK g

such that

D[X, Y] = [DX,Y] + [X,DY].

Definition (2b) of the Jacobi identity says, that [X, ] acts like a deriva-tion. A representation of a Lie algebra g on a vector space V over a fieldK is a homomorphism of Lie algebras : g EndK V. The adjoint rep-resentation ofg on the vector space g is defined by ad X(Y) = [X, Y].ad X lies in Der(g) because of the Jacobi identity.

A direct sum of two Lie algebras a and b is the vector space directsum a b with unchanged bracket law within each component and[a, b] = 0.

Let g be a finite-dimensional Lie algebra over K, X, Y g. Thesymmetric bilinear form defined by

B(X, Y) = Tr(ad X ad Y)is called Killing form. The Killing form satisfies

B([X, Y], Z) = B(X, [Y, Z])

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1. CLASSIFICATION OF COMPLEX SEMISIMPLE LIE ALGEBRAS 9

We call this property invariance. The radical of B (or generally of anybilinear form) defined by

rad B = {v g|B(v, u) = 0 u g}is an ideal ofg because of the invariance of B. B is called degenerate

if rad B = 0, otherwise it is called nondegenerate.1.1. Proposition. Letg be a finite-dimensional Lie algebra. The

following conditions are equivalent:

(1) g is semisimple(2) The Killing form ofg is nondegenerate(3) g = g1 gm withgj a simple ideal for all j. This decom-

position is unique and the only ideals of g are direct sums ofvariousgj.

1.2. Corollary. If g is semisimple, then [g, g] = g. If a is anyideal ing, thena is an ideal andg = a

a.

A Lie algebra g is called reductive if

g = [g, g] Zgwith [g, g] being semisimple and Zg abelian.

1.3. Corollary. If g is reductive, then g = [g, g] Zg with [g, g]semisimple and Zg abelian.

We now give the definitions of all matrix Lie algebras we will need.The bracket relation will always be defined the way it is done for Liealgebras of endomorphisms above.

H denotes the quaternions, a division algebra over R with basis{1, i , j , k} satisfying the following conditions:

i2 = j2 = k2 = 1ij = k,jk = i,ki = j

ji = k,kj = i,ik = jThe real part of a quaternion is given by Re(a + ib +jc + kd) = a. Wedefine some matrices used in the definitions later on. Let In denote theidentity matrix of dimension n-by-n. Let

Jn,n = 0 InIn 0 , Im,n = Im 00 In and Kn,n = 0 InIn 0 .

The following proposition enables us to check reductiveness.

1.4. Proposition. Letg be a real Lie algebra of matrices overR,C orH. Ifg is closed under conjugate transpose (i.e.: (X) = (X)t gX g) theng is reductive.

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Proof. Define an inner product X, Y = Re Tr(XY) for X, Y ing. Let a be an ideal in g and denote a the orthogonal complement ofa. The g = a a as vector space. To see that a is an ideal in g wechoose arbitrary elements X a, Y g and Z a and compute

[X, Y], Z =ReTr(XY Z Y XZ)= Re Tr(XZY XY Z)= Re Tr(X(YZ) X(ZY))= X, [Y, Z]

Since Y is in g, [Y, Z] is in a. Thus the right hand side is 0 for allZ and hence [X, Y] is in a and a is an ideal and g is reductive.

In the sequel we will use matrix Lie algebras listed below. Theseare all reductive by proposition 1.4. To check semisimplicity one mightuse corollary 1.3, to see that g is semisimple if Zg = 0.

Reductive Lie algebrasgl(n,C) = {n-by-n matrices over C}gl(n,R) = {n-by-n matrices over R}gl(n,H) = {n-by-n matrices over H}

Semisimple Lie algebras over Csl(n,C) = {X gl(n,C)| Tr X = 0} for n 2so(n,C) = {X gl(n,C)|X+ Xt = 0} for n 3sp(n,C) = {X gl(n,C)|XtJn,n + Jn,nX = 0} for n 1 Semisimple Lie algebras over R

sl(n,R

) = {X gl(n,R)| Tr X = 0} for n 2sl(n,H) = {X gl(n,H)| Re Tr X = 0} for n 1so(p, q) = {X gl(p + q,R)|XIp,q + Ip,qX = 0} for p + q 3su(p, q) = {X sl(p + q,C)|XIp,q + Ip,qX = 0} for p + q 2sp(n,R) = {X gl(2n,R)|XtJn,n + Jn,nX = 0} for n 1sp(p, q) = {X gl(p + q,H)|XIp,q + Ip,qX = 0} for p + q 1so(2n) = {X su(n, n)|XtKn,n + Kn,nX = 0} for n 2Now we want to understand the bracket relation of complex semisim-

ple Lie algebras.

1.5. Proposition. If g is any finite-dimensional Lie algebra overC and h is a nilpotent subalgebra, then there is a finite subset hsuch that

(1) g =

g where

g := {X g|(ad H (H)I)nX = 0 for all H h and some n}(2) h g0

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(3) [g, g] g+The g are called generalized weight spaces ofg relative to adh with

generalized weights . The members ofg are called generalized weightvectors. The decomposition statement (1) holds for any representationof a nilpotent Lie algebra over C on a finite-dimensional complex vec-

tor space. The proof of this decomposition uses Lies Theorem whichstates that there is a simultaneous eigenvector for any representation ofa solvable Lie algebra on a finite-dimensional vector space over an al-gebraically closed field. Statement (2) is clear since ad h is nilpotent onh. In statement (3) we set g+ = {0} if + is no generalized weight.The proof consists of an elementary calculation. As a consequence g0 isa subalgebra ofg. A nilpotent Lie subalgebra h of a finite-dimensionalcomplex Lie algebra g is a Cartan subalgebra ifh = g0. One proves thath is a Cartan subalgebra if and only ifh = Ng(h). Ifg is semisimple aCartan algebra h is maximal abelian.

These are the two important theorems concerning Cartan subalge-

bras of finite-dimensional complex Lie algebras:1.6. Theorem. Any finite-dimensional complex Lie algebra g has

a Cartan subalgebra.

1.7. Theorem. If h and h are Cartan subalgebras of a finite-dimensional complex Lie algebra g, then there exists an inner auto-morphism a Intg such that a(h) = h.

We say that h and h are conjugate via a. Because of this conjuga-tion all Cartan subalgebras of a complex Lie algebra g have the samedimension, which is called rank ofg.

This decomposition is simpler for semisimple Lie algebras. Let gbe a complex semisimple Lie algebra with Killing form B and Cartansubalgebra h. The generalized weights of the representation adh on gare called roots. The set of roots is denoted by and is called rootsystem. The decomposition

g = h

g

is called root space decomposition of g. This decomposition has anumber of nice properties:

1.8. Proposition.

Theg are 1-dimensional and are there-

fore given byg = {X g|(ad H)X = (H)X for all H h}.

[g, g] = g+. If and are in {0} and + = 0 then B(g, g) = 0. If is in {0} then B is nonsingular ong g. If then

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B|hh is nondegenerate and consequently there exists to eachroot h an element H h such that (H) = B(H, H)

for all H h. spansh. Choose g E = 0 for all so [H, E] = (H)E. If

Xg then [E, X] = B(E, X)H.

If and then (H) is a rational multiple of (H). If then (H) = 0. The action of adh ong is simultaneously diagonable. If H and H h then B(H, H) = (H)(H). The pair of vectors{E, E} can be chosen so thatB(E, E) =

1.

We define a bilinear form , on h by , = B(H, H) =(H) = (H).

1.9. Proposition. LetV be theR linear span of inh. ThenV

is a real form of the vector spaceh and the restriction of the bilinearform , to V V is a positive definite inner product. Let h0 be theR linear span of all H for thenh0 is a real form of the vectorspaceh, the members of V are exactly those linear functionals that arereal onh0. Restricting those linear functionals to operate onh0 yieldsanR isomorphism of V to h0.

Let ||2 = , and . The mapping s : h0 h0 defined by

s() = 2, ||2

is called root reflection. The root reflections are orthogonal transfor-mations which carry to .A reduced abstract root system in a finite-dimensional real vector

space V with inner product , is a finite set of nonzero elementssuch that

(1) spans V

(2) the orthogonal transformation s() = 2,||2 for carry to itself

(3) 2,||2 is an integer for , (4) implies 2 / (without this condition the abstract

root system is called nonreduced)1.10. Theorem. The root system of a complex semisimple Lie al-

gebrag with respect to a Cartan subalgebra h forms a reduced abstractroot system inh0.

Two abstract root systems in V and in V are isomorphic ifthere exists a vector space isomorphism : V V such that () =

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and2, ||2 =

2(), ()|()|2

for , . An abstract root system in V is said to be reducibleif admits a nontrivial disjoint decomposition = withevery member of orthogonal to every member of . is calledirreducible if no such decomposition exists.

1.11. Theorem. The root system of a complex semisimple Liealgebra g with respect to a Cartan subalgebra h is irreducible as anabstract root system if and only ifg is simple.

Lets take a look at some properties of abstract root systems.

1.12. Proposition. Let be an abstract root system in the vectorspace V with inner product , .

(1) If is in , then is in .(2) If in is reduced, then the only members of {0} pro-

portional to are , 2 and 0, 2 cannot occur if isreduced.(3) If and {0}, then 2,||2 {0, 1, 2, 3, 4}

and 4 only occurs in a nonreduced system for = 2.(4) If and are nonproportional members of such that || 0, then is a root or 0. If

, < 0, then + is a root or 0.(6) If , and neither + nor in {0}, then

, = 0.(7) If

and

{0

}, then the string containing

has the form + n for p n q with p 0 and q 0.There are no gaps. Furthermore p q = 2,||2 . The stringcontaining contains at most four roots.

The abstract reduced root systems with V = R2 are the following:

A1 A1 A2 B2 C2 G2

GG

yy

oo

GG

111111

ppoo

$$111

111

GG

?????????

yy ccoo

11???

????

??

GG

?????cc

yy

oo

11???

??

ffMMMM GG

111111yy

ppVVqqqq88MM

MMoo

$$111

111

xxqqqq

with A1 A1 being the only reducible one of the above.We want to introduce a notion of positivity on V, the vector space

containing an abstract reduced root system , such that

for any V\{0} either or is positive and the sum of positive elements is positive and positive multiples

of positive elements are positive.

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Later on, when we will classify real semisimple Lie algebras, we willinsiste on a special ordering of roots. Therefore the way we intro-duce positivity is by means of a lexicographic ordering. Let B ={1, 2, . . . , n} be a basis of V. An arbitrary element V de-composes to =

ni=1 aii. We say that > 0 if there is an index k,

such that ai = 0 for all 1

i

k

1 and ak > 0, otherwise < 0. It iseasily seen, that this notion of positivity preserves the above properties.We say that > if > 0.

We say a root V is simple if > 0 and does notdecompose in = 1 + 2 with 1 and 2 both positive roots. The setof simple roots is denoted by . Because of the first condition of ourpositivity we either have < or > for , . Then we obtainan ordering of simple roots by 1 < 2 < < l

1.13. Proposition. With l = dim V, there are l simple roots

{1, 2, . . . , l} =

which are linearly independent. If is a root and is decomposed by = a11 + a22 + + all, then all ai = 0 have the same sign andall ai are integers.

Let be a reduced abstract root system in an l dimensional vectorspace V and let = {1, 2, . . . , l} denote the simple roots in a fixedordering. The l-by-l matrix A = (Aij) given by

Aij =2i, j

|i|2is called the Cartan matrix of and . This matrix depends onthe ordering of the simple roots but distinct orderings lead to Cartan

matrices which are conjugate by a permutation matrix.We examine some properties of Cartan matrices.

1.14. Proposition. The Cartan matrix A = (Aij) of and hasthe following properties:

(1) Aij is inZ for all i, j(2) Aii = 2 for all i(3) Aij 0 for i = j(4) Aij = 0 if and only if Aji = 0(5) there exists a diagonal matrix D with positive diagonal entries

such that DAD1 is symmetric positive definite.

An arbitrary square matrix A satisfying the above properties iscalled abstract Cartan matrix. Two abstract Cartan matrices are iso-morphic if they are conjugate by a permutation matrix.

1.15. Proposition. A reduced abstract root system is reducible ifand only if, for some enumeration of the indices, the correspondingCartan matrix is block diagonal with more than one block.

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Via this proposition we move the notion of reducibility from reducedabstract root systems to Cartan matrices. An abstract Cartan matrixis reducible if, for some enumeration of the indices, the matrix is blockdiagonal with more than one block. Otherwise it is irreducible.

The last step in reducing the problem of classification to the essen-tial minimum are Dynkin diagrams. We associate to a reduced abstractroot system with simple roots and Cartan matrix A the followinggraph: Each simple root i is represented by a vertex, and we attachto that vertex a weight proportional to |i|2. We will omit writing theweights if they are all the same. We connect two given vertices corre-sponding to two distinct simple roots i and j by AijAji edges. Theresulting graph is called the Dynkin diagram of . It follows from ourlast proposition, that a Dynkin diagram is connected if and only if is irreducible.

Lets look at three examples, one reducible and the others irre-ducible.

A1 A1 A2 G2

Root system 2

1

GG

yy

oo

2

1

GG

111111

ppoo

$$111

111

21 ffMMMM GG

111111yy

ppVVqqqq88MM

MMoo

$$111

111

xxqqqq

Cartan matrix

2 00 2

2 1

1 2

2 31 2

Dynkin diagram

e e e e e1

e3

Because of the fact that any two Cartan subalgebras of g are con-jugate via Intg we know, that different choice of a Cartan algebra hleads to isomorphic root systems. To see that the choice of a simplesystem leads to isomorphic Cartan matrices we introduce the Weylgroup.

Let be an abstract reduced root system in a vector space V. Themapping s : V V defined by

s() := 2,

|

|2

is the reflection of V at the hyperplane orthogonal to . The groupof reflections generated by these s with is called the Weylgroup and is denoted by W = W() (if is the root system of a Liealgebra g and a Cartan subalgebra h we also write W(g, h)). As theses preserve the whole group W preserves . If = {1, . . . , l} is asimple system in , then W() is generated by the reflections si with

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i . The Weyl group acts simple transitive on the set of simplesystems in .

1.16. Theorem. Let and be two simple systems in . Thereexists one and only one element s W such that s() = .

1.17. Corollary.

Let be an abstract root system and let

+

and + be two positive systems, with corresponding simple systems and . The Cartan matrices of and are isomorphic.

Proof. By the above theorem we obtain an s W() such that = s(). We fix an enumeration of = {1, . . . , l} and choosean enumeration of = {1, . . . , l} such that j = s(j) for all j {1, . . . , l}. So we have

2i, j|i|2 =

2si, sj|si|2 =

2i, j|i|2

since s is orthogonal and hence the resulting Cartan matrices are equalafter a permutation of indices which means that they are isomorphic.

The Weyl group in another important tool in many proofs along theclassification. It is also used in the proof of the following Proposition.

1.18. Proposition. Let and be two nonisomorphic reducedroot systems with simple systems resp. . Then the Cartan matricesA of and A of are nonisomorphic.

Now we will give an outline of the classification of abstract Car-tan matrices. We will work simultaneously with Cartan matrices andtheir associated Dynkin diagrams. First we observe two operations onDynkin diagrams and their counterparts on Cartan matrices.

(1) Remove the ith vertex and all attached edges from an abstractDynkin diagram. The counterpart operation on an abstractCartan matrix is removing the ith row and column from thematrix.

(2) If the ith and jth vertices are connected by a single edge theirweights are equal. Collapse the two vertices to a single oneremoving the connecting edge, retaining all other edges. Thecounterpart operation collapses the ith and jth row and col-

umn replacing the 2-by-2 matrix from the ith

and jth

indices

2 11 2

by the 1-by-1 matrix (2)

One shows that these two operations make abstract Dynkin dia-grams out of abstract Dynkin diagrams and abstract Cartan matricesout of abstract Cartan matrices. Using the defining properties of ab-stract Cartan matrices plus operation (1) we get the following

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1.19. Proposition. LetA be an abstract Cartan matrix. If i = j,then

(1) AijAji < 4(2) Aij {0, 1, 2, 3}

An important step which uses the above proposition in the classifi-

cation is the following

1.20. Proposition. The abstract Dynkin diagram associated to thel-by-l abstract Cartan matrix A has the following properties:

(1) there are at most l pairs of verticesi < j with at least one edgeconnecting them

(2) there are no loops(3) there are at most three edges attached to one vertex.

Using these tools we obtain the following classification of irreducibleabstract Dynkin diagrams in five steps. Note that reducible abstractDynkin diagrams are not connected and can therefore be obtained byputting irreducible ones side by side.

Step 1: None of the following configurations occurs:

e e ee

e

ddde e e e

e

ee e

e

eddd

ddd

Otherwise we use operation (2) to collapse all the single-line part in the center to a single vertex leading to a violationof 1.20 (3).

Step 2: We do a raw classification by the maximal number oflines connecting two vertices.

There is a triple line. By 1.20 (3) the only possibility is(G2)

e1

e2

There is a double line, but no triple line. The graph inthe middle of the figure in step 1 shows that only one pairof vertices connected by two edges exists.

(B , C, F) e1

ep

ep+1

el

There are only single lines. In this situation we call

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e ee

ea triple point. If there is no triple point, then the absenceof loops implies that the diagram is

(A) e e e eIf there is a triple point there is only one because if thethird diagram in the figure in step 1. So the other possi-bility is

(D, E) e1

ep

e1

ep+q

er

ep+1

ddd

Step 3: Now we address the problem of possible weights going

through the three point of the previous step in reverse order: If the ith and jth vertices are connected by a single line,then Aij = Aji = 1 which implies that the weights wiand wj of these vertices are equal. Thus in the cases (A)and (D, E) all weights are equal and we may take themto be 1. In this situation we omit writing the weights inthe diagram.

In the case (B , C, F) we have Ap,p+1 = 2 and Ap+1,p =1 (Ignoring the possibility of the reverted situation isno loss of generality). The defining property 5 of abstractCartan matrices leads to

|p+1

|2 = 2

|p

|2. Taking k = 1

for k p we get k = 2 for k p + 1. In the case (G2) similar reasoning leads to |1|2 = 1 and

|2|2 = 3.Step 4: The remaining steps deal with special situations. In

this step we cover the case (B , C, F). In this case only thesediagrams are possible:

(B) e1 e2 e2 e2(C) e1 e1 e1 e2(F4) e1 e1 e2 e2

For a proof one uses the Schwarz inequality and the definingproperties of Cartan matrices.

Step 5: In the case (D, E) the only possibilities are

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(D) e e ee

e

ddd

(E) e1

epe

ep+1

ep+2

where p {3, 4, 5}. For a proof one uses the Parseval equality.

These steps lead to the following

1.21. Theorem. Up to isomorphism the connected Dynkin dia-grams are the following:

An for n 1 Bn for n 2 Cn for n 3 Dn for n 4 E6 E7 E8 F4 G2

n refers to the number of vertices of the Dynkin diagram. The restric-tions of n in the first four items are made to avoid identical diagrams.The diagrams carrying those names are listed in the following table.

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An e e e eBn e1 e2 e2 e2Cn e

1

e1

e1

e2

Dn e e ee

e

ddd

E6 e e ee

e eE7 e e e e

ee e

E8 e e e eee e e

F4 e1 e1 e2 e2G2 e1 e3

We have got a classification of reduced abstract Dynkin diagrams(resp. Cartan matrices) now which gives us a classification of reducedabstract root systems. We want to show, that an isomorphism of theroot systems of two complex semisimple Lie algebras lifts to an isomor-phism of these algebras themselves. The technique used will be to usegenerators and relations, realizing any complex semisimple Lie algebraas a quotient of a free Lie algebra by an ideal generated by some re-lations. First lets look at some properties of complex semisimple Liealgebras.

Let g be a complex semisimple Lie algebra, h a Cartan subalgebra,

its root system with simple system = {1, . . . , l}, B a nondegen-erate symmetric invariant bilinear form on g that is positive definiteon the real form ofh where the roots are real and let A = (Aij) be theCartan matrix of and . For 1 i l let

hi = 2|i|2Hi , ei a nonzero root vector for i and

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fi the nonzero root vector for i satisfying B(ei, fi) = 2|i|2 .In this situation the set X = {hi, ei, fi}li=1 generates g as a Lie algebra.The elements ofX are called standard generators ofg relative to h and. These generators satisfy the following properties within g.

(1) [hi, hj ] = 0

(2) [ei, fj] = ijhi(3) [hi, ej] = Aijej(4) [hi, fj ] = Aijfj(5) (ad ei)

Aij+1ej = 0 when i = j(6) (ad fi)

Aij+1fj = 0 when i = jThese relations are called Serre relations for g.

To build up a complex semisimple Lie algebra out of generatorsand relations we introduce the notion of free Lie algebras. A free Liealgebra on a set X is a pair (F, ) consisting of a Lie Algebra F anda function : X F with the following universal mapping property:Whenever l is a complex Lie algebra and : X

l is a function,

then there exists a unique Lie algebra homomorphism such that thediagram

F

00 e1 en1 > > e1 e2 >

> e2 en > > e2 e3 >> > >> en2 en > en2 en1 >

> en1 en > 0All negative roots follow in reversed order. The simple roots are all

ei ei+1 with 1 i n 1. Using the Killing form B we obtaina correspondence between a root ei ej and Hij h0, where Hij isthe diagonal matrix with 1 in the ith diagonal entry, 1 in the jthdiagonal entry and 0 elsewhere. This enables us to calculate the entriesof the Cartan matrix

Akl =2k, lk, k =

2k(Hl)

k(Hk).

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With k = ek ek+1 and l = el el+1 this readsAkl =

2(ek ek+1)(Hl,l+1)(ek ek+1)(Hk,k+1) .

Since

(ek ek+1)(Hl,l+1) = 2 for k = l

1 for k + 1 = l or k = l + 10 else

we see that

(An)kl =

2 11 2 . . .

. . . . . . 11 2

.

We have already seen a picture of the root system of sl(3,C) = A2 andfrom the calculations above we know that the Dynkin diagram ofAn is

ee1 e2 ee2 e3 een1 en een en+1

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CHAPTER 2

Classification of real semisimple Lie algebras

Let V be a vector space over R. We call

VC := V R Cthe complexification of V. The mapping m(c) : C C given byz cz is R-linear. Thus 1 m(c) : V RC V RC defines a scalarmultiplication with C. VRC is a vector space over C with the naturalembedding V V R C by v v 1. If{vi}iI is a basis of V overR, {vi 1}iI is a basis of VC over C. Therefore the dimensions

dimR V = dimC V

C

correspond in the above way.Let W be a vector space over C. Restricting the definition of scalars

to R leads to a vector space WR over R. If {vj}jI is a basis of W,then {vj, ivj}jI is a basis ofWR. We get (VC)R = V iV. Therefore

dimC W =1

2dimR W

R

is the correspondence of dimensions.If a complex vector space W and a real vector space V are related

byWR = V

iV

then V is called real form ofW. The conjugate linear map : VC VCthat is 1 on V and 1 on iV is called conjugation of VC with respectto the real form V.

Let g be a real Lie algebra and gC = g C its complexification.The mapping

[, ] : (g C) (g C) g Cgiven by

(X a) (Y b) ([X, Y] ab)extends the bracket in a complex bilinear way. Surely the restrictionof scalars of a complex Lie algebra gives a real Lie algebra. Therefore

both, complexification and restriction of scalars make Lie algebras outof Lie algebras.

For a real Lie algebra g we note that

[g, g]C = [gC, gC]

since by g gC we get [g, g] [gC, gC] and this also holds for theC subspace [g, g]C of gC. For the reverse let a, b C and X, Y g,

25

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26 2. CLASSIFICATION OF REAL SEMISIMPLE LIE ALGEBRAS

then [X a, Y b] = [X, Y] ab [g, g]C. Allowing arbitrary linearcombinations on the left we obtain [gC, gC] [g, g]C.

Lets look at the relation of the Killing forms. Let g0 be a real Liealgebra with Killing form Bg0 . Let g

C0 be its complexification with its

Killing form denoted by BgC0

. Fix any basis ofg0. This is also a basis

for its complexification gC0

. Therefore Tr(ad X

ad Y) is unaffected bycomplexifying and the Killing forms are related by

BgC0|g0g0 = Bg0 .

By Cartans criterion for semisimplicity (i.e.: A Lie algebra is semisim-ple if and only if its Killing form is nondegenerate) we see, that thecomplexification gC0 is semisimple if and only ifg0 is semisimple.

The situation is a little bit more complicated in the case of restric-tion of scalars. Let g be a complex Lie algebra with Killing form Bgand let gR be the real Lie algebra obtained by restriction of scalarswith Killing form BgR. Let B = {vj}jI be a basis of g. ThenB = {vj, ivj}jI is a basis of g

R

. For X g we write adg X as thematrix (ckl)k,lI with respect to the basis B. Look at the same X gR.In the basis B, adgR X is described by the same matrix replacing ckl by

akl bklbkl akl

where akl = Re ckl and bkl = Im ckl. Therefore the Killing

forms are related by

BgR = 2 Re Bg.

Again by Cartans criterion for semisimplicity we get, gR is semisimpleif and only ifg is semisimple.

We will identify two special real forms of complex semisimple Lie

algebras now. The first one will be called split real form. Let g be acomplex semisimple Lie algebra, h a Cartan subalgebra, = (g, h)the set of roots and B the Killing form ofg.

2.25. Theorem. For each it is possible to choose root vectorsX g such that the following conditions hold for all ,

[X, X] = H H as in proposition 1.8 [X, X] = N,X+ if + [X, X] = 0 if + = 0 and + /

and the constants N, satisfy

N, =

N,

N2, = 12 q(1 +p)||2where + n is the string of with p n q.

Since the above theorem shows that N2, is positive, N, is realand we obtain a real form by defining

h0 = {H h|(H) R }

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2. CLASSIFICATION OF REAL SEMISIMPLE LIE ALGEBRAS 27

g0 = h0

RX.

Every real form ofg containing such an h0 for some Cartan subalgebrah is called split real form of g and the above construction shows thatsuch a real form exists for every complex semisimple Lie algebra.

Another special real form, which exists for every complex semisim-ple Lie algebra is called compact real form. A compact real form is areal form that is a compact Lie algebra. A real Lie algebra g is compactif the analytic group Int g of inner automorphisms is compact.

The compact real form will be of greater importance in the sequelthan the split real form. We construct it using a split real form. Firstof all lets specify how a compact real form is characterized. Let u0 bea real form of a complex semisimple Lie algebra g. If the Killing formBu0 is negative definite u0 is called compact real form ofg.

2.26. Theorem. Every complex semisimple Lie algebrag containsa compact real form.

Proof. Let h be a Cartan subalgebra ofg and let X be the rootvectors as in the construction of the split real form. Define

u0 =

RiH +

R(X X) +

Ri(X + X).

Since this is clearly a vector space real form we have to check that itis closed under bracket and that its Killing form is negative definite.Lets check the occurring brackets. Assume = .

[iH, iH] = 0 [iH, (X X)] = ||2i(X + X) [iH, i(X + X)] = ||

2

(X X) [(X X), (X X)] =N,(X+ X(+)) N,(X+ X(+))

[(X X), i(X+ X)] =N,i(X++ X(+)) N,i(X++ X(+))

[i(X + X), i(X+ X)] = N,(X+ X(+)) N,(X+ X(+))

[(X X), i(X + X)] = 2iHThese computations show that u0 is closed under bracket, so we have areal form on our hands. To show its compactness we check the Killingform. We know that the Killing forms Bu0 ofu0 and Bg ofg are related

by Bu0 = Bg|u0u0. SinceBg(g, g) = 0 for , {0} and + = 0

RiH is orthogonal to

R(X X) and toRi(X + X). By the same argument

B((X X), (X X)) = 0 B((X X), i(X + X)) = 0

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B(i(X + X), i(X + X)) = 0.Since B is positive definite on

RH it is negative definite on

RiH. With just two cases left we compute B((X X), (X X)) = 2B(X, X) = 2 B(i(X + X), i(X + X)) = 2B(X, X) = 2

and obtain that Bg|u0u0 is negative definite and therefore u0 is a com-pact real form.

Let g be a Lie algebra. An automorphism : g g such that 2 =idg is called an involution. Such an involution yields a decompositioninto eigenspaces to the eigenvalues +1 and 1. An involution of areal semisimple Lie algebra g0 such that the symmetric bilinear form

B(Z, Z) := B(Z,Z)

is positive definite is called Cartan involution. For complex g these Car-tan involutions correspond to compact real forms. The first situationwhere we observe this correspondence is the following.

2.27. Proposition. Letg be a complex semisimple Lie, u0 a com-pact real form and the conjugation ofg with respect to u0. Then isa Cartan involution ofgR.

Proof. Surely 2 = idgR. The Killing forms ofg and gR are related

by BgR(Z1, Z2) = 2 Re Bg(Z1, Z2). (Therefore gR is semisimple if and

only if g is semisimple.) Decompose Z g as Z = X + iY withX, Y u0. For Z = 0 we get

Bg(Z,Z) = Bg(X+ iY,X iY)= Bg(X, X) + Bg(Y, Y)

= Bu0(X, X) + Bu0(Y, Y) < 0.It follows that

(BgR)(Z, Z) = BgR(Z,Z) = 2 Re Bg(Z,Z)

is positive definite on gR and therefore is a Cartan involution ofgR.

To study an arbitrary real form of complex semisimple Lie algebraswe will align a compact real form to it. For a real form g0 of a complexsemisimple Lie algebra g we want to find a compact real form u0 suchthat u0 = (u0 g0) (u0 ig0).

2.28. Lemma. Let and be involutions of a vector space V. LetV+ denote the eigenspace of to the eigenvalue 1 and V the eigen-space of to the eigenvalue 1. Using the similar notation for weget

=

V+ = (V+ V+) (V+ V)V = (V V+) (V V)

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Proof.

(): Let x V be an arbitrary element. x = x+ + x withx+ V+ and x V. Because of commutativity we have

(x+) =

(x+) =

(x+)

(x+)

V+

(): We start with an arbitrary x V. Decomposing we getx = x+ + x = x+,+ + x+, + x,+ + x,.

Computing

(x) = ((x+, + x+, + x,+ + x,))= (x+,+ x+, + x,+ x,)= x+,+ x+, x,+ + x,= (x+,+ + x+, x,+ x,)= ((x+,+ + x+, + x,+ + x,))= (x)

Thus and commute.

So we search for real forms with commuting involutions.

2.29. Theorem. Letg0 be a real semisimple Lie algebra, a Cartaninvolution and any involution of g0. Then there exists a Intg0such that 1 commutes with .

Proof. Since is a Cartan involution, B is an inner product ong0. Let = . For any automorphism a ofg0 we have B(aX,aY) =B(X, Y) for all X, Y

g0. Since is an automorphism and since

2 = 2 = 1 and = 1, we computeB(X,Y) = B(X, 1Y) = B(X, 11Y)

= B(X,Y) = B(X,Y)

and hence

B(X,Y) = B(X,Y).

Thus is symmetric and its square = 2 is positive definite. Thusr for < r < is a one parameter group in Int g0. Then

= 2 = = = = 2 = 1.

In terms of a basis ofg0 that diagonalizes , the matrix form of this

equation isiiij = ij

1jj for all i and j.

We see that

riiij = ijrjj

and therefore

r = r.

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Now we explicitly give the automorphism = 14 which fulfills

(1) = 14

14 =

12

= 12 1 =

1

21

= 12 =

12

= 12 =

14

14

= (1)as required.

With this result on our hands we easily get the following

2.30. Corollary. Every real semisimple Lie algebrag0 has a Car-tan involution.

Proof. Let g be the complexification ofg0. By our previous resultwe choose a compact real form u0 of g such that the involutions respectively of gR with respect to g0 respectively u0 commute. Wehave g0 = {X g|X = X}. Because of this and the commutativityof the involutions we get

X = X = X

and therefore restricts to an involution = |g0 ofg0. Furthermorewe have

B(X, Y) = Bg0(X,Y) = Bg(X,Y) =1

2(BgR)(X, Y)

for X, Y g0 and so B is positive definite on g0 and is a Cartaninvolution.

Another useful Corollary of Theorem 2.29 deals with a uniquenessproperty of Cartan involutions.

2.31. Corollary. Any two Cartan involutions of a real semisimpleLie algebrag0 are conjugate via Intg0.

Proof. Let and be Cartan involutions ofg0. By Theorem 2.29we can find an automorphism Intg0 such that 1 commuteswith . Hence we may assume without loss of generality that and commute. Therefore their decomposition in +1 and 1 eigenspacesare compatible. Assume that X g0 lies in the +1 eigenspace of andin the 1 eigenspace of . Then X = X and X = X and

0 < B(X, X) = B(X,X) = B(X, X)0 < B(X, X) =

B(X, X) = +B(X, X)

which contradicts the assumption. Therefore = .

Reinterpreting this result we see that any two compact real forms ofa complex semisimple Lie algebra g are conjugate via Int g. Each com-pact real form has a determining associated conjugation. These con-

jugations are Cartan involutions ofgR and are conjugate by a memberof Int gR. Int gR = Int g completes the argument.

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An important Corollary in view of the classification is the followingone.

2.32. Corollary. Let A be an abstract Cartan matrix. Up toisomorphism there exist one and only one compact semisimple real Liealgebra g0 with its complexification (g0)

C having a root system with A

as Cartan matrix.From the first chapter we get the existence and uniqueness ofg, we

know of the existence of a compact real form and because of the aboveargument all compact real forms are conjugate by Int g.

We will now introduce the notion of Cartan decomposition of areal semisimple Lie algebra g0 and we will see that this correspondsto Cartan involutions. A vectorspace direct sum g0 = k0 p0 ofg0 iscalled Cartan decomposition if

(1) the following bracket laws are satisfied

[k0, k0]

k0 , [k0, p0]

p0 , [p0, p0]

k0

(2) the Killing form

Bg0 is

negative definite on k0positive definite on p0.

Let X k0 and Y p0. By the first defining property(ad Xad Y)(k0) p0 and (ad Xad Y)(p0) k0.

Therefore Tr(ad Xad Y) = 0 and hence the Killing form Bg0(X, Y) =0. Since Y = Y also B(X, Y) = 0. This means that k0 and p0 areorthogonal under Bg0 and B.

We will now describe the correspondence between Cartan invo-

lutions and Cartan decompositions. First let be a Cartan involu-tion ofg0. The involution defines an eigenspace decomposition in aneigenspace k0 to the eigenvalue +1 and an eigenspace p0 to the eigen-value 1. We have a decomposition of the form g0 = k0 p0. LetX, X k0 and Y, Y p0 and notice that is an automorphism. Weget

[X, X] = [X,X] = [X, X] [k0, k0] k0[X, Y] = [X,Y] = [X, Y] = [X, Y] [k0, p0] p0[Y, Y] = [Y,Y] = [Y, Y] = [Y, Y] [p0, p0] p0

As we have seen above it follows that k0 and p0 are orthogonal under Bg0

and B. B is positive definite since is a Cartan involution and henceBg0 is negative definite on k0 and positive definite on p0. Thereforeg0 = k0 p0 is a Cartan decomposition.

Conversely starting with a Cartan decomposition g0 = k0 p0 wedefine a mapping

=

+1 on k01 on p0

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respects bracket because for X, X k0 and Y, Y p0 we have[k0, k0] k0 [X, X] = [X, X] = [X,X]

[k0, p0] p0 [X, Y] = [X, Y] = [X, Y] = [X,Y][p0, p0] k0 [Y, Y] = [Y, Y] = [Y, Y] = [Y,Y]

We know of the orthogonality of k0 and p0 under both Bg0 and B and

we know that Bg0 is negative definite on k0 and positive definite on p0.Hence B is positive definite. Therefore is a Cartan involution.

Let g0 = k0 p0 be a Cartan decomposition of g0. By bilinearityof the Killing form we see, that k0 ip0 is a compact real form ofg = (g0)C. Conversely if l0 respectively q0 is the +1 respectively 1eigenspace of an involution , then is a Cartan involution only if thereal form l0 iq0 of (g0)C is compact. For a complex semisimple Liealgebra g, gR = u0 iu0 is a Cartan decomposition ofgR.

To understand the second important decomposition, the Iwasawadecomposition, of Lie algebras we look at an example on group level.

Define

G = SL(m,C), K = SU(m,C), A = {diag(a1, . . . , am)|ai R+}and

N = {

1 n1,2 n1,m0 1

. . ....

.... . . . . . nm1,m

0 0 1

|ni,j C}.

The Iwasawa decomposition states that there is a decomposition G =KAN or more precisely the multiplication : K A N G isa diffeomorphism. To show this in our special case we take the stan-

dard basis {e1, . . . , em} ofCm

and an arbitrary g G. Applying gto the basis we obtain a basis {ge1, . . . , g em}. The Gram-Schmidt or-thogonalization process transforms this basis into an orthonormal ba-sis {v1, . . . , vm} ofCm. By the nature of this process we get a matrixk SU(m) such that k1vj = ej and k1g is upper triangular with pos-itive diagonal entries. i.e.: k1g AN. g = k(k1g) K(AN) showsthat is onto and K AN = {1} that it is one-one. Smoothness isgranted by the explicit formulae of the Gram-Schmidt Process.

Our goal is to observe the equivalent decomposition on algebra level.Untill know we used the subscript 0 to refer to real forms. We willchange this notation for some time because we need a subscript refer-

ring to linear functionals. To avoid constructions like g0,0 we will omitthis subscript for a while.So let g be a real semisimple Lie algebra with a Cartan involution

and corresponding Cartan decomposition g = k p. Let B be anondegenerate, symmetric, invariant, bilinear form on g such that

B(X, Y) = B(X,Y) and B := B(X,Y) is positive definite.

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Then B is negative definite on the compact real form k ip. ThereforeB is negative definite on a maximal abelian subspace of k ip. Bythe invariance of B and the fact, that two Cartan subalgebras of gC

are conjugate by Int gC, we conclude that for any Cartan subalgebraof gC, B is positive definite on the real subspace where all roots arereal-valued. We define orthogonality and adjoints by B, which is aninner product on g.

Before we go into the decomposition we need the following lemma.

2.33. Lemma. Letg be a real semisimple Lie algebra and a Cartaninvolution. For all X g we have (ad X) = ad X relative to theinner product B.

Proof.

B((ad X)Y, Z) = B([X,Y], Z) = B(Y, [X,Z])= B(Y, [X, Z]) = B(Y, [X, Z])= B(Y, (ad X)Z) = B((ad X)Y, Z)

Let a be a maximal abelian subspace ofp. Existence is guaranteedby finite dimensionality. {ad H|H a} is a commuting set of selfad-

joint transformations of g. To show selfadjointness we use the abovelemma. For X g we have

(ad H)X = ( ad H)X = [H,X] = [H, X] = (ad H)X.Commutativity is given by the Jacobi identity. For a let

g := {X g|(ad H)X = (H)X H a}.Ifg

= 0 and

= 0, we call a restricted root ofg, or more precisely

of (g, a), g a restricted root space with its elements called restrictedroot vectors. Let denote the set of restricted roots.

2.34. Proposition. The restricted roots and restricted root spaceshave the following properties:

(1) g = g0

g is an orthogonal direct sum.(2) [g, g] g+(3) g = g(4) (5) g0 = a m orthogonally, wherem = Zk(a)

Proof. (1) see construction(2) Let H a, X g and Y g. We compute

(ad H)[X, Y] = [H, [X, Y]]

= [[H, X], Y] + [X, [H, Y]]

= [(H)X, Y] + [X, (H)Y]

= ((H) + (H))[X, Y]

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(3) Let H a and X g. A quite similar computation does thetrick.

(ad H)X= [H,X]

= [H,X]

= [H, X]= (H)X

(4) A consequence of (3).(5) g0 = g0 by 3. Hence g0 = (k g0) (p g0). Since a p g0

and a is maximal abelian in p, a = p g0. By definitionk g0 = Zk(a).

We choose a notion positivity on a and define the set + of positiverestricted roots and n =

+ g. We collect some facts on the

occuring subalgebras.

n is a nilpotent subalgebra ofg by 2.34 (2). a is abelian by definition. [a,n] = n because for all = 0 we have [a, g] = g. [a n, a n] = n a n is a solvable subalgebra.

The Iwasawa decomposition (on Lie algebra level) states the follow-ing

2.35. Proposition. Letg be a semisimple Lie algebra. g is a vectorspace direct sumg = k a n with k, a andn as above.

Proof. an n is a direct sum because of 2.34 (1) and 2.34 (3).To show that k + a + n is a direct sum we observe some intersecions.

Since a g0 and n = + g, a n = 0. Let X k (a n).X k gives X = X and X a n gives X a n. So X a nand X = X a n, hence X lies in a. Since a p we concludeX k p = 0.

The second step of the proof is to show that the direct sum kanis all ofg. Let X be an arbitrary element ofg. We write X as

X = H+ X0 +

X

where H a, X0 m and X g for all . We write

X =+(X + X)=+(X + X) + +(X X).Since maps X + X onto itself (X + X) k and since Xand X g, (X X) g n.

An arbitrary X decomposes to

X = (X0 +

+(X + X)) + H + (

+(X X)) g k a n

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The following Lemma will enable us to find Cartan algebras in realLie algebras.

2.36. Lemma. Letg be a real semisimple Lie algebra. There existsa basis

{Xi

}of g such that the matrices representing ad g have the

following properties:

(1) the matrices of ad k are skew symmetric(2) the matrices of ad a are diagonal with real entries(3) the matrices of ad n are upper triangular with all diagonal en-

tries 0.

Proof. Recall that we decomposed g = g0

g orthogonally.Let {Xi} be an orthonormal basis ofg, which is compatible with thisdecomposition. We do a reordering of these vectors such that Xi giand Xj gj with i < j implies i j.

(1) Let X

k. Therefore X = X and (ad X) =

ad X = ad X. We have used this argument before for H a.

(2) Since each Xi is either a restricted root vector or in g0 thematrices of ad a are diagonal, necessarily real.

(3) [gi , gj ] gi+j

Define the rank of a real semisimple Lie algebra g as the dimensionof any Cartan subalgebra h ofg. This is well defined since h is a Cartansubalgebra if and only ifhC is Cartan in gC.

2.37. Proposition. Letk p be a Cartan decomposition of g. Leta be maximal abelian in p and m = Zk(a). If t is a maximal abeliansubspace ofm thenh = a t is a Cartan subalgebra of g.

Proof. We have to prove that hC is maximal abelian in gC andthat adgC h

C is simultaneously diagonable.By bilinearity of the bracket hC is abelian. To prove maximality let

Z = X + iY be in hC. If Z commutes with hC then so do X and Y.Thus we do not loose generality in using X h to test commutativity.If X commutes with hC it lies in a m. This also holds for X. Thus(X + X) k lies in m and commutes with t, hence is in t. Similarargumentation gives (X X) a. Thus X is in a t and hence hCis maximal abelian.

Using the same basis as above ad tconsists of skew symmetric ma-trices. These are diagonable over C. With the matrices in ad a alreadydiagonal we get a family of commuting matrices and hence the membersof adhC are diagonable.

Using h = atas Cartan subalgebra ofg, we build the set (gC, hC)of roots ofgC with respect to the Cartan subalgebra hC. The root space

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decomposition ofgC is given by

gC = hC

(gC).

By definition

(gC

) = {X gC

|(ad H)X = (H)X H hC

}and

g = {X g|(ad H)X = (H)X H a}.Hence

g = g

|a=

(gC)

and

mC = tC |a=0

(gC).

2.38. Corollary. If t is a maximal abelian subspace ofm = Zk(a)then the Cartan subalgebrah = a thas the property that all roots arereal valued ona it. Ifm = 0 theng is a split real form ofgC.

Proof. The values of the roots on a member H of adh are theeigenvalues of ad H. For H a, ad H is self adjoint and hence has realeigenvalues. For H t, ad H is skew adjoint and hence has imaginaryeigenvalues.

If m = 0, then t = 0 and h = a. All roots are real on a and gcontains the real subspace of a Cartan subalgebra hC

gC. Hence g is

a split real form ofgC.

We want to impose an ordering on the root system , such thatthe positive system + extends +. We form a lexicographic orderingon (a+ it), taking values on a before it. If is nonzero on a thenthe positivity of only depends on its values on a. Thus + extends+.

The following theorem will be used in some proofs in the sequel.

2.39. Theorem. LetG be a compact connected Lie group with Liealgebra g0. Any two maximal abelian subalgebras of g0 are conjugatevia Ad(G).

We will now oberserve the possible choices for all parts of the Iwa-sawa decomposition. From the Cartan decomposition we already nowthat k is unique up to conjugacy.

2.40. Lemma. Let H a with (H) = 0 for all , thenZg(H) = m a. Hence Zp(H) = a.

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Proof. Let X be in Zg(H) and decompose

X = H0 + X0 +

X

with H0 a, X0 m and X g. Then 0 = [H, X] =

(H)Xand hence (H)X = 0 for all

. By our assumption that (H)

=

0, X = 0.

Now we fix a subalgebra k ofg and look at the possible choices ofa maximal abelian subspace a ofp.

2.41. Theorem. Ifa anda are two maximal abelian subalgebras ofp then there is a member k K with Ad(k)a = a, where K is the ana-lytic subgroup of G with Lie algebrak. Consequentlyp =

kKAd(k)a.

Proof. We can easily find an H a with (H) = 0 for all since the union of the kernels of all is only a finite union ofhyperplanes in a. By lemma 2.40 such an H a gives Zp(H) = a.Similarly we find an H a such that Zp(H) = a. By compactness ofAd(K), choose a k0 K such that B(Ad(k0)H, H) B(Ad(k)H, H)for all k K. For any Z k

r B(Ad(exp rZ)Ad(k0)H, H)is a smooth function of r that is minimized for r = 0. Differentiatingand setting r = 0 we obtain

0 = B((ad Z)Ad(k0)H, H) = B(Z, [Ad(k0)H, H]).

[Ad(k0)H, H] is in k. Since B(k, p) = 0 and since B is nondegenerate,

[Ad(k0)H, H] = 0. Thus Ad(k0)H is in Zp(H) = a. Since a is abelian

a Zp(Ad(k0)H) = Ad(k0)Zp(H) = Ad(k0)a.Since a is maximal abelian in p we even get equality a = Ad(k0)a

.This proves the first statement of the theorem.

Let X p and extend the abelian subspace RX ofp to a maximalabelian subspace a. Using the first part of the proof we write a =Ad(k)a and hence X kKAd(k)a. Therefore

p =kK

Ad(k)a.

For a fixed subalgebra k we have found that all possible a are con-jugate. Now we fix k and a and observe the possible choices ofn. Bis an inner product on g and can be restricted to an inner product ona. Since we can identify a with H a we can transfer B to adenoting it , .

2.42. Proposition. Let be a restricted root and let E be anonzero restricted root vector for .

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(1) [E, E] = B(E, E)H and B(E, E) < 0.(2) RH RE RE is a Lie subalgebra of g isomorphic to

sl(2,R) and the isomorphism can be defined so that the vector

H = 2H||2 corresponds to h =

1 00 1

.

(3) If E is normalized so that B(E, E) =

2

||2 , then

k = exp

2(E + E)

is a member of the normalizer NK(a) of a in K and Ad(k)acts as the reflection s ona

.

Proof. (1) Since g = g the vector

[E, E] [g, g] g0 = a mand from

[E

, E

] = [E

, E

] =

[E

, E

]

it follows that [E, E] lies in a. For H a we computeB([E, E], H) = B(E, [E, H])

= (H)B(E, E)

= B(H, H)B(E, E)

= B(B(E, E)H, H).

Since B is nondegenerate on a and B is positive definite weget the stated results

[E

, E

] = B(E

, E

)H

B(E, E) = B(E, E) < 0.(2) Let

H =2

||2 H, E =

2

||2 E, E = E.

Then (1) shows that

[H, E] = 2E

, [H

, E

] = 2E, [E, E] = H

which is all we need.(3) We normalize the vectors such that B(E, E) =

2

||2 , which

always works because of (1). If (H) = 0, then

Ad(k)H= Ad(exp 2 (E + E))H

=(expad 2

(E + E))H

=

n=01n!

(ad 2

(E + E))nH

= H.

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For the element H the following equalities hold:

(ad 2

(E + E))H = (E E)

(ad 2

(E + E))2H = 2H.

Using this equalities we get

Ad(k)H =

n=01n!

(ad 2

(E + E))nH

=

m=01

(2m)!((ad

2(E + E))

2)mH+

m=01

(2m+1)!(ad 2 (E + E))((ad

2 (E + E))

2)mH=

m=01

(2m)!(2)mH +

m=0

1(2m+1)!

(2)m(E E)= (cos )H + (sin )(E E)= H.

This is what we stated.

2.43. Corollary. is an abstract root system in a. need notbe reduced.

Proof. We verify that satisfies the axioms for an abstract rootsystem. To see that spans a, assume (H) = 0 for some H a.Then [H, g] = 0 for all and hence [H, g] = 0. But the center Zg = 0and hence H = 0. Thus spans a.

Let sl denote the Lie subalgebra mentioned in 2.42 (2). This actsby ad on g and hence on gC. Complexifying we obtain a representationof slC

= sl(2,C) on gC. The element H = 2 H||2 which corresponds toh acts diagonably with integer eigenvalues. H acts on g by the scalar(2 H||2 ) = 2

,||2 . Hence 2

,||2 is an integer.

The last property to show is that the reflection s() of along is in for all , . Define k as in 2.42 (3), let H a and X g.Then

[H, Ad(k)X]=Ad(k)[Ad(k)1H, X] = Ad(k)[s1 (H), X]

= (s1 (H))Ad(k)X = (s)(H)Ad(k)X

and hence gs() is not 0.

2.44. Corollary. Any two choices ofn are conjugate by Ad n for

some n NK(a).We ran through all parts of the Iwasawa decomposition and see that

an Iwasawa decomposition ofg is unique up to conjugacy by Int g.An interesting aspect in classifying real semisimple Lie algebras are

the conjugacy classes of their Cartan subalgebras. Now, that we donot deal with subscripts referring to root spaces any longer we revert

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to the subscript 0 for real Lie algebras. So g will denote a complex Liealgebra again and g = (g0)C ifg0 is a real Lie algebra.

Let g0 be a real semisimple Lie algebra, a Cartan involutionand g0 = k0 p0 the corresponding Cartan decomposition. Let B beany nondegenerate, symmetric, bilinear form such that B(X,Y) =B(X, Y) and B is positive definit.

In the case of complex Lie algebras the situation is quite easy sinceall Cartan subalgebras are conjugate. This is not true in the real case.However, the following proposition holds.

2.45. Proposition. Any Cartan subalgebrah0 of a real semisimpleLie algebrag0 is conjugate via Intg0 to a stable Cartan subalgebra.

Proof. Let h0 be any Cartan subalgebra ofg0 with complexifica-tion h Cartan in g. Let the conjugation ofg with respect to g0.

Let u0 be the compact real form of g built out of the split formcorresponding to h as in 2.26. and let be the conjugation with respect

tou

0. Since ih

0 u

0 is exactly the part ofh

which lies inu

0, (h

) =h

.The conjugations and are involutions of gR and is a Cartaninvolution. As proven before = (()2)

14 IntgR = Int g is the

element that makes and = 1 commute. Since (h) = h and(h) = h also (h) = h and (h) = h. Using commutativity we compute

(g0) = (g0) = (g0)

which shows that (g0) = g0. Since h0 = h g0 we obtain (h0) = h0.Let = |g0. Clearly (h0) = h0. Since

1((u0)) = (u0) = (u0)

is a conjugation of g with respect to (u0). Taking X and Y in gR

we have

(BgR)(X, Y) = BgR(X, Y).Restricting X and Y to g0 this equals

2Bg0(X,Y) = 2(Bg0)(X, Y).Consequently is a Cartan involution of g0. Since any two Cartaninvolutions ofg0 are conjugate via Int g0 their exists a Intg0 suchthat = 1. Then (h0) is a Cartan subalgebra ofg0 and

((h0)) = 1(h0) = ((h0)) = (h0)

shows that it is stable.

Without loss of generality we restrict to the study of stable Cartansubalgebras. Let h0 be a stable Cartan subalgebra of g0. Thenh0 = t0 a0 with t0 k0 and a0 p0. As seen above all roots are realvalued on a0 it0. We call dim t0 the compact dimension and dim a0the noncompact dimension ofh0.

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We call a stable Cartan subalgebra h0 maximally noncompact ifits noncompact dimension is maximal. We call it maximally compactif its compact dimension is maximal.

In any case a0 is an abelian subspace ofp0. Therefore h0 = t0 a0 ismaximally noncompact if and only ifa0 is a maximal abelian subspaceofp0.

A similar statement is true for maximally compact Cartan algebras.To see this we need the following

2.46. Proposition. Let t0 be a maximal abelian subspace of k0.Then h0 = Zg0(t0) is a stable Cartan subalgebra of g0 of the formh0 = t0 a0 with a0 p0.

Proof. From our construction of h0 we know that it decomposesto h0 = t0 a0 where a0 = h0 p0. Therefore h0 is stable. Since all stable subalgebras of a real semisimple Lie algebra are reductive, sois h0. Hence [h0, h0] is semisimple.

We have[h0, h0] = [t0 a0, t0 a0] = [a0, a0].

Recall that [p0, p0] k0. Since t0 = h0 k0 we get[h0, h0] = [a0, a0] t0.

Thus the semisimple Lie algebra [h0, h0] is abelian and hence must be0. Consequently h0 is abelian.

h = (h0)C is maximal abelian in g. Since all elements of adg0(t0)are skew adjoint and all elements of adg0(a0) are self adjoint and t0commutes with a0, ad h0 is diagonably on g. Therefore h is a Cartansubalgebra ofg and consequently h0 is a Cartan subalgebra ofg0.

So similar to the above we see, that for a stable h0 = t0 a0, t0is an abelian subspace of k0 and h0 is maximally compact if and onlyif t0 is maximal abelian in k0. We proceed with two statements aboutconjugacy of special Cartan subalgebras.

2.47. Proposition. Among stable Cartan subalgebrash0 ofg0 themaximally noncompact ones are all conjugate viaK, and the maximallycompact ones are all conjugate via K.

2.48. Proposition. Up to conjugacy by Intg0, there are only fi-nitely many Cartan subalgebras ofg0.

Recall that we used Dynkin diagrams to classify complex semisim-ple Lie algebras. If we want to use something similar in the classifi-cation of real semisimple Lie algebras we have to refine this concept.These refined diagrams will be called Vogan diagrams and will consistof Dynkin diagrams plus additional information.

Let g0 be a real semisimple Lie algebra with complexification g. Let be a Cartan involution and g0 = k0 p0 the corresponding Cartan

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decomposition. Let B be any nondegenerate symmetric invariant bi-linear form on g0 such that B(X,Y) = B(X, Y) and B is positivedefinite. Let h0 = t0 a0 be a stable Cartan subalgebra with t0 k0and a0 p0. From Corrolary 2.38 we know that all roots of (g, h),where h = (h0)C, are real valued on a0 it0. Hence the roots take realvalues on a0 and imaginary values on t0. We call a root real if it takesreal values on all ofh0, equivalently the root vanishes on t0 and we calla root imaginary if it takes purely imaginary values on h0, equivalentlyit vanishes on a0. If a root does not vanish on either of t0 and a0 it iscalled complex.

Now let h0 = t0 a0 be a maximally compact stable Cartansubalgebra of g0 with complexification h = t a. Let = (g, h)be the set of roots. Since we choose a maximally compact Cartansubalgebra there are no real roots.

Let {H1, . . . , H n} be a basis of it0 and let {Hn+1, . . . , H m} be abasis ofa0. Then {H1, . . . , H m} is a basis of it0 a0 and also a basisofh. Let ,

. We say that > if there is an index l such that

(Hl) > (Hl) and (Hj) = (Hj) for all j < l.

For any root we define by

(H) = (1H).

Let E be a nonzero root vector for and calculate

[H,E] = [1H, E] = (

1H)E = ()(H)E

to see that is a root again. If is purely imaginary, then = .Thus g is stable and hence

g = (gk)

(g

p).

But, as dim g = 1 we either have g k or g p. We call animaginary root compact ifg k or we call it noncompact ifg p.

Since is +1 on t0 and 1 on a0 and since there are no real roots,which means no roots that vanish on t, (+) = +. Therefore permutes the simple roots. More precisely fixes the imaginary roots,which vanish an a, and it permutes the complex roots in 2-cycles sinceit flips the real parts.

Let g0 be a real semisimple Lie algebra and h0 a Cartan subalgebraofg0. Let be a Cartan involution ofg0. Let

+ be a system of positiveroots built in the above way. The Vogan diagram of (g0, h0,

+) consists

of the Dynkin diagram of +

with 2-element orbits under labeled andwith 1-element orbits corresponding to noncompact imaginary rootspainted. Observe that this triple totally determines the Vogan diagram.Let (g0, h0,

+) and (g0, h0,

+) be two isomorphic triples. Since (g0)C

is isomorphic to (g0)C their Vogan diagrams are based on the same

Dynkin diagram. But they also have isomorphic Cartan subalgebrasand positive systems and hence they have the same 2-element orbits

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and noncompact imaginary roots. Hence they have the same Vogandiagram.

In order to classify complex semisimple Lie algebras, by classify-ing Dynkin diagrams, we needed the Isomorphism Theorem 1.23 andthe Existence Theorem 1.24, which gave us a one-one correspondenceof Lie algebras and Dynkin diagrams. The same has to be done forreal semisimple Lie algebras and Vogan diagrams. First lets state ananalog for the Isomorphism Theorem. The proof will be made out ofsteps, each step decreasing the possible differences between the two Liealgebras.

2.49. Theorem. Letg0 andg0 be real semisimple Lie algebras. If

two triples(g0, h0, +) and (g0, h

0,

+) have the same Vogan diagram,theng0 andg

0 are isomorphic.

Proof. Since the Lie algebras have the same Vogan diagram, theyalso have the same Dynkin diagram. By the Isomorphism Theorem

1.23 we do not loose generality in assuming (g0)

C

= (g0)C

= g.Let u0 = k0 ip0 be the compact real form of g associated to g0and u0 = k

0 ip0 the one associated to g0. Since any two compact real

forms of a complex semisimple Lie algebra g are conjugate via Int g,there exists x Intg such that xu0 = u0. xg0 is a real form ofg that isisomorphic to g0 and has Cartan decomposition xg

0 = xk

0 xp0. Since

xk0 ixp0 = xu0 = u0, there is no loss of generality in assuming thatu0 = u0 from the outset. Then

(u0) = u0 and (u0) = (u0) = u

0 = u0.

We now use theorem 2.39 to see that the Cartan subalgebras ofg0

and g0 complexify to the same Cartan subalgebra ofg: Let h0 = t0 a0respectively h0 = t0 a0 be the Cartan subalgebras ofg0 respectivelyg0 decomposed with respect to respectively

. Since t0 ia0 andt0 ia0 are maximal abelian in u0 and u0 is compact, there exists ak Intu0 such that k(t0 ia0) = t0 ia0. kg0 is isomorphic to g0and xh0 = xt

0 xa0. Since k(t0 ia0) = kt0 ika0, there is no loss in

generality in assuming that (t0ia0) = (t0ia0). Therefore the Cartansubalgebras h0 and h

0 complexify to the same Cartan subalgebra ofg,

which we will denote by h. The space

u0 h = t0 ia0 = t0 ia0

is a maximal abelian subspace ofu0.Because the complexifications of both real Lie algebras and their

Cartan subalgebras are the same now, their root systems coincide.However their positive systems still may differ. But their exists ak Intu0 normalizing u0 h with k+ = +. If we replace g0by kg0 and argue the same way we walked through twice, we mayassume +

= + from the outset.

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What we have done so far is that we may assume without loss ofgenerality that

u0 = u0u0 h = u0 h

= and + = +

The next step is to choose normalizations of the root vectors relativeto h. Let B be the Killing form ofg. Recall the construction of a splitform of a complex semisimple Lie algebra in 2.25. We obtained rootvectors X. Out of this split form we constructed a compact real form

u0 =

R(iH) +

R(X X) +

Ri(X + X)

of g. The subalgebra u0 contains the real subspace

RiH of hwhere all roots are imaginary, which is just u0 h. Since any twocompact real forms are conjugate by Int g there exists a g Intg suchthat gu0 = u0. Then gu0 = u0 is built from g(u0 h) and the rootvectors gX. The two maximal abelian subspaces u0 h and g(u0 h)of u0 are conjugate by u Intu0. Hence ug(u0 h) = u0 h. LetY = ugX for all . Then u0 is built from ug(u0 h) = u0 hand the root vectors Y.

u0 =

R(iH) +

(Y Y) +

Ri(Y + Y)

Now we will use what we know about the Cartan involutions ofg0 and g

0 out of the Vogan diagram. Since the automorphism of

+

defined by and are the same, and have the same effect on h.Thus (H) = (H) for all H h. If is an imaginary simple root,then (Y) = Y =

(Y) if is unpainted(Y) = Y = (Y) if is painted.

Remember that the 2-element orbits under of complex simple rootsare labeled in the Vogan diagram. For we write Y = aY.Since (u0) = u0 we know that

(u0 span{Y, Y}) u0 span{Y, Y}.This means that

(R(Y Y) +Ri(Y + Y)) (R(Y Y) + Ri(Y + Y)).

Let x and y R and x + iy = z C. Thenx(Y Y) + yi(Y + Y) = (x + iy)Y (x iy)Y = zY zYSince (zYzY) = zaYzaY lies in (R(YY)+Ri(Y+Y)) it must be of the form wY wY. Thus from za = za weconclude that

a = a.

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Furthermore

aa = 1

since

aa = aaB(Y, Y) = B(aY, aY) =

B(Y, Y) = B(Y, Y) = 1.Combining these two results we get

|a| = aa = aa = 1.Since Y =

2Y = (aY) = aaY we have

aa = 1.

For each pair of complex simple roots it is therefore possible to choosesquare roots

a and

a such that

a

a = 1.

Similarly we write Y = bY and obtain the same results asabove and choose square roots.

|b| = 1b

b = 1

We can define H and H in u0 h by the following conditions: (H) = 0 = (H) for imaginary simple exp( 1

2(H)) =

a, exp(

12

(H)) =

a for complex simple and

exp( 1

2(H

)) =

b

, exp( 12

(H)) =

b

for complex simple and

The last step in the proof is to show that the equation

Ad(exp 12

(H H)) = Ad(exp 12

(H H)) holds. On all ofh and on each X where is an imaginary simple root, acts like . On these subspaces the two sides are equal.

If is complex simple, then

Ad(exp 12 (H H))Y = (e12(HH)Y)

= b

a

b Y=

ba

Y

=ab

Y

= Ad(exp 12

(H H)) YThis proves the above equation.

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For some X k we have Ad(exp 1

2(H H))X = Ad(exp 1

2(H H))X

and hence X lies in k. With Y p the only thing that changes is anadditional minus on the right side. Therefore the inclusions

Ad(exp 12 (H H))(k) kAd(exp 12 (H H))(p) p

hold. By dimensional argumentation we easily see that we get equalityon these inclusions. Since the element Ad(exp 12 (H H)) carries u0to itself, it must carry k0 = u0 k to k0 = u0 k and p0 = u0 p top0 = u0 p. Hence it must carry g0 = k0 p0 to g0 = k0 p0.

Now, that we have proved uniqueness, we address the question ofexistence. We define an abstract Vogan diagram as an abstract Dynkindiagram together with an automorphism of the diagram, which indi-cates the 1- and 2-element orbits of vertices, and a subset of the 1-element orbits, which indicates the painted vertices. Clearly, everyVogan diagram is an abstract Vogan diagram.

2.50. Theorem. If an abstract Vogan diagram is given, then thereexists a real semisimple Lie algebrag0, a Cartan involution , a max-imally compact stable Cartan subalgebra h0 = t0 a0 and a positivesystem + of (g, h) that takes t0 before ia0 such that the given dia-gram is the Vogan diagram of (g0, h0,

+).

Proof. By the Existence Theorem 1.24 there is a complex semisim-ple Lie algebra g which corresponds to the Dynkin diagram on which

the abstract Vogan diagram is based. Let h be a Cartan subalgebra ofg. Let = (g, h) be a root system with a positive system +. LetX be the root vectors of the corresponding split real form of g anddefine a compact real form u0 ofg in terms ofh and X by

u0 =

R(iH) +

R(X X) +

Ri(X + X).

The abstract Vogan diagram defines an automorphism of theDynkin diagram, which extends linearly to h and is isometric. Wewant to see that maps onto itself. Let + be a positive root.We write =

nii as a sum of simple roots and call

ni the levelof . We show that (+)

by induction on the level n of . If

the level is 1, then is simple and thereof we know that is simpleeither. Now let n > 1. Assume that for all positive withlevel < n and let be of level n. We choose a simple root j such that, j > 0. Then the reflection of on the hyperplane defined by j,

sj() = 2, j

|j|2 j =

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is a positive root of smaller level than . By induction hypothesis and j are in . Since is isometric = sj() and therefore is in .

We can transfer from h to h, retaining the same name. Define on the root vectors X for simple roots by

X = X if is unpainted and forms a 1-element orbitX if is painted and forms a 1-element orbit

X if forms a 2-element orbit.

By the Isomorphism Theorem 1.23 extends to an automorphism ofg. The uniqueness in 1.23 implies that 2 = 1.

The main step is to prove that u0 = u0. Let B be the Killing formofg. For define the constant a by X = aX. Then

aa = aaB(X, X)

= B(aX, aX)

= B(X, X)

= B(X, X)

= 1

.

So if we prove, that

a = 1 for all +this also proves the result for all . Again we prove by inductionon the level of . For simple roots, which have level 1, this is trueby definition. Let + be of level n and inductively assume thata = 1 holds for all with level < n. Choose some positive roots and such that = + . Clearly and are of smaller level than

. Remember that we have chosen X in the construction of the splitreal form such that

X = N1,[X, X]

= N1,[X, X]

= N1,aa[X, X]

= N1,N,aaX+.

Thereforea = N

1,N,aa.

By induction hypothesis aa =

1 and Theorem 2.25 tells us that

N2, =12

q(1 +p)||2 = 12

q(1 +p)||2 = N2,.Hence N1,N, = 1. This proves a = 1 and the induction iscomplete.

Let us see that

(R(X X) +Ri(X + X)) R(X X) +Ri(X + X).

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Like in the proof of Theorem 2.49 we calculate for some z = x + iywith x, y R

x(X X) + yi(X + X) = zX zXReverting argumentation of 2.49 the desired result is equivalent to thefact that

(zX zX) = zaX zaXis of the form wX wX where z, w C. But this is clear sincewe know that a = 1 for all . Since carries roots to roots wehave

(

R(iH)) =

R(iH).

This shows that u0 = u0.Let k and p be the +1 and 1 eigenspaces of in g, so that g = kp.

We have

u0 = (u0k)

(u0

p).

Define k0 = u0 k and p0 = i(u0 p). Thenu0 = k0 ip0.

Since u0 is a vector space real form ofg so is

g0 = k0 p0.Since u0 = u0 and is an involution we have the bracket relations

[k0, k0] k0[k0, p0] p0[p0, p0]

k0.

Therefore g0 is closed under brackets and is a Lie algebra real form ofg. The involution

(X) =

+X for X k0X for X p0

Hence is a Cartan involution ofg0. maps u0 h to itself. and therefore

u0 h = (u0 k h) (u0 p h)= (k0 h) (ip0 h)= (k0

h)

i(p0

h).

The abelian subspace u0 h is a real form ofh and so is h0 = (k0 h) (p0 h). The subspace h0 is contained in g0 and is therefore a stableCartan subalgebra ofg0.

A root that is real on all ofh0 has the property that = .Since (+) = +, there are no such real roots. Hence h0 is maximallycompact.

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Let us verify that + results from a lexicographic ordering thattakes i(k0 h) before p0 h. Define the following sets

{i}li=1 the set of simple roots of + in 1-element orbits{i, i}mi=1 the set of simple roots of + in 2-element orbits

{i}l+2m

i=1 the set of all simple roots of +

in the following order

{1, . . . , l, l+1, l+2, . . . , l+2m1, l+2m} ={1, . . . , l, 1, 1, . . . , m, m}

Relative to the basis {i}l+2mi=1 define the dual basis {i}l+2mi=1 by i, j =ij. We shall write j or j or j in place of i to see the originof i more easily. We define a lexicographic ordering by using innerproducts with the ordered basis

1 , . . . , l , 1+ 1 , . . . , m+ m , 1

1 , . . . , m

m

which takes i(k0 h) before p0 h. Let be in + and decompose

=l

i=1

nii +mj=1

rjj +mj=1

sjj.

Then

, i = ni 0and

, j + j = rj + sj 0.If all these inner products were 0, then all coefficients of were 0. Thus has positive inner product with the first member of our ordered basisfor which the inner product is nonzero. The lexicographic orderingyields + as a positive system. Consequently (g0, h0,

+) is a triplediscribing a real Lie algebra with the Vogan diagram we started off.

In order to classify real semisimple Lie algebras we take a look attheir complexifications.

2.51. Theorem. Letg0 be a simple Lie algebra overR and letg beits complexification. The following two situations may occur:

(1) g0 is complex, which means that g0 = sR

for some comlpex s.Theng isC isomorphic to s s.(2) g0 is not complex. Theng is simple overC.

Proof. (1) Let J be multiplication by1 in g0. We want

to show that the R linear map L : g s s given byL(X+ iY) = (X+ J Y , X JY),

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where X, Y in g0, is an isomorphism. L respects brackets since

L([X+ iY,X + iY]) = L(([X, X] [Y, Y]) + i([Y, X] + [X, Y]))= ([X, X] + [J Y , J Y ] + [X,JY] + [J Y , X],

[X, X] + [J Y , J Y ] [J Y , X] [X,JY])= [(X+ J Y , X JY), (X + JY, X JY)]= [L(X+ iY), L(X + iY)]

and it is one-one. By dimensional argumentation L is an Risomorphism. Let s be the same real Lie algebra as g0 butmultiplication by

1 is defined as multiplication by i. Tosee that L is a C isomorphism ofg with s s we compute

L(i(X+ iY)) = L(Y + iX)= (Y + JX, Y JX)= (J(X+ JY), J(X JY)).

To complete the proof we show that s is C isomorphic to s.s has a compact real form u0. The conjugation of s with

respect to u0 is R linear and respects bracket and we have toshow, that : s s is a C isomorphism. Let U and V be inu0. Then

(J(U + JV)) = (V + JU)= V JU= J(U JV)= J(U + JV)

shows this isomorphism.(2) Let bar denote conjugation of g with respect to g0. If a is a

simple ideal in g then so is a. Hence a

a and a+a are ideals in

g invariant under conjugation and hence are complexificationsof ideals in g0. Thus they are 0 or all of g. Since a = 0 wehave a + a = g.

Ifa a = 0 then g = a a. Let : g0 g be the inclusionof g0 in g and : g a the projection of g to a. = is an R homomorphism of Lie algebras. If ker is nonzero,it must be g0 since g0 is simple. In this case g0 is containedin a. Since conjugation fixes g0 we get g0 a a = 0 whichis a contradiction. So ker = 0 and is one-one. Since aaswell as g0 are of dimension

12 dimR g, is onto and hence an

isomorphism. But this would mean that g0 is complex which

contradicts our initial assumption.We conclude that a a = g and hence a = g. Therefore g

is simple as asserted.

2.52. Proposition. Letg be a complex Lie algebra which is simpleoverC. ThengR is simple overR.

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Proof. Suppose that a is an ideal in gR. By Cartans Criterion forSemisimplicity gR is semisimple. Hence [a, gR] a = [a,

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