Maximal Subalgebras of Kac–Moody Algebras#

El Mamoun Souidi* and Mostafa Zaoui

Departement de Mathematiques et d’Informatique, Faculte des Sciences,Rabat, Morocco

ABSTRACT

We use the classification of finite order automorphisms by Kac to characterize allmaximal subalgebras, regular, semisimple, reductive or not of a simple complex

Lie algebra (up to conjugacy) that we can determine from its Dynkin diagram.Using Barnea et al. [Barnea, Y., Shalev, A., Zelmanov, E. I. (1998). Gradedsubalgebras of affine Kac–Moody algebras. Israel J. Math. 104:321–334] weextend our results to the case of affine Kac–Moody algebras. We also point out

some inaccuracies in the Dynkin paper [Dynkin, E. B. (1957a). Semisimple sub-algebras of semisimple Lie algebras. Amer. Math. Soc. Transl t. 6:111–244].

Key Words: Maximal subalgebras; Dynkin diagrams; Kac–Moody algebras.

AMS Subject Classification (2000): 17B40; 17B20; 17B67.

#Communicated by J. Alev.*Correspondence: El Mamoun Souidi, Departement de Mathematiques et d’Informatique,

Faculte des Sciences, B.P 1014, Rabat, Morocco; Fax: +(212) 37775471; E-mail: souidi@fsr.ac.ma.

COMMUNICATIONS IN ALGEBRA�

Vol. 32, No. 7, pp. 2573–2588, 2004

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DOI: 10.1081/AGB-120037400 0092-7872 (Print); 1532-4125 (Online)

Copyright # 2004 by Marcel Dekker, Inc. www.dekker.com

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INTRODUCTION

In the Lie theory, the classification problem for (maximal) subalgebras orsubgroups is a classical one. A number of problems in geometry and algebra leadto this one. In particular the classical problem of the classification of primitive trans-formations which was posed in the 19th century by Lie and solved by him in smalldimensions (Lie, 1930).

Nowadays, motivations for finding all (maximal) subalgebras (resp. subgroups)of simple Lie algebras (resp. groups) come from both mathematics, for example,systems of differential equations satisfying a superposition principle, and from physics,for example, symmetry breaking, complete sets of commuting operators in aquantum mechanical system, dynamical systems, etc.

This problem has been the focus of much research which produced somebeautiful results. We first cite the famous papers of Dynkin (1957a,b) where theclassification of semisimple subalgebras (subgroups) of complex semisimple Liealgebras (groups) has been achieved. In the same spirit see also Karpelevich (1951)and Mal’cev (1944=1955). Tits (1955) has reviewed these papers.

Recently, in a series of papers, Patera, Hussin, Winternitz, and Zassenhausexamined the maximal abelian subalgebras of Lie algebras see Hussin et al. (1992)and references within. Almost a paper for each classical Lie algebra.

The notion of maximality depends on the characteristic of the groud field. InBruce (1973), it is shown that a Lie algebra of type E6 over a field k of characteristic0 has a maximal subalgebra of type G2 if and only if

ffiffiffiffiffiffiffi�7p 2 k.

Whilst, Komrakov (1990b) (and the references within) gives without proofs aclassification of the maximal subalgebras of the simple finite dimensional real Liealgebras. The same author (Komrakov, 1990a) investigates reductive subalgebrasof semisimple real Lie algebra, also without any proofs. See also Marcel (1976) forthe case of maximal solvable subalgebras of the real classical Lie algebras where a listis given of all conjugacy classes of maximal solvable subalgebras of glðn;RÞ, suð2rÞ,suðp; qÞ, so�ð2rÞ, spð2r;RÞ, spð2p; 2qÞ and a representative of each class is given interms of a matrix algebra. Furthermore a large class of maximal solvable subalgebrasfor some Lie superalgebras is studied by Shchepochkina (1994). In the literature, thereis also many papers investigating different type of maximal subalgebras of p-algebras.

In this paper we try to answer the following question: Can we determine all themaximal subalgebras of a simple Lie algebra from its Dynkin diagram? This questionis asked since Borel and De Siebenthal (1949) have showed that regular reductivemaximal subalgebras of a simple Lie algebra can be determined directly from itscompleted Dynkin diagram. We use the classification of finite order automorphismsby Kac, to characterize all maximal subalgebras, regular or not, semisimple or notand reductive or not (up to conjugacy) of a simple complex Lie algebra that wecan determine from its Dynkin diagram.

The response to the above question is partial since we do not find all maximalLie subalgebras but a large part englobing those of Karpelevich (1951) and Boreland De Siebenthal (1949) (where proofs are of geometrical nature), and someS-subalgebras given by Dynkin (1957a,b). Our procedure based on Zaoui (1987) ismuch simpler, we avoid the use of table or the classification of subroot systems asused in Dynkin (1957a,b) and Tits (1955). Many explicit examples are given.

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To extend these results in an obvious way to affine Kac–Moody algebras we useBarnea et al. (1998) where it is shown that for a simple finite dimensional algebraover C, if g0 is a maximal subalgebra in g then g0 �C½t; t�1� is a maximal gradedone in g�C½t; t�1�.

This paper is divided as follows: the first section is to set notations and recallsome results. The second section is devoted to show a necessary condition for max-imality and then investigate the automorphisms s ¼ ðEj; kÞ of order 2 and s ¼ ðEj; 1Þof order m > 2 to deduce all maximal semisimple subalgebras we seek for. As anexample, the algebra D4 is detailed. In Sec. 3, the automorphisms s ¼ ðE0 þ Ej; 1Þof order 2 are investigated to deduce that maximal reductive subalgebras are thosewhich semisimple parts are maximal of corank 1. The last section is reserved tocharacterize nonreductive Lie subalgebras and their semisimple parts by studyingautomorphisms the s ¼ ðE0 þ Ej; 1Þ of order >2.

1. PRELIMINARIES

To make this paper somewhat self contained, we recall some results from Kac(1990) and fix some notations.

All Lie algebras considered here are complex. By Z, C and R we note the ringof integers, the fields of complex and real numbers respectively. We let S to be thelattice S ¼ ZþE0 � � � � � ZþE‘ where fE0; E1; . . . ; E‘g is the canonical basis of R‘þ1.

Let g be a simple Lie algebra and s an automorphism of g of order m

(i.e., sm ¼ 1). Set e ¼ e2pi=m. Then each eigenvalue of s has the form ej withj 2 Z=mZ :¼ Zm and since s is diagonalizable we have the Zm-gradation of g i.e.,

g ¼ g0 � g1 � � � � � gm�1 ð1Þ

with ½gi; gj� � giþj where gj is the eigenspace of s for the eigenvalue ej. The subspaceg0 is a subalgebra of g and the gi for i ¼ 1 � � �m� 1 are g0-modules.

Recall that the integer k is the order of the Dynkin diagram automorphism, itis 1 for the algebras A1, Cn, n � 2, Bn, n � 3, E7, E8, F4, G2; k ¼ 2 for An, n � 2,Dn, n � 4, E6 and k ¼ 3 for D4. We also use the labels ai’s attached to the Dynkindiagrams. For a simple Lie algebra XN we denote X

ðkÞN the corresponding Kac–

Moody algebra. And by E0;E1; . . . ;E‘ the generators of g as in Kac (1990,Chap 8.3). Let now s ¼ ðs0; . . . ; s‘Þ be a sequence of nonnegative relatively primeintegers and m ¼ k

P‘i¼0 aisi then the relations

sðEjÞ ¼ e2pisj=mEj; j ¼ 0; . . . ; ‘

define (uniquely) an mth order automorphism of g. It will be denoted s ¼ ðs; kÞ andcalled Kac automorphism. Moreover up to conjugation by an automorphism,s ¼ ðs; kÞ exhausts all mth order automorphisms of g. And s ¼ ðs; kÞ is inner ifand only if k ¼ 1.

Let GðkÞ be the Dynkin diagram of gðkÞ and i1; i2; . . . ; ip its vertices for whichsi1 ¼ si2 ¼ � � � ¼ sip ¼ 0, then the subalgebra g0 ¼ fx 2 g=sx ¼ xg is isomorphic toa direct sum G0 � Z where G0 is a semisimple subalgebra whose Dynkin diagram

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is the subdiagram of GðkÞ consisting of the vertices i1; i2; . . . ; ip and Z is theð‘� pÞ-dimensional center of g0.

If now j1; j2; . . . ; jq are the vertices of GðkÞ such that sj1 ¼ sj2 ¼ � � � ¼ sjq ¼ 1then the g0-module g1 is isomorphic to a direct sum of q irreducible g0-modules withhighest weight �aj1 ;�aj2 ; . . . ;�ajq where ai’s are the roots of g.

We call a Lie subalgebra g0 of g regular if it is stable by adh where h is aCartan subalgebra of g (see for example Bourbaki, 1975). A nonregular subalgebrais also called an S-subalgebra. A subalgebra is called parabolic if it contains a Borelsubalgebra. A subalgebra is said reductive if its adjoint representation is semisimpleor equivalently g ¼ G0 � Z where G0 is semisimple and Z is the center of g.

2. MAXIMAL SEMISIMPLE LIE SUBALGEBRAS

2.1. A Necessary Condition for Maximality

Lemma 2.1. Let g be a simple Lie algebra and s ¼ ðs; kÞ be a Kac automorphismof order m inducing the gradation (1). If m is not prime then g0 is not maximalsubalgebra of g.

Proof. Assume m ¼ p � q where p and q are integers such that p > 1 and q > 1.We have the following gradation induced by s:

g ¼ g0 � g1 � � � � � gm�1

Set t ¼ sp, it is an automorphism of g of order q > 1 so it induces the Zq-gradation:

g ¼ g00 � g01 � � � � � g0q�1

where g00 ¼ fx 2 g; t � x ¼ xg is a reductive subalgebra fixed pointwise by t. Assumethat g0 g00. Since q > 1, g01 ¼ fx 2 g; tx ¼ e0 � xg 6¼ 0, (e0q ¼ 1). On the other hand, ifx 2 g00 then s � x 2 g00 because tsx ¼ spsx ¼ sspx ¼ stx ¼ sx hence g00 is stable by swhich is an automorphism of order p > 1, so induces a Zp-gradation on g00:

g00 ¼ g0 � g1 � � � � � gp�1

where g0 ¼ fx 2 g00; s � x ¼ xg is a reductive subalgebra of g00, and g1 6¼ f0g sincep > 1 on ðg00Þ. Finally for every x 2 g0 we have s � x ¼ x, so x 2 g0 and we havethe following inclusions:

g0 g0 g0 � g1 g00 g00 � g01 g:

So g0 is not maximal. &

Remark 2.2. Let s ¼ ðEj; kÞ be a Kac automorphism of prime order m ¼ kPr

i¼0 aisiand Ej an element of the canonical basis of the lattice S then:

– If m ¼ 1 then we have necessarily k ¼ 1; sj ¼ aj ¼ 1 and si ¼ 0 for all i 6¼ j.In this case g ¼ g0 and s ¼ ðEj; 1Þ.

– If k ¼ 2 or 3 then sj ¼ aj ¼ 1 and si ¼ 0 for all i 6¼ j:

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An Example of Nonmaximal Subalgebra

Let g be an algebra of type F4 and s ¼ ðE3; 1Þ. The Dynkin diagram of Fð1Þ4 is

o1�������o2�������o3¼¼¼¼¼>o4�������o2

and a0 ¼ 1; a1 ¼ 2; a2 ¼ 3; a3 ¼ 4; a4 ¼ 2. Take s ¼ ðE3; 1Þ with

E3 ¼ ðs0; s1; s2; s3; s4Þ ¼ ð0; 0; 0; 1; 0Þ:

It is an automorphism of order m ¼ a0s0 þ a1s1 þ a2s2 þ a3s3 þ a4s4 ¼ 4it induces the gradation g ¼ g0 � g1 � g2 � g3 where g0 ffi A3 � A1 whose Dynkindiagram is

o�������o�������o o

It is obtained by removing the 4th vertex of label a3 ¼ 4. We have g2 6¼ 0 (other-wise g0 � g1 and g0 � g3 would be ideals of g) hence g0 is contained strictly in theproper subalgebra g0 � g2 so g0 is not maximal.

2.2. Automorphisms r= (ðj, k) of Order 2 (k = 1 or 2)

Proposition 2.3. If s ¼ ðEj; kÞ is a Kac automorphism of order m¼ 2 inducing theZ2-gradation g ¼ g0 � g1 then g0 is maximal in g.

Proof. If g0 is a proper subalgebra of g containing strictly g0 then g0=g0 would bea proper submodule of the irreducible g0-module g=g0 that is isomorphic to g1, acontradiction. &

Remark 2.4. (i) In the last proof we do not ask any more about the structure ofg0 but only to be a Lie subalgebra. So g0 is maximal among all Lie subalgebras(reductive or not) of g.

(ii) A subalgebra g0 of g which contains g0 and g1 or gm�1 is equal to g.

(iii) For s ¼ ðEj; kÞ, g0 is semisimple.

Example of Maximal Subalgebra

Let g be a Lie algebra of type C2 and E1, E2, F1, F2, H1;H2 its Chevalleygenerators and H1;H2 a basis of a Cartan subalgebra h and nþ, n� its Borelsubalgebras then

g ¼ n� � h� nþ

¼ C½F1; ½F1;F2�� �C½F1;F2� �CF1 �CF2 �CH1 �CH2

�C½E1; ½E1;E2�� �C½E1;E2� �CE1 �CE2

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If we set E0 ¼ ½F1; ½F1;F2�� and F0 ¼ ½E1; ½E1;E2�� then g is generated by E0, E1 andE2 since E0 is a lowest vector for the adjoint representation of g and g ¼ UðnþÞðE0Þwhere UðnþÞ is the universal enveloping algebra of nþ. The automorphism s ¼ ðE1; 1Þis of order 2 because a1 ¼ 2, and

sðE0Þ ¼ 1 � E0; sðE1Þ ¼ �1 � E1; sðE2Þ ¼ 1 �E2:

The decomposition of g induced by s is g ¼ g0 � g1 where g0 is the semisimplesubalgebra fixed by s hence it is generated by eigenvectors of s associated to theeigenvalue 1; namely E0;F0;E2;F2. Set H0 ¼ ½E0;F0� then:

g0 ¼ C½F1; ½F1;F2�� �CF2 � h�C½E1; ½E1;E2�� �CE2

¼ ðCF0 �CH0 �CE0Þ � ðCF2 �CH2 �CE2Þffi A1 � A1

g1 ¼ CE1 �C½E1;E2� �CF1 �C½F1;F2�¼ CE1 �C½E1;E2� �C½E0; ½E2;E1�� �C½E2; ½E0; ½E2;E1���

In the second expression F1 and ½F1;F2� are expressed as elements of Uðnþ0 ÞðE1Þwhere nþ0 is the upper Borel subalgebra of g0.

In Table 1 we give, for each simple Lie algebra, the maximal semisimple sub-algebras fixed pointwise by an automorphism of order m ¼ 2. (The isomorphismsA1 ffi B1 ffi C1; D2 ffi A1 � A1; D3 ffi A3; B0 ¼ f0g are understood).

2.3. Maximal Subalgebras of r= (ðj, 1) of Order m > 2

In this section we show that if s ¼ ðEj; 1Þ is of prime order m > 2, then thesubalgebra g0 is maximal in g. First, recall that g0 is regular and any subalgebra of

Table 1. Maximal g0 for two order s ¼ ðEj; kÞ.k ¼ 1 k ¼ 2

g g0 g g0

Br Dp � Br�p; 2 � p � r A2 A1

Cr Cp � Cr�p; 1 � p � r � 1 A2r Br

Dr Dp �Dr�p; 2 � p � r � 2 A2r�1 Cr ;Dr

E6 A1 � A5 Drþ1 Bp � Br�p; 0 � p � r

E7 A7 E6 C4;F4

A1 �D6

E8 A1 � E7

D8

G2 A1 � A1

F4 B4

A1 � A3

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g containing g0 is also regular. Furthermore any subalgebra g0 of g which contains g0has the following form:

g0 ¼ h��M

a2D0ga�

where, if D and D0 are root systems of g and g0, then D0 is a closed subset of Dcontaining D0. The automorphism s ¼ ðEj; 1Þ induces the gradation: g ¼ g0 � g1� � � � � gm�1 where g1 is an irreducible g0-module. About other g0-modulesgqðq > 1Þ we have

Lemma 2.5. Let s ¼ ðEj; 1Þ be a Kac automorphism of order m > 2, inducing thegradation g ¼ g0 � g1 � � � � � gm�1 then

(i) gq is irreducible and gq ¼ ½g1; gq�1� for all q such that 0 < q � m� 1.(ii) Every subalgebra containing strictly g0, contains also a g0-module gq,

ð0 < q � m� 1Þ.(iii) gm�q ¼ oðgqÞ where o is the Cartan involution of g. Furthermore every

subalgebra g0 which contains g0 and gm�1 is conjugated to g by o.(iv) ½g2; g2� 6¼ f0g for g ¼ E8 and s ¼ ðE4; 1Þ.

Proof. Let E0; . . . ;E‘ be a set of generators of g as in Sec. 1.

(i) We know g1 is irreducible, say with lowest weight vector Ej, and assume asinduction hypothesis that gq is irreducible for a q such that 1 < q < m� 1. Let thenE be its highest weight vector (i.e., for all i 2 f0; 1; . . . ; ‘g � fjg, E satisfies½Ei;E� ¼ 0Þ then ½Ej;E� 6¼ 0, else E would be a highest weight vector for the adjointrepresentation of g which is unique because of the irreducibility, then E 2 g0, but thiscannot happen by hypothesis. On the other hand, let nþ0 and n�0 be the Borel sub-algebras of g0, then:

for all x 2 Uðnþ0 Þ; x � ½Ej;E� ¼ ½xEj;E� ð2Þ

for all y 2 Uðn�0 Þ; y � ½Ej;E� ¼ ½Ej; yE� ð3Þ

Let E0 2 gqþ1; since E0;E1; . . . ;E‘ generates g, the vector E0 can be written asE0 ¼ adEin � � � adEi1ðE0Þ where Ej appears ðqþ 1Þ (modm) times. Then E0 can bewritten as x1 � ½Ej; x2ðE0Þ� where x1 2 Uðnþ0 Þ and x2ðE0Þ 2 gq. But E is a highestvector of gq, so there exists x0 2 Uðnþ0 Þ and y0 2 Uðnþ0 Þ such that x0x2ðE0Þ ¼ E

and y0ðEÞ ¼ x2ðE0Þ: Using (2) and (3) we deduce E0 ¼ x1ð½Ej; x2ðE0Þ�Þ ¼x1ð½Ej; y0ðEÞ�Þ ¼ x1y0ð½Ej;E�Þ. Thus every element in gqþ1 belongs to Uðg0Þ�½Ej;E�; hence gqþ1 is irreducible since it does not contain any invariant propersubspace by the action of g0. Furthermore, from (2) and (3) we get gqþ1 ¼ ½g1; gq�.

(ii) Let g0 be a subalgebra of g which contains g0 and a linear combinationof 2 vectors Ea and Eb associated to the distinct roots a and b such that Ea 2 gqand Eb 2 gp, ðp; q > 0Þ. The set faigi2f0;1;...;‘g�fjg is a basis of h�, then there

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exists an i 2 f0; 1; . . . ; ‘g � fjg such that ha;Hii 6¼ hb;Hii. We can assume withoutloosing generality ha;Hii < hb;Hii and consider the ai-string through a and b, thereexist nonnegative integers a; a0; b and b0 such that: ha;Hii ¼ a0 � a; hb;Hii ¼ b0 � b.Set Ea

i � Ea :¼ ½Ei; ½Ei; ½� � � ½Ei;Ea� � � ���� and so on. Then Eai � Ea 6¼ 0 and

Eaþ1i � Ea ¼ 0; Eb

i � Eb 6¼ 0 and Ebþ1i � Eb ¼ 0. Fa0

i � Ea 6¼ 0 and Fa0þ1i � Ea ¼ 0;

Fb0i � Eb 6¼ 0 and Fb0þ1

i � Eb ¼ 0. We have a0 � a < b0 � b: Consider the combinationc1Ea þ c2Eb in g0 with Ea 2 gq and Eb 2 gp. Since g0 contains g0, it contains all theelements Ep

i � ðc1Ea þ c2EbÞ and Fpi � ðc1Ea þ c2EbÞ where p is a nonnegative integer.

The different possible cases are:

– a < b, in this case Ebi ðc1Ea þ c2EbÞ ¼ c1E

bi Ea þ c2E

bi Eb ¼ c2E

bi Eb 2 gp with

Ebi Eb 6¼ 0 and Eb

i Eb 2 gp which is irreducible, so gp � g0.– a ¼ b, in this case we have a0 < b0 and Fb0

i � ðc1Ea þ c2EbÞ ¼ c1Fb0i Ea

þ c2Fb0i Eb ¼ c2F

b0i Eb 2 gp with Fb0

i Eb 6¼ 0 and Fb0i Eb 2 gp so gp � g0.

– a > b, in this case we have Eai ðc1Ea þ c2EbÞ ¼ c1E

ai Ea þ c2E

ai Eb ¼

c1Eai Ea 2 gq with Ea

i Ea 6¼ 0 and Eai Ea 2 gq but gq is irreducible, hence gq � g0.

If the linear combination is made by more than 2 vectors, we successivelyeliminate them as above. Hence in all cases a subalgebra g0 which contains properlyg0, contains also a g0-module gq for some 0 < q � m� 1.

(iii) Let e ¼ e2pi=m. Since sHj ¼ Hj and sEj ¼ eEj we have sHj ¼ s½Ej;Fj� ¼½sEj; sFj� ¼ e½Ej; sFj�, Fj is an eigenvector of s, it is clear that sFj ¼ e�1Fj, thenFj 2 gm�1. Furthermore ½Fi;Ej� ¼ 0 for all i 2 f0; . . . ; ‘g � fjg since Ej is a lowestvector of g1 then ½Ei;Fj� ¼ o½Fi;Ej� ¼ 0 for all i 6¼ j hence Fj is a highest vectorof gm�1 and gm�1 ¼ Uðn�0 Þ � Fj ¼ oðg1Þ. Using now (i) we get gm�q ¼ oðgqÞ. If g0 isa subalgebra of g containing g0 and gm�1, it is conjugated by the automorphism oto the algebra oðg0Þ but

oðg0Þ � oðg0 � gm�1Þ ¼ oðg0Þ � oðgm�1Þ ¼ g0 � g1

thus oðg0Þ ¼ g.

(iv) If g ¼ E8 and s ¼ ðE4; 1Þ, in this case s is of order 5 and

g ¼ g0 � g1 � g2 � g3 � g4:

Let a and b be the roots of E8 such that

a ¼ a1 þ a2 þ 2a3 þ 2a4 þ 2a5 þ a6 þ a7 þ a8

and

b ¼ a2 þ a3 þ 2a4 þ 2a5 þ a6 þ a8;

their sum is

aþ b ¼ a1 þ 2a2 þ 3a3 þ 4a4 þ 4a5 þ 2a6 þ a7 þ 2a8

is also a root. Let Ea, Eb 2 g2 the root spaces of a and b then ½Ea;Eb� ¼ Eaþb 6¼ 0,it follows that ½g2; g2� 6¼ f0g and Eaþb 2 g4. &

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Now, we can prove the following

Proposition 2.6. Let g be a simple Lie algebra and s ¼ ðEj; 1Þ a Kac automorphismof prime order m ¼ aj. Then s induces on g the gradation:

g ¼ g0 � g1 � � � � � gm�1

where g0 is a maximal subalgebra of g. Furthermore g0 is semisimple.

Proof. Let g0 be a subalgebra of g containing strictly g0, by (Lemma 2.5, ii) itcontains a g0-module gq for some q such that 1 � q � m� 1.

If m ¼ 3 we have q ¼ 1 or q ¼ 2 ¼ m� 1, in the two cases g0 ¼ g, and g0 ismaximal.

If m ¼ 5 in this case g ffi E8 and s ¼ ðE4; 1Þ. If g0 contains g2 (resp. g3), it contains½g2; g2� (resp. ½g3; g3� ¼ o½g2; g2�) which is nontrivial by (Lemma 2.5, iv). Thus thereexists a nonzero element of g4 ¼ gm�1 (resp. g1) which is in g0 and we have g4 � g0,(resp. g1 � g0) so g0 ¼ g and g0 is maximal. &

In the Table 2 we list simple Lie algebras and their maximal semisimple sub-algebras fixed pointwise by the (inner) automorphism of type s ¼ ðEj; 1Þ with orderm > 2.

Remark 2.7. (i) The subalgebra g0 is maximal among all Lie subalgebras of g.

(ii) Each maximal subalgebra g0 considered above is fixed by an automorphismof type ðEj; kÞ i.e., its Dynkin diagram is obtained from that of gðkÞ by removing onevertex; so g0 is semisimple subalgebra of g with the same rank as g.

Remark 2.8. In the Tables 1 and 2 we have listed for each simple Lie algebras allregular semisimple subalgebras (up to conjugacy). Comparing our results with (Dynkin,1957a, Table 12, p. 150) we become aware of the following inaccuracies: A3 � A1 isnot maximal subalgebra in F4 as it is shown in the Subsec. 2.1. A3 � A3 � A1 is notmaximal subalgebra in E7 and A3 �D5, A5 � A2 � A1 and A7 � A1 are not maximalsubalgebras in E8. Theses inaccuracies are because of Theorem 5.5, p. 148 in(Dynkin, 1957a) which does not hold in the reductive case.

The fact that the automorphism s is inner (or equivalently g0 is regular) is crucialfor the proof of Lemma 2.5. If the automorphism s were not inner, as it is for the

Table 2. Maximal semisimple g0 for s ¼ ðEj; 1Þ of order prime m > 2.

g g0 m g g0 m

E6 A2 � A2 � A2 3 E8 A4 � A4 5

E7 A2 � A5 3 A8 3E8 A2 � E6 3 F4 A2 � A2 3

G2 A2 3

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case g ¼ D4 and s ¼ ðE0; 3Þ or s ¼ ðE2; 3Þ, for example, Proposition 2.6) falls as wewill see below, that is, the subalgebra g0 is not maximal.

2.4. D4 and its Automorphisms of Order 2 and 3

Let g ¼ D4 with its Chevalley generators Ei, Hi, Fi, i ¼ 1; . . . ; 4. The algebra g

has automorphisms of order 2 and 3.Let t be the order two automorphism of D4 defined by tE1 ¼ E1;

tE2 ¼ E2; tE3 ¼ E4; tE4 ¼ E3 it fixes a simple subalgebra gt0 of type B3 having forgenerators e1 ¼ E1; e2 ¼ E2; e3 ¼ E3 þ E4;h1 ¼ H1;h2 ¼ H2; h3 ¼ H3þ H4; f1 ¼ F1;f2 ¼ F2; f3 ¼ F3 þ F4. t induces the following gradation

g ¼ gt0 � gt1 ð4Þ

gt1 is an irreducible gt0-module, of lowest weight vector et, having for weight the shortminimal root of gt0 ¼ B3. Explicitly gt1 has for basis the following vectors:Z ¼ H3 � H4 2 gt1 since t � Z ¼ �Z, and we have

½Z;E2� ¼ 0; ½Z;F2� ¼ 0; ½Z;E1� ¼ 0; ½Z;F1� ¼ 0; ½Z; e3� ¼ E3 � E4;

½Z; ½E2; e3�� ¼ ½E2;E3 � E4�; ½Z; ½E1; ½E2; e3��� ¼ ½E1; ½E2;E3 � E4��;½Z; f3� ¼ F3 � F4; ½Z; ½F2; f3�� ¼ ½F2;F3 � F4�;½Z; ½F1; ½F2; f3��� ¼ ½F1; ½F2;F3 � F4�� ¼ et:

This shows that gt0 is a maximal subalgebra of g. It is clear, from Dynkindiagram of D

ð2Þ4 (Fig. 1) that the subalgebra of D4 defined by the vertex 0, 1, 2 is also

a subalgebra of type B3 which is conjugated to gt0 by an inner automorphism of gt0that we can determine explicitly. Let now s be the automorphism of order 3 ofg ¼ D4 defined by s � E1 ¼ E3; s � E2 ¼ E2; s � E3 ¼ E4; s � E4 ¼ E1, it fixes a sub-algebra gs0 of type G2 having for generators: e1 ¼E1þE3þE4; h1 ¼H1þH3þH4;f1 ¼ F1 þ F3 þ F4; e2 ¼ E2;h2 ¼ H2; f2 ¼ F2. The gradation induced by s is

g ¼ gs0 � gs1 � gs2 ð5Þ

We know that gs1 is an irreducible gs0-module. An explicit construction of gs1 (and soof gs2) can be made as for t.

Set h3 ¼ H1 þ e2H3 þ eH4 and h4 ¼ H1 þ eH3 þ e2H4 we verify easily thatsh3 ¼ eh3; sh4 ¼ e2h4, Uðgs0Þ � h3 ¼ gs1, and Uðgs0Þ � h4 ¼ gs2. So we have the followingbasis for gs1 : h3; ½½e1; ½e1;e2��;h3�; ½½e1;e2�;h3�; ½e1;h3�; ½½f1; ½f1;f2��;h3�; ½½f1;f2�;h3�,and ½f1;h3�.

Figure 1. Dynkin diagram of Dð2Þ4 .

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We are essentially interested in showing that gs0 ffi G2 is not maximal in g ffi D4.By remark 2.4 (ii), we know that g is generated by gs0 and any nonzero vector of gs1(or of gs2). Hence any proper subalgebra which contains strictly gs0, cannot be gener-ated only by gs0 and elements of gs1 � gs2 which are not eigenvectors of s. Hence let Z3and Z4 the vectors defined by ðe� e2ÞZ3 ¼ h3 þ h4 and Z4 ¼ e2h3 þ eh4 thenZ3 ¼ H3 � H4 and Z4 ¼ �H1 þ 2H3 � H4 are linearly independent and generatetwo gs0-modules g1 :¼ Uðgs0Þ � Z3 and g2 :¼ Uðgs0Þ � Z4 which are isomorphic to gs1and gs2 as gs-modules. Hence

g ¼ gs0 � g1 � g2 ð6ÞExplicitly g1 has for basis:

Z3 ¼ H3 �H4; ½e1; Z3� ¼ 2ðE3 � E4Þ; ð½e2; Z3� ¼ 0Þ;½½e1; e2�; Z3� ¼ 2½E2;E3 � E4�; ½½e1; ½e1; e2��; Z3� ¼ �4½E1; ½E2;E3 � E4��;½f1; Z3� ¼ 2ðF3 � F4Þ; ð½f2;h� ¼ 0Þ; ½½f1; f2�; Z3� ¼ 2½F2;F3 � F4�;½½f1; ½f1; f2��; Z3� ¼ �4½F1; ½F2;F3 � F4��:

In the same way we can make explicit a basis for g2. Remark that g1 ¼ gt1,comparing gradations (5) and (6) of g then gs0 � g2 ¼ gt0. But g2 is an irreduciblemodule whose all weights are of multiplicity one, so gs0 is maximal in gt0: Summarizingthe above in:

Proposition 2.9. The algebra D4 has automorphisms of order two and three, whichfixe pointwise subalgebras of type B3 and G2 respectively and we have the followinginclusions of maximal subalgebras : G2B3D4.

3. MAXIMAL REDUCTIVE SUBALGEBRAS

In this section we study maximal reductive subalgebras. We will see that thiskind of subalgebras are those which semisimple parts are maximal of corank 1. Con-sider the triangular decomposition of a simple Lie algebra g ¼ n� � h� nþ withrespect to a Cartan subalgebra h. Let s ¼ ðE0 þ Ej; 1Þ be an automorphism of orderm ¼ 2 of g. We have necessary a0 ¼ aj ¼ 1, s0 ¼ sj ¼ 1 and si ¼ 0 for all i 6¼ j. Letg ¼ g0 � g1 be the Z2-gradation induced by s. The subalgebra g0 is reductive of rankN ¼ rankðgÞ and g0 ¼ G0 �C:H where G0 is a semisimple subalgebra of rank N � 1,whose Dynkin diagram is obtained by removing the vertices 0 and j of that of gð1Þ

and H is a vector in h defined (up to multiplicative scalar) by hai;Hi ¼ 0 fori 6¼ 0; j where the ai’s are the simple roots of g. Set

G0 ¼ n�0 � h0 � nþ0

the triangular decomposition of G0. Hence that of g0 is g0 ¼ n�0 � h� nþ0 . On theother hand g1 is a direct sum of two irreducible g0-modules say M0 and Mj

g1 ¼ M0 �Mj

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of lowest weight vectors E0 and Ej, respectively. Each element in M0 (resp. Mj) hasthe form

adEi1 � adEi2 � � � adEit � � � ðE0Þ; it 6¼ 0; j

respectively

adEi1 � adEi2 � � � adEit � � � ðEjÞ; it 6¼ 0; j

It is clear that Mj � nþ. Keeping the above notations we get

Lemma 3.1. The Kac automorphism s ¼ ðE0 þ Ej; 1Þ with aj ¼ 1 induces aZ2-gradation g ¼ g0 � g1 where g0 is reductive. If G0 is the semisimple partof g0 then g0 ¼ G0 �CH ¼ n�0 � h� nþ0 , where H 2 h is such that ½H;G0� ¼ 0.Furthermore nþ0 �Mj ¼ nþ.

Proof. Only the last assertion needs proof. Since Mj � nþ and nþ0 � nþ thennþ0 �Mj � nþ. Let now v be an element of the canonical basis of nþ ¼ L

a2Dþ ga;v belongs to some ga, with a 2 Dþ. Then v can be written as v ¼ adEi1 �adEi2 � � � adEit�1

ðEitÞ satisfying furthermore, for each h 2 h:

adh � ðvÞ ¼ ½h; v� ¼ ðai1 þ ai2 þ � � � þ ait�1þ aitÞðhÞ � v ¼ aðhÞv

For a ¼ PNi¼1 kiai 2 Dþ we have ki � ai for all i such that 1 � i � N , hence kj � 1

and aj appears at most once (since aj ¼ 1) in the expression: a ¼ ai1 þ ai2þ � � � þ ait�1

þ ait so Ej appears at most once in the expression

v ¼ adEi1 � adEi2 � � � adEit�1ðEitÞ:

We distinguish the following cases:

– If it0 6¼ 0; j for 1 � t0 � t then v 2 nþ0 :– If Eit ¼ Ej and it0 6¼ 0 for 1 � t0 � t then v 2 Mj:– If Eit0 ¼ Ej with 1 � t0 � t, we have the following two possibilities.

(i) v ¼ adEjðzÞ where z ¼ adEi1 � adEi2 � � � adEit�1ðEitÞ with it0 6¼ 0; j (which is

equivalent to the fact that z 2 nþ0 ); in this case

v ¼ ½Ej; z� ¼ �½z;Ej� ¼ �adzðEjÞ 2 Mj:

(ii) v ¼ adEi1 � adEi2 � � � adEit0adEjðzÞ where i1; . . . ; it0 6¼ 0; j and z 2 nþ0 . Wehave

adEit0 � adEjðzÞ ¼ ½Eit0 ; ½Ej; z�� ¼ �½Ej; ½z;Eit0 �� � ½z; ½Eit0 ;Ej��¼ ad½z;Eit0 �ðEjÞ � adz � adEit0 ðEjÞ

which is inMj. If we repeat this process sufficiently we get v 2 Mj. Hence each v 2 nþ

is in nþ0 or in Mj thus nþ0 �Mj ¼ nþ. &

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It remains to study the maximality of g0 in g. We will show later that it is notmaximal as a Lie subalgebra, however it is maximal among reductive subalgebrasof g.

Proposition 3.2. Let g be a simple Lie algebra not of type Dl, l � 4. The auto-morphism s ¼ ðE0 þ Ej; 1Þ with aj ¼ 1 induces a Z2-gradation g ¼ g0 � g1 where g0is a reductive subalgebra of maximal rank and is maximal among all reductive sub-algebras of g. Moreover if G0 is the semisimple part of g0, then G0 is a maximalsemisimple subalgebra among all the semisimple subalgebras of g of rankrankðgÞ � 1.

Proof. We keep the notations as above. s induces the gradation g ¼ g0 �M0 �Mj.Since nþ � g0 �Mj by lemma 3.1 and M0 \ ðg0 �MjÞ ¼ f0g then M0 \ nþ ¼ f0gand since h � g0 then M0 � n�. On the other hand g0 is reductive, so oðg0Þ ¼ g0 thusoðMjÞ ¼ M0 and n�0 �M0 ¼ n�. Now every reductive subalgebra g0 which containsproperly g0, contains a nonzero element of the irreducible g0-module M0

(resp. Mj), so it contains also M0 (resp. Mj) and oðM0Þ ¼ Mj (resp. oðMjÞ ¼ M0);then it contains g0 �M0 �Mj ¼ g � g0. So g0 is maximal among reductive sub-algebras of g.

For the maximality of G0, we have nþ0 �Mj ¼ nþ and n�0 �M0 ¼ n� withM0 ¼ oðMjÞ. Since g is not of type Dl, l � 4 then M0 and Mj are not isomorphicas G0-modules. Hence every subalgebra G0

0 which contains properly G0 containsG0 �Mj or G0 �M0, so it contains nþ or n�. If G0

0 is semisimple, it contains nþ

and n�, therefore ½nþ; n�� which contains h then G00 ¼ g and G0 is maximal among

semisimple subalgebras of g. &

This proposition is false for Dl, l � 4, in fact take Dl ¼ rð2lÞ, it has a Kacautomorphism of order two: s ¼ ðE0 þ E1; 1Þ whose fixed subalgebra is the directsum of rð2ðl� 1ÞÞ and a one-dimensional center. But rð2ðl�1ÞÞ rð2l�1Þ rð2lÞ(inclusions are strict) so the semisimple part G0 of g0 is not maximal among the semi-simple subalgebra of g.

In Table 3 we list all subalgebras G0 that are semisimple, regular, andmaximal among all semisimple subalgebras of g of rank rankðgÞ � 1.

In the remaining section we are interested in nonreductive maximal subalgebrasof g.

Table 3. Maximal regular semisimple G0 of rank rankðgÞ � 1.

g G0

Ar ðr � 2Þ Ar�p � Ap�1

Br ðr � 3Þ Br�1

Cr ðr � 2Þ Ar�1

Dr ðr � 4Þ Ar�1

E6 D5

E7 E6

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4. MAXIMAL NONREDUCTIVE SUBALGEBRAS

4.1. Automorphism r= (ð0 + ðj; 1) of Order 2

We first consider, as a particular case, automorphism s ¼ ðE0 þ Ej; 1Þ withaj ¼ 1. This automorphism induces a Z2-gradation on g ¼ g0 �M0 �Mj. ConsiderL ¼ g0 �Mj, then we have L ¼ n�0 � h� nþ0 �Mj ¼ n�0 � h� nþ. Furthermore L

is a Lie subalgebra, in fact: ½g0; g0� � g0 � g0 �Mj; ½g0;Mj� � Mj � g0 �Mj;½Mj;Mj� � nþ � g0 �Mj, that is L is closed for Lie brackets, hence it is a subalgebraof g. Moreover L contains h� nþ, so L is parabolic.

We now show that L is maximal in g. Let L0 be a subalgebra of g containingproperly L; there exists an eigenvector x corresponding to a weight of g0-moduleM0, and x 2 L0. M0 is irreducible, so it is contained in L0. But g ¼ g0 �M0 �Mj � L0 � g, thus L0 ¼ g. So L is a maximal subalgebra in g. Remark also thatg0 �M0 is a maximal subalgebra, conjugated to L by the Cartan involutiono : oðg0 �M0Þ ¼ oðg0Þ � oðM0Þ ¼ g0 �Mj. Then we have:

Proposition 4.1. Let s ¼ ðE0 þ Ej; 1Þ be an automorphism such that aj ¼ 1 inducinga Z2-gradation g ¼ g0 � g1 where g1 is a sum of two irreducible g0-modules M0

and Mj. Then g0 �Mj and g0 �M0 are two maximal parabolic subalgebras of g

conjugated by the Cartan involution.

Remark 4.2. The subalgebra g0, which is maximal among all reductive subalgebrasof g, is not maximal as Lie subalgebra of g, since it is properly contained in g0 �Mj

and g0 �M0, which are maximal subalgebras in g. The semisimple parts G0 of theseparabolic subalgebras are those listed in the Table 3.

4.2. Automorphisms r= (ð0 + ðj; 1) of Order >2

We go on investigating other nonreductive regular subalgebras of a simple Liealgebra g. Let sðE0 þ Ej; 1Þ be an inner automorphism of order m > 2, which induceson g the Zm-gradation g ¼ g0 � g1 � � � � � gm�1 with g1 ¼ M0 �Mj. We will be inter-ested in the subalgebra L generated by g0 and Mj. In this case L is not equal tog0 �Mj. We show:

Proposition 4.3. Let s ¼ ðE0 þ Ej; 1Þ be an automorphism of g of order m ¼ 1þ ajinducing the Zm-gradation g ¼ g0 � g1 � � � � � gm�1 where g1 is a direct sum of twoirreducible g0-modulesM0 andMj. Let Ej a lowest vector ofMj. Then the subalgebraL generated by g0 and Ej is a maximal (parabolic) subalgebra of g.

According to results recalled in Sec. 1, the subalgebra g0 is one dimensionalcenter.

Before giving the proof, we look at the following example of g ¼ C2. The Dynkindiagram of C

ð1Þ2 is

o1¼¼¼¼>o

2<¼¼¼¼o

1

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Let E1;E2;F1 and F2 the Chevalley generators of g and H1;H2 a basis of a Cartansubalgebra h. We have the following decomposition of g

g ¼ n� � h� nþ

¼ C½F1; ½F1;F2�� �C½F1;F2� �CF1 �CF2 �CH1 �CH2

�C½E1; ½E1;E2�� �C½E1;E2� �CE1 �CE2

Let F0 ¼ ½E1; ½E1;E2�� and E0 ¼ ½F1; ½F1;F2��. We consider the automorphisms ¼ ðE0 þ Ej; 1Þ of finite order m ¼ 1þ 2 ¼ 3. g has the Z3-gradation g ¼ g0 �g1 � g2 where g0 ¼ CF2 � h�CE2 and

g1 ¼ M0 �M1 ¼ fC½F1; ½F1;F2��g � fC½E1;E2� �CE1gg2 ¼ fC½E1; ½E1;E2��g � fC½F1;F2� �CF1g

Let L be the Lie algebra generated by g0 and M1

L ¼ n�0 � h� nþ ¼ fCF2g � h� fC½E1; ½E1;E2�� �C½E1;E2� �CE1 �CE2g

Let L0 be a subalgebra of g which contains properly L. Then it contains one of thethree elements: F1, ½F2;F1� or ½F1; ½F1;F2�� ¼ E0: If E0 2 L0 then L0 contains E0;E1

and E2 which generate g so L0 ¼ g. If F1 2 L0 then L0 contains F0;F1 and F2 whichgenerate g and here again we have L0 ¼ g. If ½F1;F2� 2 L0 then L0 containsF1 ¼ ½E2; ½F1;F2�� and then L0 ¼ g. Thus L is maximal in g.

Proof of Proposition 4.3. Recall that a positive root can always be writtena ¼ ai1 þ ai2 þ � � � þ ait , (ai’s are the simple roots) where all the partial sumsai1 þ ai2 þ � � � þ ait0 , ðt0 � tÞ are roots. Moreover if aj appears in the expression ofa ¼ ai1 þ ai2 þ � � � þ ait it is possible to rearrange the ait0 to have j ¼ it.

Let L be the subalgebra generated by g0 and Mj, or it is the same thing to say, byg0 and Ej. Let L

0 be a subalgebra that contains properly L. Then L0 contains a non-zero vector, say F , of n� which is not in g0. Assume F 2 g�a with a ¼ ai1 þ ai2þ � � � þ ait�1

þ aj and where all the partial sums ait0 þ � � � þ ait�1þ aj, ðt0 � tÞ are

roots of g then

adHðFÞ ¼ �ðai1 þ ai2 þ � � � þ ait�1þ ajÞðHÞ � F

adH � adEi1ðFÞ ¼ �ðai2 þ � � � þ ait�1þ ajÞðHÞ � F

Repeating this process with Ei2 ;Ei3 ; . . . ;Eit�1, we obtain

adEi2 � adEi3 � � � adEit�1ðFÞ 2 g�aj ¼ CFj

but Ei2 ;Ei3 ; . . . ;Eit�1are vectors in L0, it follows that Fj 2 L0. Hence L0 contains

F1;F2; . . . ;F‘; it then contains n� thus L0 ¼ g and L is maximal in g. &

Remark 4.4. The maximal subalgebras studied in this section cover all non-reductive maximal subalgebras of a simple finite dimensional Lie algebras.

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ACKNOWLEDGMENT

The first author is very grateful to Professors J. Alev and Mme M. P. Malliavinfor hospitality in their team at Paris VI University.

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Received October 2002Revised March 2004

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