Lie Algebras

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http://france.elsevier.com/direct/CRASS1/

Lie Algebras

Lie algebras generated by 3-forms

Rudolf Philippe Rohr

University of Geneva, Section of Mathematics, 2-4, rue du Lièvre, CH-1211 Geneva 4, Switzerland Received 4 November 2004; accepted 21 December 2005

Available online 3 February 2006 Presented by Michel Duflo

Abstract

LetU be a real vector space,B an inner product onU andT ∈ ∧U^∗ a 3-form. The 3-formT defines two natural maps, [·,·]U:∧²U→Uandσ:U→ ∧²U^∗∼=so(U, B)given by[x, y]U=2B(T (x, y,·))andσ (x)=T (x,·,·). We show that[·,·]U is a Lie bracket if and only ifg_T ≡Im(σ )is a Lie subalgebra of so(U, B). To cite this article: R.P. Rohr, C. R. Acad. Sci. Paris, Ser. I 342 (2006).

Résumé

Algèbres de Lie engendrées par des 3-formes. SoitUun espace vectoriel réel,Bun produit euclidien surUetT∈ ∧U^∗une 3-forme. La 3-formeT permet de définir deux applications,[·,·]U:∧²U→Uetσ:U→ ∧²U^∗∼=so(U, B)telles que[x, y]U= 2B(T (x, y,·))etσ (x)=T (x,·,·). On va démontrer que[·,·]U est un crochet de Lie si et seulement sig_T ≡Im(σ )est une sous-algèbre de Lie de so(U, B). Pour citer cet article : R.P. Rohr, C. R. Acad. Sci. Paris, Ser. I 342 (2006).

Version française abrégée

Soitgune algèbre de Lie quadratique et(·,·)sa forme bilinéaire symétrique invariante. On définit la 3-forme de CartanT ∈ ∧³g^∗parT (x, y, z)=(x,[y, z]), x, y, z∈g(voir [2]). Cette 3-forme définit une classe dans cohomologie deg. De plus sigest simple elle engendreH³(g). On peut changer de point de vue. SoitUun espace vectoriel réel (de dimension finie) muni d’une forme bilinéaire symétrique non dégénéréeB etT ∈ ∧³U^∗une 3-forme. On considère le crochet antisymétrique suivant,

[·,·]U:U∧U→U, x∧y→2B

T (x, y,·) ,

avecB:U^∗→Ul’isomorphisme induit parB. Si[·,·]U satisfait l’identité de Jacobi, alors :Udevient une algèbre de Lie avec[·,·]U comme crochet ;Bdevient une forme invariante surU;T est une 3-forme de Cartan. Considérons l’homomorphisme d’espaces vectoriels suivant,

E-mail address: [email protected] (R.P. Rohr).

doi:10.1016/j.crma.2006.01.006

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σ:U→ ∧²U^∗∼=so(U, B), x→σ (x)≡T (x,·,·).

On noteg_T ≡Im(σ )l’image de cette homomorphisme. Cette définition est inspirée de la Définition 3.1 de [1]. Le résultat principal de cette Note est le théorème suivant :

Théorème 0.1. Supposons queBest définie positive. Alors,[·,·]U est un crochet de Lie si et seulement sigT est une sous algèbre de Lie de so(U, B). De plus :

(i) σ est un homomorphisme d’algèbre de Lie, i.e.[σ (x), σ (y)] =σ ([x, y]U); (ii) Ker(σ )est un idéal abélien ;

(iii) T est une 3-forme de Cartan pourB(i.e.T (x, y, z)=B([x, y]U, z)).

La preuve de ce théorème dans le sens direct est un exercice facile (voir Remarque 1). Par contre, la preuve de la réciproque nécessite les étapes suivantes :

(i) On démontre que si deux éléments (σ (x)etσ (y)) degT commutent alors[x, y]U =0 (voir Lemme 2.1). De ce lemme on déduit :

(a) l’algèbre de LiegT est semi-simple (voir Lemme 2.2) ;

(b) sa décomposition en somme directe d’algèbres de Lie simples est donnée parg_T =

ig_T_i avecT =

iT_i oùT_i∈ ∧³U_i^∗etU=

iU_i⊕Ker(σ )est la décomposition orthogonale correspondante (voir Lemme 2.4).

(ii) On montre que σ (x)·y ≡ [σ (x), y] = [x, y]U définit une structure g_T-module sur U (voir Section 2). Puis on démontre queU est homomorphe au module adjoint de g_T et queσ :U→g_T est un homomorphisme de g_T-modules (voir Section 3).

1. Introduction

Letgbe a quadratic Lie algebra, with(·,·):g×g→R its invariant bilinear form. We can define a 3-formT ∈ ∧³g^∗ byT (x, y, z)=([x, y], z), x, y, z∈g. This 3-form, called a Cartan 3-form (see [2]), defines a class in the cohomology ofg. In particular, forgsimple,T generatesH³(g). We can change the point of view. LetU be a real vector space (of finite dimension), equipped with a nondegenerate symmetric bilinear formB, and letT ∈ ∧³U^∗be a 3-form. We define a skew-symmetric bracket

[·,·]U:U∧U→U, x∧y→2B

T (x, y,·)

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whereB:U^∗→Uis the isomorphism induced byB. If[·,·]Uis a Lie bracket (i.e.[·,·]U satisfies the Jacobi identity), then the bilinear formBis invariant with respect to the adjoint action, i.e.∀x, y, z∈U, B([x, y]U, z)+B(y,[x, z]U)= 0 andT is the Cartan 3-form. Furthermore, we define the vector space homomorphism

σ:U→ ∧²U^∗∼=so(U, B),

x→σ (x)≡T (x,·,·). (2)

We denote its image byg_T ≡Im(σ )⊆so(U, B). This definition is inspired by Definition 3.1 of [1]. The main result of this Note is the following theorem:

Theorem 1.1. Assume thatBis positive definite. Then,[·,·]U is a Lie bracket if and only ifg_T is a Lie subalgebra of so(U, B). Moreover, in this case we have:

(i) σ is a Lie algebra homomorphism, i.e.[σ (x), σ (y)] =σ ([x, y]U), (ii) Ker(σ )is an Abelian ideal,

(iii) T is the Cartan 3-form forB (i.e.T (x, y, z)=B([x, y]U, z)).

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Remark 1. Assume that[·,·]Uis a Lie bracket. Then, proving thatgT is a Lie subalgebra of so(U, B)is an elementary exercise: let Cl(U, B)be the Clifford algebra and denote[σ (x), y] =σ (x)y−yσ (x)the commutator in Cl(U, B) (here we use the Chevalley isomorphism Cl(U, B)∼= ∧U^∗, see Theorem II.1.6, p. 41 of [3]). We have[x, y]U = 2BT (x, y,·)= [σ (x), y], which impliesσ (x)∼=ad_x. Moreover, ad_[_x,y_]_U = [ad_x,ad_y]implies thatg_T ⊆so(U, B)is a Lie subalgebra andσ ([x, y]U)= [σ (x), σ (y)], i.e.σis a Lie algebra homomorphism.

The above argument works for any signature of the bilinear form B. Our proof of the reciprocal, presented in Sections 2 and 3, depends on the positivity ofB. We conjecture that the theorem above hold true if the bilinear form is only nondegenerate.

2. Properties ofgT

In this section we assume that g_T is a Lie subalgebra of so(U, B)and thatB is positive definite. Below we list some useful properties of the mapσ and of the Lie subalgebrag_T.

Lemma 2.1. Letx, y∈Usuch that[σ (x), σ (y)] =0. Then,[x, y]U=0.

Proof. Let {e_i}i=1..n by an orthonormal basis of U. By Theorem 16 of [5] (page 293), we have ∀x, y ∈ U, [σ (x), σ (y)] =2_n

i=1(ι(e_i)ι(x)T )∧(ι(e_i)ι(y)T )(hereι is the left contraction operator). If[σ (x), σ (y)] =0, we have that_n

i=1(ι(e_i)ι(x)T )∧(ι(e_i)ι(y)T )=0. In particular, we obtain 0=ι(x)ι(y)

n i=1

ι(e_i)ι(x)T

∧

ι(e_i)ι(y)T

= − n

i=1

T (y, x, e_i)2

,

which implies[x, y]U=0. 2

Remark 2. From the above lemma we deduce that the Lie algebra g_T is either trivial or nonabelian. Indeed, if [σ (x), σ (y)] =0 for allx, y∈U, thenT (x, y,·)=0 andσ=0.

Lemma 2.2. The Lie algebrag_T is semisimple.

Proof. gT is a reductive Lie algebra, since it is a Lie subalgebra of so(U, B)andB is positive definite. Then, we havegT =Z(g_T)⊕ [gT,gT]. Ifσ (x)∈Z(g_T)(the center ofgT), Lemma 2.1 impliesι(x)ι(y)T =0,∀y ∈U and σ (x)=0. 2

We are interested in the decomposition of gT into a direct sum of simple Lie algebras. For that, we need the following definition.

Definition 2.3. The 3-formT is called reducible if there exists a nontrivial orthogonal decompositionU=U₁⊕U₂⊕ Ker(σ ), and nontrivial elementsT_i∈ ∧³U_i^∗,i=1,2, such thatT =T₁+T₂. Otherwise we say thatT is irreducible.

Lemma 2.4. The Lie algebrag_T is simple if and only if the 3-formT is irreducible. Moreover, the decomposition of g_T into simple Lie subalgebras is given by

gT =

i

gT_i,

where T_i ∈ ∧³U_i^∗ are the irreducible components of T =

iT_i, andU =Ker(σ )⊕

iU_i is the corresponding orthogonal decomposition.

Proof. The proof in the⇒)direction is obvious. We will give the proof in the⇐)direction. Assume thatg_T decom- poses into a direct sum of Lie algebrasg_T =g₁⊕g₂. This decomposition defines a decomposition ofUinto a direct sum of vector spacesU=U₁⊕U₂⊕Ker(σ ), whereU_i=σ⁻¹(g_i)∩Ker(σ )^⊥,i=1,2. This decomposition together

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with Lemma 2.1 implies that∀x∈U₁and∀y∈U₂,T (x, y,·)=0. Then,T =T₁+T₂, withT_i ≡T|_∧³U_i^∗,i=1,2, and we haveg_i=g_T_i,i=1,2. Moreover,U₁⊥U₂. Indeed, we have thatB(B(σ (x)), y∧z)=B(x,[y, z]U), which implies(Ker(Bσ ))^⊥=Im([·,·]U). Then, for eachx∈U1there existswx∈ ∧²U1such thatB(ι(wx)T )=x (hereι is the left contraction operator), and we have∀x∈U₁and∀y∈U₂thatι(y)ι(w_x)T=B(x, y)=0, i.e.U₁⊥U₂. 2

The vector spaceU is ag_T-module, withg_T-action given byσ (x)·y≡ [σ (x), y], where the bracket is the com- mutator in the Clifford algebra Cl(U, B)and we identify Cl(U, B)and∧U^∗with the Chevalley isomorphism.

Lemma 2.5. Theg_T-moduleUhas the following properties:

(i) σ (x)·y= [x, y]U,∀x, y∈U, (ii) Uis a faithful module,

(iii) [σ (x), σ (y)] =0⇒σ (x)·y=0, (iv) σ (x)·y= −σ (y)·x,∀x, y∈U,

(v) Uis irreducible if and only ifg_T is simple.

Proof. The point (i) follows from the fact that in the Clifford algebra Cl(U, B)we have[x,•] =2ι(x)•(see Theo- rem 5, page 284 of [5]). The points (ii) and (iv) follow from the point (i), the point (iii) follows from the point (i) and Lemma 2.1 and the point (v) follows from the point (i) and Lemma 2.4. 2

3. The homomorphism theorem

In this section, we prove a theorem that allows us to complete the proof of Theorem 1.1.

Theorem 3.1. Letgbe a simple Lie algebra(his its Cartan subalgebra),U a faithful g-module, andσ:U→ga vector space isomorphism. Assume that the following properties hold true:

P1 ∀v, w∈U,σ (v)·w= −σ (w)·v, P2 ∀v, w∈σ⁻¹(h),σ (v)·w=0.

Thenσ is an isomorphism ofg-modules, i.e.[σ (x), σ (y)] =σ (σ (x)·y), ∀x, y∈U, and thenx∧y→σ (x)·yis a Lie bracket onU.

Proof. First we complexifygandU. Letg^C=h^C⊕

α∈Rg_αbe the root space decomposition,Rbe the set of roots ofg^Cwith respect toh^C. This decomposition induces a decomposition ofU^Cbyσ, i.e.U^C=U⁰⊕

α∈RU^α, where U⁰=σ⁻¹(h^C)andU^α=σ⁻¹(g_α).

The structure of theg^C-moduleU^Cinduces another decomposition,U^C=

λ∈PUλ, wherePis the set of weights ofUandUλ⊆Uis the subspace of weightλ. The subspace of weight zero is denoted byU0≡Uλ=0.

(i) The property P2 impliesU⁰⊆U₀.

(ii) We prove that every root ofg^Cis a weight ofU, i.e.R⊆P. Leth∈U⁰andv^α∈U^α. Then, property P1, the point (i) above and Lemma 20.1, page 107 of [4] imply thatσ (h)·v^α∈U_α. We shall show that for every rootα there exists an elementh_α∈U⁰such asσ (h_α)·v^α=0. Indeed, if it is not the case, then there isβ=αsuch that for somev^β∈U^β,σ (v^β)·v^α=0 (because the moduleUis faithful). We have then∀h∈U⁰,

σ (h)·σ v^β

·v^α

=β(h)σv^β·v^α

= −σ (h)·σ v^α

·v^β= −

σ (h), σ v^α

·v^β−σ v^α

·

σ (h)·v^β

=α(h) σ v^β

·v^α

of weightβ

+σ v^α

· σ

v^β

·h

of weightα+β

.

This is a contradiction. Indeed if the last term vanishes, thenα=β, and if it is not the case then a nonvanishing eigenvector of weightα+βis proportional to an eigenvector of weightβ.

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(iii) The point (i) says thatU⁰⊆U₀and the point (ii) says thatR⊆P. Then, for dimensional reasons,U⁰=U₀and R=P. This implies that theg-moduleUis isomorphic to the adjoint module, i.e. there exists an isomorphism ofg-modulesΦ:g→U such thatΦ([σ (x), σ (y)])=σ (x)·Φ(σ (y)),Φ(σ (U⁰))=U₀andΦ(σ (U^α))=U_α. Moreover, the skew-symmetric mapx∧y→σ (x)·Φ(σ (y))defines a Lie bracket onU. In the end of the proof we will show thatΦ=λσ⁻¹, whereλ∈C^∗.

(iv) We will prove thatU_α=U^α for all rootsα. Letαbe a root. Then,∀h∈U⁰we haveα(h)=0 iffσ (h)·v^α=0.

Indeed, from the point (ii) we know that there existh_α∈U⁰such thatσ (h_α)·v^α=0. We have∀h∈U⁰that α(h)σ (h_α)·v^α=σ (h)·σ (h_α)·v^α=α(h_α)σ (h)·v^α.

We know that there exist ah∈U⁰such thatα(h)=0. This implies thatα(h_α)=0. Ifα(h)=0 thenσ (h)·v^α= 0, and ifσ (h)·v^α=0 thenα(h)=0. Moreover, from these equations we deduce that there exist ax_α∈U_αsuch that for allh∈U⁰we have

σ (h)·v^α=α(h)xα.

From this identity and becauseσ (h)·xα=α(h)xα, we deduce that there existkα∈U⁰such thatv^α=xα+kα. We shall prove thatk_α=0. Indeed, for all rootsαandβwe haveσ (v^α)·v^β= −σ (v^β)·v^α, which implies

σ v^α

·x_β+σ v^β

·x_α

of weightα+β

=α(k_β)x_α

of weightα

+β(k_α)x_β

of weightβ

.

Thenβ(k_α)=0 for every rootβ, which impliesk_α=0, i.e.v^α=x_α. ThenU_α=U^α, and for allh∈U₀,v^α∈U^α we have[σ (h), σ (v^α)] =σ (σ (h)·v^α).

(v) Letv^α∈U^α,v^β∈U^βandv^α⁺^β∈U^α⁺^β such that[σ (v^α), σ (v^α)] =σ (v^α⁺^β). We shall show thatσ (v^α)·v^β= v^α⁺^β. Indeed, for allh∈U₀, we have

σ v^α⁺^β

·h

−α+β(h)v^α+β

=σ v^α

·σ v^β

·h

−β(h)σv^α·v^β

−σ v^β

·σ v^α

·h

−α(h)σv^α·v^β

.

Thenσ (v^α)·v^β=v^α⁺^β, i.e.[σ (v^α), σ (v^α)] =σ (σ (v^α)·v^β).

(vi) From the point (iii) we know thatU₀=U⁰, then∀h, h∈U₀,[σ (h), σ (h)] =σ (σ (h)·h). From the point (iv) we have∀α∈R,∀v^α∈U^α and∀h∈U₀that [σ (h), σ (v_α)] =σ (σ (h)·v_α). And from the point (v) we have

∀α, β∈R,∀v^α∈U^αand∀v^β∈U^β that[σ (v^α), σ (v^α)] =σ (σ (v^α)·v^β). We can then conclude∀x, y∈Uthat [σ (x), σ (y)] =σ (σ (x)·y), i.e.Φ=λσ⁻¹,λ∈C^∗. 2

Now we can finalize the prove of Theorem 1.1. First we decompose the Lie algebrag_T into a direct sum of simple Lie algebras,g_T =

ig_T_i, whereT_i∈ ∧³U_i^∗are the irreducible components ofT (see Lemma 2.4). Second we apply the Theorem 3.1 to eachgT_i. Finally we conclude that[·,·]U is a Lie bracket,σ is a Lie algebra homomorphism and with Lemma 2.1 that Ker(σ )is an Abelian ideal.

Acknowledgements

I am very grateful to A. Alekseev, J. Greenstein, E. Meinrenken and E. Petracci for their comments and very useful suggestions. This work was supported in part by the Swiss National Science Foundation.

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