Solvable Lie Algebras

2.5 Semisimple Lie Algebras

2.5.1 Solvable Lie Algebras

We begin with a Lie-algebraic condition for nilpotence of a linear transformation.

Lemma 2.5.1.Let V be a finite-dimensional complex vector space and let A∈ End(V). Suppose there exist Xi,Yi ∈End(V) such that A=∑^ki=1[Xi,Yi] and [A,X_i] =0 for all i. Then A is nilpotent.

Proof. LetΣbe the spectrum ofA, and let{P_λ}λ∈Σbe the resolution of the identity forA(see Lemma B.1.1). ThenP_λX_i=X_iP_λ=P_λX_iP_λ for alli, so

P_λ[Xi,Yi]P_λ=P_λX_iP_λY_iP_λ−P_λY_iP_λX_iP_λ= [P_λX_iP_λ,P_λY_iP_λ].

Hence tr(P_λ[Xi,Yi]P_λ) =0 for alli, so we obtain tr(P_λA) =0 for allλ∈Σ. However, tr(P_λA) =λdimV_λ, where

V_λ={v∈V : (A−λ)^kv=0 for somek}.

It follows thatV_λ =0 for allλ6=0, so thatAis nilpotent. ut Definition 2.5.2.A finite-dimensional representation (π,V)of a Lie algebra gis completely reducibleif everyg-invariant subspaceW ⊂Vhas ag-invariant comple- mentary subspaceU. ThusW∩U={0}andV=W⊕U.

Theorem 2.5.3.Let V be a finite-dimensional complex vector space. Supposegis a Lie subalgebra ofEnd(V)such that V is completely reducible as a representation of g. Letz={X∈g:[X,Y] =0 for all Y∈g}be the center ofg. Then

1. every A∈zis a semisimple linear transformation;

2.[g,g]∩z=0;

3.g/zhas no nonzero abelian ideal.

Proof. Complete reducibility implies thatV=^L_iV_i, where eachV_iis invariant and irreducible under the action ofg. IfZ∈zthen the restriction ofZtoV_icommutes with the action ofg, hence is a scalar by Schur’s lemma (Lemma 4.1.4). This proves (1). Then (2) follows from (1) and Lemma 2.5.1.

2.5 Semisimple Lie Algebras 109

To prove (3), leta⊂g/zbe an abelian ideal. Thena=h/z, wherehis an ideal in gsuch that[h,h]⊂z. But by (2) this implies that[h,h] =0, sohis an abelian ideal ing. LetBbe the associative subalgebra of End(V)generated by[h,g]. By Lemma 2.5.1 we know that the elements of[h,g]are nilpotent endomorphisms ofV. Since [h,g]⊂his abelian, it follows that the elements ofBare nilpotent endomorphisms.

If we can prove thatB=0, thenh⊂zand hencea=0, establishing (3).

We now turn to the proof thatB=0. LetAbe the associative subalgebra of End(V)generated byg. We claim that

AB⊂BA+B. (2.38)

Indeed, for X,Y ∈gandZ∈hwe have [X,[Y,Z]]∈[g,h] by the Jacobi identity, sincehis an ideal. Hence

X[Y,Z] = [Y,Z]X+ [X,[Y,Z]]∈BA+B. (2.39) Letb∈Band suppose thatX b∈BA+B. Then by (2.39) we have

X[Y,Z]b= [Y,Z]X b+ [X,[Y,Z]]b∈[Y,Z]BA+B⊂BA+B.

Now (2.38) follows from this last relation by induction on the degree (in terms of the generators fromgand[h,g]) of the elements inAandB.

We next show that

(AB)^k⊂B^kA+B^k (2.40)

for every positive integerk. This is true fork=1 by (2.38). Assuming that it holds fork, we use (2.38) to get the inclusions

(AB)^k+1= (AB)^k(AB)⊂(B^kA+B^k)(AB)⊂B^kAB

⊂B^k(BA+B)⊂B^k+1A+B^k+1. Hence (2.40) holds for allk.

We now complete the proof as follows. SinceB^k=0 for ksufficiently large, the same is true for(AB)^k by (2.40). Suppose(AB)^k+1=0 for some k≥1. Set C= (AB)^k. ThenC²=0. SetW =CV. Since AC⊂C, the subspace W is A- invariant. Hence by complete reducibility ofV relative to the action ofg, there is an A-invariant complementary subspaceUsuch thatV =W⊕U. NowCW=C²V=0 andCU⊂CV=W. ButCU⊂Ualso, soCU⊂U∩W={0}. HenceCV=0. Thus C=0. It follows (by downward induction onk) thatAB=0. SinceI∈A, we con-

clude thatB=0. ut

For a Lie algebra g we define thederived algebra D(g) = [g,g]and we set D^k+1(g) =D(D^k(g))fork=1,2, . . .. One shows by induction onkthatD^k(g)is invariant under all derivations ofg. In particular,D^k(g)is an ideal ingfor eachk, andD^k(g)/cD^k+1(g)is abelian.

Definition 2.5.4.gissolvableif there exists an integerk≥1 such thatD^kg=0.

It is clear from the definition that a Lie subalgebra of a solvable Lie algebra is also solvable. Also, ifπ:g //his a surjective Lie algebra homomorphism, then

π(D^k(g)) =D^k(h).

Hence the solvability ofgimplies the solvability ofh. Furthermore, ifgis a nonzero solvable Lie algebra and we chooseksuch thatD^k(g)6=0 andD^k+1(g) =0, then D^k(g)is an abelian ideal ingthat is invariant under all derivations ofg.

Remark 2.5.5.The archetypical example of a solvable Lie algebra is then×nupper- triangular matricesb_n. Indeed, we haveD(bn) =n⁺_n, the Lie algebra ofn×nupper- triangular matrices with zeros on the main diagonal. Ifn⁺_n,r is the Lie subalgebra of n⁺_n consisting of matricesX= [x_{i j}]such thatx_{i j}=0 for j−i≤r−1, thenn⁺_n =n⁺_n,1 and[n⁺_n,n⁺_n,r]⊂n⁺_n,r+1forr=1,2, . . .. HenceD^k(bn)⊂n⁺_n,k, and soD^k(bn) =0 for k>n.

Corollary 2.5.6.Supposeg⊂End(V)is a solvable Lie algebra and that V is com- pletely reducible as a g-module. Thengis abelian. In particular, if V is an irre- ducibleg-module, thendimV =1.

Proof. Letz be the center of g. If z6=g, then g/zwould be a nonzero solvable Lie algebra and hence would contain a nonzero abelian ideal. But this would con- tradict part (3) of Theorem 2.5.3, so we must havez=g. Given thatgis abelian andV is completely reducible, we can find a basis forV consisting of simultaneous eigenvectors for all the transformationsX∈g; thusV is the direct sum of invariant one-dimensional subspaces. This implies the last statement of the corollary. ut We can now obtain Cartan’s trace-form criterion for solvability of a Lie algebra.

Theorem 2.5.7.Let V be a finite-dimensional complex vector space. Letg⊂End(V) be a Lie subalgebra such thattr(XY) =0for all X,Y ∈g. Thengis solvable.

Proof. We use induction on dimg. A one-dimensional Lie algebra is solvable. Also, if[g,g]is solvable, then so isg, sinceD^k+1(g) =D^k([g,g]). Thus by induction we need to consider only the caseg= [g,g].

Take any maximal proper Lie subalgebrah⊂g. Thenhis solvable, by induction.

Hence the natural representation ofhong/hhas a one-dimensional invariant sub- space, by Corollary 2.5.6. This means that there exist 06=Y ∈gandµ∈h^∗such that

[X,Y]≡µ(X)Y (modh)

for allX∈h. But this commutation relation implies thatCY+his a Lie subalgebra of g. Since h was chosen as a maximal subalgebra, we must haveCY+h=g.

Furthermore,µ6=0 because we are assumingg= [g,g].

Given the structure ofgas above, we next determine the structure of an arbitrary irreducibleg-module(π,W). By Corollary 2.5.6 again, there existw₀∈W andσ∈ h^∗such that

π(X)w₀=σ(X)w₀ for allX∈h.

2.5 Semisimple Lie Algebras 111

Setw_k=π(Y)^kw₀andW_k=Cw_k+···+Cw₀. We claim that forX∈h,

π(X)w_k≡(σ(X) +kµ(X))w_k (modW_k−1) (2.41) (whereW₋1={0}). Indeed, this is true fork=0 by definition. If it holds forkthen π(h)W_k⊂W_kand

π(X)wk+1=π(X)π(Y)wk=π(Y)π(X)wk+π([X,Y])wk

≡(σ(X) + (k+1)µ(X))w_k+1 (modW_k).

Thus (2.41) holds for allk. Letmbe the smallest integer such thatWm=Wm+1. Then Wmis invariant underg, and henceWm=W by irreducibility. Thus dimW =m+1 and

tr(π(X)) =

∑

^m

k=0

σ(X) +kµ(X) = (m+1)

σ(X) +m 2µ(X) for allX∈h. However,g= [g,g], so tr(π(X)) =0. Thus

σ(X) =−m

2µ(X) for allX∈h. From (2.41) again we get

tr(π(X)²) =

∑

^m

k=0

k−m

2 2

µ(X)² for allX∈h. (2.42) We finally apply these results to the given representation ofgonV. Take a com- position series{0}=V₀⊂V₁⊂ ··· ⊂V_r=V, where each subspaceV_jis invariant undergandW_i=V_i/V_i−1is an irreducibleg-module. Write dimW_i=m_i+1. Then (2.42) implies that

tr_V(X²) =µ(X)²

∑

^r

i=1 m_i k=0

∑

k−1

2m_i2

for allX∈h. But by assumption, tr_V(X²) =0 and there existsX∈hwithµ(X)6=0.

This forcesm_i=0 fori=1, . . . ,r. Hence dimW_i=1 for eachi. Sinceg= [g,g], this implies thatgV_i⊂V_i−1. If we take a basis forV consisting of a nonzero vector from eachW_i, then the matrices forgrelative to this basis are strictly upper triangular.

Hencegis solvable, by Remark 2.5.5. ut

Recall that a finite-dimensional Lie algebra issimpleif it is not abelian and has no proper ideals.

Corollary 2.5.8.Letgbe a Lie subalgebra ofEnd(V)that has no nonzero abelian ideals. Then the bilinear formtr(XY)ongis nondegenerate, andg=g₁⊕ ··· ⊕g_r (Lie algebra direct sum), where eachg_iis a simple Lie algebra.

Proof. Letr={X∈g: tr(XY) =0 for allY∈g}be the radical of the trace form.

Thenris an ideal ing, and by Cartan’s criterionris a solvable Lie algebra. Suppose

r6=0. Thenrcontains a nonzero abelian idealathat is invariant under all derivations ofr. Henceais an abelian ideal ing, which is a contradiction. Thus the trace form is nondegenerate.

To prove the second assertion, letg₁⊂gbe an irreducible subspace for the adjoint representation ofgand define

g^⊥₁ ={X∈g: tr(XY) =0 for allY ∈g₁}.

Then g^⊥₁ is an ideal in g, and g₁∩g^⊥₁ is solvable by Cartan’s criterion. Hence g₁∩g^⊥₁ =0 by the same argument as before. Thus[g1,g^⊥₁] =0, so we have the decomposition

g=g₁⊕g^⊥₁ (direct sum of Lie algebras) .

In particular,g₁is irreducible as an adg₁-module. It cannot be abelian, so it is a

simple Lie algebra. Now use induction on dimg. ut

Corollary 2.5.9.Let V be a finite-dimensional complex vector space. Supposegis a Lie subalgebra ofEnd(V)such that V is completely reducible as a representation ofg. Letz={X∈g:[X,Y] =0 for all Y∈g}be the center ofg. Then the derived Lie algebra[g,g]is semisimple, andg= [g,g]⊕z.

Proof. Theorem 2.5.3 implies thatg/zhas no nonzero abelian ideals; hence g/z is semisimple (Corollary 2.5.8). Since g/z is a direct sum of simple algebras, it satisfies[g/z,g/z] =g/z. Letp:g //g/zbe the natural surjection. Ifu,v∈gthen p([u,v]) = [p(u),p(v)]. Since pis surjective, it follows thatg/zis spanned by the elementsp([u,v])foru,v∈g. Thusp([g,g]) =g/z. Now Theorem 2.5.3 (2) implies that the restriction of pto[g,g]gives a Lie algebra isomorphism withg/zand that dim([g,g]) +dimz=dimg. Henceg= [g,g]⊕z. ut

Letgbe a finite-dimensional complex Lie algebra.

Definition 2.5.10.TheKilling formofgis the bilinear formB(X,Y) =tr(adXadY) forX,Y∈g.

Recall thatgissemisimpleif it is the direct sum of simple Lie algebras. We now obtainCartan’s criterionfor semisimplicity.

Theorem 2.5.11.The Lie algebragis semisimple if and only if its Killing form is nondegenerate.

Proof. Assume thatgis semisimple. Since the adjoint representation of a simple Lie algebra is faithful, the same is true for a semisimple Lie algebra. Hence a semisimple Lie algebragis isomorphic to a Lie subalgebra of End(g). Let

g=g₁⊕ ··· ⊕g_r

(Lie algebra direct sum), where eachg_i is a simple Lie algebra. Ifmis an abelian ideal ing, thenm∩g_i is an abelian ideal ing_i, for eachi, and hence is zero. Thus m=0. HenceBis nondegenerate by Corollary 2.5.8.

2.5 Semisimple Lie Algebras 113

Conversely, suppose the Killing form is nondegenerate. Then the adjoint repre- sentation is faithful. To show thatgis semisimple, it suffices by Corollary 2.5.8 to show thatghas no nonzero abelian ideals.

Supposeais an ideal ing,X ∈a, andY ∈g. Then adXadY mapsgintoaand leavesainvariant. Hence

B(X,Y) =tr(adX|^aadY|^a). (2.43) Ifais an abelian ideal, then adX|^a=0. SinceBis nondegenerate, (2.43) implies

thatX=0. Thusa=0. ut

Corollary 2.5.12.Suppose gis a semisimple Lie algebra and D∈Der(g). Then there exists X∈gsuch that D=adX .

Proof. The derivation propertyD([Y,Z]) = [D(Y),Z] + [Y,D(Z)]can be expressed as the commutation relation

[D,adY] =adD(Y) for allY∈g. (2.44) Consider the linear functionalY 7→tr(DadY)ong. Since the Killing form is non- degenerate, there existsX ∈gsuch that tr(DadY) =B(X,Y)for allY ∈g. Take Y,Z∈gand use the invariance ofBto obtain

B(adX(Y),Z) =B(X,[Y,Z]) =tr(Dad[Y,Z]) =tr(D[adY,adZ])

=tr(DadYadZ)−tr(DadZadY) =tr([D,adY]adZ). Hence (2.44) and the nondegeneracy ofBgive adX=D. ut

For the next result we need the following formula, valid for any elementsY,Zin a Lie algebrag, anyD∈Der(g), and any scalarsλ,µ:

D−(λ+µ)k[Y,Z] =

∑

r

k r

[(D−λ)^rY,(D−µ)^k−rZ]. (2.45) (The proof is by induction on kusing the derivation property and the inclusion–

exclusion identity for binomial coefficients.)

Corollary 2.5.13.Letgbe a semisimple Lie algebra. If X∈gandadX=S+N is the additive Jordan decomposition inEnd(g)(with S semisimple, N nilpotent, and [S,N] =0), then there exist Xs,Xn∈gsuch thatadXs=S andadXn=N.

Proof. Letλ ∈Cand set

g_λ(X) =^[

k≥1

Ker(adX−λ)^k

(the generalizedλeigenspace of adX). The Jordan decomposition of adXthen gives a direct-sum decomposition

g=^M

λ

g_λ(X),

andSacts byλong_λ(X). TakingD=adX,Y∈g_λ(X),Z∈g_µ(X), andksufficiently large in (2.45), we see that

[g_λ(X),g_µ(X)]⊂g_λ+µ(X). (2.46) HenceSis a derivation ofg. By Corollary 2.5.12 there existsXs∈gsuch that adXs=

S. SetX_n=X−X_s. ut