WARNING: UNCORRECTED NOTES 1. Day 4 Solvable and nilpotent Lie algebras
• (a) Definitions
• (b) Lie’s theorem
• (c) Engel’s theorem
• (d) Characterization by the Killing form
1.1. Definitions. Note that if h,h0 ⊂g are ideals, then so is [h,h0], by the Jacobi identity. We define a sequence of important ideals:
Thedescending central series(suite centrale), defined inductively: C0(g) = g;Ci(g) = [g, Ci−1g].
Thederived series, again defined inductively: D0(g) =g,Di(g) = [Di−1(g), Di−1(g)].
Definition 1.1. The Lie algebragis nilpotent (resp. solvable) ifCi(g) = 0 fori >>0 (resp. ifDi(g) = 0 for i >>0).
Obviously,Ci(g)⊃Di(g), so any nilpotent Lie algebra is solvable.
Exercise 1.2. Let b,n⊂gl(n) be the upper triangular, resp. strictly upper triangular, subalgebras. Show that b is solvable but not nilpotent, and n is nilpotent.
Lemma 1.3. Let h⊂g be an ideal. Then g is solvable if and only if hand g/h are solvable.
More generally, any subalgebra of a solvable Lie algebra is solvable.
Finally, if h, andh0 are solvable ideals ofg, then so ish+h0.
Proof. Since Di(h)⊂Di(g) and since Di(g) maps onto Di(g/h), one direc- tion is clear; the last statement also follows. Now suppose g and g/h are solvable, and say Dn(g/h) = 0. ThenDn(g) ⊂h, and Dn+i(g)⊂Di(h) for i≥0.
For the final statement, we recall the isomorphism (h+h0)/h0 −→h/(h∼ ∩h0).
The right hand side is a quotient of a solvable Lie algebra, hence solvable;
but h0 is solvable, hence the first statement implies that so is h+h0. This
completes the proof.
This shows that the following definition is meaningful.
Definition 1.4. The radical of a Lie algebragis the maximal solvable ideal.
A Lie algebra gis semisimple if its radical is trivial.
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Lemma 1.5. Let h⊂g be an ideal, with g nilpotent. Then h and g/h are nilpotent.
Moreover, the center of any non-zero nilpotent Lie algebra is non-trivial.
Proof. The first part follows as in the solvable case. Supposeg is nilpotent and non-zero. Then there exists i such that Ci(g) 6= 0, Ci+1(g) = 0. But 0 = Ci+1(g) = [g, Ci(g)] implies that Ci(g) is contained in the center of
g.
1.2. Lie’s theorem. In what follows, K is an algebraically closed field of characteristic zero, and gis a (finite-dimensional) Lie algebra over K.
Theorem 1.6. The following are equivalent:
(i) g is solvable.
(ii) Every irreducible representation of g is1-dimensional.
(ii’) Every representation ofg has a (non-zero) eigenvector.
(iii) Every representation(r, V) of g has ag-stable flag, i.e. r(g) is con- tained in the algebra of upper-triangular matrices of End(V) (for some basis). [trigonalisable]
Proof. Conditions (ii) and (ii’) are clearly equivalent. Moreover, (ii’) implies (iii) by induction. (Also (iii) implies (ii’).)
To prove that (iii) implies (i), we apply (iii) to the adjoint representation of gon itself. Thus gcontains a g-stable complete flag; if dimg=N, say
g=g0 ⊃g1 ⊃ · · · ⊃gN ={0}
with dimgi/gi+1 = 1 for all i, and each gi is stable for the adjoint rep- resentation of g. The condition of being ad(g)-stable is equivalent to the condition that eachgi is an ideal. Thus each gi/gi+1 is a 1-dimensional Lie algebra, and is therefore abelian. It follows that g has a filtration by Lie subalgebras gi such that the successive quotients are abelian, which is one of the equivalent characterizations of solvability.
The hard step is the implication (i) ⇒ (ii’). We prove this by induction on dimg. Suppose dimg = 1; then (ii’) follows because K is algebraically closed. Now suppose dimg > 1, so dimg/[g,g] ≥ 1. Let (r, V) be a rep- resentation of g. Let h ⊃ [g,g] be of codimension 1 in g. Any subspace of g containing [g,g] is a subalgebra; so by induction, the representation r restricted to h has a non-zero eigenvector v ∈ V. Let λ : h→K be the eigenvalue ofv, and let Vλ ⊂V be the λ-eigenspace ofh. The main step is to prove thatVλ is ag-stable subspace ofV. Assume this for the moment, and let Y ∈g,Y /∈h. Then Y has an eigenvector, sayv0, in Vλ, because K is algebraically closed. Thus v0 is an eigenvector for all ofg.
It remains to show that, if v ∈ Vλ, then Y v ∈ Vλ, in other words, that XY v=λ(X)Y v for allX ∈h. But
XY v=Y Xv+ [X, Y]v=λ)XY v+ [X, Y]v
so we need to show that [X, Y]v = 0. Moreover, h is an ideal in g, so [X, Y]∈h, and [X, Y]v=λ([X, Y])v. Thus we have to show that
λ([X, Y]) = 0,∀X∈h, Y ∈g.
Let 06=w∈Vλ and letWk be the span of w, Y w, . . . , Yk−1w. Let n >0 be the smallest integer such that w, Y w, . . . , Ynw are linearly dependent;
then
dimWn=n; Wn+i=Wn∀i≥0.
. Write W = Wn; obviously Y W ⊂ W. We show by induction that, for all X ∈ h, X fixes Wk and has upper triangular matrix relative to the basisw, Y w, . . . , Yk−1wwith diagonal entries λ(X). Fork= 1 this is clear;
assume it’s true for k. Use the formula
XYkw=XY·Yk−1w=Y X·Yk−1w−[Y, X]Yk−1w+Y X·Yk−1w (modW)k By induction,
X·Yk−1w≡λ(X)Yk−1w (modWk−1) so that
XYkw=λ(X)Ykw+Y(Wk−1) +Wk=λ(X)Ykw (modWk).
ThusT rW(X) =n·λ(X) for allX ∈h. In particular T rW([X, Y]) =n·λ([X, Y])
for X ∈ h, Y ∈ g. But both X and Y stabilize W, so [X, Y] is the com- mutator of two endomorphisms of W, and therefore has trace 0. It follows that
nλ([X, Y]) = 0
and since char(K) = 0, this implies λ([X, Y]) = 0, as required.
1.3. Engel’s theorem.
Lemma 1.7. Let g be a nilpotent Lie algebra. Then for any X ∈g, adX is a nilpotent endomorphism of g: there exists n >0 such that adnX = 0.
Proof. This is easy. Say Cn(g) = 0. Then for all Xi ∈ g, i = 1, . . . , n, Qn
i=1adXi = 0. Indeed,adXn−i(Ci(g))⊂Ci+1(g) by definition.
Engel’s theorem is the converse.
Theorem 1.8 (Engel’s theorem). Let g be a (finite-dimensional) Lie alge- bra. Suppose, for all X∈gthere exists n=n(X) such thatadnX = 0. Then g is a nilpotent Lie algebra.
Corollary 1.9. The subalgebra n⊂gl(n) is nilpotent.
Proof. It suffices to show that for every X ∈ n, adX is nilpotent. More generally, we can show that, ifXis any nilpotent endomorphism ofV (finite- dimensional) then adX :gl(V) → gl(V) is nilpotent.
Define the endomorphismsrX and `X ofgl(V) by rX(Y) =Y X, `X(Y) =XY.
These obviously commute and both rX and `X are nilpotent. The sum and difference of two commuting nilpotent endomorphisms of a vector space is again nilpotent (use the binomial theorem). Thus adX = `X −rX is
nilpotent.
The proof of Engel’s theorem is based on the following result that is a strengthening of Lie’s theorem for nilpotent Lie algebras. Note that we do not requireK to be algebraically closed.
Theorem 1.10. Let g ⊂ gl(V) be a Lie subalgebra consisting of nilpotent endomorphisms, with V 6= 0. Then there exists a nonzero v ∈V such that gv= 0.
Proof. Induct on N = dimg. When N = 1 this is clear. Let h ⊂ g be a proper subalgebra. It follows from the proof of the Corollary that h acts on gl(V) by nilpotent endomorphisms, and thus also on the vector space g/h. By induction, there is a nonzeroY ∈g/hsuch thatadX(Y) = 0 for all X ∈ h. In other words, [X, Y]∈ hfor all X ∈h, but note that Y /∈ h. In other words, the normalizer ofhing is strictly bigger than h.
In the preceding argument, we can take h to be a maximal proper sub- algebra of g. Its normalizer is strictly bigger than h, and therefore must equal g. Thus any maximal proper subalgebra of g is an ideal. Suppose dimg/h > 1. Any nonzero Z ∈ g/h generates a 1-dimensional subalgebra of g/h, and its inverse image ing is a proper subalgebra containing h. This contradicts the maximality of h, so h is of codimension 1. Choose Z ∈ g, Z /∈h; so thatg=KZ +h.
By induction,Vh ⊂V is nontrivial. Since his an ideal in g,Vh is stable underg: ifv ∈Vh then for allY ∈h
Y Zv =ZY v+ [Y, Z]v= 0
because [Y, Z] ∈ h. Now Z is nilpotent on V and stabilizes Vh, hence is nilpotent on Vh and has a non-zero kernel. This completes the proof.
Corollary 1.11. Let g ⊂gl(V) be a Lie subalgebra consisting of nilpotent endomorphisms, with dimV =M >0. Then there is a complete flag
V =V0 ⊃V1⊃ · · · ⊃VM ={0}
with dimVi/Vi+1 = 1 for i≤M−1 such that, for all i, g(Vi)⊂Vi+1.
Proof. By induction again. TakeVM−1 =Kv for any nonzero v ∈ Vg and let V1 =V /VM−1. Then dimV1 = dimV −1 and by induction contains a
complete flag. Pull back toV.
Proof of Engel’s theorem. By hypothesis,ad(g)⊂gl(g) consists of nilpotent endomorphisms. It follows that there isZ ∈gsuch that [X, Z] =adX(Z) = 0 for all X∈g. In other words, the centerzg of gis non-zero. On the other hand, ad(zg) = 0 by definition, so ad(g) = ad(g/zg) consists of nilpotent elements. Since dimg/zg<dimg, we can apply induction (the case dimg= 1 is trivial) and conclude thatg/zgis a nilpotent Lie algebra. It follows that there existsnsuch that Cn(g)⊂zg, and so Cn+1(g)⊂[g, zg] = 0.
1.4. Characterization by the Killing form. We recall the statement of the Jordan decomposition for endomorphisms. Asemisimpleendomorphism is a diagonalizable endomorphism.
Theorem 1.12. Let K be any field.
(a) Let V be a finite-dimensional vector space over K, X ∈ End(V).
Then there is a unique pair of elements Xs, Xn ∈End(V) such that X = Xs+Xn, Xs is semisimple, Xn is nilpotent, and [Xs, Xn] = 0.
(b) There are polynomials ps, pn ∈K[T](depending on X), with ps(0) = pn(0) = 0, such that Xs = ps(X), Xn = pn(X). In particular, Xs and Xn
commute with any endomorphism Y such that [X, Y] = 0.
(c) If B ⊂A⊂V and X(B)⊂A thenXs(B)⊂A, Xn(B)⊂A.
Corollary 1.13. LetX ∈gl(V). SupposeXis semisimple (resp. nilpotent).
Then so is adX ∈End(gl(V)). In particular, for anyX,adXs+adXn is the Jordan decomposition for adX.
Proof. We have seen this for nilpotent X. If X is semisimple, say X = diag(a1, . . . , an) for some basis, then an easy (and important) computation shows that
adX(Eij) = (ai−aj)Eij
where Eij is the elementary matrix with 1 at the (ij) place and zero else- where (in the chosen basis).
Thus for any X, adX = adXs +adXn is a decomposition as a sum of a semisimple and a nilpotent matrix. The two obviously commute, hence they
are the Jordan components.
Henceforward, K is assumed of characteristic zero.
Theorem 1.14 (Cartan’s criterion for solvability). Let g ⊂ gl(V) be a subalgebra with V finite-dimensional. Suppose that T r(XY) = 0 for all X∈[g,g] and all Y ∈g. Then g is solvable.
In particular, suppose the Killing form Bad(X, Y) = 0 for all X ∈[g,g], Y ∈g. Then g is solvable.
The proof of the first statement requires a lemma. First I derive the second statement. Indeed, let g0 = ad(g) ⊂ gl(g). It follows from the hypothesis and the first part that g0 is solvable. But g0 =g/Z whereZ is the center of g; thusg is also solvable.
To prove the theorem, we use the following lemma.
Lemma 1.15. Let A⊂B ⊂gl(V), withdimV finite. Let M ={X∈gl(V) | [X, B]⊂A}
Suppose X ∈M satisfies T r(XY) = 0 for allY ∈M. Then X is nilpotent.
Proof. LetX=Xs+Xnbe the Jordan decomposition, withXs=diag(a1, . . . , an) for some basis. Let E denote the Q-vector subspace of K spanned by the eigenvalues ai; it is a finite-dimensional Q-vector space. We will show that Hom(E,Q) = 0; this implies thatE= 0.
So let f ∈Hom(E,Q), and consider Y =diag(f(a1), . . . , f(an)) (in the same basis). Then adY(Eij) = (f(ai)−f(aj))Eij. Claim there exists a polynomial g∈K[T] such that, for all i, j,
g(ai−aj) =f(ai)−f(aj).
One can certainly find such a polynomial (by Lagrange interpolation) pro- vided
ai−aj =ak−a` ⇒f(ai)−f(aj) =f(ak)−f(a`).
But this is clear because f is linear. Then
adY =g(adXs) =g◦ps(adX)
where Xs =ps(X),ps(0) = 0, because adXs is the semisimple part of adX. ThusadY is a polynomial inadX without constant term. SinceadX(B)⊂A, it follows that adY(B)⊂A, so thatY ∈M. Thus
0 =T r(XY) =X
f(ai)ai. Recall thatf(ai)∈Qfor all i, so thatP
aif(ai) is aQ-linear combination of elements ofE, and hence is in E. Apply f to both sides: we get
0 =f(0) =X
f(ai)f(ai)
is a sum of squares of rational numbers; hence each f(ai) = 0. Since the ai
spanE,f ≡0.
Proof of Cartan’s criterion. In order to prove that g is solvable, it suffices to prove that [g,g] is nilpotent; and by Engel’s theorem it suffices to show that allX ∈[g,g] are nilpotent endomorphisms. We apply the lemma with A= [g,g], B =g, so that
M ={X ∈gl(V)|[X,g]⊂[g,g]}.
Clearly B ⊂M. Our hypothesis is that T r(XY) = 0 for all X∈[g,g], Y ∈ g. The lemma implies that X is nilpotent provided T r(XZ) = 0 for all
X∈[g,g], Z ∈M. However, if U, V ∈g,Z ∈M, a simple calculation using the invariance of the trace shows that
T r([U, V]Z) =T r(U[V, Z]) =T r([V, Z]U).
Since Z ∈ M, [V, Z] ∈ [g,g] and by the hypothesis, T r([V, Z]U) = 0 for any U ∈ g. So T r([g, g]M) = 0 and the lemma applies to show that any
X∈[g,g] is nilpotent.
Corollary 1.16. More generally, let φ:g → End(V) be a representation, and hthe radical of Bφ. Then φ(h) is solvable.
Proof. The radical h of Bφ is an ideal of g: if X, Z ∈ g, Y ∈ h, then
−T r(φ(adX(Y))◦φ(Z)) =T r(φ([Y, X]◦φ(Z)) =T r([φ(Y), φ(X)]◦φ(Z)) = T r(φ(Y)◦[φ(X), φ(Z)]) = 0. It follows that Bφ |h =Bφ|h. It now follows
from Cartan’s criterion thatφ(h) is solvable.
