WARNING: UNCORRECTED NOTES 1. Day 5
• (a) Semisimple Lie algebras (Killing form)
• (b) Complete reducibility (Weyl’s theorem)
• (c) Reductive Lie algebras
• (d) Jordan decomposition
1.1. Semisimple Lie algebras (Killing form). The theory of semisimple Lie algebras is one of the high points of mathematics. The next two weeks presents the Cartan-Weyl structure theory and classification of semisimple Lie algebras. First, however, they need to be defined. Letgbe a Lie algebra over K. We let rad(g) denote the largest solvable ideal in g (this is well defined by the properties of solvable Lie algebras).
Definition 1.1. The Lie algebra g, of dimension > 1, is simple if it has no non-trivial ideals. It is semisimple if rad(g) = 0, and reductive if rad(g) coincides with the center ofg.
Exercise 1.2. Show that (a) every simple algebra is semisimple; (b)g/rad(g) is semisimple (or0) for anyg; (c) the direct sum of semisimple Lie algebras is semisimple.
We now assume K to be algebraically closed of characteristic zero.
Lemma 1.3. Let (φ, V) be an irreducible representation of g. Then the restriction of φ to rad(g) is given by a character λ:rad(g) → K. In par- ticular, φ([g, rad(g)]) = 0.
Proof. By Lie’s theorem, V contains an eigenvector v for rad(g). Let λ : rad(g)→ K denote its eigenvalue,Vλ theλ-eigenspace. If we show thatVλ is stable under g, then it must coincide with V. If X ∈g,Y ∈rad(g), and w∈Vλ, then
Y Xw=XY w+ [Y, X]w=λ(Y)Xw+λ([Y, X])w
because [Y, X]∈rad(g), so we need to show thatλ([Y, X]) = 0. The proof
is analogous to the proof of Lie’s theorem.
Corollary 1.4. Supposeghas a representation (φ, V) for which the bilinear form Bφ is nondegenerate. Then g is reductive.
Proof. The ideal [g, rad(g)] acts trivially on every irreducible subquotient of V. Say 0 ⊂ V1 ⊂V2· · · ⊂ Vn =V is a sequence with each Vi/Vi−1 irre- ducible. Thenφ(X)(Vi)⊂Vi−1for allX∈[g, rad(g)], and sinceφ(Y)Vi ⊂Vi
1
for all Y ∈ g, we have φ(X)φ(Y)(Vi) ⊂ Vi−1 for all i whenever X ∈ [g, rad(g)]. Thus
Bφ(X, Y) =T r(φ(X)φ(Y)) = 0∀X∈[g, rad(g)], Y ∈g.
In other words [g, radg] is contained in the kernel of Bφ. Since Bφ is non- degenerate, [g, rad(g)] = 0, and thus rad(g) equals the center of g, which
completes the proof.
Applying this to the standard representations of classical Lie algebras, we see that they are all reductive. Moreover, the centers of sp(2n) and so(n) are trivial, thus they are semisimple. (This should be an exercise in TD.) Theorem 1.5 (Cartan). The K-Lie algebra g is semisimple if and only if its Killing form is nondegenerate.
Proof. IfB=Bad is nondegenerate theng is reductive. But the center acts as 0 on the adjoint representation, hence it must be zero. Conversely, ifgis semisimple, its center is trivial, and therefore ad:g → End(g) is injective.
Lethbe the kernel ofB. Note that his an ideal ofg. Indeed, ifX∈hand Y, Z ∈g, then B(adY(X), Z) +B(X, adY(Z)) = 0, but B(X, adY(Z)) = 0 by definition, thus adY(X) ∈ h. We have seen that the restriction of the Killing form of g to h is the Killing form of h; thus the Killing form of h is trivial. It follows from Cartan’s criterion that h is solvable, hence zero,
sincerad(g) = 0.
Theorem 1.6. Any semisimple Lie algebragis a direct sum of simple ideals.
In particular, g= [g,g].
Proof. Lethbe an ideal ofg,h⊥its dual with respect toBad. The argument in the proof above shows thath⊥is again an ideal, hence so ish∩h⊥. So the restriction ofBad toh∩h⊥is the Killing form of the latter, but it’s obviously trivial, so h∩h⊥ is a solvable ideal, hence is zero. By induction, we obtain the simple decomposition. For the second assertion, we can thus assume g to be simple (and non-trivial), so [g,g] is either 0 org. But if [g,g] = 0 then
Bad = 0, which is impossible; thusg= [g,g].
1.2. Complete reducibility (Weyl’s theorem). We continue to assume K algebraically closed of characteristic zero.
Theorem 1.7. (Weyl) Let g be a Lie algebra over K. Then g is semisim- ple if and only if every finite-dimensional representation of g is completely reducible.
Remark 1.8. Weyl’s proof over Cused the “unitary trick”. The algebraic proof is due to Casimir.
Proof. First the easy direction. If every finite-dimensional representation of g is completely reducible, then the adjoint representation in particular is completely reducible. Write g = ⊕igi where each gi is an irreducible summand of the adjoint representation. Then each gi is an ideal and is
either simple (any ad(gi)-stable summand is ad(g)-stable) or of dimension 1 and therefore abelian. But a 1-dimensional Lie algebra has reducible but indecomposable representations, so there are no 1-dimensional factors.
So assumegis semisimple and let (φ, V) be a finite-dimensional represen- tation. First suppose V has an invariant submodule W of codimension 1.
Then V /W is the trivial representation. Indeed, any Lie algebra homomor- phism from g toK =End(V /W) is trivial, becauseg = [g,g]. Thus V fits in an exact sequence
0 →W →V → K → 0.
Induct on the dimension of W to show we may assume W is irreducible.
Indeed, if W ⊃W0, then we have an exact sequence 0→ W/W0 →V /W0 → K → 0.
which splits by induction. Say
V /W0=W/W0⊕W /W˜ 0
for some ˜W ⊂ V. Then ˜W /W0 −→K, which (by induction) gives yet an-∼ other split exact sequence
W˜ −→W∼ 0⊕W00
with dimW00= 1 mapping onto the originalK. It follows thatW00∩W = 0, soV =W ⊕W00.
Thus we may assume W irreducible. Use the following properties of the Casimir operatorCg:
• For any faithful representation (φ, V),T r(φ(Cg)) = dimV.
• Cg acts as 0 on the trivial representation.
• (Schur’s Lemma)Cg acts as a scalar cW on any irreducible W. The last two have already been seen. The first holds (thanks to Lemma
??) providedBφis nondegenerate. But we have seen (Corollary??) that, for a faithful representationφ, the radical ofBφis solvable; sincegis semisimple, the radical is trivial.
We may assume (φ, V) is faithful; otherwise we replace g by its image in End(V). Let φ0 denote the action of g on W. Then 0 6= T r(φ(Cg)) = T r(φ0(Cg) =cWdimW˙ becauseCg acts as 0 onV /W. It follows thatCghas two distinct eigenvalues onV, namely 0 andcW (because the trace ofCg is the same on V and on W. Thus the kernel of Cg is a complement to W in V.
This completes the case where V /W is of dimension 1. In general, let W ⊂V, withV /W 6= 0. ConsiderH=HomK(V, W) as a representation of g, and let
A={h∈H |h|W ∈K·idW}; B={h∈H |h|W = 0} ⊂A
These are g-invariant subspaces and dimA/B = 1. So by the first part, there is an invariant decomposition A =B ⊕D with D the one-diensional trivial representation. Leth∈Dbe a generator; thush|W =αidW for some
α6= 0, and we may assume α= 1. The fact that Dis trivial implies that h is a homomorphism of g-modules. Then ker(h)∩W = 0 and dim ker(h) = dimV −dimV, soV =W ⊕ker(h) is an invariant decomposition.
1.3. Jordan decomposition.
Theorem 1.9. Letgbe a semisimple Lie algebra over an algebraically closed field of characteristic 0. Let X ∈ g. Then there is a unique decomposition X =Xs, Xn∈g such that, for any representation φ of g, φ(X) =φ(Xs) + φ(Xn) is the Jordan decomposition of φ(X). Moreover, [Xs, Xn] = 0.
Proof. Step 1: We show that if g ⊂ End(V) is a semisimple Lie algebra, then for any X ∈g, Xs and Xn are also ing. Recall that any subspace of End(V) fixed by ad(X) is also fixed by ad(Xs) and ad(Xn); in particular ad(Xs) andad(Xn) both fixg, hence belong to the normalizer N(g) of gin End(V). Moreover, gis clearly an ideal in N(g).
Let N0 be the set of Y ∈ N(g) such that, for all g-invariant W ⊂ V, Y(W) ⊂ W and T r(Y |W) = 0. Since g = [g,g], g ⊂ N0 and is an ideal in N0 (since it’s already an ideal in N(g)). Moreover, any subspace W of V fixed by X is fixed by Xs and Xn, and since T rW(Xn) = 0 for any representation W and T rW(X) = 0, it follows that T rW(Xs) = 0. Thus Xs, Xn ∈ N0. Write N0 = g ⊕M as a sum of g-modules, using Weyl’s theorem. Since [g, N0] ⊂g (an ideal), and [g, M]⊂ M (it’s a submodule) the adjoint action of g on M takes M to g∩M = 0, hence M is a trivial module.
Now let W be any irreducibleg-submodule ofV. If Y ∈M, thenY is in particular in N0, hence fixes W: butY commutes withg, hence by Schur’s lemma Y acts as a scalar on W. On the other hand, T rW(Y) = 0 because Y ∈ N0, so Y acts as 0 on W. Since V is the direct sum of irreducible g-submodules by Weyl’s theorem,Y acts as 0 onV, henceM = 0. It follows thatN0 =g, and so Xs, Xn∈g.
Step 2: Apply this to V = g with the adjoint representation; note that g → End(g) is injective. Then step 1 gives elements Xs, Xn∈ g withX = Xs+Xn; and since this is the Jordan decomposition of ad(X), it is unique and [Xs, Xn] = 0.
Step 3: Now show thatφ(X) =φ(Xs) +φ(Xn) is the Jordan decomposi- tion of φ(X) for any (φ, V). The adjoint representation of X on g induces the adjoint representation of φ(X) on φ(g). Since ad(Xs) is semisimple and ad(Xn) is nilpotent, the same holds for ad(φ(Xs)) andad(φ(Xn)), act- ing on φ(g). Thus adφ(g)(φ(Xs)) = adφ(g)(φ(X))s and adφ(g)(φ(Xn)) = adφ(g)(φ(X))n inφ(g) by Step 2 applied toφ(g). But viewingU =φ(X) as an element of End(V), we know that
adgl(V)(U)s=adgl(V)(Us), adgl(V)(U)n=adgl(V)(Un)
by the Jordan decomposition in gl(V). This remains true upon restriction to any U-invariant submodule ofgl(V), in particular
adφ(g)(φ(X))s=adφ(g)(phi(X)s), adφ(g)(φ(X))n=adφ(g)(phi(X)n).
Combining these, we have
adφ(g)(φ(Xs)) =adφ(g)(φ(X))s=adφ(g)(φ(X)s)
and likewise for Xn. Since adφ(g) is injective on φ(g), this implies that φ(Xs) =φ(X)s (and likewise forXn), which concludes the proof.
Definition 1.10. Let g be a semisimple Lie algebra over an algebraically closed field of characteristic 0. SayX ∈gis semisimple (resp. nilpotent) if X=Xs (resp. X=Xn).
