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Complete Reducibility of Representations Complete reducibility of representations MAT 552 April 6, 2020 Complete reducibility of representations Today’s topic is the following fundamental theorem. Theorem Every finite-dimensional representation of a semisimple Lie algebra is completely reducible. It is enough to consider complex Lie algebras (because g is semisimple iff gC is semisimple). Let g be a semisimple complex Lie algebra. Let V be a finite-dimensional complex representation of g. Goal: V is a direct sum of irreducible representations. I Proof using compact groups. I Proof using homological algebra. Compact real forms Let g be a complex semisimple Lie algebra. 1. There is a real subalgebra k such that g = k ⊗R C, and k is the Lie algebra of a compact Lie group K. 2. If G is a connected complex Lie group with Lie(G) = g, then one can choose K ⊆ G. The Lie algebra k is called the compact real form of g. It is unique up to conjugation. The Lie group K is called the compact real form of G. Example For g = sl(n, C), the compact real form is k = su(n). Weyl’s unitary trick Weyl used compact real forms to prove complete reducibility: I Let g be a semisimple complex Lie algebra. I Then g is the complexification of k = Lie(K), where K is a compact, connected, and simply-connected Lie group. I The categories of finite-dimensional complex representations of K, k, and g are equivalent. I Since K is compact, every finite-dimensional representation of K is completely reducible. I Therefore every finite-dimensional representation of g is also completely reducible. This argument is known as Weyl’s unitary trick. A proof using homological algebra In the rest of the lecture, I will present a more elementary proof that uses only basic homological algebra. The idea is that the obstruction to complete reducibility is given by a certain type of cohomology, and one can show that this cohomology is vanishes. To prove the vanishing, we use the Casimir element, which is a special central element in the universal enveloping algebra of g. Recall that the universal enveloping algebra Ug is the quotient of the tensor algebra C ⊕ g ⊕ (g ⊗ g) ⊕ (g ⊗ g ⊗ g) ⊕ · · · by the relations x ⊗ y − y ⊗ x − [x, y]. The Casimir element Let g be a Lie algebra, and B a non-degenerate invariant symmetric bilinear form on g. Let x1,..., xn ∈ g be a basis, and x 1,..., x n ∈ g the dual basis with respect to B. So i B(x , xj ) = δi,j . Lemma The element n X i CB = xi x ∈ Ug i=1 does not depend on the choice of basis, and is central in Ug. CB is called the Casimir element determined by B. Example: sl(2, C) Let g = sl(2, C), with bilinear form B(x, y) = tr(xy). Let e∗, h∗, f ∗ be the basis dual to e, h, f . Then 1 h∗ = h, 2 because tr(h2) = 2, tr(he) = tr(hf ) = 0. Similarly, e∗ = f and f ∗ = e. Therefore the Casimir element is 1 C = h2 + ef + fe. B 2 We showed sometime earlier that this is central. The Casimir element: proof Lemma The element n X i CB = xi x ∈ Ug i=1 does not depend on the choice of basis, and is central in Ug. B induces an isomorphism g ∼= g∗, x 7→ B(x, −). Thus g ⊗ g ∼= g ⊗ g∗ ∼= End(g), x ⊗ y 7→ B(y, −) x. This is an isomorphism of representations: the element [z, x] ⊗ y + x ⊗ [z, y] is sent (by ad-invariance of B) to the endomorphism B(y, −)[z, x] + B([z, y], −) x = B(y, −)[z, x] − B(y, [z, −]) x. The Casimir element: proof Under the isomorphism g ⊗ g ∼= g ⊗ g∗ ∼= End(g), x ⊗ y 7→ B(y, −) x, the element n X i I = xi ⊗ x i=1 goes to the identity operator in End(g). Therefore: I I does not depend on the choice of basis. I I commutes with the action of g (because id does). Since g ⊗ g → Ug is a morphism of representations, we see P i that CB = xi x does not depend on the choice of basis and is central. Application of the Casimir element Lemma Let V be a nontrivial irreducible representation of a semisimple Lie algebra g. Then there is a central element CV ∈ Z(Ug) that acts I by a nonzero constant on V , I and by zero on the trivial representation. In fact, the Casimir element defined by the Killing form always works. But this is more difficult to prove. Proof: special case Let BV (x, y) = tr(ρ(x)ρ(y)). This is an invariant bilinear form. Suppose first that BV is nondegenerate. I Let CV be the Casimir element defined by BV . I CV obviously acts by zero in the trivial representation (because all elements of g act by zero). I By Schur’s lemma, CV acts as a constant in the irreducible representation V , say ρ(CV ) = λ idV . I On the other hand, X i tr ρ(CV ) = tr(ρ(xi )ρ(x )) = dim g, i i because tr(ρ(xi )ρ(x )) = BV (xi , x ) = 1. I Thus λ = dim g/ dim V 6= 0. Proof: general case In general, let I = ker BV ⊆ g. This is an ideal in g. I I 6= g: otherwise, tr(ρ(x)ρ(y) = 0 for all x, y ∈ g, which would imply that ρ(g) ⊆ End(V ) is solvable; but quotients of semisimple Lie algebras are semisimple. 0 0 I From last time, g = I ⊕ g for a nonzero ideal g . 0 0 I g is semisimple, and the restriction of BV to g is nondegenerate. 0 I Let CV be the Casimir element of g corresponding to BV . 0 I CV is central in Ug (because I and g commute). I The argument on the previous slide shows that CV acts in V as the nonzero constant dim g0/ dim V . Main theorem Theorem Every finite-dimensional representation of a semisimple Lie algebra is completely reducible. We use the functors Exti (W , V ) from homological algebra: I Let V , W be (finite-dimensional) representations of g. I We consider V , W as modules over the algebra Ug. I We have a sequence of abelian groups i i Ext (W , V ) = ExtUg(W , V ), i = 0, 1, 2,... 0 I Elements of Ext (W , V ) = Hom(W , V ) are morphisms. Main theorem 1 I Elements of Ext (W , V ) correspond to extensions 0 → V → E → W → 0, taken up to isomorphism. 1 I If Ext (W , V ) = 0, then every such extension splits: E ∼= V ⊕ W I Conclusion: complete reducibility is equivalent to Ext1(W , V ) = 0 for all V , W . I This is what we are going to prove. Step 1 of the proof Lemma 1 If V is irreducible, then Ext (C, V ) = 0. In other words, every extension 0 → V → E → C → 0 splits. There are two cases: V trivial, and V nontrivial. Suppose that V = C is trivial I E is a two-dimensional representation such that ρ(x) is strictly upper triangular for all x ∈ g. I Therefore ρ(g) is nilpotent. I Since g is semisimple, ρ(g) = {0}. ∼ I This says that E = C ⊕ C. Step 1 of the proof Still consider the extension 0 → V → E → C → 0. Suppose that V is nontrivial. I CV ∈ Ug that acts as zero in C, and as a nonzero constant λ in V . I On E, the eigenvalues of CV are λ with multiplicity dim V , and 0 with multiplicity 1. I Thus E = V ⊕ W , where W is the 0-eigenspace of CV . I Since CV is central, W is a subrepresentation: ρ(CV )e = 0 =⇒ ρ(CV )ρ(x)e = ρ(x)ρ(CV )e = 0 I p : E → C takes W isomorphically to C. ∼ I Therefore E = V ⊕ C. Step 2 of the proof Lemma 1 Ext (C, V ) = 0 for any representation V . A short exact sequence of representations 0 → V1 → V → V2 → 0 gives a long exact sequence of Ext-groups. In particular, 1 1 1 · · · → Ext (C, V1) → Ext (C, V ) → Ext (C, V2) → · · · 1 1 is exact. Consequently, Ext (C, V1) = Ext (C, V2) = 0 implies 1 that Ext (C, V ) = 0. 1 Since we know that Ext (C, V ) = 0 for every irreducible representation, we get the result by induction on dim V . Step 3 of the proof Lemma Ext1(W , V ) = 0 for any two representations V , W. Consider an arbitrary extension 0 → V → E → W → 0. We are going to prove that it splits: E ∼= V ⊕ W (1) Apply the functor HomC(W , −); this produces another short exact sequence of representations 0 → HomC(W , V ) → HomC(W , E) → HomC(W , W ) → 0. Step 3 of the proof (2) Recall that we have g Hom(C, HomC(A, B)) = HomC(A, B) = Hom(A, B), where (−)g = Hom(C, −) is the functor of g-invariants. (3) If we apply this to 0 → HomC(W , V ) → HomC(W , E) → HomC(W , W ) → 0. 1 and recall that Ext (C, HomC(W , V )) = 0, we see that 0 → Hom(W , V ) → Hom(W , E) → Hom(W , W ) → 0 is still exact. Step 3 of the proof (4) The fact that 0 → Hom(W , V ) → Hom(W , E) → Hom(W , W ) → 0 s 7→ p ◦ s is exact means that there is a morphism s : W → E whose composition with p : E → W equals idW .
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