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 . (5) This gives a splitting of the extension, so E ∼= V ⊕ W . Conclusion We have shown Ext1(W , V ) = 0 for all representations V , W . Thus all (ﬁn.-dim.) representations are completely reducible. Application: structure of reductive Lie algebras

Suppose g is reductive: g/z(g) is semisimple. Then

g = z(g) ⊕ g0,

where g0 is a semisimple ideal. (We stated this earlier.)

I Consider the adjoint representation of g on itself.

I It makes g into a representation of ad(g) ∼= g/z(g).

I The center z(g) is a trivial subrepresentation. 0 I By complete reducibility, g = z(g) ⊕ g . 0 I g is a subrepresentation for ad(g), hence an ideal in g. 0 I g → g/z(g) is an isomorphism. 0 I Therefore g is semisimple as a Lie algebra.