Solutions for Humphreys

Introduction to Lie Algebras and Representation Theory Selected Solutions Aban Husain

Chapter 1

Basic Concepts 1.1

Definitions and First Examples

11. Verify that the commutator of two derivations of an F-Algebra is again a derivation, whereas the ordinary product need not be. Let A be an F-algebra, a, b ∈ A and δ, δ 0 ∈ Der(A). [δ, δ 0 ](ab) = δδ 0 (ab) − δ 0 δ(ab) = δ(δ 0 (a)b + aδ 0 (b)) − δ 0 (δ(a)b + aδ(b)) = (δδ 0 )(a)b + δ 0 (a)δ(b) + δ(a)δ 0 (b) + a(δδ 0 )(b) − (δ 0 δ)(a)b − δ(a)δ 0 (b) − δ 0 (a)δ(b) − a(δ 0 δ)(b) = (δδ 0 − δ 0 δ)(a)b + a(δδ 0 − δ 0 δ)(b) = [δ, δ 0 ](a)b + a[δ, δ 0 ](b) The commutator is clearly a derivation. (δδ 0 )(ab) = δ(δ 0 (a)b + aδ 0 (b)) = (δδ 0 )(a)b + δ 0 (a)δ(b) + δ(a)δ 0 (b) + a(δδ 0 )(b) So the ordinary product is only a derivation if δ 0 (a)δ(b) + δ(a)δ 0 (b) = 0 for all a, b ∈ A. 12. Let L be a Lie algebra over an algebraically closed field and let x ∈ L. Prove that the subspace of L spanned by the eigenvectors of ad x is a subalgebra. As the bracket is bilinear, it suffices to show that the commutator of two eigenvectors is contained in the subspace to show that the subspace spanned by the eigenvectors is closed under the Lie bracket. Let y, z be

1

eigenvectors of ad x with eigenvalues λ and µ. ad x([y, z]) = [x, [y, z]] = −[z, [x, y]] + [y, [x, z]] = −λ[z, y] + µ[y, z] = (λ + µ)[y, z] So the subspace spanned by the eigenvectors of ad x is a subalgebra.

1.2

Ideals and Homomorphisms

10. Let σ be the automorphism of sl(2, F) defined as σ = exp(ad x)◦exp(ad−y)◦ exp(ad x). Verify that σ(x) = −y, σ(y) = −x and σ(h) = −h.       0 1 1 0 0 0 Given the basis x = , h = , and y = for 0 0 0 −1 1 0     0 0 0 0 −2 0    −1 0 0 so 0 0 1 and ad y = sl(2, F) we have ad x = 0 2 0 0 0 0     1 0 0 1 −2 −1 1 0. 1  and exp(ad y) =  1 exp(ad x) = 0 1 −1 −2 1 0 0 1 σ(x) = exp(ad x) ◦ exp(ad−y) ◦ exp(ad x)(x) = exp(ad x) ◦ exp(ad−y)(x) = exp(ad x)(x + h − y) = x − 2x + h + x − h − y = −y σ(x) = exp(ad x) ◦ exp(ad−y) ◦ exp(ad x)(h) = exp(ad x) ◦ exp(ad−y)(−2x + h) = exp(ad x)(−2x − h) = −2x + 2x − h = −h σ(x) = exp(ad x) ◦ exp(ad−y) ◦ exp(ad x)(y) = exp(ad x) ◦ exp(ad−y)(−x + h + y) = exp(ad x)(−x) = −x 11. If L = sl(n, F), g ∈ GL(n, F), prove the map of L to itself defined by x 7→ −gxt g −1 belongs to Aut L. When n = 2, g = identity matrix, prove 2

that the automorphism is inner. For any x ∈ sl(n, F), Tr(−gxt g−1) = Tr(−xt ) = −Tr(x) = 0, so the map x 7→ −gxt g −1 is a linear map from L to L. As g ∈ GL(n, F), the kernel is trivial and for any x ∈ sl(n, F), the element (−g −1 xg)t ∈ sl(n, F) maps onto it. For x, y ∈ sl(n, F) [−gxt g −1 , −gy t g −1 ] = gxt y t g −1 − gy t xt g −1 = −g(xy)t g −1 + g(yx)t g −1 = −g(xy − yx)t g −1 = −g[x, y]t g −1 Thus the map belongs to Aut L. When n = 2 and g is the identity matrix, x 7→ −xt . Using the basis for sl(2, F) from the previous question, we have x 7→ −y, y 7→ −x, and h 7→ −h. Therefore the map is σ = exp(ad x) ◦ exp(ad−y) ◦ exp(ad x) and is clearly an inner automorphism.

1.3

Solvable and Nilpotent Lie Algebras

10. Let L be a Lie algebra, K an ideal of L such that L/K is nilpotent and such that ad x K is nilpotent for all x ∈ L. Prove that L is nilpotent. n For each x ∈ L there exists some n such that (ad x) (y) ∈ K as L/K is nilpotent. As ad x K is nilpotent for each x ∈ L, there exists some m such that (ad x)m (ad x)n (y) = (ad x)m+n (y) = 0 for each x ∈ L and y ∈ L. Therefore each x ∈ L is ad-nilpotent and so, by Engel’s Theorem, L is nilpotent.

3

Chapter 2

Semisimple Lie Algebras 2.1

Theorems of Lie and Cartan

5. If x, y ∈ End V commute, prove that (x + y)s = xs + ys and (x + y)n = xn + yn . If x, y commute and are semisimple, let Vi be the λi eigenspace of x. Then for v ∈ Vi , x(y(v)) = y(x(v)) = y(λi v) = λi (y(v)), so y(Vi ) ⊆ Vi . Therefore y Vi is semisimple and for some basis of Vi can be written as a diagonal matrix, while for any basis of Vi , x Vi can be written as a diagonal matrix as x acts as a scalar. As such, x + y is semisimple. For nilpotent x and y that commute, there m such that
kand Pn+mexists n n+m−k xn = 0 and y m = 0. Then (x + y)n+m = k=0 n+m x y = 0 and k clearly x + y is nilpotent. Let x = xs + xn and y = ys + yn be the Jordan decomposition of x and y. The semisimple and nilpotent parts of each commute with all endomorphisms that x and y commute with. Therefore xs +ys is semisimple, xn + yn nilpotent and the two commute. By uniqueness of the Jordan decomposition, we must have (x + y)s = xs + ys and (x + y)n = xn + yn . 8. It suffices to check the hypothesis of Cartan’s criterion for x, y ranging over a basis of L. For L with basis (x, y, z) and commutators [x, y] = z, [x, z] = y [y, z] = 0, verify solvability using Cartan’s criterion. For the  three elements, we have   0 0 0 0 ad x = 0 0 1, ad y =  0 0 1 0 −1 4

  0 0 0 0 0, ad z = −1 0 0 0

 0 0 0 0. 0 0

[L, L] has basis (y, z) and so: Tr((ad Tr((ad Tr((ad Tr((ad Tr((ad

y)(ad y)(ad y)(ad z)(ad z)(ad

y)) = 0 z)) = 0 x)) = 0 x)) = 0 z)) = 0

Therefore L is solvable.

2.2

Killing Form

6. Let char F = p 6= 0. Prove that L is semisimple if its Killing form is nondegenerate. Let I be any abelian ideal in L, and S the radical of the Killing form. For x ∈ I, y ∈ L we have adxady : L → I and (adxady)2 : I → 0, so for each x ∈ I and y ∈ L, adxady is nilpotent and so κ(x, y) = 0. Therefore I ⊆ S, but as the Killing form is nondegenerate, S = 0 and L contains no abelian ideals. Thus L is semisimple. 8. Let L = L1 ⊕ · · · ⊕ Lt be the decomposition of a semisimple Lie algebra L into simple ideals. Show that the semisimple and nilpotent parts of x ∈ L are the sums of the semisimple and nilpotent parts in the various Li of the components of x. Let x ∈ L and x = x1 + · · · + xt with xi ∈ Li , and xi = si + ni be the Jordan decomposition of xi ∈ Li . As each si is semisimple and [si , sj ] = 0 for all i 6= j, s = s1 + · · · + st is semisimple. Similarly, n = n1 + · · · + nt is nilpotent. [s, n] = [s1 , n1 ] + · · · + [st , nt ] = 0 Therefore, x = s + n is the Jordan decomposition of x.

2.3

Complete Reducibility of Representations

8. If L is a Lie algebra, then L acts (via ad) on (L ⊗ L)∗ which may be identified with the space of all bilinear forms β on L. Prove that β is associative if and only if L.β = 0.

5

For x, y, z ∈ L (x.β)(y ⊗ z) = −β(x.(y ⊗ z)) = −β(x.y ⊗ z + y ⊗ x.z) = −β([x, y] ⊗ z + y ⊗ [x, z]) = β([y, x] ⊗ z) − β(y ⊗ [x, z]) If β is associatve then clearly L.β = 0 and if L.β = 0 then β([y, x] ⊗ z) = β(y ⊗ [x, z]) and so β is associative. 9. Let L0 be a semisimple subalgebra of a semisimple Lie algebra L. If x ∈ L0 , its Jordan decomposition in L0 is also its Jordan decomposition in L. Let σ : L0 → gl(L) with σ(x) = adL x. For x ∈ L0 , let x = xs + xn be its Jordan decomposition in L0 . Then adL x = adL xs + adL xn and by uniqueness of the Jordan decomposition, x = xs + xn is the Jordan decomposition in L.

2.4

Representations of sl(2, F)

2.5

Root Space Decomposition

6

Chapter 3

Root Systems 3.1

Axiomatics

3.2

Simple Roots and Weyl Groups

3.3

Classification

3.4

Construction of Root Systems and Automorphisms

3.5

Abstract Theory of Weights

7

Chapter 4

Isomorphism and Conjugacy Theorems 4.1

Isomorphism Theorem

4.2

Cartan Subalgebras

4.3

Conjugacy Theorems

8

Chapter 5

Existence Theorem 5.1

Universal Enveloping Algebras

5.2

Generators and Relations

5.3

The Simple Algebras

9

Chapter 6

Representation Theory 6.1

Weights and Maximal Vectors

6.2

Finite Dimensional Modules

6.3

Multiplicity Formula

6.4

Characters

6.5

Formulas of Weyl, Konstant, and Steinberg

10