Proof of theorem from proposition. Let 0 ’ E ’ E be an exact

sequence of g modules, and we may assume that E = 0. We want to ¬nd an

invariant complement to E in E. De¬ne W to be the subspace of Homk (E, E )

whose restriction to E is a scalar times the identity, and let V ‚ W be the

subspace consisting of those linear transformations whose restrictions to E is

zero. Each of these is a submodule of End(E). We get a sequence

0’V ’W ’k’0

and hence a complementary line of invariant elements in W . In particular, we

can ¬nd an element, T which is invariant, maps E ’ E , and whose restriction

to E is non-zero. Then ker T is an invariant complementary subspace. QED

As an illustration of construction of the Casimir operator consider g = sl(2)

with

10 01 00

h= , e= , f= .

0 ’1 00 10

Then

tr(ad h)2 = 8

tr(ad e)(ad f ) = 4

so the dual basis to the basis h, e, f is h/8, f /4, e/4, or, if we divide the metric

by 4, the dual basis is h/2, f, e and so the Casimir operator C is

12 1

h + ef + f e = h2 + h + 2f e.

2 2

This coincides with the C that we used in Chapter II.

72 CHAPTER 4. ENGEL-LIE-CARTAN-WEYL

Chapter 5

Conjugacy of Cartan

subalgebras.

It is a standard theorem in linear algebra that any unitary matrix can be di-

agonalized (by conjugation by unitary matrices). On the other hand, it is easy

to check that the subgroup T ‚ U (n) consisting of all unitary matrices is a

maximal commutative subgroup: any matrix which commutes with all diagonal

unitary matrices must itself be diagonal; indeed if A is a diagonal matrix with

distinct entries along the diagonal, any matrix which commutes with A must be

diagonal. Notice that T is a product of circles, i.e. a torus.

This theorem has an immediate generalization to compact Lie groups: Let

G be a compact Lie group, and let T and T be two maximal tori. (So T and T

are connected commutative subgroups (hence necessarily tori) and each is not

strictly contained in a larger connected commutative subgroup). Then there

exists an element a ∈ G such that aT a’1 = T . To prove this, choose one

parameter subgroups of T and T which are dense in each. That is, choose x

and x in the Lie algebra g of G such that the curve t ’ exp tx is dense in T

and the curve t ’ exp tx is dense in T . If we could ¬nd a ∈ G such that the

a(exp tx )a’1 = exp t Ada x

commute with all the exp sx, then a(exp tx )a’1 would commute with all ele-

ments of T , hence belong to T , and by continuity, aT a’1 ‚ T and hence = T .

So we would like to ¬nd and a ∈ G such that

[Ada x , x] = 0.

Put a positive de¬nite scalar product ( , ) on g, the Lie algebra of G which is

invariant under the adjoint action of G. This is always possible by choosing any

positive de¬nite scalar product and then averaging it over G.

Choose a ∈ G such that (Ada x , x) is a maximum. Let

y := Ada x .

73

74 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

We wish to show that

[y, x] = 0.

For any z ∈ g we have

d

([z, y], x) = (Adexp tz y, x)|t=0 = 0

dt

by the maximality. But

([z, y], x) = (z, [y, x])

by the invariance of ( , ), hence [y, x] is orthogonal to all g hence 0. QED

We want to give an algebraic proof of the analogue of this theorem for Lie

algebras over the complex numbers. In contrast to the elementary proof given

above for compact groups, the proof in the general Lie algebra case will be

quite involved, and the ¬‚avor of the proof will by quite di¬erent for the solvable

and semi-simple cases. Nevertheless, some of the ingredients of the above proof

(choosing “generic elements” analogous to the choice of x and x for example)

will make their appearance. The proofs in this chapter follow Humphreys.

5.1 Derivations.

Let δ be a derivation of the Lie algebra g. this means that

δ([y, z]) = [δ(y), z] + [y, δ(z)] ∀ y, z ∈ g.

Then, for a, b ∈ C

(δ ’ a ’ b)[y, z] = [(δ ’ a)y, z] + [y, (δ ’ b)z]

(δ ’ a ’ b)2 [y, z] = [(δ ’ a)2 y, z] + 2[(δ ’ a)y, (δ ’ b)z] + [y, (δ ’ b)2 z]

(δ ’ a ’ b)3 [y, z] = [(δ ’ a)3 y, z] + 3[(δ ’ a)2 y, (δ ’ b)z )] +

3[(δ ’ a)y, (δ ’ b)2 z] + [y, (δ ’ b)3 z]

. .

. .

. .

n

(δ ’ a ’ b)n [y, z] = [(δ ’ a)k y, (δ ’ b)n’k z].

k

Consequences:

• Let ga = ga (δ) denote the generalized eigenspace corresponding to the

eigenvalue a, so (δ ’ a)k = 0 on ga for large enough k. Then

[ga , gb ] ‚ g[a+b] . (5.1)

• Let s = s(δ) denote the diagonizable (semi-simple) part of δ, so that

s(δ) = a on ga . Then, for y ∈ ga , z ∈ gb

s(δ)[y, z] = (a + b)[y, z] = [s(δ)y, z] + [y, s(δ)z]

so s and hence also n = n(δ), the nilpotent part of δ are both derivations.

5.1. DERIVATIONS. 75

• [δ, ad x] = ad(δx)]. Indeed, [δ, ad x](u) = δ([x, u]) ’ [x, δ(u)] = [δ(x), u].

In particular, the space of inner derivations, Inn g is an ideal in Der g.

• If g is semisimple then Inn g = Der g. Indeed, split o¬ an invariant comple-

ment to Inn g in Der g (possible by Weyl™s theorem on complete reducibil-

ity). For any δ in this invariant complement, we must have [δ, ad x] = 0

since [δ, ad x] = ad δx. This says that δx is in the center of g. Hence

δx = 0 ∀x hence δ = 0.

• Hence any x ∈ g can be uniquely written as x = s + n, s ∈ g, n ∈ g

where ad s is semisimple and ad n is nilpotent. This is known as the

decomposition into semi-simple and nilpotent parts for a semi-simple Lie

algebra.

• (Back to general g.) Let k be a subalgebra containing g0 (ad x) for some

x ∈ g. Then x belongs g0 (ad x) hence to k, hence ad x preserves Ng (k)

(by Jacobi™s identity). We have

x ∈ g0 (ad x) ‚ k ‚ Ng (k) ‚ g

all of these subspaces being invariant under ad x. Therefore, the character-

istic polynomial of ad x restricted to Ng (k) is a factor of the charactristic

polynomial of ad x acting on g. But all the zeros of this characteristic

polynomial are accounted for by the generalized zero eigenspace g0 (ad x)

which is a subspace of k. This means that ad x acts on Ng (k)/k without

zero eigenvalue.

On the other hand, ad x acts trivially on this quotient space since x ∈ k

and hence [Ng k, x] ‚ k by the de¬nition of the normalizer. Hence

Ng (k) = k. (5.2)

We now come to the key lemma.

Lemma 3 Let k ‚ g be a subalgebra. Let z ∈ k be such that g0 (ad z) does not

strictly contain any g0 (ad x), x ∈ k. Suppose that

k ‚ g0 (ad z).

Then

g0 (ad z) ‚ g0 (ad y) ∀ y ∈ k.

Proof. Choose z as in the lemma, and let x be an arbitrary element of k.

By hypothesis, x ∈ g0 (ad z) and we know that [g0 (ad z), g0 (ad z)] ‚ g0 (ad z).

Therefore [x, g0 (ad z)] ‚ g0 (ad z) and hence

ad(z + cx)g0 (ad z) ‚ g0 (ad z)

for all constants c. Thus ad(z + cx) acts on the quotient space g/g0 (ad z). We

can factor the characteristic polynomial of ad(z + cx) acting on g as