76 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

where f is the characteristic polynomial of ad(z + cx) on g0 (ad z) and g is the

characteristic polynomial of ad(z + cx) on g/g0 (ad z). Write

f (T, c) = T r + f1 (c)T r’1 + · · · fr (c) r = dim g0 (ad z)

g(T, c) = T n’r + g1 (c)T n’r’1 + · · · + gn’r (c) n = dim g.

The fi and the gi are polynomials of degree at most i in c. Since 0 is not an

eigenvalue of ad z on g/g0 (ad z), we see that gn’r (0) = 0. So we can ¬nd r + 1

values of c for which gn’r (c) = 0, and hence for these values,

g0 (ad(z + cx)) ‚ g0 (ad z).

By the minimality, this forces

g0 (ad(z + cx)) = g0 (ad z)

for these values of c. This means that f (T, c) = T r for these values of c, so each

of the polynomials f1 , . . . , fr has r + 1 distinct roots, and hence is identically

zero. Hence

g0 (ad(z + cx)) ⊃ g0 (ad z)

for all c. Take c = 1, x = y ’ z to conclude the truth of the lemma.

5.2 Cartan subalgebras.

A Cartan subalgebra (CSA) is de¬ned to be a nilpotent subalgebra which is its

own normalizer. A Borel subalgebra (BSA) is de¬ned to be a maximal solvable

subalgebra. The goal is to prove

Theorem 10 Any two CSA™s are conjugate. Any two BSA™s are conjugate.

Here the word conjugate means the following: De¬ne

N (g) = {x| ∃y ∈ g, a = 0, with x ∈ ga (ad y)}.

Notice that every element of N (g) is ad nilpotent and that N (g) is stable

under Aut(g). As any x ∈ N (g) is nilpotent, exp ad x is well de¬ned as an

automorphism of g, and we let

E(g)

denote the group generated by these elements. It is a normal subgroup of the

group of automorphisms. Conjugacy means that there is a φ ∈ E(g) with

φ(h1 ) = h2 where h1 and h2 are CSA™s. Similarly for BSA™s.

As a ¬rst step we give an alternative characterization of a CSA.

Proposition 13 h is a CSA if and only if h = g0 (ad z) where g0 (ad z) con-

tains no proper subalgebra of the form g0 (ad x).

5.3. SOLVABLE CASE. 77

Proof. Suppose h = g0 (ad z) which is minimal in the sense of the proposition.

Then we know by (5.2) that h is its own normalizer. Also, by the lemma,

h ‚ g0 (ad x) ∀x ∈ h. Hence ad x acts nilpotently on h for all x ∈ h. Hence, by

Engel™s theorem, h is nilpotent and hence is a CSA.

Suppose that h is a CSA. Since h is nilpotent, we have h ‚ g0 (ad x), ∀ x ∈

h. Choose a minimal z. By the lemma,

g0 (ad z) ‚ g0 (ad x) ∀x ∈ h.

Thus h acts nilpotently on g0 (ad z)/h. If this space were not zero, we could

¬nd a non-zero common eigenvector with eigenvalue zero by Engel™s theorem.

This means that there is a y ∈ h with [y, h] ‚ h contradicting the fact h is its

own normalizer. QED

Lemma 4 If φ : g ’ g is a surjective homomorphism and h is a CSA of g

then φ(h) is a CSA of g .

Clearly φ(h) is nilpotent. Let k = Ker φ and identify g = g/k so φ(h) = h + k.

If x + k normalizes h + k then x normalizes h + k. But h = g0 (ad z) for some

minimal such z, and as an algebra containing a g0 (ad z), h+k is self-normalizing.

So x ∈ h + k. QED

Lemma 5 φ : g ’ g be surjective, as above, and h a CSA of g . Any CSA

h of m := φ’1 (h ) is a CSA of g.

h is nilpotent by assumption. We must show it is its own normalizer in g. By

the preceding lemma, φ(h) is a Cartan subalgebra of h . But φ(h) is nilpotent

and hence would have a common eigenvector with eigenvalue zero in h /φ(h),

contradicting the selfnormalizing property of φ(h) unless φ(h) = h . So φ(h) =

h . If x ∈ g normalizes h, then φ(x) normalizes h . Hence φ(x) ∈ h so x ∈ m

so x ∈ h. QED

5.3 Solvable case.

In this case a Borel subalgebra is all of g so we must prove conjugacy for CSA™s.

In case g is nilpotent, we know that any CSA is all of g, since g = g0 (ad z) for

any z ∈ g. So we may proceed by induction on dim g. Let h1 and h2 be Cartan

subalgebras of g. We want to show that they are conjugate. Choose an abelian

ideal a of smallest possible positive dimension and let g = g/a. By Lemma 4

the images h1 and h2 of h1 and h2 in g are CSA™s of g and hence there is

a σ ∈ E(g ) with σ (h1 ) = h2 . We claim that we can lift this to a σ ∈ E(g).

That is, we claim

Lemma 6 Let φ : g ’ g be a surjective homomorphism. If σ ∈ E(g ) then

78 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

there exists a σ ∈ E(g) such that the diagram

φ

g ’’’ g

’’

¦ ¦

¦ ¦

σ σ

g ’’’ g

’’

φ

commutes.

Proof of lemma. It is enough to prove this on generators. Suppose that

x ∈ ga (y ) and choose y ∈ g, φ(y) = y so φ(ga (y)) = ga (y ), and hence we

can ¬nd an x ∈ N (g) mapping on to x . Then exp ad x is the desired σ in the

above diagram if σ = exp ad x . QED

Back to the proof of the conjugacy theorem in the solvable case. Let m1 :=

φ (h1 ), m2 := φ’1 (h2 ). We have a σ with σ(m1 ) = m2 so σ(h1 ) and h2 are

’1

both CSA™s of m2 . If m2 = g we are done by induction. So the one new case

is where

g = a + h1 = a + h2 .

Write

h2 = g0 (ad x)

for some x ∈ g. Since a is an ideal, it is stable under ad x and we can split it

into its 0 and non-zero generalized eigenspaces:

a = a0 (ad x) • a— (ad x).

Since a is abelian, ad of every element of a acts trivially on each summand, and

since h2 = g0 (ad x) and a is an ideal, this decomposition is stable under h2 ,

hence under all of g. By our choice of a as a minimal abelian ideal, one or the

other of these summands must vanish. If a = a0 (ad x) we would have a ‚ h2

so g = h2 and g is nilpotent. There is nothing to prove. So the only case to

consider is a = a— (ad x). Since h2 = g0 (ad x) we have

a = g— (ad x).

Since g = h1 + a, write

x = y + z, y ∈ h1 , z ∈ g— (ad x).

Since ad x is invertible on g— (ad x), write z = [x, z ], z ∈ a— (ad x). Since a is

an abelian ideal, (ad z )2 = 0, so exp(ad z ) = 1 + ad z . So

exp(ad z )x = x ’ z = y.

So h := g0 (ad y) is a CSA (of g), and since y ∈ h1 we have h1 ‚ g0 (ad y) = h

and hence h1 = h. So exp ad z conjugates h2 into h1 . Writing z as sum of

its generalized eigencomponents, and using the fact that all the elements of a

commute, we can write the exponential as a product of the exponentials of the

summands. QED