then s = id. Indeed, if s = id, write s in a minimal fashion as a product of

simple re¬‚ections. By what we have just proved, it must send some simple root

into a negative root. So W permutes the Weyl chambers without ¬xed points.

We now show that W acts transitively on the Weyl chambers:

Let γ ∈ E be a regular element. We claim

∃ s ∈ W with (s(γ), ±) > 0 ∀ ± ∈ ∆.

Indeed, choose s ∈ W so that (s(γ), δ) is as large as possible. Then

(s(γ), δ) ≥ (s± s(γ), δ)

= (s(γ), s± δ)

= (s(γ), δ) ’ (s(γ), ±) so

(s(γ), ±) ≥ 0 ∀ ± ∈ ∆.

We can™t have equality in this last inequality since s(γ) is not orthogonal to any

root. This proves that W acts transitively on all Weyl chambers and hence on

all bases.

We next claim that every root belongs to at least one base. Choose a (non-

regular) γ ⊥ ±, but γ ∈ Pβ , β = ±. Then choose γ close enough to γ so that

(γ, ±) > 0 and (γ, ±) < |(γ, β)| ∀ β = ±. Then in ¦+ (γ) the element ± must be

indecomposable. If β is any root, we have shown that there is an s ∈ W with

s β = ±i ∈ ∆. By (5.13) this implies that every re¬‚ection sβ in W is conjugate

’1

∈ W . Since W is

by an element of W to a simple re¬‚ection: sβ = s si s

generated by the sβ , this shows that W = W .

5.8 Length.

De¬ne the length of an element of W as the minimal word length in its expression

as a product of simple roots. De¬ne n(s) to be the number of positive roots

made negative by s. We know that n(s) = (s) if (s) = 0 or 1. We claim that

(s) = n(s)

in general.

Proof by induction on (s). Write s = s1 · · · si in reduced form and let

± = ±i . We have s± ∈ ¦’ . Then n(ssi ) = n(s) ’ 1 since si leaves all positive

roots positive except ±. Also (ssi ) = (s) ’ 1. So apply induction. QED

Let C = C(∆) be the Weyl chamber associated to the base ∆. Let C denote

its closure.

Lemma 13 If », µ ∈ C and s ∈ W satis¬es s» = µ then s is a product of

simple re¬‚ections which ¬x ». In particular, » = µ. So C is a fundamental

domain for the action of W on E.

5.9. CONJUGACY OF BOREL SUBALGEBRAS 89

Proof. By induction on (s). If (s) = 0 then s = id and the assertion is clear

with the empty product. So we may assume that n(s) > 0, so s sends some

positive root to a negative root, and hence must send some simple root to a

negative root. So let ± ∈ ∆ be such that s± ∈ ¦’ . Since µ ∈ C, we have

(µ, β) ≥ 0, ∀β ∈ ¦+ and hence (µ, s±) ¤ 0. So

0 ≥ (µ, s±)

= (s’1 µ, ±)

= (», ±)

≥ 0.

So (», ±) = 0 so s± » = » and hence ss± » = µ. But n(ss± ) = n(s) ’ 1 since

s± = ’± and s± permutes all the other positive roots. So (ss± ) = (s) ’ 1

and we can apply induction to conclude that s = (ss± )s± is a product of simple

re¬‚ections which ¬x ».

5.9 Conjugacy of Borel subalgebras

We need to prove this for semi-simple algebras since the radical is contained in

every maximal solvable subalgebra.

De¬ne a standard Borel subalgebra (relative to a choice of CSA h and a

system of simple roots, ∆) to be

b(∆) := h • gβ .

β∈¦+ (∆)

De¬ne the corresponding nilpotent Lie algebra by

n+ (∆) := gβ .

β∈¦+

Since each s± can be realized as (exp e± )(exp ’f± )(exp e± ) every element of W

can be realized as an element of E(g). Hence all standard Borel subalgebras

relative to a given Cartan subalgebra are conjugate.

Notice that if x normalizes a Borel subalgebra, b, then

[b + Cx, b + Cx] ‚ b

and so b + Cx is a solvable subalgebra containing b and hence must coincide

with b:

Ng (b) = b.

In particular, if x ∈ b then its semi-simple and nilpotent parts lie in b.

From now on, ¬x a standard BSA, b. We want to prove that any other BSA,

b is conjugate to b. We may assume that the theorem is known for Lie algebras

of smaller dimension, or for b with b © b of greater dimension, since if dim

90 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

b © b = dim b, so that b ⊃ b, we must have b = b by maximality. Therefore

we can proceed by downward induction on the dimension of the intersection

b©b.

Suppose b © b = 0. Let n be the set of nilpotent elements in b © b . So

n = n+ © b .

Also [b © b , b © b ] ‚ n+ © b = n so n is a nilpotent ideal in b © b .

Suppose that n = 0. Then since g contains no solvable ideals,

k := Ng (n ) = g.

Consider the action of n on b/(b © b ). By Engel, there exists a y ∈ b © b with

[x, y] ∈ b © b ∀x ∈ n . But [x, y] ∈ [b, b] ‚ n+ and so [x, y] ∈ n . So y ∈ k.

Thus y ∈ k © b, y ∈ b © b . Similarly, we can interchange the roles of b and

b in the above argument, replacing n+ by the nilpotent subalgebra [b , b ] of

b , to conclude that there exists a y ∈ k © b , y ∈ b © b . In other words, the

inclusions

k©b⊃b©b, k©b ⊃b©b

are strict.

Both b © k and b © k are solvable subalgebras of k. Let c, c be BSA™s

containing them. By induction, there is a σ ∈ E(k) ‚ E(g) with σ(c ) = c. Now

let b be a BSA containing c. We have

b ©b⊃c©b⊃k©b⊃b ©b

with the last inclusion strict. So by induction there is a „ ∈ E(g) with „ (b ) = b.

Hence

„ σ(c ) ‚ b.

Then

b © „ σ(b ) ⊃ „ σ(c ) © „ σ(b ) ⊃ „ σ(b © k) ⊃ „ σ(b © b )

with the last inclusion strict. So by induction we can further conjugate „ σb

into b.

So we must now deal with the case that n = 0, but we will still assume

that b © b = 0. Since any Borel subalgebra contains both the semi-simple and

nilpotent parts of any of its elements, we conclude that b © b consists entirely

of semi-simple elements, and so is a toral subalgebra, call it t. If x ∈ b, t ∈ t =

b © b and [x, t] ∈ t, then we must have [x, t] = 0, since all elements of [b, b] are

nilpotent. So

Nb (t) = Cb (t).

Let c be a CSA of Cb (t). Since a Cartan subalgebra is its own normalizer, we

have t ‚ c. So we have

t ‚ c ‚ Cb (t) = Nb (t) ‚ Nb (c).

Let t ∈ t, n ∈ Nb (c). Then [t, n] ∈ c and successive brackets by t will eventually

yield 0, since c is nilpotent. Thus (ad t)k n = 0 for some k, and since t is semi-

simple, [t, n] = 0. Thus n ∈ Cb (t) and hence n ∈ c since c is its own normalizer

5.9. CONJUGACY OF BOREL SUBALGEBRAS 91

in Cb (t). Thus c is a CSA of b. We can now apply the conjugacy theorem for

CSA™s of solvable algebras to conjugate c into h.

So we may assume from now on that t ‚ h. If t = h, then decomposing

b into root spaces under h, we ¬nd that the non-zero root spaces must consist

entirely of negative roots, and there must be at least one such, since b = h.

But then we can ¬nd a „± which conjugates this into a positive root, preserving

h, and then „± (b ) © b has larger dimension and we can further conjugate into

b.

So we may assume that

t‚h

is strict.

If