exp(ad e± )(h) = h ’ ±(h)e±

for any h ∈ h. Now [f± , e± ] = ’h± so

(ad f± )2 (e± ) = [f± , ’h± ] = [h± , f± ] = ’2f± .

So

1

(id ’ ad f± + (ad f± )2 )(h ’ ±(h)e± )

exp(’ ad f± )(exp ad e± )h =

2

= h ’ ±(h)e± ’ ±(h)f± ’ ±(h)h± + ±(h)f±

= h ’ ±(h)h± ’ ±(h)e± .

If we now apply exp ad e± to this last expression and use the fact that ±(h± ) = 2,

we get the right hand side of (5.11).

5.6 Bases.

∆ ‚ ¦ is a called a Base if it is a basis of E (so #∆ = = dimR E = dimC h)

and every β ∈ ¦ can be written as ±∈∆ k± ±, k± ∈ Z with either all the

coe¬cients k± ≥ 0 or all ¤ 0. Roots are accordingly called positive or negative

and we de¬ne the height of a root by

ht β := k± .

±

µ i¬ » ’ µ is a sum of

Given a base, we get partial order on E by de¬ning »

positive roots or zero. We have

(±, β) ¤ 0, ±, β ∈ ∆ (5.12)

since otherwise (±, β) > 0 and

2(β, ±)

s± (β) = β ’ ±

(±, ±)

is a root with the coe¬cient of β = 1 > 0 and the coe¬cient of ± < 0, contradict-

ing the de¬nition which says that roots must have all coe¬cients non-negative

or non-positive.

To construct a base, choose a γ ∈ E, | (γ, β) = 0 ∀ β ∈ ¦. Such an element

is called regular. Then every root has positive or negative scalar product with

γ, dividing the set of roots into two subsets:

¦ = ¦+ ∪ ¦’ , ¦’ = ’¦+ .

86 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

A root β ∈ ¦+ is called decomposable if β = β1 + β2 , β1 , β2 ∈ ¦+ , indecom-

posable otherwise. Let ∆(γ) consist of the indecomposable elements of ¦+ (γ).

Theorem 12 ∆(γ) is a base, and every base is of the form ∆(γ) for some γ.

Proof. Every β ∈ ¦+ can be written as a non-negative integer combination of

∆(γ) for otherwise choose one that can not be so written with (γ, β) as small

as possible. In particular, β is not indecomposable. Write β = β1 + β2 , βi ∈

¦+ . Then β ∈ ∆(γ), (γ, β) = (γ, β1 ) + (γ, β2 ) and hence (γ, β1 ) < (γ, β) and

(γ, β2 ) < (γ, β). By our choice of β this means β1 and β2 are non-negative

integer combinations of elements of ∆(γ) and hence so is β, contradiction.

Now (5.12) holds for ∆ = ∆(γ) for if not, ±’β is a root, so either ±’β ∈ ¦+

so ± = ± ’ β + β is decomposable or β ’ ± ∈ ¦+ and β is decomposable.

This implies that ∆(γ) is linearly independent: for suppose ± c± ± = 0 and

let p± be the positive coe¬cients and ’qβ the negative ones, so

p± ± = qβ β

± β

all coe¬cients positive. Let be this common vector. Then ( , ) = p± qβ (±, β) ¤

0 so = 0 which is impossible unless all the coe¬cients vanish, since all scalar

products with γ are strictly positive. Since the elements of ¦ span E this shows

that ∆(γ) is a basis of E and hence a base.

Now let us show that every base is of the desired form: For any base ∆,

let ¦+ = ¦+ (∆) denote the set of those roots which are non-negative integral

combinations of the elements of ∆ and let ¦’ = ¦’ (∆) denote the ones which

are non-positive integral combinations of elements of ∆. De¬ne δ± , ± ∈ ∆ to

be the projection of ± onto the orthogonal complement of the space spanned by

the other elements of the base. Then

(δ± , ± ) = 0, ± = ± , (δ± , ±) = (δ± , δ± ) > 0

so γ = r± δ± , r± > 0 satis¬es

(γ, ±) > 0 ∀ ± ∈ ∆

hence

¦+ (∆) ‚ ¦+ (γ)

and

¦’ (∆) ‚ ¦’ (γ)

hence

¦’ (∆) = ¦’ (γ).

¦+ (∆) = ¦+ (γ) and

Since every element of ¦+ can be written as a sum of elements of ∆ with non-

negative integer coe¬cients, the only indecomposable elements can be the ∆,

so ∆(γ) ‚ ∆ but then they must be equal since they have the same cardinality

= dim E. QED

5.7. WEYL CHAMBERS. 87

5.7 Weyl chambers.

De¬ne Pβ := β ⊥ . Then E ’ Pβ is the union of Weyl chambers each con-

sisting of regular γ™s with the same ¦+ . So the Weyl chambers are in one to

one correspondence with the bases, and the Weyl group permutes them.

Fix a base, ∆. Our goal in this section is to prove that the re¬‚ections

s± , ± ∈ ∆ generate the Weyl group, W , and that W acts simply transitively

on the Weyl chambers.

Each s± , ± ∈ ∆ sends ± ’ ’±. But acting on » = cβ β, the re¬‚ection s±

does not change the coe¬cient of any other element of the base. If » ∈ ¦+ and

» = ±, we must have cβ > 0 for some β = ± in the base ∆. Then the coe¬cient

of β in the expansion of s± (») is positive, and hence all its coe¬cients must be

non-negative. So s± (») ∈ ¦+ . In short, the only element of ¦+ sent into ¦’ is

±. So if

1

β then s± δ = δ ’ ±.

δ :=

2 +

β∈¦

If β ∈ ¦+ , β ∈ ∆, then we can not have (β, ±) ¤ 0 ∀± ∈ ∆ for then β ∪ ∆

would be linearly independent. So β ’ ± is a root for some ± ∈ ∆, and since we

have changed only one coe¬cient, it must be a positive root. Hence any β ∈ ¦

can be written as

β = ±1 + · · · + ±p ±i ∈ ∆

where all the partial sums are positive roots.

Let γ be any vector in a Euclidean space, and let sγ denote re¬‚ection in the

hyperplane orthogonal to γ. Let R be any orthogonal transformation. Then

sRγ = Rsγ R’1 (5.13)

as follows immediately from the de¬nition.

Let ±1 , . . . , ±i ∈ ∆, and, for short, let us write si := s±i .

Lemma 12 If s1 · · · si’1 ±i < 0 then ∃j < i, j ≥ 1 so that

s1 · · · si = s1 · · · sj’1 sj+1 · · · si’1 .

Proof. Set βi’1 := ±i , βj := sj+1 · · · si’1 ±i , j < i ’ 1. Since βi’1 ∈ ¦+

and β0 ∈ ¦’ there must be some j for which βj ∈ ¦+ and sj βj = βj’1 ∈ ¦’

implying that that βj = ±j so by (5.13) with R = sj+1 · · · si’1 we conclude that

sj = (sj+1 · · · si’1 )si (sj+1 · · · si’1 )’1

or

sj sj+1 · · · si = sj+1 · · · si’1

implying the lemma. QED

As a consequence, if s = s1 · · · st is a shortest expression for s, then, since

st ±t ∈ ¦’ , we must have s±t ∈ ¦’ .

88 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

Keeping ∆ ¬xed in the ensuing discussion, we will call the elments of ∆ sim-

ple roots, and the corresponding re¬‚ections simple re¬‚ections. Let W denote

the subgroup of W generated by the simple re¬‚ections, s± , ± ∈ ∆. (Eventually