So x, y, t± span a solvable three dimensional algebra. Acting as ad on g,

it is superdiagonizable, by Lie™s theorem, and hence ad t± , which is in the

commutator algebra of this subalgebra is nilpotent. Since it is ad semi-

simple by de¬nition of h, it must lie in the center, which is impossible.

• Choose e± ∈ g± , f± ∈ g’± with

2

κ(e± , f± ) = .

κ(t± , t± )

Set

2

h± := t± .

κ(t± , t± )

Then e± , f± , h± span a subalgebra isomorphic to sl(2). Call it sl(2)± . We

shall soon see that this notation is justi¬ed, i.e that g± is one dimensional

and hence that sl(2)± is well de¬ned, independent of any “choices” of

e± , f± but depends only on ±.

• Consider the action of sl(2)± on the subalgebra m := h• gn± where n ∈

Z. The zero eigenvectors of h± consist of h ‚ m. One of these corresponds

to the adjoint representation of sl(2)± ‚ m. The orthocomplement of

h± ∈ h gives dim h’1 trivial representations of sl(2)± . This must exhaust

all the even maximal weight representations, as we have accounted for all

the zero weights of sl(2)± acting on g. In particular, dim g± = 1 and no

integer multiple of ± other than ’± is a root. Now consider the subalgebra

p := h • gc± , c ∈ C. This is a module for sl(2)± . Hence all such c™s

must be multiples of 1/2. But 1/2 can not occur, since the double of a

root is not a root. Hence the ±± are the only multiples of ± which are

roots.

Now consider β ∈ ¦, β = ±±. Let

k := gβ+j± .

Each non-zero summand is one dimensional, and k is an sl(2)± module. Also

β + i± = 0 for any i, and evaluation on h± gives β(h± ) + 2i. All weights di¬er

by multiples of 2 and so k is irreducible. Let q be the maximal integer so that

β + q± ∈ ¦, and r the maximal integer so that β ’ r± ∈ ¦. Then the entire

string

β ’ r±, β ’ (r ’ 1)±, . . . β + q±

5.5. ROOTS. 83

are roots, and

β(h± ) ’ 2r = ’(β(h± ) + 2q)

or

β(h± ) = r ’ q ∈ Z.

These integers are called the Cartan integers.

We can transfer the bilinear form κ from h to h— by de¬ning

(γ, δ) = κ(tγ , tδ ).

So

β(h± ) = κ(tβ , h± )

2κ(tβ , t± )

=

κ(t± , t± )

2(β, ±)

= .

(±, ±)

So

2(β, ±)

= r ’ q ∈ Z.

(±, ±)

Choose a basis ±1 , . . . , ± of h— consisting of roots. This is possible because

the roots span h— . Any root β can be written uniquely as linear combination

β = c1 ±1 + · · · + c ±

where the ci are complex numbers. We claim that in fact the ci are rational

numbers. Indeed, taking the scalar product relative to ( , ) of this equation

with the ±i gives the equations

(β, ±i ) = c1 (±1 , ±i ) + · · · + c (± , ±i ).

Multiplying the i-th equation by 2/(±i , ±i ) gives a set of equations for the

coe¬cients ci where all the coe¬cients are rational numbers as are the left hand

sides. Solving these equations for the ci shows that the ci are rational.

Let E be the real vector space spanned by the ± ∈ ¦. Then ( , ) restricts

to a real scalar product on E. Also, for any » = 0 ∈ E,

(», ») :=: κ(t» , t» )

:= tr(ad t» )2

±(t» )2

=

±∈¦

> 0.

So the scalar product ( , ) on E is positive de¬nite. E is a Euclidean space.

In the string of roots, β is q steps down from the top, so q steps up from the

bottom is also a root, so

β ’ (r ’ q)±

84 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

is a root, or

2(β, ±)

β’ ± ∈ ¦.

(±, ±)

But

2(β, ±)

β’ ± = s± (β)

(±, ±)

where s± denotes Euclidean re¬‚ection in the hyperplane perpendicular to ±. In

other words, for every ± ∈ ¦

s± : ¦ ’ ¦. (5.6)

The subgroup of the orthogonal group of E generated by these re¬‚ections

is called the Weyl group and is denoted by W . We have thus associated to

every semi-simple Lie algebra, and to every choice of Cartan subalgebra a ¬nite

subgroup of the orthogonal group generated by re¬‚ections. (This subgroup

is ¬nite, because all the generating re¬‚ections, s± , and hence the group they

generate, preserve the ¬nite set of all roots, which span the space.) Once we

will have completed the proof of the conjugacy theorem for Cartan subalgebras

of a semi-simple algebra, then we will know that the Weyl group is determined,

up to isomorphism, by the semi-simple algebra, and does not depend on the

choice of Cartan subalgebra.

We de¬ne

2(β, ±)

β, ± := .

(±, ±)

So

β, ± = β(h± ) (5.7)

= r’q ∈Z (5.8)

and

s± (β) = β ’ β, ± ±. (5.9)

So far, we have de¬ned the re¬‚ection s± purely in terms of the root struction

on E, which is the real subspace of h— generated by the roots. But in fact,

s± , and hence the entire Weyl group arises as (an) automorphism(s) of g which

preserve h. Indeed, we know that e± , f± , h± span a subalgebra sl(2)± isomorphic

to sl(2). Now exp ad e± and exp ad(’f± ) are elements of E(g). Consider

„± := (exp ad e± )(exp ad(’f± ))(exp ad e± ) ∈ E(g). (5.10)

We claim that

Proposition 16 The automorphism „± preserves h and on h it is given by

„± (h) = h ’ ±(h)h± . (5.11)

In particular, the transformation induced by „± on E is s± .

5.6. BASES. 85

Proof. It su¬ces to prove (5.11). If ±(h) = 0, then both ad e± and ad f± vanish

on h so „± (h) = h and (5.11) is true. Now h± and ker ± span h. So we need

only check (5.11) for h± where it says that „ (h± ) = ’h± . But we have already

veri¬ed this for the algebra sl(2). QED

We can also verify (5.11) directly. We have