D = Dg = {» ∈ h— |(», φ) ≥ 0 ∀φ ∈ ¦+ }

R

and

Dr = {» ∈ h— |(», φ) ≥ 0 ∀φ ∈ ¦+ }

R r

so

D ‚ Dr

and we have chosen a cross-section C of Wr in W as

C = {w ∈ W |wD ‚ Dr },

10.6. EIGENVALUES OF THE DIRAC OPERATOR. 173

so

W = Wr · C, Dr = wD.

w∈C

We let L = Lg ‚ h— denote the lattice of g integral linear forms on h, i.e.

R

(µ, φ)

L = {µ ∈ h— |2 ∈ Z ∀φ ∈ ∆}.

(φ, φ)

We let

1

ρ = ρg = φ

2

φ∈∆+

and

1

ρr = φ.

2

φ∈∆+

r

We set

Lr = the lattice spanned by L and ρr ,

and

Λ := L © D, Λr := Lr © Dr .

For any r module Z we let “(Z) denote its set of weights, and we shall

assume that

“(Z) ‚ Lr .

For such a representation de¬ne

mZ := max (γ + ρr , γ + ρr ). (10.20)

γ∈“(Z)

For any µ ∈ Λr we let Zµ denote the irreducible module with highest weight µ.

Proposition 30 Let

“max (Z) := {µ ∈ “(Z)|(µ + ρr , µ + ρr ) = mZ }.

Let µ ∈ “max (Z). Then

1. µ ∈ Λr .

2. If z = 0 is a weight vector with weight µ then z is a highest weight vector,

and hence the submodule U (r)z is irreducible and equivalent to Zµ .

3. Let

Ymax := Zµ

µ∈“max (Z)

and

Y := U (r)Ymax .

Then mZ ’ (ρr , ρr ) is the maximal eigenvalue of Casr on Z and Y is the

corresponding eigenspace.

174 CHAPTER 10. THE KOSTANT DIRAC OPERATOR

Proof. We ¬rst show that

µ ∈ “max ’ µ + ρr ∈ Λr .

Suppose not, so there exists a w = 1, w ∈ Wr such that

wµ + wρr ∈ Λr .

But w changes the sign of some of the positive roots (the number of such changes

being equal the length of w in terms of the generating re¬‚ections), and so ρr ’wρr

is a non-trivial sum of positive roots. Therefore

(wµ + wρr , ρr ’ wρr ) ≥ 0, (ρr ’ wρr , ρr ’ wρr ) > 0

and

wµ + ρr = (wµ + wρr ) + (ρr ’ wρr )

satis¬es

(wµ + ρr , wµ + ρr ) > (wµ + wρr , wµ + wρr ) = (µ + ρr , µ + ρ) = mZ

contradicting the de¬nition of mZ . Now suppose that z is a weight vector

with weight µ which is not a highest weight vector. Then there will be some

irreducible component of Z containing z and having some weight µ such that

µ ’ µ is a non trivial sum of positive roots. We have

µ + ρr = (µ ’ µ) + (µ + ρr )

so by the same argument we conclude that

(µ + ρr , µ + ρr ) > mZ

since µ + ρr ∈ Λr , and again this is impossible. Hence z is a highest weight

vector implying that µ ∈ Λr . This proves 1) and 2).

We have already veri¬ed that the eigenvalue of the Casimir Casr on any Zγ

is (γ + ρr , γ + ρr ) ’ (ρr , ρr ). This proves 3).

Consider the irreducible representation Vρ of g corresponding to ρ = ρg . By

the same arguments, any weight γ = ρ of Vρ lying in D must satisfy (γ, γ) <

(ρ, ρ) and hence any weight γ of Vρ satisfying (γ, γ) = (ρ, ρ) must be of the form

γ = wρ

for a unique w ∈ W. But

wρ = ρ ’ φ = ρ ’ φJ

φ∈Jw

where

Jw := w(’¦+ ) © ¦+ .

10.6. EIGENVALUES OF THE DIRAC OPERATOR. 175

We know that all the weights of Vρ are of the form ρ ’ φJ as J ranges over all

subsets of ¦+ . So

(ρ, ρ) ≥ (ρ ’ φJ , ρ ’ φJ ) (10.21)

where we have strict inequality unless J = Jw for some w ∈ W .

Now let » ∈ Λ, let V» be the corresponding irreducible module with highest

weight » and let γ be a weight of V» . As usual, let J denote a subset of the

positive roots, J ‚ ¦+ . We claim that

Proposition 31 We have

(» + ρ, » + ρ) ≥ (γ + ρ ’ φJ , γ + ρ ’ φJ ) (10.22)

with strict inequality unless there exists a w ∈ W such that

γ = w», and J = Jw

in which case the w is unique.

Proof. Choose w such that

w’1 (γ + ρ ’ φJ ) ∈ Λ.

Since w’1 (γ) is a weight of V» , » ’ w’1 (γ) is a sum (possibly empty) of positive

roots. Also w’1 (ρ ’ φJ ) is a weight of Vρ and hence ρ ’ w’1 (ρ ’ φJ ) is a sum

(possibly empty) of positive roots. Since

» + ρ = (» ’ w’1 γ) + (ρ ’ w’1 (ρ ’ φJ ) + w’1 (γ + ρ ’ φJ )),

we conclude that

(» + ρ, » + ρ) ≥ (w’1 (γ + ρ ’ φJ ), w’1 (γ + ρ ’ φJ )) = (γ + ρ ’ φJ , γ + ρ ’ φJ )