12

DAVID C. VELLA

The proof of thi s i s exactl y the same a s in the cas e J = A and i s

omitted. _

Proposition 3.2. Let V be a finit e dimensional Pj-module. Let v be a weight

vector of weight X e A and suppose v i s fixed by R^B) = U. Let V be the

Pj-submodule of V generated by v. Then

a) If a c A(V') then u £ X in the partial order on A induced by J; i.e.

X - a i s a nonnegative integral combination of the a* e J.

b) The dimension of the weight space V/ i s one.

c) For any a e J, X,a e

Z+.

d) V' contains a unique maximal proper submodule V" and V

/

/V

/ /

i s

irreducible.

Proof. To show a) and b) i t suffice s to show V ' G V, where V = kv © ( 2 V,,).

pX u

Define a map f:Pj - V' by f(g) = g»v. (f i s a variet y morphism because V' i s

rational.) Consider f I applied to u0tu., with tu-. c B = T»U, and

l(U"nLj)»B

2 X 1

u

2

e U~ n Lj = Uj

0

. (U~ denotes the unipotent radical of the "opposite"

Borel subgroup B~.) We get f(u

2

tu

1

) = U g t u ^ v = u

2

t» v = X(t)u2*v. But u

2

has

coordinates in root groups U

a

only for a e -fcj, so f(u

2

tu

1

) e V by Proposition

3.1. Since Uj

0

B i s dense in P j we have f(Pj) Q V.

To prove c) le t a be a simple root in J. Then Sa(X) e A(V') by part a)

of Proposition 3.1, s o by part a) above,

s

a

M * *•• Since

Sa(X) = X - X,aa, we get X,a e

Z+.

Finally, for d), le t V" be the sum of al l proper submodules. Since ever y

proper submodule intersect s V/ trivially , so does V" and thu s V" i s proper.

It i s clearl y the unique maximal proper submodule of V .

D