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
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
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