PROJECTIVE MODULES 11

Then /(A) G ob Cgr with ^ / ( A ) = J(A). Moreover /(A) is projective in CgT. By Frobenius reciprocity

dim

F

Homu(g)(I(\),£gr(v)) = dim

F

Homu(Go)(P(X),£gr(n) \u(G0))

= [£9r(v) \u(G0y. L(\)].

For A = fi the above expression is non-zero. This implies that /(A) has a graded projective

indecomposable summand corresponding to the irreducible Cgr(X). We will denote this graded

projective cover by Vgr(X).

Theore m 1.2.5. For each X G t we have TCgr{X) = £(A) and TVgriX) = V(X).

Proof. For each A G T observe that

$ 3

&mF

Homu{6)(I(X),£gr(ii)) = ^ [ £

5

r O )

\u(G0): L(x)}

net

M€T

E[^)k(G

0

)^(A)]

= ] T d i m

F

# o r a t f

( g )

( / ( A ) , £(//)).

net

The first sum indicates the number of graded projective indecomposable summands in /(A), while

the last sum indicates the number of projective indecomposable summands in /(A). Since FVgr(X)

is a projective U(g) module it follows that the last sum must be greater than or equal to the first.

Hence equality holds and so by the Krull-Schmidt theorem and the fact that V(X) is a projective

summand of ^"P

pr

(A) we have FVgr{X) = V(X). From the computation above for each //, A G T we

also have

[Cgr(n) \u(Go): L(X)] = [£(/*) \u(Goy. 1(A)].

It follows that

d i m

F

£

5 r

( A ) = dim

F

£(A) .

Consequently, JrCgr(X) = C(X) because C(X) is a homomorphic image of TCgr(X).u