20 PRAMOD N. ACHAR

(2) If i 0, then

Homi(A(λ),A(λ))

= 0, and Hom(A(λ),A(λ))

∼

= k.

(3) If i 0 and n ≥ 0, then

Homi(A(λ),A(λ)

n ) = 0.

(4) If μ ∈ Λ+ and λ = μ, then RHom(A(w0μ), A(λ)) = 0.

Proof. (1) Fix λ. In the proof, we will assume that μ is dominant and that

λ ≤ μ, and we will show that RHom(A(wμ), A(λ)) = 0 for all w ∈ W . We

proceed by induction with respect to . Assume that for all ν μ, we already

know that RHom(A(ν), A(λ)) = 0. We know that RHom(ON ⊗ M(μ),A(λ)) = 0

by Lemma 5.5(1). On the other hand, applying RHom(−, A(λ)) to (5.7), we have

RHom(ON ⊗ M(μ),A(λ)) ∈

RHom(A(μ), A(λ)) ∗ RHom(A(νk), A(λ)) ∗ · · · ∗ RHom(A(ν1), A(λ)).

All terms on the right-hand side with νi μ vanish by assumption and can be

omitted. The remaining terms are those with νi ∈ Wμ. By Lemma 5.3 and the

inductive assumption, there is some integer n 0 such that

RHom(A(νi), A(λ))

∼

= RHom(A(μ), A(λ)) n if νi ∈ Wμ.

Therefore, the expression above simpliﬁes to

(5.8) RHom(ON ⊗ M(μ),A(λ)) ∈

RHom(A(μ), A(λ)) ∗ RHom(A(μ), A(λ)) n1 ∗ · · · ∗ RHom(A(μ), A(λ)) nm .

By Lemma 2.2(1), we conclude that RHom(A(μ), A(λ)) = 0.

(2) The ﬁrst assertion of this part is contained in Lemma 5.2. The second

assertion will be proved together with part (3) in the next paragraph.

(3) This proof is similar to that of part (1). We know from Lemma 5.5(2) that

RHom(ON ⊗ M(λ),A(λ))

∼

=

k. We may again carry out the calculations leading

to (5.8), this time with μ = λ. Since n1,...,nm 0, Lemma 2.2(2) tells us that

Hom(A(λ),A(λ))

∼

=

k, and that for i 0,

Homi(A(λ),A(λ))

is concentrated in

strictly positive degrees. In other words, for n ≥ 0,

Homi(A(λ),A(λ)

n ) = 0.

(4) If μ ≥ λ, then this is an instance of part (1). On the other hand, if μ λ,

then we apply Serre–Grothendieck duality and Lemma 5.1:

RHom(A(w0μ), A(λ))

∼

=

RHom(DA(λ), DA(w0μ))

∼

=

RHom(A(−λ), A(−w0μ)).

Now, −w0μ and −w0λ are both dominant, and −w0μ −w0λ. In particular, we

have −λ − w0μ, so RHom(A(−λ), A(−w0μ)) = 0 by part (1) again.

Lemma 5.7. Let C be the category of ﬁnitely-generated graded B-equivariant

modules over the graded ring k[u]. Then

DbC

is generated as a triangulated category

by objects of form k[u] ⊗ V n , where V is a ﬁnite-dimensional B-representation.

Proof. In this proof, we will say that an object M ∈ C is free if it is a direct

sum of objects of the form k[u] ⊗ V n . Let R be the functor which forgets the B-

action (but retains the grading). Clearly, R takes free objects of C to free (graded)

k[u]-modules. However, a module M ∈ C may have the property that R(M) is a

free module while M itself is not. Let us call a module Mweakly free if R(M) is

free.

It is easy to see that C has “enough” free objects, i.e., that every module is

a quotient of a free module. Therefore, every module M has (possibly inﬁnite)

resolution by free modules · · · → F1 → F0 → M → 0. Hilbert’s syzygy theorem,

in the form found in, say, [CLO, Corollary 3.19], asserts that there is some n such

20