(2) If i 0, then
= 0, and Hom(A(λ),A(λ))

= k.
(3) If i 0 and n 0, then
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

k, and that for i 0,
is concentrated in
strictly positive degrees. In other words, for n 0,
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
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
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
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