PERVERSE COHERENT SHEAVES IN GOOD CHARACTERISTIC 19
Since N is an affine variety, kills no nonzero object of
DbCohG×Gm
(N ), so it
suffices to check that the object RΓ(Rπ∗(i∗O
˜
N
α

p∗S
(Q)))

= RΓ(Rp∗(i∗O
˜
N
α

p∗S (Q))) vanishes. By the projection formula. we have Rp∗(i∗O
˜
N
α
p∗S (Q))

=
S (k[uα] Q), so to prove (5.4), we must check that RΓ(S (k[uα] Q)) = 0, or
(5.6) RindB(k[uα]
G
Q) = 0.
But RindBα
P
(k[uα] Q)

= k[uα] ⊗L RindBα
P
Q, so (5.6) follows from (5.1).
Lemma 5.4. For any μ Λ+, we have Rπ∗p∗S (M(μ))

=
ON M(μ). More-
over, there are weights ν1,...,νk such that
(5.7) ON M(μ) A(ν1) · · · A(νk) A(μ)
where either νi μ or νi but νi = μ for each i.
As a consequence, ON M(μ)
PCohG×Gm
(N ), and there is a surjective map
M(μ) A(μ).
Proof. Since π∗(ON M(μ))

=
p∗S (M(μ)), the projection formula implies
that Rπ∗p∗S (M(μ))

=
Rπ∗O
˜
N
⊗L (ON M(μ)). But Rπ∗O
˜
N

=
ON by [BK,
Theorem 5.3.2], so Rπ∗p∗S (M(μ))

=
ON M(μ).
Next, there is a surjective map of B-representations resB G M(μ) kμ. The
kernel of this map has a filtration whose subquotients are various kν1 , . . . , kνk , where
either νi μ or νi and ν μ. Applying the functor
Rπ∗p∗S
, we see that (5.7)
holds. It now follows from Lemma 5.2 that
Rπ∗p∗S
(M(μ))
PCohG×Gm
(N ). In
particular, there is a distinguished triangle
K ON M(μ) A(μ)
with K A(ν1) · · ·∗ A(νk). Since all three terms belong to
PCohG×Gm
(N ), this is
actually a short exact sequence in that category, and the map ON M(μ) A(μ)
is surjective.
Lemma 5.5. Let λ, μ
Λ+.
(1) If λ μ, then RHom(ON M(μ),A(λ)) = 0.
(2) If λ = μ, then RHom(ON M(μ),A(λ))

= k.
Proof. We have RHom(ON M(μ),A(λ))

= RHom(M(μ), RΓ(A(λ))) by
adjunction. We will work with the latter object. Using (4.3), we have
RHomG(M(μ), RΓ(A(λ)))

= RHomB(resB
G
M(μ), k[u] kλ).
Of course, all weights of M(μ) are μ, and all weights of k[u]⊗kλ are λ. Part (1)
then follows from Lemma 4.1.
For part (2), let J k[u] be the ideal spanned by all homogeneous elements
of strictly negative degree. Thus, k[u]

= (J kμ). Since all weights of
J are μ, Lemma 4.1 again tells us that RHom(resB
G
M(μ),J kμ) = 0. We
conclude that
RHomG(M(μ), RΓ(A(μ) n ))

=
RHomB(resB
G
M(μ), n )

= RHomG(M(μ), RindB
G
n )

= RHomG(M(μ), N(μ) n ),
and this clearly vanishes for n = 0 and is 1-dimensional when n = 0.
Proposition 5.6. Let λ Λ+. We have:
(1) If μ Λ and μ λ, then RHom(A(μ), A(λ)) = 0.
19
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