PERVERSE COHERENT SHEAVES IN GOOD CHARACTERISTIC 19

Since N is an aﬃne variety, RΓ 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 ∈ Wμ 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 ∈ Wμ 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] ⊗ kμ

∼

= kμ ⊕ (J ⊗ kμ). Since all weights of

J ⊗ kμ 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), kμ n )

∼

= RHomG(M(μ), RindB

G

kμ 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