PERVERSE COHERENT SHEAVES IN GOOD CHARACTERISTIC 11

with

(3.4) Y ∈ Δs ∗ Δs m1 ∗ · · · ∗ Δs mk and I ∈ Ik+1 ∗ · · · ∗ Il ⊂

(s)D.

Recall that Hom(Δs[k] m , I) = Hom(Δs[k] m , I[1]) = 0 for any k, m ∈ Z, by

Deﬁnition 2.6(2). It follows that Hom(X[−d],I) = Hom(X[−d]),I[1]) = 0. From

the long exact sequence obtained by applying Hom(Δs, −) and Hom(X[−d], −)

to (3.3), we obtain natural isomorphisms

Hom(Δs,Y )

∼

→ Hom(Δs,Y ) and Hom(X[−d],Y )

∼

→ Hom(X[−d],Y ).

The ﬁrst of these shows that g : Δs → Y factors in a unique way through Y . Let

g : Δs → Y be the induced map. Now, the fact that g◦f = 0 means that g ◦f is in

the kernel of Hom(X[−d],Y ) → Hom(X[−d],Y ), so the second isomorphism above

shows that g ◦ f = 0. However, g does not necessarily have the other properties

claimed in the proposition.

To repair this, we use the “rearrangement” method again. By Lemma 2.8(3),

we have Hom(Δs m , Δs[1] n ) = 0 if n m, so it follows that

Δs n ∗ Δs m ⊂ Δs m ∗ Δs n if n m.

Using this to rearrange terms in (3.4), we may ﬁrst assume without loss of generality

that m1

≤ m2 ≤ · · · ≤ mk, and then we may write

Y ∈ Δs m1 ∗ · · · ∗ Δs mj ∗ Δs ∗ Δs mj+1 ∗ · · · ∗ Δs mk

where m1 ≤ · · · ≤ mj ≤ 0 mj+1 ≤ · · · ≤ mk. In other words, we have a

distinguished triangle

J → Y → Y →

with J ∈ Δs m1 ∗ · · · ∗ Δs mj and Y ∈ Δs ∗ Δs mj+1 ∗ · · · ∗ Δs mk . Let

g : Δs → Y be the natural map. Clearly, g ◦ f = 0. It remains only to check

that g is injective. Recall that a map in A is injective if and only if its cone in D

actually lies in the abelian category A (and in that case, the cone is the cokernel).

By construction, the cone of g lies in Δs mj+1∗···∗Δs mk ⊂ A, as desired.

Proposition 3.7. Let k ≥ 0 and d ≥ 1. If n k + d, then

Homd(

˜

Δ

(k)

s

n , Δs) =

Homd(∇s,

˜

∇

s

(k)

−n ) = 0.

Proof. We proceed by induction on d. Suppose ﬁrst that d = 1. Note that

every composition factor of Δs−n is some Σt m with (t, m)

Δ

(s, −n). The

assumption that n k+1 means that (s, −n)

Δ

(s, −k), so in view of Lemma 2.1,

the result follows from Propostion 3.4(2).

Now, suppose d 1. Given f ∈

Homd(

˜

Δ

(k)

s

n , Δs), choose an embedding

g : Δs → Y as in Lemma 3.6, and let Z = Y/g(Δs). Consider the exact sequence

Homd−1(

˜

Δ

(k)

s

n , Z) →

Homd(

˜

Δ

(k)

s

n , Δs) →

Homd(

˜

Δ

(k)

s

n , Y ).

For any m 0, we have n − m k + d − 1, so it follows from the inductive

assumption that

Homd−1(

˜

Δ

(k)

s

n , Δs m )

Homd−1(

˜

Δ

(k)

s

n − m , Δs) = 0 if m 0.

Recall from Lemma 3.6 that the standard subquotients of Z are all various Δs m

with m 0. It follows that

Homd−1(

˜

Δ

(k)

s

n , Z) = 0, so the map

Homd(

˜

Δ

(k)

s

n , Δs) →

Homd(

˜

Δ

(k)

s

n , Y )

11