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
Previous Page Next Page