12 PRAMOD N. ACHAR
is injective. The morphism f is in the kernel of this map (because g f = 0), so
f = 0. Thus,
Homd(
˜
Δ
(k)
s
n , Δs) = 0, as desired.
Proposition 3.8. If t s, or else if t = s and n m k + d, then
Homd(
˜
Δ
(k)
s
n , Σt m ) =
Homd(Σt
n ,
˜

s
(k)
m ) = 0.
Proof. If t s, this follows from Definition 2.6(2) and the fact that
˜
Δ
(k)
s
n
is s-quasistandard. If t = s, consider the exact sequence
Homd(
˜
Δ
(k)
s
n , Δs m )
Homd(
˜
Δ
(k)
s
n , Σs m )
Homd+1(
˜
Δ
(k)
s
n , Rs m ).
The first term vanishes by Proposition 3.7, and the last again because
˜
Δ
(k)
s
n is
s-quasistandard and Rs m (s)A. Therefore,
Homd(
˜
Δ
(k)
s
n , Σs m ) = 0.
Corollary 3.9. If t = s, or else if t = s and n m k + d, then we have
Homd(
˜
Δ
(k)
s
n ,
∇t
m ) =
Homd(Δs
n ,
˜

s
(k)
m ) = 0.
3.3. Subcategories associated to convex sets. The next step towards our
theorem is to show that certain Serre subcategories of A with finitely many simple
objects have enough projectives, and that these projectives have desirable Hom-
vanishing properties in D.
Definition 3.10. A subset Ξ S × Z is said to be convex if the following two
conditions hold:
(1) For each (s, n) Ξ, we have Δs n
ΞA
and
∇s
n
ΞA.
(2) For any s S, the set of integers {n Z | (s, n) Ξ} is either empty or
an interval {a0,a0 + 1,...,a0 + k}.
Definition 3.11. Let Ξ S × Z be a finite convex set. Let (s, n) Ξ, and let
as = min{m Z | (s, m) Ξ}. Let
Δs
Ξ
=
˜
Δ
(n−as).
s
The object Δs Ξ n is said to be Ξ-standard, or the Ξ-standard cover of Σs n .
A Ξ-standard filtration of an object X ΞA is a filtration each of whose
subquotients is a Ξ-standard object.
Lemma 3.12. Every finite subset of S × Z is contained in a finite convex set.
Proof. Given a finite set Ξ S × Z, let F0(Ξ) S be the set of s S such
that one of the following conditions holds:
There is an n Z with (s, n) Ξ but either Δs n /
ΞA
or
∇s
n /
ΞA.
There are integers a b c such that (s, a), (s, c) Ξ but (s, b) / Ξ.
Let F (Ξ) be the lower closure of F0(Ξ), i.e.,
F (Ξ) = {s S | there is some s0 F0(Ξ) such that s s0}.
It is obvious that F0(Ξ) is finite, and then it follows from (2.1) that F (Ξ) is finite
as well. Of course, Ξ is convex if and only if F (Ξ) is empty.
We prove the lemma by induction on the size of F (Ξ). If F (Ξ) = ∅, let
s F (Ξ) be a maximal element (with respect to ≤). Then s F0(Ξ). Let us put
a0 = min{n | (s, n) Ξ}, b0 = max{n | (s, n) Ξ}.
12
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