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