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Strong generation and ghosts

The proof of Brown Representability is a powerful way to construct a “resolution” in a triangulated category. We wish to be able to run it successfully without the condition on the existence of coproducts.

It turns out that a good substitute for compact generation is the notion of strong generation. To define this, we need some intermediate notions.

Defintion. Let $\mathcal S$ be a subcategory of a triangulated category $\mathcal T$. We write $\langle \mathcal S \rangle_0$ for the smallest additive subcategory closed under shifts and containing the objects of $\mathcal S$.

The smallest subcategory closed under retracts and containing $\mathcal S$ is denoted $\overline{\mathcal S}$.

Given another subcategory $\mathcal S^\prime$, we write $\mathcal S \ast \mathcal S^\prime$ for the full subcategory consisting of objects $X$ of $\mathcal T$ for which there is triangle \(S \to X \to S^\prime \to S[1]\) with $S \in \mathcal S$ and $S^\prime \in \mathcal S^\prime$.

For $n>1$, we set \(\langle \mathcal S \rangle_n = \overline{\langle \mathcal S \rangle}_{n-1} \ast \overline{\langle \mathcal S \rangle}_0\)

We say that an object $G$ strongly generates $\mathcal T$ if there is some $n$ with \(\overline{\langle G \rangle}_n = \mathcal T\)

Lemma. The operation $\ast$ is associative.

Proof. (Expand to view)

This is a consequence of the Octahedral Axiom, TR4. We will omit the details.

From this lemma we see that \(\overline{\langle G \rangle}_n = \overline{\underbrace{ {\overline{\langle G \rangle}_0 \ast \cdots \ast \overline{\langle G \rangle}_0}}_n}\)

We introduce another notion which we will see is related.

Definition. Let $\mathcal S$ be a subcategory of $\mathcal T$. We say that a map $f: X \to Y$ is $\mathcal S$-ghost if \(\operatorname{Hom}(S,f) = 0\) for all objects of $\mathcal S$.

Example. Take $R$, a ring, as an object of $D(\operatorname{Mod} R)$. Then, a map $f: C \to D$ is $R$-ghost if and only if the induced map \(H^0(f) : H^0(C) \to H^0(D)\) is zero.

Lemma. If $X \to Y$ is ghost for all shifts of objects from $\mathcal S$, then it is $\overline{\langle \mathcal S \rangle}_0$ -ghost.

Proof. (Expand to view)

Note that $$ \operatorname{Hom}\left( \bigoplus S_i, f) = \bigoplus \operatorname{Hom}(S_i,f) $$ for finite sums. If $T$ is retract of an object $S$ and $\operatorname{Hom}(S,f)=0$, then $\operatorname{Hom}(T,f)$ is a retract of $0$ and hence is $0$.

Lemma (Ghost Lemma). Let $\mathcal S$ be a subcategory of a triangulated category $\mathcal T$. If $f_{i+1}: X_i \to X_{i+1}$ is a sequence of $\langle \mathcal S \rangle_0$-ghosts for $0 \leq i \leq n$, then the composition $f_n \circ \cdots \circ f_1$ is $\overline{\langle \mathcal S \rangle}_{n+1}$-ghost.

Proof. (Expand to view)

We prove this via induction on $n$. The case of $n=0$ is clear. For the induction step, we must prove that if $f: X \to Y$ is $\overline{\langle \mathcal S \rangle}_n$-ghost and $g: Y \to Z$ is $\langle \mathcal S \rangle_0$-ghost, then $gf$ is $\overline{\langle \mathcal S \rangle}_{n+1}$-ghost. In general, if a map $h$ is $\mathcal R$-ghost for some subcategory $\mathcal R$, then it is automatically $\overline{\mathcal R}$-ghost. So we reduce to checking $gf$ is $\langle \mathcal S \rangle_{n+1}$- ghost. For an object $U$ from $\langle \mathcal S \rangle_{n+1}$, we have a triangle $$ W \to U \to S \to W[1] $$ with $W \in \overline{\langle \mathcal S \rangle}_n$ and $S \in \overline{\langle \mathcal S \rangle}_0$. Given a map $U \to X$, the composition $$ W \to U \to X \overset{f}{\to} Y $$ vanishes as $f$ is $\overline{\langle \mathcal S \rangle}_n$-ghost. Thus, there exists a $S \to Y$

making the diagram commute. But as $g: Y \to Z$ is $\langle \mathcal S \rangle_0$-ghost, the composition $$ S \to Y \to Z $$ is $0$. Thus, $$ U \to X \overset{f}{\to} Y \overset{g}{\to} Z $$ is also $0$, and so, $gf$ is $\overline{\langle \mathcal S \rangle}_{n+1}$-ghost.

Note that $\mathcal S$-ghosts form an ideal in $\mathcal T$.

Corollary. Assume that there is a subcategory $\mathcal S$ of $\mathcal T$ with \(\overline{\langle \mathcal S \rangle}_n = \mathcal T\) Then, the ideal of $\langle \mathcal S \rangle_0$-ghosts is uniformly nilpotent with degree $n$.

Finally, if we can produce $\langle \mathcal S \rangle_0$ approximations, then we have a converse to the previous corollary.

Lemma. Suppose we have a subcategory $\mathcal S$ such that

  • the ideal of $\langle \mathcal S \rangle_0$-ghosts is uniformly nilpotent of degree $n$ and
  • for any object $X$ of $\mathcal T$ there exists a triangle \(S \to X \to Y \to S[1]\) with $S \in \overline{\langle \mathcal S \rangle}_0$ and $X \to Y$ $\langle \mathcal S \rangle_0$-ghost.

Then, \(\overline{\langle \mathcal S \rangle}_n = \mathcal T\)

Proof. (Expand to view)

Using induction, we can produce a sequence of ghosts $$ X_0 = X \to X_1 \to \cdots \to X_{n-1} \to X_n $$ with $$ \operatorname{cone}(X_i \to X_{i+1}) \in \langle \mathcal S \rangle_0 $$ Let $Y$ be the cone over the composition $X \to X_n$. Then from TR4 and the facts that each cone over $X_i \to X_{i+1}$ lies in $\langle \mathcal S \rangle_0$, one sees that $$ Y \in \langle \mathcal S \rangle_n $$ As it is composition of $n$ ghosts, the map $X \to X_n$ is zero. Thus, $$ Y \cong X_n \oplus X[1] $$ So $$ X \in \overline{\langle \mathcal S \rangle}_n $$