## Strong generation and saturation

Definition. Let $\mathcal T$ be a triangulated category linear over a field $k$. We say $\mathcal T$ is Ext-finite if for any pair of objects $X$ and $Y$ we have $$\dim_k \bigoplus_{j \in \mathbb{Z}} \operatorname{Hom}(X,Y[j]) < \infty$$ An additive functor $H : \mathcal T \to \operatorname{Vect} k$ is Ext-finite if for any object $X$ we have $$\dim_k \bigoplus_{j \in \mathbb{Z}} H(X[j]) < \infty$$

Lemma. Let $\mathcal T$ be an Ext-finite triangulated category and let $X$ and $Y$ be objects. Then there exists a triangle $$Z \to X \to X^\prime \to Z[1]$$ where $Z \in \langle Y \rangle_0$ and $X \to X^\prime$ is $\langle Y \rangle_0$-ghost.

Proof. (Expand to view)

We introduce some notation. For $$\operatorname{Hom}^\ast(Y,X) \otimes_k Y$$ we really mean $$\bigoplus_{i=1}^d Y[n_i]$$ where $d$ is the dimension of $\bigoplus \operatorname{Hom}(Y[j],X)$ and $n_1,\ldots,n_d$ are the homological degrees of a basis. This object is independent of the basis up to isomorphism. There is a natural evaluation map $$\operatorname{Hom}^\ast(Y,X) \otimes_k Y \to X$$ whose action of the factor $Y[n_i]$ is the corresponding map in the basis for $\bigoplus \operatorname{Hom}(Y[j],X)$. We set $$Z = \operatorname{Hom}^\ast(Y,X) \otimes_k Y$$ and use the triangle $$Z \to X \to X^\prime \to Z[1]$$ where $X^\prime$ is the cone over $Z \to X$. By construction, any map $Y[l] \to X$ will lift to $Y[l] \to Z$. Just take the factor in $Z$ corresponding to $Y[l] \to X$. Then, the composition $Y[l] \to X \to X^\prime$ will be $0$. So $X \to X^\prime$ is $\langle Y \rangle_0$-ghost.

Assume we have a cohomological functor $H : \mathcal T^{op} \to \operatorname{Vect} k$ which is Ext-finite. We want to employ the “resolution” procedure from Brown representability to $H$ with a single object (and its shifts) playing the role of the compact generators.

We start with $$X_0 := \bigoplus_{j \in \mathbb{Z}} H(G[j]) \otimes_k G[j]$$ Since the summands of $X_0$ are indexed by elements $H(G[j])$, there is a natural element of $H(X_0)$ given by using the element from the index. From Yoneda, we get a natural transformation $$\xi_0 : h^{X_0} \to H$$

Inductively, we set $$K_i(G[j]) := \ker\left( \xi_i(G[j]): \operatorname{Hom}(G[j],X_i) \to H(G[j]) \right)$$ and $$Y_i := \bigoplus_{j \in \mathbb{Z}} K_i(G[j]) \otimes_k G[j]$$ We have natural map $Y_i \to X_i$ and we let $$X_{i+1} = \operatorname{cone}(Y_i \to X_i)$$

Note that from Ext-finiteness of $\mathcal T$ and $H$ all this constructions yield well-defined objects of $\mathcal T$.

As in the proof of Brown Representability, we get a natural transformation $$\xi_{i+1} : h^{X_i} \to H$$ such that the diagram

commutes.

Since we do not have anything beyond finite coproducts, we cannot pass to the homotopy colimit to achieve a representing object on the thick subcategory generated by $G$ so we need to prove that after finite number of steps we can do this.

Lemma. On $\overline{\langle G \rangle}_0$, $H$ is isomorphic to a retract of $h^{X_1}$.

Proof. (Expand to view)

For any object $E \in \overline{\langle G \rangle}_0$, we have a commutative diagram

where the diagonal map vanishes from the construction $X_1$. Thus, we get a map $$H(E) \to \operatorname{Hom}(E,X_1)$$ which splits $$\operatorname{Hom}(E,X_1) \to H(E)$$ This construction is natural in $E$.

So on $\overline{\langle G \rangle}_0$, we have represented our object, at least allowing for retracts. To achieve this we needed that $$\operatorname{Hom}(S,X_i) \to H(X_i) \to 0$$ is exact and the composition $$\ker \xi_i(S) \to \operatorname{Hom}(S,X_i) \to \operatorname{Hom}(S,X_{i+1})$$ is zero.

We will show that we can achieve this for $\overline{\langle G \rangle}_n$ by composing a sequence of the maps $X_i \to X_{i+1}$.

Proposition. For each $n \geq 1$, there is an $M > 0$ such that $H$ is isomorphic to a retract of $X_M$ on $\overline{\langle G \rangle}_n$.

Proof. (Expand to view)

We will prove the following statement. For each $n$ there is a $N$ such that for any $E$ in $\overline{\langle G \rangle}_n$ we have that * the sequence $$\operatorname{Hom}(E,X_N) \to H(E) \to 0$$ is exact and * the composition $$\ker \xi_N(E) \to \operatorname{Hom}(E,X_N) \to \operatorname{Hom}(E,X_{N+2})$$ is $0$. Then we get $H$ is isomorphic to a retract of $h^{X_{N+2}}$ on $\overline{\langle G \rangle}_n$. The base case of the induction is the previous lemma. Assume it is true for $n$. Note it suffices to prove it for $\langle G \rangle_0 \ast \langle G \rangle_n$ since the two conditions are preserved under passage to sums, shifts, and summands. Take $E$ lying in a triangle $$F \to E \to E^\prime \to F[1]$$ with $F \in \langle G \rangle_0$ and $E^\prime \in \langle G \rangle_n$. Then for each $m$, we have the commutative diagram

We can take $m$ large enough so that $$\operatorname{coker} \xi_m(E^\prime) = \operatorname{coker} \xi_m(F) = 0$$ A diagram chase gives a injective map $$\operatorname{coker} \xi_m(E) \to \operatorname{ker} \xi_m(E^\prime[-1])$$ Taking $m$ larger if necessary we have $$\operatorname{coker} \xi_m(E) = 0$$ so $$\operatorname{Hom}(E,X_m) \to H(E) \to 0$$ is exact. Finally, $m^\prime > m$ so that $$\operatorname{ker} \xi_m(E) \to \operatorname{Hom}(E,X_{m^\prime})$$ and $$\operatorname{ker} \xi_m(F) \to \operatorname{Hom}(F,X_{m^\prime})$$ are both $0$ maps.

If we know that for some large enough $n$ we have $$\overline{\langle G \rangle}_n = \mathcal T$$ then we get the following corollary due to Bondal and Van den Bergh

Corollary. If $\mathcal T$ is a strongly generated Ext-finite triangulated category and $H$ is an Ext-finite (co)homological functor, then $H$ is representable by a retract of an object of $\mathcal T$.

Note that being a strongly generated triangulated category is a condition preserved by taking opposites.