## 13.35 Generators of triangulated categories

In this section we briefly introduce a few of the different notions of a generator for a triangulated category. Our terminology is taken from [BvdB] (except that we use “saturated” for what they call “épaisse”, see Definition 13.6.1).

Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. Denote $\langle E \rangle _1$ the strictly full subcategory of $\mathcal{D}$ consisting of objects in $\mathcal{D}$ isomorphic to direct summands of finite direct sums

\[ \bigoplus \nolimits _{i = 1, \ldots , r} E[n_ i] \]

of shifts of $E$. For $n > 1$ let $\langle E \rangle _ n$ denote the full subcategory of $\mathcal{D}$ consisting of objects of $\mathcal{D}$ isomorphic to direct summands of objects $X$ which fit into a distinguished triangle

\[ A \to X \to B \to A[1] \]

where $A$ is an object of $\langle E \rangle _1$ and $B$ an object of $\langle E \rangle _{n - 1}$. Each of the categories $\langle E \rangle _ n$ is a strictly full additive subcategory of $\mathcal{D}$ preserved under shifts and under taking summands. But, $\langle E \rangle _ n$ is not necessarily closed under “taking cones”, hence not necessarily a triangulated subcategory.

Lemma 13.35.1. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. The subcategory

\[ \langle E \rangle = \bigcup \nolimits _ n \langle E \rangle _ n \]

is a strictly full, saturated, triangulated subcategory of $\mathcal{D}$ and it is the smallest such subcategory of $\mathcal{D}$ containing the object $E$.

**Proof.**
To prove this it suffices to show: if $A \in \langle E \rangle _ a$ and $B \in \langle E \rangle _ b$ and if $A \to X \to B \to A[1]$ is a distinguished triangle, then $X \in \langle E \rangle _{a + b}$. We omit the details.
$\square$

Definition 13.35.2. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$.

We say $E$ is a *classical generator* of $\mathcal{D}$ if the smallest strictly full, saturated, triangulated subcategory of $\mathcal{D}$ containing $E$ is equal to $\mathcal{D}$, in other words, if $\langle E \rangle = \mathcal{D}$.

We say $E$ is a *strong generator* of $\mathcal{D}$ if $\langle E \rangle _ n = \mathcal{D}$ for some $n \geq 1$.

We say $E$ is a *weak generator* or a *generator* of $\mathcal{D}$ if for any nonzero object $K$ of $\mathcal{D}$ there exists an integer $n$ and a nonzero map $E \to K[n]$.

This definition can be generalized to the case of a family of objects.

Lemma 13.35.3. Let $\mathcal{D}$ be a triangulated category. Let $E, K$ be objects of $\mathcal{D}$. The following are equivalent

$\mathop{\mathrm{Hom}}\nolimits (E, K[i]) = 0$ for all $i \in \mathbf{Z}$,

$\mathop{\mathrm{Hom}}\nolimits (E', K) = 0$ for all $E' \in \langle E \rangle $.

**Proof.**
The implication (2) $\Rightarrow $ (1) is immediate. Conversely, assume (1). Then $\mathop{\mathrm{Hom}}\nolimits (X, K) = 0$ for all $X$ in $\langle E \rangle _1$. Arguing by induction on $n$ and using Lemma 13.4.2 we see that $\mathop{\mathrm{Hom}}\nolimits (X, K) = 0$ for all $X$ in $\langle E \rangle _ n$.
$\square$

Lemma 13.35.4. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. If $E$ is a classical generator of $\mathcal{D}$, then $E$ is a generator.

**Proof.**
Assume $E$ is a classical generator. Let $K$ be an object of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits (E, K[i]) = 0$ for all $i \in \mathbf{Z}$. By Lemma 13.35.3 $\mathop{\mathrm{Hom}}\nolimits (E', K) = 0$ for all $E'$ in $\langle E \rangle $. However, since $\mathcal{D} = \langle E \rangle $ we conclude that $\text{id}_ K = 0$, i.e., $K = 0$.
$\square$

## Comments (0)