## 13.36 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, and our definition of $add(\mathcal{A})$ is different).

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 $E$. It is clear that in the notation of Section 13.35 we have

$\langle E \rangle _1 = smd(add(E[-\infty , \infty ]))$

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 $\langle E \rangle _{n - 1}$. In the notation of Section 13.35 we have

$\langle E \rangle _ n = smd(\langle E \rangle _1 \star \langle E \rangle _{n - 1})$

Each of the categories $\langle E \rangle _ n$ is a strictly full additive (by Lemma 13.35.3) subcategory of $\mathcal{D}$ preserved under shifts and under taking summands. But, $\langle E \rangle _ n$ is not necessarily closed under “taking cones” or “extensions”, hence not necessarily a triangulated subcategory. This will be true for the subcategory

$\langle E \rangle = \bigcup \nolimits _ n \langle E \rangle _ n$

as will be shown in the lemmas below.

Lemma 13.36.1. Let $\mathcal{T}$ be a triangulated category. Let $E$ be an object of $\mathcal{T}$. For $n \geq 1$ we have

$\langle E \rangle _ n = smd(\langle E \rangle _1 \star \ldots \star \langle E \rangle _1) = smd({\langle E \rangle _1}^{\star n}) = \bigcup \nolimits _{m \geq 1} smd(add(E[-m, m])^{\star n})$

For $n, n' \geq 1$ we have $\langle E \rangle _{n + n'} = smd(\langle E \rangle _ n \star \langle E \rangle _{n'})$.

Proof. The left equality in the displayed formula follows from Lemmas 13.35.1 and 13.35.2 and induction. The middle equality is a matter of notation. Since $\langle E \rangle _1 = smd(add(E[-\infty , \infty ])])$ and since $E[-\infty , \infty ] = \bigcup _{m \geq 1} E[-m, m]$ we see from Remark 13.35.6 and Lemma 13.35.2 that we get the equality on the right. Then the final statement follows from the remark and the corresponding statement of Lemma 13.35.4. $\square$

Lemma 13.36.2. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. The subcategory

$\langle E \rangle = \bigcup \nolimits _ n \langle E \rangle _ n = \bigcup \nolimits _{n, m \geq 1} smd(add(E[-m, m])^{\star n})$

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. The equality on the right follows from Lemma 13.36.1. It is clear that $\langle E \rangle = \bigcup \langle E \rangle _ n$ contains $E$, is preserved under shifts, direct sums, direct summands. 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 $X \in \langle E \rangle _{a + b}$ by Lemma 13.36.1. Hence $\bigcup \langle E \rangle _ n$ is also preserved under extensions and it follows that it is a triangulated subcategory.

Finally, let $\mathcal{D}' \subset \mathcal{D}$ be a strictly full, saturated, triangulated subcategory of $\mathcal{D}$ containing $E$. Then $\mathcal{D}'[-\infty , \infty ] \subset \mathcal{D}'$, $add(\mathcal{D}) \subset \mathcal{D}'$, $smd(\mathcal{D}') \subset \mathcal{D}'$, and $\mathcal{D}' \star \mathcal{D}' \subset \mathcal{D}'$. In other words, all the operations we used to construct $\langle E \rangle$ out of $E$ preserve $\mathcal{D}'$. Hence $\langle E \rangle \subset \mathcal{D}'$ and this finishes the proof. $\square$

Definition 13.36.3. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$.

1. 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 $\langle E \rangle = \mathcal{D}$.

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

3. 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 13.36.4. Let $\mathcal{D}$ be a triangulated category. Let $E, K$ be objects of $\mathcal{D}$. The following are equivalent

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

2. $\mathop{\mathrm{Hom}}\nolimits (E', K) = 0$ 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 13.36.5. 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 13.36.4 $\mathop{\mathrm{Hom}}\nolimits (E', K) = 0$ for all $E'$ in $\langle E \rangle$. However, since $\mathcal{D} = \langle E \rangle$ we conclude that $\text{id}_ K = 0$, i.e., $K = 0$. $\square$

Lemma 13.36.6. Let $\mathcal{D}$ be a triangulated category which has a strong generator. Let $E$ be an object of $\mathcal{D}$. If $E$ is a classical generator of $\mathcal{D}$, then $E$ is a strong generator.

Proof. Let $E'$ be an object of $\mathcal{D}$ such that $\mathcal{D} = \langle E' \rangle _ n$. Since $\mathcal{D} = \langle E \rangle$ we see that $E' \in \langle E \rangle _ m$ for some $m \geq 1$ by Lemma 13.36.2. Then $\langle E' \rangle _1 \subset \langle E \rangle _ m$ hence

$\mathcal{D} = \langle E' \rangle _ n = smd( \langle E' \rangle _1 \star \ldots \star \langle E' \rangle _1) \subset smd( \langle E \rangle _ m \star \ldots \star \langle E \rangle _ m) = \langle E \rangle _{nm}$

as desired. Here we used Lemma 13.36.1. $\square$

Remark 13.36.7. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. Let $T$ be a property of objects of $\mathcal{D}$. Suppose that

1. if $K_ i \in D(A)$, $i = 1, \ldots , r$ with $T(K_ i)$ for $i = 1, \ldots , r$, then $T(\bigoplus K_ i)$,

2. if $K \to L \to M \to K[1]$ is a distinguished triangle and $T$ holds for two, then $T$ holds for the third object,

3. if $T(K \oplus L)$ then $T(K)$ and $T(L)$, and

4. $T(E[n])$ holds for all $n$.

Then $T$ holds for all objects of $\langle E \rangle$.

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