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Let $D$ be a triangulated category (the triangulated category in my mind is $D^{b}(X)$, that is the derived category of bounded complex of coherent sheaves on a smooth projective variety), $A \subset D$ is a subset of $D$. By $\langle A \rangle$ we mean the triangulated category generated by $A$ (i.e. the smallest triangulated category contain $A$). Let $$ A ^{\perp} : = \{ b \in D\mid \forall a \in A, \forall i \in \mathbb{Z}, \operatorname{Hom}(a, b[i])=0\}.$$

Then I was wondering if the following statement is true:

$\langle A \rangle = D\quad \Longleftrightarrow \quad A^{\perp}= \{0\}.$

I saw this result used implicitly in several places when $D = D^{b}(X)$, but even in this case, I still do not know the validity of the statement.

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Do you mean $\forall b\in A$ instead of $\forall b\in D$? –  Mahdi Majidi-Zolbanin Mar 28 '13 at 0:21
Thank you for pointing out, I have edited the post. –  Li Yutong Mar 28 '13 at 0:43
The implication $\implies$ is an easy exercise. In the other direction, it is easy of you assume the existence of a right adjoint to the inclusion $\langle A \rangle \to D$. Are there any extra conditions on your $A$? –  name Mar 28 '13 at 1:09
I think you probably want to assume also that $<A>$ is thick, that is closed under direct summands, otherwise there's no chance for your statement to work. –  Benjamin Antieau Mar 28 '13 at 1:26
As soon as the inclusion admits a right adjoint, there is a canonical equivalence of categories $\left<A\right>^\perp \stackrel{\sim}{\to} D/\left<A\right>$. Also $\left<A\right>$ is automatically thick under this assumption. See section 1.2 of Beilinson-Vologodsky for a reference. –  Adeel Dec 23 '13 at 14:51
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5 Answers

up vote 7 down vote accepted

Edited based on Sasha's answer:

I will assume that we are interested in the thick subcategory generated by $A$.

Under this assumption, the desired statement is closely related to a theorem of Neeman and Ravenel. See for instance the paper by Bondal and van den Bergh. The theorem of Neeman and Ravenel says that a subset $A$ of compact objects in a compactly generated triangulated category $D$ generates $D$ if and only if it classically generates $D^c$. Here, to say that $A$ classically generates $D^c$ means that $D^c$ is the smallest thick (closed under summands and equivalences) subcategory of $D^c$ containing $A$, while to say that $A$ generates $D$ means that if $Hom(a,b[i])=0$ for all $i\in\mathbb{Z}$ and $a\in A$, then $b=0$. Recall that an object $a$ is compact if $Hom(a,-)$ commutes with coproducts. This really only makes sense in the big derived category $D$, not in the small derived category $D^c$.

In the setting of (quasi-compact and quasi-separated) schemes, $D=D_{qc}(X)$, the derived category of complexes of $O_X$-modules with quasi-coherent cohomology, and $D^c=D_{perf}(X)$, the triangulated category of perfect complexes. When $X$ is smooth and projective, as in the original question, $D_{perf}(X)=D^b(X)$.

Thus, to finish answering the question, if the right-orthogonal to $A$ in $D$ vanishes, then the right-orthogonal to $A$ in $D^c$ vanishes (this is where compactness and compact generation is used). In this case $A$ generates $D$ by definition, which by the theorem implies that $A$ classically generates $D^c$, as desired.

In my original answer, I mistakenly went the other way, concluding that the right-orthogonal to $A$ in $D$ vanishes from the assumption that the right-orthogonal to $A$ in $D^c$ vanishes. This is just not true in general, as I interpret Sasha's example to show. For instance, consider $D^b(\mathbb{Z})$, the bounded derived category of the integers (I realize it's not projective, but the projective case is similar). Then, any complex right-orthogonal to $\mathbb{Z}/p$ for all $p$ must be zero. This can be seen because any complex has closed support in $Spec\mathbb{Z}$. So, if it contains no closed point it contains no point at all. But, if the support of a perfect complex is $0$, the complex is quasi-isomorphic to $0$. Thus, if $A=\{\mathbb{Z}/p\}$, then $A^\perp=0$ in $D^b(\mathbb{Z})$. It is pretty clear that the complex $\mathbb{Z}$ cannot be generated from these. One reason is as follows: if it could be, then it support would not contain the generic point. But, it does.

To summarize, if $X$ is a scheme and if $\langle A\rangle$ denotes the smallest thick subcategory of $D_{perf}(X)$ containing $A$, then $\langle A\rangle=D_{perf}(X)$ if and only if the right-orthogonal to $A$ in $D_{qc}(X)$ vanishes. This is not what was asked in the original post. But, it is at least good to know.

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Ben, my counterexample seems to contradict to what you claimed. –  Sasha Mar 28 '13 at 3:29
@Sasha: you are absolutely right. Thanks for pointing it out. I updated my answer. –  Benjamin Antieau Mar 28 '13 at 3:58
Thank you so much!! I am not good at triangulated category, so I try to summarize your answer as well as the others: If $<A>$ is a thick triangulated category, then the statement is true. Is this understanding correct or not? One of example I came across is $X=\mathbb{P}^n$ and $A=\{ \mathcal{O}(i)\mid i \in \mathbb{Z}\}$, is this fulfill your criteria? –  Li Yutong Mar 29 '13 at 0:14
@Li Yutong: the statement is true if you assume that the right orthogonal vanishes in the big derived category, $D_{qc}(X)$, not just in $D^b(X)$. It is also true if the subcategory is admissible, as Beren explained. This is the case for $\langle A\rangle$ when $A$ is the collection of line bundles on $\mathbb{P}^n$. For a detailed look at the admissible case, I cannot suggest more strongly that you look at Bondal's paper "Rrepresentation of associative algebras and coherent sheaves" and the paper of Bondal and Kapranov. –  Benjamin Antieau Mar 29 '13 at 21:23
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A nice counterexample is the following. Let $A$ be the set of all structure sheaves of points. Then $A^\perp = 0$ while $\langle A\rangle$ is definitely not $D$.

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$\Leftarrow$ is false: take $D=D^b(\mathbb Z)$ the derived category of bounded complexes of abelian groups and $A=\{\mathbb Z^2\}$. Then $\mathbb Z\notin \langle A\rangle$ but $A^{\perp}=0$. Indeed, $\langle A\rangle$ is formed by bounded complexes of free abelian groups of finite even rank, hence by Euler characteristic arguments $\mathbb Z\notin \langle A\rangle$.

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I think one should just take the thick subcategory generated by $A$. –  Benjamin Antieau Mar 28 '13 at 1:39
Well, I guess Li Yutong asked about the triangulated subcategory generated by $A$ because it is this category, and not the thick subcategory generated by $A$, what arises in his research. It's not a matter of what one should take, but of what one comes up with. –  Fernando Muro Mar 28 '13 at 1:49
@Muro: fair enough. When I said "should" I meant "in order to obtain a correct statement." –  Benjamin Antieau Mar 28 '13 at 1:58
@Antileau: the correct statement is that $\Rightarrow$ is true but $\Leftarrow$ is false. –  Fernando Muro Mar 28 '13 at 2:00
Dear Fernanado Muro, thank you for point out this nice example! –  Li Yutong Mar 29 '13 at 0:16
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This problem is very clear if we consider derived module categories of rings. For simplicity, let us consider a finite dimensional algebra $A$ over a field $k$. Consider the regular left module $_AA$ and the bounded derived category $D^b(A)$ of finitely generated $A$-modules. Then the category classically generated by $_AA$ and all its summands is the homotopy category of perfect complexes, which can be regarded as a full triangulated subcategory of $D^b(A)$, and it is a proper subcategory if the global dimension is infinity. On the other hand, $D^b(A)$ is generated by $A$.

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Let $\mathcal{T}$ be a triangulated category and let $\mathcal{S} \subset \mathcal{T}$ be a triangulated subcategory. The left orthogonal of $\mathcal{S}$ is the subcategory

$\phantom{}^\perp S = \lbrace Y \in \mathcal{T} \mid [X,Y] = 0 \text{ for all } X \in S\rbrace$

while the right orthogonal of $\mathcal{S}$ is the subcategory

$\mathcal{S}^\perp = \lbrace X \in \mathcal{T} | [X,Y] = 0 \text{ for all } Y \in \mathcal{S}\rbrace$.

Both these subcategories are thick triangulated subcategories of $\mathcal{T}$.

We say that $\mathcal{S}$ is right admissible if the inclusion functor $\mathcal{S} \hookrightarrow \mathcal{T}$ has a right adjoint. This is equivalent to the following condition: for every object $X$ in $\mathcal{T}$ there exists an exact triangle $W \rightarrow X \rightarrow Y \rightarrow W[1]$ with $W \in \mathcal{S}$ and $Y \in \mathcal{S}^\perp$.

Thus, if $S$ is a right admissible subcategory then $S^\perp = (0)$ implies that for every object $X \in \mathcal{T}$ there is an object $W \in \mathcal{S}$ and an isomorphism $W \rightarrow X$. Thus, $\mathcal{S}^\perp = (0)$ if and only $\mathcal{S} = \mathcal{T}$ when $\mathcal{S}$ is a right admissible subcategory. (Obviously, $\mathcal{T}^\perp = (0)$.)

(Note: it is easy to check that $A^\perp$ as defined by you is the same as $\langle A\rangle^\perp$ as defined by me.)

Edit to answer comment: $\langle A \rangle$ won't be admissible in general. However, admissible subcategories are often studied in the literature on $\mathcal{T} = D^b(X)$; it is quite likely that the subcategories the original poster is coming across in the literature are admissible.

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But why should $\langle A \rangle$ be admissible in general? Or even in the case of $D^b(X)$ with $X$ smooth projective? –  name Mar 28 '13 at 1:24
In general it is not. For instance, if $Z$ is a proper closed subset of $X$ (with $X$ connected say), then the subcategory of complexes supported on $Z$ is a non-admissible subcategory in $D^b(X)$. –  Benjamin Antieau Mar 28 '13 at 1:40
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