5
$\begingroup$

Let $\mathcal{T}$ be a triangulated category which has arbitraty direct sums. An object $E\in \mathcal{T}$ is called compact if the functor Hom$(E,-)$ commutes with arbitrary direct sums.

A triangulated category $\mathcal{T}$ is called compactly generated if there is a set $\mathcal{S}$ of objects of $\mathcal{T}$ which consists of compact objects and satisfies $$ \text{Hom}(G[n],X)=0, ~\forall G\in \mathcal{S}, \forall n\in \mathbb{Z} \text{ implies } X=0. $$

In particular, an object $E$ is a compact generator of $\mathcal{T}$ if $\{E\}$could play the role of $\mathcal{S}$ in the above definition.

It is well-known that for a quasi-compact, separated scheme $X$, the derived category of complexes of quasi-coherent $\mathcal{O}_X$-modules $D(X)$ has a compact generator.

My question is: is there a triangulated category $\mathcal{T}$ which is compactly generated but does not have a compact generator?

$\endgroup$

2 Answers 2

11
$\begingroup$

Examples can be found amongst the derived categories of algebraic stacks, see Hall--Rydh: Algebraic groups and compact generation of their derived categories of representations, in particular theorem A. The easiest example should be $\mathrm{B}\mathbb{G}_{\mathrm{m}}$.

$\endgroup$
7
$\begingroup$

Possibly the stack example in pbelman's answer is of this form, but an elementary way to construct examples is by taking infinite products.

Let $\{\mathcal{C}_i\}_{i\in I}$ be an infinite collection of nonzero compactly generated triangulated categories, and $\mathcal{C}=\prod_{i\in I}\mathcal{C}_i$.

It is easy to see that an object $(X_i)_{i\in I}$ of $\mathcal{C}$ is compact if and only if $X_i$ is compact in $\mathcal{C}_i$ for every $i$ and $X_i=0$ for all but finitely many $i$. Such objects clearly generate $\mathcal{C}$, but a generating set must involve generators of every $\mathcal{C}_i$, so no single compact object of $\mathcal{C}$ is a generator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.