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?

up vote 11 down vote accepted

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 $\mathrm{B}\mathbb{G}_{\mathrm{m}}$.

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.

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