Let $X$ be a variety (or more generally a quasi-compact, separated scheme) and $D(X)$ be the derived category of complexes of $\mathcal{O}_X$-modules with quasi-coherent cohomologies. Let $\mathcal{E}$ be a compact generator of $D(X)$. It is well-known that $\mathcal{E}$ is a perfect complex (Generators and representability Theorm 3.1.1). Moreover it is also well-known that $\mathcal{E}$ is quasi-isomorphic to a strictly perfect complex $\mathcal{S}$, i.e. each $\mathcal{S}^n$ is a locally free $\mathcal{O}_X$-module with finite rank and $\mathcal{S}^n=0$ for $|n|\gg 0$. So without loss of generality we can assume $\mathcal{E}$ itself is strictly perfect.

Now we consider the dual of $\mathcal{E}$, $\mathcal{E}^{\vee}$, i.e. $\mathcal{E}^{\vee,n}=\mathcal{H}om(\mathcal{E}^{-n},\mathcal{O}_X)$ and the differential is defined naturally. In fact $\mathcal{E}^{\vee}$ could be considered as $\mathcal{RH}om(\mathcal{E},\mathcal{O}_X)$, see Stack Project Lemma 21.35.9.

My question is: is $\mathcal{E}^{\vee}$ also a compact generator of the derived category $D(X)$? Why?

  • 1
    $\begingroup$ I think that if you believe that: 1. there's a theory of symmetric monoidal stable infinity categories 2. compact <=> dualizable (for nice enough scheme - in particular your example) Then the dual of a compact generator has to be dualizable and hence compact. I think a similar formal argument can show that it if one is a right generator (right orthogonal is the whole category) than the other is a left generator (left orthogonal is the whole category). $\endgroup$ Jan 26, 2017 at 9:26

1 Answer 1


Let me try to sketch an argument (though I am not quite sure in details).

A theorem of Neeman implies that a compact object $M$ in a compactly generated triangulated category $T$ is a generator if and only if the smallest full dense subcategegory $T$ containing $M$ equals with the subcategory of compact objects of $T$. Now, the latter condition remains true when you apply duality (since compactness is preserved by duality; isn't it in your case?).

  • $\begingroup$ I am still not sure why the smallest full dense subcategory $T$ containing $M$ is preserved by duality. $\endgroup$ Feb 8, 2017 at 15:17
  • $\begingroup$ Isn't the restrition of the duality functor to compact objects an fully faithful (contravariant) functor? $\endgroup$ Feb 9, 2017 at 19:18

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.