5
$\begingroup$

Recall that a triangulated subcategory $\mathcal{A}$ of a triangulated category $\mathcal{B}$ is called admissible if the inclusion functor has both left and right adjoints.

Is it true that all admissible subcategories of $D^b(\mathbb{P}^n_k)$ (the bounded derived category of coherent sheaves on $\mathbb{P}^n_k$, for a field $k$) are generated by exceptional collections?

Perhaps the right setting for the question is to consider admissible subcategories $\mathcal{A}$ of a $k$-linear triangulated category $\mathcal{B}$ generated by a (strong) exceptional collection rather than just $D^b(\mathbb{P}^n_k)$.

$\endgroup$
4
  • 2
    $\begingroup$ This is definitely not true for a category generated by an exceptional collection (without strongness assumption though): an elementary geometric counterexample can be found in the paper arxiv.org/abs/1304.0903 $\endgroup$ Nov 9, 2013 at 20:04
  • $\begingroup$ @AntonFonarev: Thanks a lot for the reference. $\endgroup$
    – naf
    Nov 10, 2013 at 6:46
  • $\begingroup$ For $n=1$ this is folklore, whilst for $n=2$ it was settled last year by Pirozhkov in arxiv.org/abs/2006.07643. For $n\geq 3$ it's open. If @naf thinks this should be an accepted answer I can turn this comment into an answer. $\endgroup$
    – pbelmans
    Apr 9, 2021 at 13:22
  • $\begingroup$ @pbelmans: Thanks, and yes, I'd be happy to accept this as an answer. $\endgroup$
    – naf
    Apr 9, 2021 at 13:47

1 Answer 1

2
$\begingroup$

Reposting my comment as a slightly extended answer.

For $n=1$ this is folklore (it is I think a pleasant exercise using global dimension 1 and the description of coherent sheaves on curves), whilst for $n=2$ it was settled last year by Pirozhkov in his preprint Admissible subcategories of del Pezzo surfaces. He also gives the analogous statement for del Pezzo surfaces.

For $n\geq 3$ it's still open as far as I know.

$\endgroup$

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.