This question is mainly a reference request about the order of a Brauer class on a smooth projective variety over $\mathbb{C}$. Namely, let $X$ be a smooth complex projective variety and $\alpha$ be a Brauer class on $X$.

I have read in a few papers on the ArXiv that the order of $\alpha$ divides the rank of any $\alpha$-twisted sheaf. In particular, the existence of a $\alpha$-twisted line bundle would imply that $\alpha$ is trivial.

On the other hand, let $ p : Y \longrightarrow X$ be a non-trivial Brauer-Severi variety associated to the class $\alpha \in Br(X)$. It seems that one can define a $p^*\alpha$-twisted line bundle $\mathcal{O}_{Y/X}(1)$ and it seems this twisted line bundle is not a line bundle (so that the class $p^* \alpha$ is non trivial).

I am not sure how to reconcile these two claims. Is there something obvious I am missing?

Thanks a lot!

  • 3
    $\begingroup$ The class $\alpha$ may be nontrivial, but the class $p^*\alpha$ is trivial. $\endgroup$ – Jason Starr Jan 8 '17 at 11:31

There are many functors from the category of twisted sheaves to the category of untwisted sheaves. Typically, if $E$ is a locally free $\alpha$-twisted sheaf then $\mathcal{H}om(E,-)$ is such a functor. It identifies the category of $\alpha$-twisted sheaves with the category of $\mathcal{E}nd(E)$-modules.

In many cases, by a rank of an $\alpha$-twisted sheaf $F$ people understand the rank of the corresponding sheaf $\mathcal{H}om(E,F)$. In this sense, existence of a rank 1 twisted sheaf is equivalent to the vanishing of the Brauer class.

Sometimes, however, one can divide this by the rank of $\mathcal{E}nd(E)$. This "divided rank" may be equal to 1 without $\alpha$ being trivial. For instance, the divided rank of $E$ itself is equal to 1.

  • $\begingroup$ thanks a lot for your answer! So just to be sure I understood, in the case of a Severi Brauer variety, the "rank" of $\mathcal{O}_{Y/X}(1)$ would be the rank of $E$ (where $E$ is the twisted vector bundle which defines $Y$). Furthermore the class $p^* \alpha$ is non-trivial and its order is the rank of $E$. Is that correct? $\endgroup$ – Libli Jan 8 '17 at 13:32
  • 2
    $\begingroup$ The class $p^*\alpha$ is trivial. $\endgroup$ – Jason Starr Jan 8 '17 at 14:08
  • $\begingroup$ @JasonStarr : so if the class $p^* \alpha$ is trivial, this means that $\mathcal{O}_{Y/X}(1)$ is a true line bundle? It seems quite strange... $\endgroup$ – Libli Jan 8 '17 at 17:15
  • 2
    $\begingroup$ @Libli: As Jason said, the class $p^*\alpha$ is trivial. On the other hand, the category of (untwisted) sheaves on $Y$ that restrict as a multiplicity of $O(1)$ to each fiber over $X$ is equivalent to the category of $\alpha$-twisted sheaves on $X$. All this, I believe, is explained in a paper of Bernardara "A semiorthogonal decomposition for Brauer-Severi schemes". $\endgroup$ – Sasha Jan 8 '17 at 17:34

OK so I think I understand what I missed from the beginning. If I take a particular $\alpha$-twisted sheaf, say $F$, then the object $F$ itself, as a twisted sheaf, depends on the Cech cocycle I use to represent $\alpha$. In particular, the Cech cocycle used to define $\mathcal{O}_{Y/X}(1)$ on a Severi Brauer variety is non-trivial.

On the other hand, the category of $\alpha$-twisted sheaves only depends (up to equivalence) on the cohomology class $\alpha \in H^2(X,\mathcal{O}^{\star}_X)_{et}$.

Hence, though $\mathcal{O}_{Y/X}(1)$ is not a line bundle, one still has that the category of $p^* \alpha$-twisted sheaves on $Y$ is equivalent to the category of untwisted sheaves on $Y$!

  • 1
    $\begingroup$ I do not believe that is the correct intepretation. $\endgroup$ – Jason Starr Jan 9 '17 at 8:07
  • $\begingroup$ @JasonStarr : and the correct interpretation would be? $\endgroup$ – Libli Jan 10 '17 at 21:39

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.