# Twisted line bundles Brauer class

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!

• The class $\alpha$ may be nontrivial, but the class $p^*\alpha$ is trivial. – 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.

• 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? – Libli Jan 8 '17 at 13:32
• The class $p^*\alpha$ is trivial. – Jason Starr Jan 8 '17 at 14:08
• @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... – Libli Jan 8 '17 at 17:15
• @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". – 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 $Y$!

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