Timeline for Modules over an Azumaya algebra and modules over the associated Brauer-Severi variety
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 8, 2015 at 11:01 | comment | added | Bernie | @Jason : Ah, I see. I wasn't thinking "etale" enough :-). Thanks a lot for your help. | |
| Sep 7, 2015 at 15:07 | comment | added | Jason Starr | If you begin with a sheaf $N$ that is (etale locally) isomorphic to $(I^\vee)\otimes_{\mathcal{O}_X}f^*L$, then $G(N)$ is $f_*(\textit{Hom}_{\mathcal{O}_X}(I^\vee,I^\vee)\otimes_{\mathcal{O}_X} f^*L)$, which by the projection formula is $\mathcal{A}\otimes_{\mathcal{O}_Y} L$. Thus $F(G(N))$ is $I^\vee\otimes_{f^*\mathcal{A}} f^*(\mathcal{A}\otimes_{\mathcal{O}_Y} L) \cong I^\vee\otimes_{\mathcal{O}_X}f^*L$. | |
| Sep 7, 2015 at 14:51 | comment | added | Bernie | @Jason. Thanks, what you wrote is what I looking for :-). Denoting your first functor by $F$ and the second one by $G$, I can show that $G\circ F=id$ using the projection formula. But if I try to compute $F\circ G$ using base change and the formula $f_{*}Hom_X(f^{*}E,N)=Hom_Y(E,f_{*}N)$ I end up with a sheaf of the form $f^{*}f_{*}(N)$ where $N$ is a bundle on $X$ with the desired properties. Why is this isomorphic to $N$? Maybe a silly question, but usually the natural morphism $f^{*}f_{*}N\rightarrow N$ is far from being an isomorphism. | |
| Sep 4, 2015 at 18:16 | comment | added | Jason Starr | I am afraid that I do not understand what you are asking. The functors, $M \mapsto I^\vee\otimes_{f^*\mathcal{A}} f^*M$ and $N\mapsto f_*\textit{Hom}_{\mathcal{O}_X}(I^\vee,N)$ establish an equivalence between the category of $\mathcal{A}$-modules that are locally free as $\mathcal{O}_Y$-modules and the category of locally free $\mathcal{O}_X$-modules whose restriction to all geometric fiber is isomorphic to a direct summand of $(I^\vee)^{\oplus n}$ for some choice of $n$. Once you have the functors, checking that they are inverse can be done after base change to $Y'$, where it is easy. | |
| Sep 4, 2015 at 15:07 | comment | added | Bernie | @Jason : My problem is not with the sheaf $I$ (all about $I$ is in Quillen's article Higher algebriac K-theory I), but with sheaves of $End(I)$ modules. For examples, assume $M$ has rank one as an $A$-module and is locally free, then the corresponding $O_X$-module, is locally free of rank $r$ on $X$. Does every rank $r$ bundle appaer this way? Does every rank $r$ bundle on $X$ give an $A$-module of rank one after tensoring with the dual of $I$ and pushing down to $S$? I don't see how to find properties of the bundles on $X$, maybe all possible bundles can appear? | |
| Sep 4, 2015 at 12:22 | comment | added | Jason Starr | ... Thus, by descent and Skolem-Noether, there is a canonical isomorphism of $f^*\mathcal{A}$ with $\text{End}(I)$ (or its opposite ring, as you write, I am not quite sure). For restriction to fibers, surely sasha meant "geometric fibers". Thus, base change again to $Y'$ and check there. | |
| Sep 4, 2015 at 12:18 | comment | added | Jason Starr | Just to point out, the sheaf $I$ that you mention is the unique locally free sheaf of rank $r$ sitting in the short exact sequence $$0\to \Omega_{X/Y}\to I\to \mathcal{O}_X \to 0,$$ coming from the canonical generator of $R^1f_*\Omega_{X/Y}$. If you make any base change to $Y'$ such that $\mathcal{A}' = \text{End}(E)$, then $X'$ is $\mathbb{P}_Y(E)$, and $I'$ is $(f')^*E^\vee\otimes \mathcal{O}_{\mathbb{P}(E)}(-1)$. Thus the pullback of $\text{End}(E)$ equals the pullback of $\text{End}(I)$. | |
| Sep 4, 2015 at 12:01 | history | asked | Bernie | CC BY-SA 3.0 |