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Feb 10, 2012 at 9:13 vote accept Sasha
Feb 9, 2012 at 19:00 answer added user19475 timeline score: 2
Feb 9, 2012 at 17:15 comment added Sasha @Dror: I don't know details, but my general feeling says that your argument does not contradict my statement? In my understanding, Serre-Hohcschild follows from interpretation of the unstructured global sections $Sh_G (Y) \to Ab$ as composition of $\Gamma: Sh_G (Y) \to G-mod$ with invariants $inv: G-mod \to Ab$; Where usually we already understand that $Sh_G (Y)$ is equivalent to $Sh(G\Y)$, for the $G\Y$ we have in mind. I might be wrong.
Feb 9, 2012 at 16:34 comment added Dror Speiser Hey Sasha! I don't think this has to be an equivalence of categories. If $X$ is a smooth projective variety, then the Hochschild-Serre spectral sequence gives us the sequence: $0\rightarrow Pic(X)\rightarrow Pic(\bar{X})^\Gamma\rightarrow Br(k)$, where it is known that the last map is not always zero. Using that $Pic$ describes invertible sheaves, it seems that sometimes the category $Sh(\bar{X}_{et})^\Gamma$ is bigger. A similar situation occurs when you go up one notch and consider $O_X$-algebras. There you can have equivariant Azumaya algebras that are not base-change.
Feb 9, 2012 at 13:47 history asked Sasha CC BY-SA 3.0