Timeline for Is the Brauer group functor a Zariski sheaf?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2018 at 0:19 | history | edited | Minseon Shin | CC BY-SA 4.0 |
removed misleading claims
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| Sep 27, 2015 at 14:55 | comment | added | მამუკა ჯიბლაძე | This answer must be relevant | |
| Sep 25, 2015 at 16:07 | comment | added | Qiaochu Yuan | The Brauer functor takes values in symmetric monoidal $2$-groupoids, at least; maybe to get the best descent properties you need to derive everything (so derived Azumaya algebras, etc). | |
| Sep 25, 2015 at 14:58 | vote | accept | Minseon Shin | ||
| Sep 25, 2015 at 11:46 | comment | added | David Roberts♦ | Yes, it should be something like a group 2-stack, but it's not entirely clear how. | |
| Sep 25, 2015 at 8:54 | comment | added | მამუკა ჯიბლაძე | In fact I just realized there is one more level, this 2-groupoid carries a multiplication making it a group-up-to-... in 2-groupoids | |
| Sep 25, 2015 at 8:45 | comment | added | მამუკა ჯიბლაძე | @DavidRoberts So is it a 2-stack (rather than just a (1-)stack) then? If one assigns to an affine open the 2-groupoid of Azumaya algebras, Morita equivalences and their natural transformations (over that open), there seems to be enough higher structure to formulate that, no? | |
| Sep 25, 2015 at 7:42 | comment | added | David Roberts♦ | @მამუკაჯიბლაძე - $Br$ is really the decategorification of a 2-functor (valued in something complicated and not groupoidal, as far as I've worked with it), as you say. The extent to which it is a stack is tricky: it satisfies a pseudo or lax descent for covers of separated schemes by two affines, by result of Gabber. | |
| Sep 25, 2015 at 7:09 | answer | added | Martin Bright | timeline score: 17 | |
| Sep 25, 2015 at 6:11 | comment | added | მამუკა ჯიბლაძე | It might be 2-descent: you need data of the form (Azumaya algebras $A_i$ over (rings corresponding to affine) $U_i$, $A_i|_{U_{ij}}$-$A_j|_{U_{ij}}$-bimodules $B_{ij}$ over $U_{ij}:=U_i\cap U_j$ together with coherent isomorphisms $B_{ii}\cong A_i$ over $U_i$ and $B_{ij}|_{U_{ijk}}\otimes_{A_j|_{U_{ijk}}}B_{jk}|_{U_{ijk}}$ $\cong$ $B_{ik}|_{U_{ijk}}$ over $U_{ijk}:=U_i\cap U_j\cap U_k$) for all (not necessarily distinct) $i$, $j$, $k$. In other words it might be a 2-stack rather than a sheaf. | |
| Sep 25, 2015 at 4:24 | history | asked | Minseon Shin | CC BY-SA 3.0 |