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This question arose from my answer to To what extent does a torsor determine a group: it turns out that I do not know one thing about it.

Let $G$, $G'$ be groups in some nice enough category (you may assume a topos, if you feel like that). Can one find a nice intrinsic simplification of the condition "there exists a $G$-$G'$-bitorsor"?

It is clear that a necessary condition for the existence of a bitorsor is that $G$ and $G'$ are locally isomorphic, i. e. there is an object $B$ with global support such that the groups $B\times G\to B$ and $B\times G'\to B$ over $B$ are isomorphic over $B$. Is this also sufficient?

Can one do better? By this I mean not quantifying over objects but rather concocting some condition out of $G$ and $G'$ alone?

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    $\begingroup$ Assuming $G$ and $G'$ are locally isomorphic, $G'$ determines a class $c\in\mathrm{H}^1(S,\operatorname{\underline{Aut}}(G))$ (say we work in a topos $S$). There exists a $(G,G')$-bitorsor if and only if $c$ is in the image of the map $\mathrm{H}^1(S,G)\to\mathrm{H}^1(S,\operatorname{\underline{Aut}}(G))$ deduced from the conjugation map $G\to\operatorname{\underline{Aut}}(G)$. Of course, this is a rather tautological answer, so I am not sure this is what is required. $\endgroup$ Jan 8, 2018 at 7:51
  • $\begingroup$ @LaurentMoret-Bailly Thanks, - although, yes, I would prefer something more explicit, this is still rather informative. So if you care to make this an answer I would wait for a while and if there will be nothing better I would accept this one too. $\endgroup$ Jan 8, 2018 at 8:19
  • $\begingroup$ What do I mean by something more explicit: there is an exact sequence$$0\to\operatorname{Center}(G)\to G\to\operatorname{\underline{Aut}}(G)\to\operatorname{\underline{Out}}(G)\to1$$(with the middle part a crossed module); I wonder whether there is some trick to form an obstruction for existence of a bitorsor in terms of some $\operatorname{\underline{Out}}(G)$-torsor, or a torsor over this crossed module, or some (higher degree) cohomology class with coefficients in the center or something like that... $\endgroup$ Jan 8, 2018 at 8:28
  • $\begingroup$ What does "global support" mean? $\endgroup$ Jan 9, 2018 at 6:25
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    $\begingroup$ @QiaochuYuan In a topos, just that the map to the terminal is epi. In worse categories, one has to be more careful of course - from this map being regular epi to the condition that the forgetful functor ${\mathbf C}/X\to{\mathbf C}$ is monadic, things like that $\endgroup$ Jan 9, 2018 at 6:31

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[I don't have enough 'points' to comment; below is really just a comment.]

If you consider instead your 'nice enough' category to be locales, then having a bi-torsor between two open localic groups G and G' implies that their toposes of sheaves are equivalent. Quite different localic groups can define the same topos, so my instinct is that there is no easy or natural way to determine (Morita) equivalence by inspecting the groups other than to tautologically give the definition. (Refer to Remark C5.2.14(d) for a reference to a concrete example, for the groupoid case at least.)

I was intrigued by the question because you seemed to be interested in working on things to do with torsors in more general categorical contexts. Since I have done some work on torsors (effectively Hilsum-Skandalis maps) in a cartesian category, which I think is the most general possible context, I hope you don't mind my providing a link to that work:

http://www.christophertownsend.org/Documents/HilsumSkandalisFrobenius.pdf

The general gist of the question seems to be: what can we say about how group(oids) are related/constructed given information about how their categories of equivariant sheaves are related; I find this to be an interesting avenue and only know partial answers.

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  • $\begingroup$ Hi Chris, glad to see you on MO! And thanks for the interesting link. The Morita business seems certainly very relevant for the question. While I am still long way from having digested the information there, may I ask you this: a comment by Laurent Moret-Bailly to the question suggests that there might be an obstruction of cohomological nature to the existence of a bitorsor between locally isomorphic groups. Is something like that visible in the context of your work? Do general locally isomorphic groups enter your picture? $\endgroup$ Jan 9, 2018 at 19:01
  • $\begingroup$ One more thing just came to my mind, about possibly more explicit kind of obstruction. In principle one would expect that there exists a $G$-$G'$-bitorsor iff $G=\operatorname{\underline{Aut}}(X)$ and $G'=\operatorname{\underline{Aut}}(X')$ for some locally isomorphic $X$ and $X'$ - the candidate bitorsor then being $\operatorname{\underline{Iso}}(X,X')$. However it is also clear that there are more general situations: say, both $X$ and $X'$ carry some structure that all Aut's and Iso's are required to preserve. Do you know still more general cases - bitorsors not realizable in any such way? $\endgroup$ Jan 9, 2018 at 19:09
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    $\begingroup$ Hi - I can comment now! I'll try and take a look at what you wrote when I get a moment. Cheers, Christopher $\endgroup$ Jan 11, 2018 at 9:38

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