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Let $k$ be a field, and suppose $G$ is a group-scheme over $k$ (I am happy to assume that $k=\mathbb{Q}$ and that $G$ is affine). A $G$-torsor over $k$ is a non-empty $k$-scheme $T$ equipped with an action $a:G\times T\to T$, such that $(a,\pi_2):G\times T\to T\times T$ is an isomorphism. There is an induced morphism $$ \begin{align*} T\times T\times T&\to T,\\ (t_1,t_2,t_3)&\mapsto a\bigg(\pi_2\big((a,\pi_2)^{-1}(t_1,t_2)\big),t_3\bigg)=``t_1t_2^{-1}t_3". \end{align*} $$ I would like to know to what extent the map $T\times T\times T\to T$ determines $G$. I mean the following:

For which pairs of groups $G$ and $G'$ does there exist a $G$-torsor $T$, a $G'$-torsor $T'$, and an isomorphism $T\cong T'$ compatible with the maps $T\times T\times T\to T$ and $T'\times T'\times T'\to T'$?

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  • $\begingroup$ This seems to be equivalent to asking for a $G$-torsor $T$ and an isomorphism $G_T \cong G'_T$ of group schemes over $T$. $\endgroup$ – S. Carnahan Jan 6 '18 at 20:04
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Let $b:T\times T\times T\to T$ be the map in question. We can view it as a morphism of functors (actually fpqc sheaves) on $k$-schemes: $$\begin{array}{rcl} T\times T & \longrightarrow & \operatorname{\underline{Aut}}(T)\\ (t_1,t_2) & \longmapsto & \left(\,t_3\mapsto b(t_1,t_2,t_3)\,\right) \end{array}$$ where $\underline{\mathrm{Aut}}$ stands for the sheaf of $k$-automorphisms. Thus, this map sends $(t_1,t_2)$ to the automorphism of $T$ given by the action of ``$t_1t_2^{-1}$''. So you can recover $G$ as the subsheaf of $\operatorname{\underline{Aut}}(T)$ which is the image of this map.

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Under fairly general exactness assumptions on the ambient category, one can perform the construction of a group from $(T,m)$, if $T$ is an object equipped with a heap structure $m:T^3\to T$, i. e. an associative Maltsev operation. That is, $m$ must just satisfy (diagrammatic versions of) $m(x,x,y)=m(y,x,x)=y$ and $m(m(x,y,z),t,u)=m(x,y,m(z,t,u))$.

This is done exactly as one does construct a vector space from an affine space: one gets a group as a quotient $T\times T\twoheadrightarrow G$ (which becomes the division map of the torsor $T$ over $G$, $(x,y)\mapsto x/y$ where $x/y$ is the unique element of $G$ with $(x/y)y=x$).

The quotient is by identifying $(x,y)$ with $(m(x,y,z),z)$ for all $x,y,z$. That is, one forms the coequalizer $T^3\rightrightarrows T^2\twoheadrightarrow G$ of appropriate morphisms.

The action of $G$ on $T$ is determined by $(x/y)\cdot z=m(x,y,z)$, and gives a $G$-torsor. It is an easy exercise to do this "without elements". Moreover any other group over which $T$ is a torsor and such that $m(x,y,z)$ is $(x/y)\cdot z$ for that structure too, is isomorphic to $G$, by an isomorphism compatible with the actions on $T$.

In fact switching sides one also has another group $G'$ such that $T$ is a $G$-$G'$-bitorsor.

This even works without global support assumption: defining support $S$ of $T$ as the coequalizer $T\times T\rightrightarrows T\twoheadrightarrow S$ of the projections $T\times T\rightrightarrows T$, one gets a group and a torsor structure in the slice category over $S$.

There are way too many references to choose from, spread across decades, so I think all this must be considered folklore.

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  • $\begingroup$ I am confused about one point. As you say, $T$ is a $G-G'$-bitorsor. Doesn't this mean $T$ is both a torsor for $G$ and a torsor for $G'$? $\endgroup$ – Julian Rosen Jan 7 '18 at 1:41
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    $\begingroup$ @Julian a left G torsor and right G' torsor, in a compatible way. $\endgroup$ – David Roberts Jan 7 '18 at 2:00
  • $\begingroup$ I think I see. The right action of $G'$ can be turned into a left action by composing with inversion, and my confusion came from the fact that this makes $T$ into both a (left) $G$-torsor and a (left) $G'$-torsor. If I understand correctly, this is not a contradiction because the heap structure coming from the left $G'$-action is not the heap structure $T$ starts with. $\endgroup$ – Julian Rosen Jan 7 '18 at 4:27
  • $\begingroup$ @JulianRosen Correct. That one is $m'(x,y,z)=m(z,y,x)$ $\endgroup$ – მამუკა ჯიბლაძე Jan 7 '18 at 6:24
  • $\begingroup$ Btw this implies that $G$ and $G'$ become isomorphic over $T$, since for any given $e\in T$ there is an isomorphism between $m$ and $m'$ sending $x$ to $m(e,x,e)$; however over $T$ in fact $T$ trivializes both over $G$ and $G'$, so this is trivial anyway. I believe there must be examples when $G$ and $G'$ are not isomorphic. $\endgroup$ – მამუკა ჯიბლაძე Jan 8 '18 at 3:43

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