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Given a site $C$ and an object $U$, let $G$ be a sheaf of groups on this site and let $F$ be $G$-torsor, see the Stacks Project for the general definition.

By restriction on the over category $C/U$ (also a site with obvious topology), we get a $G|_U$-torsor $F|_U$ on $C/U$.

I want to first confirm if it is true that the automorphism group of the trivial $G|_U$-torsor is identified with the group $G(U)$, the set (group) of global sections of the sheaf $G|_U$. And second, is there a nice description (in nice cases) of the automorphism group of the $G|_U$-torsor $F|_U$ on $C/U$ for a general $G$-torsor $F$? In my application, the group is highly non-commutative.

$\mathbf{Edit}:$ I think the answer to the first question is affirmative. For the second, I think in nice cases, $G(U)$ should be a subgroup of the automorphism group of $G|_U$-torsor $F|_U$. E.g. I think any $GL_n$-torsor (corresponds to a rank $n$ vector bundle on $U$) if the site is a category of schemes with e.g. Zariski topology should be a bitorsor (I hope this to be correct)—it has actions by $GL_n$ from left and right, and the two actions commute, Then any element in $GL_n(U)$ will give an automorphism of $F|_U$ by multiplying it on the right. And most probably in this case, $G(U)$ is exactly the automorphism group of $G|_U$-torsor $F|_U$ (rather than just a subgroup of the latter): this is basically because that by considering on each section set $F(V)$ as $V\to U$ ranges within a local trivializing cover of $F$ over $U$ (using that the right action is simply transitive), any two automorphisms differ (on the right) by a unique element in $G(V)$. These $G(V)$'s glue—by uniqueness—to a unique global section of $G|_U$.

I hope some expert can give a definitive answer and tell if the above reasoning is correct. Thanks in advance!

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  • $\begingroup$ It seems that for authentic bitorsor, the above reasoning should be true. While a $GL_n$-torsor has no natural right $GL_n$-action that commutes with the left action. $\endgroup$ – Lao-tzu Jan 30 at 11:40
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If you parametrize your torsor by a cocycle in $H^1(-,G)$, the automorphism group is the twisted form of $G$ given by the same cocycle, where $G$ acts on itself by conjugations (while the torsor itself is the twisted form of $G$ acting on itself by shifts).

In terms of the article from the comment below, $Aut(P)=G\wedge^G P$, where $G$ acts on $G$ by inner conjugation (so it is an inner form of $G$ as in Remark 1.8)

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  • $\begingroup$ Is your twisted form the same as this article page 3? What do you mean by "acting on itself by shifts"? I also don't understand your claim "the automorphism group is the twisted form of $G$ given by the same cocycle". Could you explain, or maybe expand your answer with more details? $\endgroup$ – Lao-tzu Feb 1 at 11:01
  • $\begingroup$ Yes, exactly that. Action by conjugation is also explained there, see 1.4.2 $\endgroup$ – Victor Petrov Feb 1 at 16:21

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