Consider an action of a smooth linear algebraic group $G$ on a variety $X$ over an arbitrary field $k$, and the quotient stack $[X/G]$. Let $p$ be a $k$-point of $X$. If the action is transitive (i.e. $G(\bar{k})$ acts transitively on $X(\bar{k})$) then we have an isomorphism $[X/G]\cong BG_p$, where $G_p$ is the scheme-theoretic stabilizer of $p$, given as follows. In one direction, send a $G$-torsor $P$ with $G$-equivariant map $\pi: P\to X$ to $\pi^{-1}(p)$ and on the other direction send a $G_p$-torsor $Q$ to $Q\times^{G_p}G$.

This cannot be right, because if $q$ is another $k$-point of $X$ then this implies $G_p\cong G_q$, but I think we only have $(G_p)_{\bar{k}}\cong (G_q)_{\bar{k}}$ (it can happen that $p$ and $q$ are conjugate over $\bar{k}$, but not over $k$). If $[X/G]\cong BG_p$ is wrong, how do I correct it?

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    $\begingroup$ By not stating it? $\endgroup$ – მამუკა ჯიბლაძე Apr 21 '18 at 19:33
  • $\begingroup$ I do not need to use it, but I keep reading people saying $[X/G]=BG_p$ as a consequence of the action being transitive and the stabilizer at a certain point being $G_p$. $\endgroup$ – Kabim Apr 21 '18 at 21:10
  • $\begingroup$ Sorry, that was a tasteless joke $\endgroup$ – მამუკა ჯიბლაძე Apr 22 '18 at 4:40
  • $\begingroup$ Ahah sorry for not getting that :) $\endgroup$ – Kabim Apr 22 '18 at 5:13

$BG_p$ and $BG_q$ are isomorphic if $G_p$ and $G_q$ are the stabilizers of points $p$ and $q$ in a homogeneous space. Indeed, the set of elements of $G$ that map $p$ to $q$ is a $G_p$-torsor, and $G_q$ is the inner twist of $G_p$ by this torsor.

$BG_p$ is isomorphic to $BG_q$ because given any $G_p$-torsor, the space of $G_p$-equivariant isomorphisms between it and the set of elements of $G$ that map $p$ to $q$ is a $G_q$-torsor, and this map can be inverted by doing the same construction in reverse.


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