Suppose $X,X'$ are two objects (say, genus 1 curves) over a field $k$ such that over the algebraic closure $X\_{\overline k} \cong X'\_{\overline k}$ and moreover, $Aut\_{\overline k}(X)$ is abelian.


Then, one can check easily that $Aut\_k(X) = Aut\_k(X')$. This is not true if the automorphism group is not abelian. Is there a conceptual reason as to why this should be true?



A conceptual explanation for a simple fact about twists of objects with abelian automorphism group
The Galois group $G$ acts on the classifying space $BAut\_{\bar k}(X)$. The set of isomorphism classes of possible choices of $X'$ is $\pi\_0$ of the space $(BAut\_{\bar k}(X)^{hG}$ of homotopy fixed points for this action, and $Aut\_k(X')$ is the fundamental group of the same space. If $Aut\_{\bar k}(X)$ is abelian, then the space $BAut\_{\bar k}(X)$ is a group and $G$ acts by group homomorphisms, so $(BAut\_{\bar k}(X)^{hG}$ is also a group, and therefore its fundamental groups at different points are isomorphic.


