Let $X$ be a curve defined over a number field $K$, and let $G\_K$ be the absolute Galois group of $K$. Let $\text{Aut}(X)$ be the group of $\overline{K}$-defined automorphisms of $X$, and consider the Galois cohomology set $H^1(G\_K, \text{Aut}(X))$.


Denote by $\text{Twist}(X/K)$ the set of twists of $X$ up to $K$-defined automorphisms.


According to Silverman, Chapter X.2, Theorem 2.2, there is a bijection
$$\text{Twist}(X/K)\longrightarrow H^1(G\_K, \text{Aut}(X)).$$


It is well known that there is an equivalence between a Galois cohomology set, such as $H^1(G\_K, \text{Aut}(X))$ and the set of $\text{Aut}(X)$-torsors under $G\_K$.


1. Consider $X$ as the trivial twist of itself. In what way is $X$ an $\text{Aut}(X)$ torsor? I mean, in order for $X$ to be an $\text{Aut}(X)$-torsor, we should have a well defined, bijective, action map $X\times \text{Aut}(X)\longrightarrow X$, yet $\text{Aut}(X)$ doesn't even act on $X$, but rather on $\overline{X}:=X\otimes \overline{K}$.
2. Is there a version of the bijection between twists and cohomology classes which holds in higher dimensions as well? A reference would be appreciated.



Equivalence between twists of a curve and torsors of its automorphism group
Actually, this has little to do with varieties and is just some general phenomenon in topos theory. For simplicity, let $\mathscr C$ be a (small) category with finite limits, endowed with a Grothendieck pretopology. For a sheaf of groups $\mathscr G$ on $\mathscr C$, write $H^1(\mathscr C,\mathscr G)$ for the set of isomorphism classes of $\mathscr G$-torsors (this has an easy interpretation in terms of cocycles, but I will work only with the torsor point of view). For a sheaf of sets $\mathscr F$ on $\mathscr C$, write $\operatorname{Twist}(\mathscr F)$ for the set of isomorphism classes of twists of $\mathscr F$, i.e. sheaves $\mathscr G$ on $\mathscr C$ for which there exists a covering $\{U\_i \to \*\}\_{i \in I}$ of the terminal object $\*$ of $\mathscr C$ such that $\mathscr G|\_{U\_i} \cong \mathscr F|\_{U\_i}$ for all $i \in I$.


Define a map $\operatorname{Twist}(\mathscr F) \to H^1(\mathscr C,\mathscr Aut(\mathscr F))$ by taking a twist $\mathscr G$ of $\mathscr F$ to the sheaf $\mathscr Isom(\mathscr G,\mathscr F)$, which is naturally an $\mathscr Aut(\mathscr F)$-torsor by the map
\begin{align\*}
\mu \colon \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) &\to \mathscr Isom(\mathscr G,\mathscr F)\\
(f,g) &\mapsto f \circ g.
\end{align\*}
Indeed, this map is clearly a left action of the sheaf of groups $\mathscr Aut(\mathscr F)$ on the sheaf of sets $\mathscr Isom(\mathscr G,\mathscr F)$, and the map
\begin{align\*}
(\mu,\text{pr}\_2) \colon \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) &\to \mathscr Isom(\mathscr G,\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F)\label{1}\tag{1}\\
(f,g) & \mapsto (fg,g)
\end{align\*}
is an isomorphism (over opens $U$ where $\mathscr G|\_U \ncong \mathscr F|\_U$, there is nothing to check!).


**Lemma.** *The map $\operatorname{Twist}(\mathscr F) \to H^1(\mathscr C,\mathscr Aut(\mathscr F))$ is a bijection.*


*Proof.* Injectivity is a little more conceptual than what Silverman shows: let $\mathscr G$ and $\mathscr H$ be twists of $\mathscr F$ such that $\mathscr Isom(\mathscr G,\mathscr F)$ and $\mathscr Isom(\mathscr H,\mathscr F)$ are isomorphic as $\mathscr Aut(\mathscr F)$-torsors. This means that there exists an isomorphism $\alpha \colon \mathscr Isom(\mathscr G,\mathscr F) \stackrel\sim\to \mathscr Isom(\mathscr H,\mathscr F)$ such that the diagram
$$\begin{array}{ccc}\mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr G,\mathscr F) & \to & \mathscr Isom(\mathscr G,\mathscr F) \\ \downarrow & & \downarrow \\ \mathscr Aut(\mathscr F) \times \mathscr Isom(\mathscr H,\mathscr F) & \to & \mathscr Isom(\mathscr H,\mathscr F)\end{array}\tag{2}\label{2}$$
commutes, where the horizontal maps are the actions and the vertical maps are given by $\alpha$. In particular, we get a map
$$\psi \colon \mathscr Isom(\mathscr G,\mathscr F) \stackrel\sim\to \mathscr Isom(\mathscr G,\mathscr H)$$
taking a local section $f \colon \mathscr G|\_U \stackrel\sim\to \mathscr F|\_U$ to the composition $\mathscr G|\_U \stackrel f\to \mathscr F|\_U \stackrel{\alpha(f)^{-1}}\longrightarrow \mathscr H|\_U$. We will show that $\psi$ descends to a map $\* \to \mathscr Isom(\mathscr G,\mathscr H)$, i.e. induces an isomorphism of sheaves $\mathscr G \stackrel\sim\to \mathscr H$.


For ease of notation, write $\mathscr T$ for the $\mathscr Aut(\mathscr F)$-torsor $\mathscr Isom(\mathscr G,\mathscr F)$. Then the map $\psi$ gives two maps $\psi\_1, \psi\_2 \colon \mathscr T \times \mathscr T \rightrightarrows \mathscr Isom(\mathscr G,\mathscr H)$ by $\psi\_i = \psi \circ \text{pr}\_i$, and it suffices to show that these agree (since $\mathscr T \times \mathscr T \rightrightarrows \mathscr T \to \*$ is a coequaliser). But \eqref{1} gives an isomorphism $\mathscr Aut(\mathscr F) \times \mathscr T \stackrel\sim\to \mathscr T \times \mathscr T$, and the compositions $\mathscr Aut(\mathscr F) \times \mathscr T \rightrightarrows \mathscr Isom(\mathscr G,\mathscr H)$ are given by $(f,g) \mapsto \alpha(fg)^{-1}fg$ and $(f,g) \mapsto \alpha(g)^{-1}g$ respectively. By \eqref{2}, the first map takes $(f,g)$ to $(f\alpha(g))^{-1}fg = \alpha(g)^{-1}g$, which is what we needed to show.


Surjectivity, as in Silverman, relies on descent theory (glueing). Let $\mathscr U = \{U\_i \to \*\}\_{i \in I}$ be a covering over which $\mathscr T$ has sections $g\_i$ for all $i \in I$, and define a glueing datum for a sheaf of sets $\mathscr G$ as follows: over $U\_i$ take the sheaf $\mathscr G\_i = \mathscr F|\_{U\_i}$, and for each $i,j \in I$ take the isomorphism $\phi\_{ij} \colon \mathscr G\_j|\_{U\_i \cap U\_j} \stackrel\sim\to \mathscr G\_i|\_{U\_i \cap U\_j}$ as the unique section $\phi\_{ij}$ of $\mathscr Aut(\mathscr F)(U\_i \cap U\_j)$ such that $\phi\_{ij} g\_j = g\_i$. This satisfies the cocycle condition, turning $(\{\mathscr G\_i\},\{\phi\_{ij}\})$ into a glueing datum. Then there exists a sheaf $\mathscr G$ on $\mathscr C$ such that $\mathscr G|\_{U\_i} = \mathscr G\_i$ for all $i \in I$; see [Stacks, Tag [04TR](https://stacks.math.columbia.edu/tag/04TR)]. It is easy to check that the $\mathscr Aut(\mathscr F)$-torsor $\mathscr Isom(\mathscr G,\mathscr F)$ is canonically isomorphic to $\mathscr T$. $\square$


Here is an even more high-brow argument using stacks of groupoids (= 'sheaves of 1-truncated homotopy types). Let $B(\mathscr Aut(\mathscr F))$ be the stackification of the prestack $U \mapsto B(\mathscr Aut(\mathscr F|\_U))$, and let $\mathbf{Twist}$ be the stack taking $U \in \mathscr C$ to the category of twists of $\mathscr F|\_U$ (with only invertible morphisms). The thing we're trying to understand is $\operatorname{Twist}(\mathscr F) = \pi\_0(\mathbf{Twist}(\mathscr F)(\*))$ (the objects up to isomorphism of the category of global sections of the stack $\mathbf{Twist}(\mathscr F)$).


There is a natural map $B(\mathscr Aut(\mathscr F|\_U)) \to \mathbf{Twist}(\mathscr F)(U)$ mapping the unique object of $B(\mathscr Aut(\mathscr F|\_U))$ to the trivial twist $\mathscr F|\_U$ of $\mathscr F|\_U$. The construction of stackification [Stacks, Tag [02ZN](https://stacks.math.columbia.edu/tag/02ZN)] shows that the associated morphism of stacks $B(\mathscr Aut(\mathscr F)) \to \mathbf{Twist}(\mathscr F)$ is an equivalence, giving a bijection $\operatorname{Twist}(\mathscr F) \stackrel\sim\to \pi\_0(B\mathscr Aut(\mathscr F)(\*))$. But for any group object $\mathscr G$ in a topos, the category $(B\mathscr G)(\*)$ of global sections of $B\mathscr G$ is the category of $\mathscr G$-torsors.


In particular, we've upgraded our bijection of 'objects up to a suitable notion of isomorphism' to an equivalence of groupoids!




---


For the scheme case, there is an extra difficulty that a twist $\mathscr G$ of a representable sheaf $\mathscr F$ (i.e. representable by a scheme $X$) need not be representable (by a scheme $Y$). More or less by definition (of algebraic space), the twist $\mathscr G$ is at least an algebraic space $Y$, as it is an étale sheaf quotient of the scheme $X\_{\bar K}$. If $X$ is projective, it has an ample line bundle, which you can use to produce an ample line bundle on $Y$, so $Y$ is projective as well (hence a scheme).


(Proof sketch: an ample line bundle $\mathscr L$ on $X$ gives an ample line bundle on $Y\_L$ after some finite Galois extension $K \to L$, and you can apply the norm map $N\_{L/K} \colon \mathbf{Pic}\_Y(L) \to \mathbf{Pic}\_Y(K)$ to get a line bundle on $Y$. It is ample since it is a sum of Galois conjugates of an ample line bundle. Exercise: find some details to worry about.)


