Suppose $k$ is a field and $K$ is a Galois extension of $k$. If $X$ is a variety defined over $k$, let $\text{Aut}(X/k)$ be the group of $k$ automorphisms of $X$, i.e. isomorphisms from $X$ to $X$ in the category of $k$-varieties, $\require{AMScd}$ \begin{CD} X @>f>> X\\ @V V V @VVV\\ \text{Spec}\,k @>>\text{Id}> \text{Spec}\,k \end{CD} Then a twist of $X$ is a variety $X'$ over $k$, such that there is a $K$-isomorphism $\phi$ $\require{AMScd}$ \begin{CD} X' \times_{\text{Spec}\,k} \text{Spec}\,K @>\phi>> X\times_{\text{Spec}\,k} \text{Spec}\,K\\ @V V V @VVV\\ \text{Spec}\,K @>>\text{Id}> \text{Spec}\,K \end{CD} Then from literatures (e.g. Chapter X of The Arithmetic of Elliptic Curves by Silverman), there is a map associated to $\phi$ defined by \begin{equation} \xi: \text{Gal}(K/k) \rightarrow \text{Aut}(X/k),~\sigma \rightarrow \phi^\sigma \circ \phi^{-1} \end{equation} Question 1: What is $\phi^{\sigma}$? Is it given by the composition \begin{equation} X' \times_k K \xrightarrow{\sigma} X' \times_k K \xrightarrow{\phi} X \times_k K \end{equation} where the first morphism is given by $\sigma$ acting on the second factor of $X' \times_k K $? If so, how to see $\phi^\sigma \circ \phi^{-1}$ is a $k$-automorphism of $X$?

Question 2: In Silverman's book, it says that $\xi$ measures $\phi$'s failure to be defined over $k$, could anyone explain the intuitions behind this statement? any interesting examples?

  • $\begingroup$ $\phi^\sigma$ should be what you get by applying $\sigma$ to the coefficients of the formula(s) describing $\phi$. It is definitely not a composition. $\endgroup$ – Lubin Apr 22 '18 at 17:51
  • $\begingroup$ @Lubin Thank you. $\endgroup$ – Wenzhe Apr 22 '18 at 19:50

The morphism $\phi^\sigma$ is defined as follow : the automorphism $\sigma$ defines an automorphism still denoted $\sigma : \operatorname{Spec}K\to\operatorname{Spec}K$. By pull-back, it defines for any $X$ an automorphism $$\sigma:X\times_k \operatorname{Spec}K\to X\times_k\operatorname{Spec}K$$

This defines a right action of $\operatorname{Gal}(K/k)$ on any $X_K=X\times_k\operatorname{Spec}K$. Now we have $\phi^\sigma=\sigma\circ\phi\circ\sigma^{-1}$. Note that this morphism is a $K$ morphism. This is because the diagram : $$\require{AMScd} \begin{CD} X'\times_k\operatorname{Spec}K@>\sigma^{-1}>>X'\times_k\operatorname{Spec}K@>\phi>>X\times_k\operatorname{Spec}K@>\sigma>>X\times_k\operatorname{Spec}K\\ @VVV@VVV@VVV@VVV\\ \operatorname{Spec}K@>\sigma^{-1}>>\operatorname{Spec}K@=\operatorname{Spec}K@>\sigma>>\operatorname{Spec}K \end{CD}$$ commutes and that the bottom row is obviously the identity.

(Note : if $G$ acts on objects $A$ and $B$ (on the right), the formula $\phi^\sigma=\sigma\circ\phi\circ\sigma^{-1}$ is the usual (right) action on $Hom(A,B)$. Note also that if $X=\operatorname{Spec}k[x_1,...,x_n]/(I), X'=\operatorname{Spec}k[y_1,...,y_m]/(J)$ and $\phi:X'_K\to X_K$ is given by $x_i\mapsto P_i(y_1,...,y_m)$, then $\phi^\sigma$ is given by $x_i\mapsto \sigma P_i(y_1,...,y_m)$ as Lubin said.)

However, it seems that this is an error in Silverman's book or an abuse of notation : $\xi_\sigma:=\phi^\sigma\circ\phi^{-1}$ is definitively not a automorphism of $C/k$, only of $C_K/K$. In fact this morphism is not invariant by Galois and we even have the cocycle equality : $\xi_{\sigma\tau}=(\xi_\sigma)^\tau\xi_\tau$.

As for question 2), $\phi$ is defined over $k$ iff it is invariant by Galois (this is obvious with the local description), in other words iff $\phi^\sigma=\phi$ for every $\sigma$, this is in turn equivalent to the cocycle $(\xi_\sigma)$ being trivial.

Have you look at the examples in Silverman's book ?

  • $\begingroup$ Thank you. I will check the examples in Silverman's book. $\endgroup$ – Wenzhe Apr 23 '18 at 18:53

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