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R.P.
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How Galois group acts on Tate-Shafarevich group?

Let $L/K$ be a quadratic number field extension. Let $\operatorname{Sha}(E/L)$ be Tate-Shafarevich group of elliptic curve $E/L$. How $\sigma \in \operatorname{Gal}(L/K)$ acts on $\operatorname{Sha}(E/L)$?

I think the action is given by $[C/L]^{\sigma}=[C^{\sigma}]$ for $[C]\in \operatorname{Sha}(E/L)$, where $C^{\sigma} $ denotes a curve whose coefficients are transported by $\sigma$.

Once I could prove $[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, this is clearly a group action easily.

To prove $[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, what we should do is to prove $C/L$ is $E/L$-torsor,in other words, there is a simply transitive algebraic group action $E/L×C^{\sigma} \to C^{\sigma}$ defined over $K$.

$f: C \cong C^{\sigma}$ over algebraic closure, thus we can define a map $E/L×C^{\sigma} \to C^{\sigma}$ by composition of $id×f^{-1}$ and $φ:E/L×C \to C$ (this is simply transitive and defined over $K$) and $f$.

This composition is indeed transitive, but I cannot prove this is defined over $K$.

Is my definition of action of $\sigma \in \operatorname{Gal}(L/K)$ on $\operatorname{Sha}(E/L)$ correct ? If so, why the last composite defined over $K$ ?

Duality
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