$\DeclareMathOperator\Sha{Sha}\DeclareMathOperator\Gal{Gal}$Let $L/K$ be a quadratic extension of number field $K$.
Let $\sigma$ be a generator of $\Gal(L/K)$.

Let $E/K$ be an elliptic curve defined over $K$ and $\Sha(E/K)$ be its Tate–Shafarevich group.
$\Gal(L/K)$ acts on $\Sha(E/L)$ naturally (cf. https://mathoverflow.net/questions/448399/how-galois-group-acts-on-tate-shafarevich-group).

There is a canonical isomorphism $\tau: E(L)\cong E_D(L), (x,y)\mapsto (x,y/\sqrt{D})$ .

> My goal is to prove $\tau$ induces **$(1-\sigma)\Sha(E/L)\cong
　\operatorname{trace}\Sha(E_D/L)$**. This isomorphism appears in p219 of
> [Yu - On Tate-Shafarevich groups over galois extensions](https://doi.org/10.1007/BF02772219).

 

 To prove this isomorphism, it is enough to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$.


 There exists an  isomorphism $\phi: \Sha(E/L)\cong \Sha(E_D/L)$ induced by $\tau$ though I cannot write down the map between them.



We can check $\sigma \tau=-\tau \sigma$  because we can calculate its coordinates explicitly, but I cannot calculate both $\sigma \phi$ and $\phi \sigma$, so I'm having difficulty to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$. How can I overcome this trouble and prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$ ? 


**Another approach of proving $(1-\sigma)\Sha(E/L)\cong \operatorname{trace}\Sha(E_D/L)$ is also appreciated.**