Defining the action of elliptic curve $E$ on a principal homogeneous space of $E$ I recently learnt (using Milne's book about elliptic curves) about principal homogeneous space for elliptic curves. For completeness I state the definition here: Let $E$ be an elliptic curve over a field $k$. A principal homogeneous space for $E$ is a curve $W$ over $k$ together with a right action of $E$ given by a regular map $W\times E \to W$ via
$$ (w, P) \mapsto w + P $$
such that
$$ (w, P) \mapsto (w, w + P) : W \times E \to W \times W $$
is an isomorphism of algebraic varieties.
Then there are a few theorems about these PH spaces and basically the PH spaces are the twists of $E$. However, if I have a twist $E'$ of $E$, I am a bit confused as to how $E$ acts on $E'$.
I thought of a simple example for this: Let $k = \mathbb{Q}$ and
$$ E: y^2 = x^3 + ax + b$$
$$ E': y^2 = x^3 + 4 a x + 8 b $$
and then $\phi : E' \to E$ by
$$ \phi(x,y) = (\frac{x}{2}, \frac{y}{2 \sqrt 2}) $$
is an isomorphism over $\mathbb{Q}(\sqrt 2)$.
To define the action of $E$ on $E'$, the obvious way to me is as follows: If $P$ is a point on $E$, and $P'$ a point on $E'$, then we can define $P' + P = P' + \phi^{-1}(P)$.
However, I'm not sure if this is the correct way of doing it. For example, if $P' = O'$ (where $O'$ is the identity point of $E'$) then this formulas gives $O' + P = O' + \phi^{-1}(P) = \phi^{-1}(P)$. But $\phi$ and $\phi^{-1}$ are not neccessarily given by polynomials with coefficients in $\mathbb{Q}$, so $\phi^{-1}(P)$ might not be in $E'(\mathbb{Q})$, which means this does not give an action on $E'(\mathbb{Q})$. Which would mean that the way I tried to define the action is wrong.
Thank you for the help.
 A: Let $E'/K$ be a twist of an elliptic curve $E/K$, we want to equip $E'$ with the structure of a principal homogeneous space for $E/K$ (FYI, you will also often see these called $K$-torsors for $(E, +)$ for extra jargon points). The structure you are looking for is not quite what you have written down, but it should appear in the results that are proven in the book.
Long story short, you have defined your action wrong, you're after: Fix an isomorphism $\phi : E' \to E$ over $\bar{K}$. Given $P' \in E'$ and $P \in E$  we define
$$ P' + P = \phi^{-1}(\phi(P') + P)$$
hence as you were worried about $P' + O = \phi^{-1}(\phi(P') + O) = P'$.
For $\sigma \in G_K$, let $\phi^{\sigma} = \sigma \phi \sigma^{-1}$ (you can also view this as $\sigma$ acting on the coefficients of $\phi$.
To see that the action is defined over $K$ notice that $\phi^{\sigma}\phi^{-1}$ is a translation on $E$, by say $Q_\sigma$. So that $\phi^{\sigma}(R) = \phi(R) + Q_\sigma$ for $R \in E'$.
\begin{align*}
\sigma(P' + P) &= (\phi^{-1})^{\sigma}(\sigma(\phi(P') + P)) \\
&= (\phi^{-1})^{\sigma}(\phi^\sigma (\sigma(P')) + \sigma(P)) \\
&= (\phi^{-1})^{\sigma}(\phi(\sigma(P')) + Q_\sigma  + \sigma(P)) \\
&= (\phi^{-1})(\phi(\sigma(P')) + \sigma(P)) \\
&= \sigma(P') + \sigma(P)
\end{align*}

The more general idea is as follows, for an object $X/K$ let $Y/K$ be a twist and fix $\phi : Y \to X$ an isomorphism over $\bar{K}$. Then $\xi_\sigma =  \phi^{\sigma}\phi^{-1}$ defines an injection $Twist(X/K) \to H^1(K, Aut(X))$ - when $X = E$ is a smooth projective curve this is a bijection (Silverman X Thrm 2.2) (it is actually true for any quasiprojective variety).
