Let $E$ be an elliptic curve over a field $\mathbb{Q}_p$, and $\phi$ is a group homomorphism from $G_{\bar{\mathbb{Q}_p}/\mathbb{Q}_p}$ to the group $\{1,-1\}$, then how to define the elliptic curve $E'$ which is $E$ twisted by $\phi$? Does the representation of Tate module of $E'$ is the representation of tate module of $E$ tensor the character?
Thanks!
Let $K$ be the subfield of $\overline{\Bbb{Q}}_p$ fixed by $\ker(\phi)$, then $Gal(K/\Bbb{Q}_p)=G_{\Bbb{Q}_p}/\ker(\phi) \cong \{1,-1\}$ thus $[K:\Bbb{Q}_p]=2$ and $K=\Bbb{Q}_p(\sqrt{d})$.
Since we are in characteristic $0$ we have a Weierstrass model $E/\Bbb{Q}_p : y^2=x^3+ax+b$ then $E_d/\Bbb{Q}_p : dY^2=X^3+aX+b$ is your quadratic twist.
On $E$ or $E_d$ the inverse is $$[-1](x,y)=(x,-y)$$
The isomorphism $E\to E_d$ is $(x,y)\mapsto (x,y/\sqrt{d})$ thus for $\sigma\in G_{\Bbb{Q}_p}$ $$\sigma(x,y/\sqrt{d})=(\sigma(x),\sigma(y)\phi(\sigma)/\sqrt{d})=[\phi(\sigma)]( \sigma(x),\sigma(y)/\sqrt{d}))$$ which means, with $\rho_{E,\ell}$ the representation of $G_{\Bbb{Q}_p}$ to $Aut(E[\ell^\infty])\cong GL_2(\Bbb{Z}_\ell)$ and for $a\in \Bbb{Z}_\ell$, $\Phi(\sigma) a=\phi(\sigma)a$ $$\rho_{E_d,\ell}(\sigma)= \phi(\sigma)\rho_{E,\ell}(\sigma)=\rho_{E,\ell}\otimes \Phi \ \ (\sigma)$$ (tensor product of $\Bbb{Z}_\ell$ representations)