Galois action on Tate twist

Question

This came up in Greenberg's paper (see chapter 2 in article no. 28 here) on the Iwasawa theory of elliptic curves. It's a small point, but I'd like to see more of the details. Fix $p$ and let $E$ be an elliptic curve over a number field $K$ with good ordinary reduction at a prime $v\mid p$. Then the absolute Galois group $G_{K_v}$ acts on the kernel $A\cong \mathbb{Q}_p/\mathbb{Z}_p$ of the surjective reduction map $E[p^\infty]\rightarrow \tilde E[p^\infty]$ by a character $\varphi:G_{K_v}\rightarrow \mathbb{Z}_p^\times$ since $\operatorname{Aut}(A)\cong \mathbb{Z}_p^\times$. Greenberg mentions that the action of $G_{K_v}$ on the Tate twist $\hat A(1):= \hom(A,\mu_{p^\infty})$ is given by $\chi\varphi^{-1}:G_{K_v}\rightarrow \mathbb{Z}_p^\times$, where $\chi:G_{K_v}\rightarrow \mathbb{Z}_p$ is the cyclotomic character coming from the action of $G_{K_v}$ on roots of unity. Why is this (the bold statement) true?

Breaking things down, I know that, given two representations $\varphi: G\rightarrow \operatorname{Aut}(V)$ and $\chi:G\rightarrow \operatorname{Aut}(W)$, the representation $\rho$ on $\hom(V,W)$ is given by defining $\rho(g)f$, for $f\in \hom(V,W)$, to be the function \begin{align}\tag{1} v\mapsto \chi(g)\big(f(\varphi(g)^{-1}(v))\big). \end{align} So, intuitively, I can see where the $\chi\varphi^{-1}$ is coming from. But I guess I'm struggling a bit to unpack how (1) translates to the above in the case of 1-dimensional representations. That is, given characters $\varphi,\chi: G\rightarrow F^\times$, coming from two group actions on $A$ and $B$, say, how does (1) reduce down to the character $\chi\varphi^{-1}:G\rightarrow F^\times$ coming from the action on $\hom(A,B)$?

Answer 1

You almost answered your own question.

There is a bijection between (...) one-dimensional representations and (...) characters, where (...) stands for some adjectives: e.g. continuous, smooth, etc.

Anyway, this bijection is given as follows: if $\chi: G\rightarrow F^\times$ is a character, then the one dimensional representation $(\rho, V)$ associated to $\chi$ is given by $\rho(g)(v) = \chi(g)\cdot v$, for any $v\in V$, where $V$ is a one-dimensional vector space over $F$.

Now it's just a matter of writing things down.

We have two characters $\phi,\chi:G\rightarrow F^\times$, and hence two one-dimensional reprsentations $(\rho_\phi, A)$ and $(\rho_\chi, B)$. The above says that $\rho_\phi(a) = \phi(g)\cdot a$ and $\rho_\chi(b) = \chi(g)\cdot b$ for any $a \in A, b \in B$.

Therefore, for any $f \in \operatorname{hom}(A, B)$ and any $g\in G$, your formula (1) for $\rho(g)(f)$ translates to: $$\rho(g)(f): v \mapsto \rho_\chi(g)(f(\rho_\phi(g)^{-1}(v))) = \chi(g)\cdot f(\phi(g)^{-1}\cdot v)= (\chi\phi^{-1})(g)\cdot f(v).$$ This means that $\rho(g)(f) = \chi\phi^{-1}(g) \cdot f$. Since it is true for all $f \in \operatorname{hom}(A, B)$ and all $g\in G$, we see that the one-dimensional representation $\operatorname{hom}(A, B)$ corresponds to the charactor $\chi\phi^{-1}$.