# Galois action on Tate twist

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 $$\hom(A,B)$$?

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 $$\chi\phi^{-1}$$.

• Thanks so much, that really clears things up. Commented Feb 1, 2020 at 0:53