Why is the following map $\mu : G_K \to M_d(\Bbb C_K) : g \mapsto a_{ij}(g)$, where $g(w_j) = \sum_{i} a_{ij}(g)w_i$, continuous?

Question

Let $K$ be a p-adic field and let $\Bbb C_K$ be the completion of its algebraic closure. Let $G_K := \operatorname{Gal}(\overline{K}/K)$. Fixing a basis $\lbrace w_1, \dots, w_d\rbrace$ of a $\Bbb C_K$-vector space $W$ we have a map

$$\mu := \mu_{(w_j)} : G_K \to M_d(\Bbb C_K), g \mapsto a_{ij}(g)$$

where the $a_{ij}(g)$ are given by $g(w_j) = \sum_{i}a_{ij}(g)w_i$ for all $j$.

The claim is that this $\mu$ defines a continuous map, but I can't see why this is the case. I assume the topology on $M_d(\Bbb C_K)$ is the product topology (let's say after identifying $M_d(\Bbb C_K)$ with $\Bbb C_K^{d^2}$).

Answer 1

Scratch that: a $\Bbb C_K$-representation of $G_K$ is an action $G_K \times W \to W$, with $W$ a $\Bbb C_K$-vector space which is continuous and semilinear. An action of $G_K$ on a $\Bbb C_K$-vector space is just a continuous homomorphism $G_K \to \operatorname{End}(W) = M_d(\Bbb C_K)$ (after fixing a basis of $W$) so this gives the continuity.

Answer 2

If I'm not wrong, I think you meant $W = \Bbb{C}_K^d$.

Any $g\in G_K= Gal(\overline{K}/K)$ extends to a continuous automorphism of $\Bbb{C}_K$.

For a fixed $w\in W$, $g(w)$ is defined the obvious way. $g\to g(w)$ is continuous because for all $n$, $w = u_n+p^n v_n$ with $u_n\in \overline{K}$, $v_n\in O_{\Bbb{C}_K}$, $g(p^nv_n)\in p^n O_{\Bbb{C}_K}$, and $g(u_n)$ is one of the finitely many conjugates of $u_n$.

$w\to g(w)$ is not $\Bbb{C}_K$ linear, it is only $K$-linear. So it defines an homomorphism $G_K\to GL_{K-vector space}(W)$.

$(g,w)\to g(w)$ is continuous because if $w'$ is close enough to $w$ then $u_n'=u_n$.

Your function is taking a basis $w_1,\ldots,w_d$ of $\Bbb{C}_K$, putting them in a matrix $M$, and letting $\mu(g)=\pmatrix{g(w_1)\\ \vdots \\ g(w_d)}M^{-1}$, clearly continuous $G_K\to \Bbb{C}_K^{d\times d}$. Not an homomorphism at all.