I have a question regarding a (probably simple) fact. However I am lacking some basic topological knowledge.
Let $G$ be a locally pro finite group, i.e. ever open neighborhood of $1_G$ contains a compact open subgroup, and $\psi:G \to \mathbb{C}^{\times}$ a group homomorphism. The following are equivalent:
- $\psi$ is continuous
- the kernel of $\psi$ is open
So I guess, that 2 $\Rightarrow$ 1: follows from the fact, that if $\psi^{-1}(1)$ is open, since every element of $\mathbb{C}^{\times}$ is a scalar multiple of $1$ and the translation is open, $\psi^{-1}(U)$, where $U$ is a open neighborhood of $1$ is a union of open sets and therefore open.
For 1 $\Rightarrow$ 2: I have read, that if $N$ is a open neighborhood to 1 in $\mathbb{C}^{\times}$, $\psi^{-1}(N)$ is open an contains, since $G$ is locally profinite, a compact open subgroup $K$ of $G$. If $N$ is chosen sufficiently small, it contains no non-trivial subgroup of $\mathbb{C}^\times$, and so $K\subseteq \ker \psi$.
However, I don't know how $K\subseteq \ker \psi$ implies, that $\ker \psi$ is open.
Thanks in advance for explanations.
To show that $K\subseteq \ker \psi$ implies that $\ker \psi$ is open:
$K$ is an open subgroup of the subgroup $\ker \psi$. So we can write $$ \ker\psi = \cup_{g \in \ker\psi} Kg $$ Which shows that $\ker \psi$ is open.
This is generalised in this brief note, for any topological group $G$ having a basis of neighbourhoods of the identity consisting of subgroups (in particular locally profinite groups), and for $\mathbb{C}^{\times}$ replaced with any topological group with no small subgroups (which includes all Lie groups, so $GL_n(\mathbb{C})$).