Why is the kernel of a Galois representation an open subgroup? Assume that $E$ is a completion of a number ﬁeld. Then either $E = \mathbb{R}$ or $\mathbb{C}$, or $E$ is a ﬁnite extension of $\mathbb{Q}_l$ for a suitable prime number $l$.
If $E = \mathbb{R}$ or $\mathbb{C}$, and $\rho : G_K \rightarrow \mathrm{Aut}_E(V )$ is a representation, then why is it the case that

$\rho$ is continuous if and only if $\ker(\rho)$ is an open normal subgroup of $G_K$?

Here, $G_K =\mathrm{Gal}(K^\text{sep}/K)$ where $K$ is a number field, $V$ is a vector space over $E$, and $\mathbb R, \mathbb C$ have the usual topology.
 A: Suppose that $\ker(\rho)$ is open. If $U$ is an open subset of $\mathrm{Aut}_E(V)$ and $g\in \rho^{-1}(U)$, then $g\ker(\rho)$ is an open subset of $\rho^{-1}(U)$. Hence, $\rho^{-1}(U)$ is open, so $\rho$ is continuous.
For the converse, we use the fact that the group $\mathrm{GL}_n(\mathbb C)$ (and by extension $\mathrm{GL}_n(\mathbb R)$) has the following property:

There exists a neighbourhood $U$ of the identity $I$ which contains no non-trivial subgroup of $\mathrm{GL}_n(\mathbb C)$.

We say that $\mathrm{GL}_n(\mathbb C)$ has no small subgroups.
In order to show that $\ker\rho$ is open, it suffices to show that $\ker\rho$ contains an open neighbourhood of the identity. Let $U$ be as above, and let $V = \rho^{-1}(U)$. Then $V$ is an open neighbourhood of the identity. Since $$G_K=\mathrm{Gal}(K^{\text{sep}}/K)\cong\displaystyle\varprojlim_L\mathrm{Gal}(L/K)$$ has a basis consisting of open neighbourhoods of the identity, it follows that $V$ contains an open subgroup, $H$.
Now, $\rho(H)$ is a subgroup of $U$, and is therefore trivial. Hence $H\subset \ker\rho$ is an open neighbourhood of the identity.
