A finite dimensional representation of a totally disconnected group is continuous if and only if it is smooth? Let $G$ be a topological group of totally disconnected type, i.e. a Hausdorff group for which every neighborhood of the identity contains a compact open subgroup.  Let $V$ be a finite dimensional complex vector space.
A representation $\pi$ of $G$ with underlying space $V$ is called smooth if stabilizer of any vector is an open subgroup of $G$.  Equivalently, $G \times V \rightarrow V, (g,v) \mapsto \pi(g)v$ is continuous, with $V$ having the discrete topology.  
On the other hand, $\pi$ is called continuous if $G \times V \rightarrow V$ is continuous with $V$ having the Euclidean topology.  Equivalently, $\pi: G \rightarrow \textrm{GL}(V), g \mapsto \pi(g)$ is a continuous function.
Is it true that a finite dimensional representation of $G$ is continuous if and only if it is mooth?
 A: Yes.  First suppose that $\pi$ is continuous.  There exists an open neighborhood of the identity $U$ in $\textrm{GL}(V)$ which does not contain any nontrivial subgroups.  See this answer.
It follows that the kernel of $\pi: G \rightarrow \textrm{GL}(V)$ is open in $G$: choose an open subgroup $H$ of $G$ contained in the preimage $\pi^{-1}(U)$.  Then $\pi(H) \subseteq \pi(\pi^{-1}(U)) \subseteq U$.  Since $\pi(H)$ is a subgroup of $\textrm{GL}(V)$, we have $\pi(H) = 1$.  This shows that $H \subseteq \textrm{Ker } \pi$.  Since $\textrm{Ker } \pi$ contains an open subgroup of $G$, it is an open subgroup of $G$.
Now if $v \in V$, then the stabilizer of $v$ contains the open subgroup $\textrm{Ker } \pi$, so the stabilizer of $v$ is open.  This shows that $\pi$ is smooth.  
Now assume that $\pi$ is smooth.  Let $v_1, ... , v_n$ be a basis of $V$, and let $H_i$ be the stabilizer of $v_i$.  By hypothesis, each $H_i$ is an open subgroup of $G$.  Then 
$$H = H_1 \cap \cdots \cap H_n$$
is also an open subgroup of $G$ which is contained in the kernel of $\pi$.  Therefore, the kernel of $\pi$ is an open subgroup of $G$.  Then the homomorphism
$$\pi: G \rightarrow \textrm{GL}(V)$$
with $V$ having the Euclidean topology, is continuous, because for example the induced map on the quotient space $G/\textrm{Ker } \pi \rightarrow \textrm{GL}(V)$ is continuous, because $G/\textrm{Ker } \pi$ is discrete.  
