Terminology: Smooth Representations of Locally Profinite Groups.

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Let $G$ be a locally profinite group.

A smooth representation is a complex representation ($V,\rho$) of $G$ such that the stabiliser of any $v \in V$ is open.

One can show that (as $\text{GL}_n(\mathbb{C})$ is a lie group and has NSS), a (finite dimensional) representation of $G$ is continuous if and only if $\ker(\rho)$ is open.

Therefore, in finite dimensions, continuous representations are smooth.

Furthermore, as $$ \ker(\rho) = \cap_v \text{Stab}_G(v), $$ and the intersection on the right can be taken as finite for finite dimensional $V$, smooth also implies continuous. So these are equivalent for finite dimensions.

What about infinite dimensions? Does either imply the other?

What is the reason for this terminology? I only ask because I am conditioned to thinking that these implications must be smooth implies continuous, and not necessarily the other way around!

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I suppose continuous here means the map $P:G \times V \rightarrow V$ is continuous, given V the discrete topology. Then smooth certainly implies continuous, literally by definition ( check the inverse image of a single vector under P is open)

But I don’t think the other side is right as it should depend on the group.

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Already $G=\mathbb Z_p$ acting on $L^2(\mathbb Z_p)$ by translation is continuous, but it is easy to make not-locally-constant functions in $L^2(\mathbb Z_p)$.

Also, it is misleading to say that smooth repn spaces "have no topology" or "have the discrete topology". Rather, they have the colimit topology from being expressed as the ascending union of their finite-dimensional subspaces. Yes, every linear map from such a space is continuous... which is why incorrect remarks about the topology do not lead directly to disaster. :)

So, in the best case, for every compact open $K$ in $G$, the subspace $V^K$ of $K$-fixed vectors is finite-dimensional, and $V=\bigcup V^K$. This is not so for $V=L^2(\mathbb Z_p)$, but is correct for $V$ the $K$-finite vectors. Things like that.