Let G be a compact group (or even profinite - Galois group). Let $V$ be a vector space over the field ${\mathbb F}\_p$ with $p$ elements, $p$ a finite prime, such that $V$ is a contable product of ${\mathbb F}\_p$ with the product topology. Let s be an irreducible continuous representation of G on $V$. Must s be finite dimensional? If not, what conditions can we add to ensure this? For example: G is profinite/abelian/solvable/finitely generated...



Irreducible representations of compact groups
Yes, $V$ is finite-dimensional. More generally, whenever a compact group $G$ act continuously on $V\neq 0$, then there is a finite-codimensional invariant closed subspace $W\neq V$. To see this, let $G$ act on the Pontryagin dual $\hat{V}$. Then this action is continuous and $\hat{V}$ is discrete. Let $v$ be a nonzero element of $\hat{V}$: then its stabilizer $G\_v$ is an open subgroup of $G$, hence has finite index. So $Gv$ is finite, and hence generates a nonzero invariant finite-dimensional subspace of $\hat{V}$. By duality, it corresponds to a closed invariant subspace $W\neq V$.


This even shows that in full generality, every compact $\mathbf{F}\_p[G]$-module is profinite as a $\mathbf{F}\_p[G]$-module.


Edit: I answer your two questions in the comments:


1) yes, the argument also works when $V$ is an arbitrary profinite abelian
group and $G$ a compact group: $V$ is then profinite as $G$-module (same argument, using that $\hat{V}$ is locally finite).


2) *is it true that there always exists an irreducible subrepresentation?*: no: indeed pick $R=\mathbf{F}\_p[[t]]$, $V=R$ (additive group) and $G=R^\times=R\smallsetminus tR$ (multiplicative group).
Since $R^\times$ generates $R$ as an $R$-algebra, a $G$-submodule of $V=R$ is the same as an ideal of $R$. The ideals of $R$ are $(0)$ and the $(t^n)$, so $(0)$ is the only finite one. Hence there is no irreducible submodule.


