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...

  • $\begingroup$ For representations over complex numbers and compact groups - all irreps are finite-dimensional - see nice answers on mathoverflow.net/questions/119402/… $\endgroup$ Jan 26, 2014 at 12:13
  • $\begingroup$ Dear Pablo, can your explain what you mean by continuous representation in this context? What morphism, precisely, is continuous? $\endgroup$
    – Joël
    Jan 26, 2014 at 18:22
  • $\begingroup$ I call a representation continuous if the action map from G X V to V is continuous. $\endgroup$
    – Pablo
    Jan 26, 2014 at 19:33
  • $\begingroup$ I don't understand this phrase "or even profinite." Every profinite group canonically carries a compact Hausdorff topology. $\endgroup$ Jan 26, 2014 at 21:23
  • $\begingroup$ @qiaochu: I think he means he's ready to assume the stronger assumption that $G$ is profinite. $\endgroup$
    – YCor
    Jan 26, 2014 at 21:50

2 Answers 2


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.

  • $\begingroup$ As I see this, the crucial fact here is that the vector space V is an abelian profinite group (this is what Pontryagin requires). It is thus natural to ask what is the appropriate gneralization to the case of an action on an abelian profinite group. What will be the conclusion in this case? A finite/finitely-generated G-invariant subgroup, or maybe something else? $\endgroup$
    – Pablo
    Jan 26, 2014 at 16:40
  • $\begingroup$ Is it true that there must exist (if we take a not necesserily irreducible representation) a nontrivial irreducible subrepresentation? $\endgroup$
    – Pablo
    Jan 26, 2014 at 18:32
  • $\begingroup$ Your counterexample works beacuse G is divisible by p (in the supernatural profinite sense)? If not, we can use the fact that the representation on the dual is discrete and thus factors through some finite quotient of G (since G is profinite) and then Maschke's theorem can be applied. Is this true? $\endgroup$
    – Pablo
    Jan 26, 2014 at 20:29
  • $\begingroup$ Can we say something interesting in the case of an action on a nonabelian profinite group? For example, a product of countably many isomorphic simple groups? (I replaced the product of cyclic groups by a product of simple ones keeping the same profinite group G...) $\endgroup$
    – Pablo
    Jan 26, 2014 at 20:31
  • $\begingroup$ @Pablo: yes, it's even much more rigid, since then the automorphism group is generated by permutations of the factors and automorphisms of the factors (i.e., $Aut(S^I)$ is the full permutational wreath product of $Aut(S)$ and $Sym(I)$). Then every action of a compact group $G$ will preserve a decomposition corresponding to a partition of $I$ into finite subsets. Namely, there exists a partition $I=\bigcup I_j$ with each $I_j$ finite such that $S^{I_j}$ is $G$-invariant for all $j$. (...) $\endgroup$
    – YCor
    Jan 26, 2014 at 21:23

I am, not an expert on the represntation theory of profinite groups. However,if all their normal subgroups are closed, then the image of any represntation is profinite. Therefore, over $\mathbb{C}$ it will be finite.

Now, Benjamin Klopsch proved that the Nottingham group has this property. Andrei Jaikin, found an elegant argument genralizing it for all just infinite pro-$p$ groups. Recently, Nik Nikolov and Dan Segal studied this in much bigger generalization, see http://arxiv.org/abs/1310.3359.

  • $\begingroup$ Thanks for your answer. Unfortunately, it doesn't seem to answer my question. I'm interested in continuous representations over F_p, when the vector space is a countable product of F_p endowed with the product topology and a continuous action of G on it. $\endgroup$
    – Pablo
    Jan 26, 2014 at 10:57
  • $\begingroup$ Sorry, I missed the positive characteristis. However, do notice that your space have another topology when you think about an ordered basis, so you need to be careful. $\endgroup$ Jan 26, 2014 at 11:12

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