MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering if there is any application for induced representations of profinite groups, for example in Galois theory or number theory? Is it a good research idea? Do you know any paper discussing this problem or related problem or at least unitary representations of profinite groups?

share|cite|improve this question

I claim that this is not more difficult/general then studying the induced representation of a finite group. (Edit: I am talking about complex representations only.)

Every (continuous) representation of a compact group is unitarizable, hence it is sufficient to study the unitary representations in this case. They also decompose into irreducible ones. The irreducible ones are finite-dimensional. This can be found under Peter-Weyl theorem.

For a profinite group $G$, an irreducible representation $\sigma$ has moreover an open kernel, i.e., the kernel is a finite index, normal subgroup.

By Frobenius reciprocity, $\sigma$ is contained in $Ind_{kern( \sigma)}^G 1$.

Because $kern( \sigma)$ is a finite index, normal subgroup, the representation $Ind_{kern( \sigma)}^G 1$ decomposes in the same way as does the right regular representation of $H=kern( \sigma) \backslash G$, i.e., every irreducible representation of $H$ occurs with multiplicity equals the dimension (again Peter-Weyl or its finite-group analogue).

share|cite|improve this answer
Thank you Marc for sharing your insight. However what you described is more about irreducible representations of profinite groups. My question was about the possible applications of induced representations from closed (or open) subgroups of profinite groups which they can be reducible. If I understood correctly your answer implies that if $H$ is (open) subgroup of a profinite group $G$ and its index in $G$ is not finite, then $ind_H^G 1$ is not irreducible, which is interesting too. – Vahid Shirbisheh Jan 9 '13 at 11:06
induction from open subgroups equals induction for finite groups. – Marc Palm Jan 9 '13 at 14:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.