6
$\begingroup$

It is a standard fact in the representation theory of finite groups that for $G,H$ finite groups, all of the irreducible representations of $G \times H$ are the external tensor product of irreps of $G$ and $H$. Today I was talking to a friend about profinite groups and it got me thinking: "Is (some version of) this result still true?" The fact that so many results from the finite case carry over makes me think that this could be true, but I have no idea how to go about proving it. The standard proof for finite groups uses a counting argument to show that they are all of this form, so certainly some higher-level techniques will be required.

Since we're considering profinite groups, we will definitely want to restrict ourselves to continuous representations on topological vector spaces. If the statement is not true in this generality, are there adjectives we can add that make it true? What if our representations are unitary, or the profinite groups are (topologically) finitely-generated? Any results, no matter the number of hypotheses, would be of interest to me.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ It should be true if you are assuming representations are on Hilbert spaces. Then since the group is compact you can assume it's unitary and also all irreducible reps are finite dimensional. Now since the image of a profinite group under a cts finite dimensional rep is a profinite lie group it is finite. You can write $G\times H$ as the inverse limit of $G_i\times H_i$ with $G_i$ finite images of $G$ and the same for $H$ and so every irrep factors through some $G_i\times H_i$. Now apply the result for finite groups. $\endgroup$
    – Benjamin Steinberg
    Jul 15, 2021 at 13:14
  • $\begingroup$ I believe every irreducible representation of a compact group on a Banach space is also finite dimensional but I'm not sure how weak the assumptions on a topological vector space can be. $\endgroup$
    – Benjamin Steinberg
    Jul 15, 2021 at 13:19
  • 2
    $\begingroup$ For $K\times G$ with $K$ compact and $G$ locally compact, this does hold for unitary Hilbert space repns. For general $G\times G'$ with locally compact $G,G'$ it can fail: see "Type I groups"... and "factor repns". I don't know much about that failure myself. $\endgroup$
    – paul garrett
    Jul 15, 2021 at 15:28
  • 2
    $\begingroup$ @YemonChoi, when I first learned about this possible failure, I was aghast and traumatized. :) Alain Robert's book (London Math Soc) does introduce these things, without being exhaustive/exhausting. :) While I'm thinking about it, I seem to recall that Robert says that a free non-abelian group on two generators (with discrete topology) behaves badly in this regard... so counter-examples to the optimistic conjecture are not amazingly pathological? ... but I do not understand such things. :) $\endgroup$
    – paul garrett
    Jul 17, 2021 at 21:27
  • 1
    $\begingroup$ @paulgarrett Just for the sake of completeness: on consulting Folland's book, a "nice" counterexample -- which I think is also the one in Robert's book -- is to take the representation of $F_2\times F_2$ on $\ell^2(F_2)$ defined by the left and right regular representation of $F_2$ acting together on the same space. This representation turns out to be irreducible, but it cannot be "factorized" as the tensor product of two other unitary irreps $\endgroup$
    – Yemon Choi
    Sep 7, 2021 at 1:08

1 Answer 1

Reset to default
7
$\begingroup$

This is not even true for finite groups, in this generality, and not even in characteristic $0$. Consider, for example, the group $Q_8 \times C_3$, where $Q_8$ is the quaternion group and $C_3$ is cyclic of order $3$, and consider $\mathbb{Q}$-representations of this direct product. The standard representation $\rho$ of $Q_8$ is not realisable over $\mathbb{Q}$, only $\rho\oplus \rho$ is. $C_3$ has an irreducible $\mathbb{Q}$-representation $\chi$, given by the sum of the two non-trivial irreducible complex characters of $C_3$. Now, $\rho\otimes \chi$ is realisable over $\mathbb{Q}$ and defines a simple $\mathbb{Q}[G]$-module, but it is not of the form $V\otimes W$ for any $\mathbb{Q}[Q_8]$-module $V$ and $\mathbb{Q}[C_3]$-module $W$.

If you wanted to restrict to finite dimensional representations over $\mathbb{C}$, then the statement will be true also for profinite groups, because any continuous complex finite dimensional representation of a profinite group will factor through a finite quotient.

Improve this answer
$\endgroup$
6
  • $\begingroup$ The complex part is essentially my answer in the comments except you need to use that all irreducible representations are finite dimensional since the regular representation of a compact group is faithful. I think it is implicit from the question mentioning a counting argument for finite groups and talking about topological vector space and unitary representations that the OP was working over the complex numbers $\endgroup$
    – Benjamin Steinberg
    Jul 15, 2021 at 14:02
  • $\begingroup$ Anyway +1. The question could be more specific on the hypotheses $\endgroup$
    – Benjamin Steinberg
    Jul 15, 2021 at 14:44
  • $\begingroup$ @BenjaminSteinberg Yes, we posted more or less simultaneously. I did not take the "topological vector space" part of the question to implicitly assume that we are working over the complex numbers. For example number theorists routinely consider continuous representations of profinite groups (e.g. of absolute Galois groups) over all sorts of rings. $\endgroup$
    – Alex B.
    Jul 15, 2021 at 14:55
  • $\begingroup$ Yes. The OP should have been clearer. As you point out even for finite groups you need a splitting field. But it is important that reps are finite dimensional to get it to factor through a finite quotient $\endgroup$
    – Benjamin Steinberg
    Jul 15, 2021 at 16:34
  • $\begingroup$ @BenjaminSteinberg Oh yeah, sorry, I was not disputing that. Now corrected -- thanks! $\endgroup$
    – Alex B.
    Jul 15, 2021 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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