Which compact groups have finitely many irreducible representations of each dimension?

Question

If my understanding is correct, this is true of sufficiently nice nonabelian Lie groups (see Ben Webster's answer below), and any finite group. On the other hand, this is false for any infinite compact abelian group by Pontryagin duality (see Kevin Buzzard's comment below), and by extension for any group with such a group as a quotient (perhaps even as a quotient of a finite-index subgroup; see Ben Webster's comment below).

So: are there any nice conditions weaker than being a Lie group which guarantee that a compact group only has finitely many irreducible representations of each dimension? (Nice necessary conditions would also be interesting.)

Answer 1

Some of the comments to the question have already indicated that finite abelianisation has something to do with it. If $G$ is a finitely generated profinite group, the following are equivalent:

1. $G$ has a finite number of isomorphism classes of complex irreducible representations of dimension $n$, for each $n$ (in this case $G$ is usually called (representation) rigid).
2. $H/[H,H]$ is finite for every open subgroup $H$ of $G$ (in this case, $G$ is said to have the property FAb).

This result is contained in Proposition 2 in this paper by Bass, Lubotzky, Magid, and Mozes.

Answer 2

EDIT: Having written up this answer, I realized Qiaochu was probably more interested in the case of $G$ not a Lie group. Oh well, I'll let the answer stand, even if it doesn't fully address the question.

A compact Lie group has finitely many simple representations of any given dimension if and only if its Lie algebra is semi-simple.

$(\Leftarrow)$: First check that it holds for a group iff it holds for the connected component of the identity (every irrep is in the induction of an irrep from that connected component). So now, assume the group is connected.

Then, note that if you have a quotient which is a torus (which is true iff the Lie algebra is not semi-simple), you're shot, because you can pull back all the irreps of the torus.

This shows you must have semi-simple Lie algebra.

$(\Rightarrow)$: Now, assume your group has semi-simple Lie algebra. You might as well pass to the universal cover, since this just makes more irreps. So really, you just have to prove the result for a semi-simple Lie algebra.

Now, use the Weyl dimension formula

$d_{\lambda}=\frac{\prod_{\alpha}(\lambda+\rho,\alpha)}{\prod_{\alpha}(\rho,\alpha)}$

to see that any rep with dimension below $n$ has its inner product with any simple root $\leq n$, and so is confined to a compact box.

Answer 3

This is an addendum to the existing answers, but might still be of interest even if this is an old question. (I apologize for bumping with an answer rather than by reformatting an existing answer; please note that the absence of MathJax is deliberate, and I request that people don't prettify the text.)

Among some people working on non-abelian harmonic analysis, compact groups with the property mentioned in the title of the question are known as tall compact groups, and seem to have been studied in some papers in the 1970s. The paper

M. F. Hutchinson, Tall profinite groups. Bull. Austral. Math. Soc. 18 (1978), no. 3, 421–428 MR 80h:20044 | free at CUP website?

contains the following:

THEOREM 2.5. For a profinite group G to be tall it is both necessary and sufficient that G satisfy the following two conditions: (i) no open normal subgroup of G has an infinite abelian quotient; (ii) G contains only finitely many open normal subgroups of index n for each positive integer n.

As a corollary one obtains the result mentioned in A. Stasinski's answer. Note that there exist tall profinite groups which are solvable, hence rather different from the setting of Lie groups.

(I only learned of Hutchinson's paper, and the terminology "tall", a few years ago — well after Qiaochu raised this question.)

Answer 4

Apart from profinite groups, you can take infinite (Tychonoff) product of $SU(2)$ with itself; this has infinitely many irreps of dimension two. But the product $\prod _{n=2}^{\infty} SU(n)$ has only finitely many irreps in each dimension