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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.)

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So for you SU(2) x S^1 isn't "sufficiently nice"?? – Kevin Buzzard Apr 30 '10 at 17:12
How about "has finite abelianization"? – S. Carnahan Apr 30 '10 at 17:19
@Scott: is it clear that finite abelianisation is necc and suff for a compact group to only have finitely many 1-dimensional reps? Hmm, I guess there are non-Hausdorff examples, unless we use the convention that compact implies Hausdorff. But using that convention it's still not clear to me that an infinite compact Hausdorff abelian group necessarily admits infinitely many 1-dimensional representations. Is this true? Aah---yes, it is true, by Pontrjagin duality! – Kevin Buzzard Apr 30 '10 at 18:29
The profinite dihedral group (the extension of Z/2 by Z-hat) has finite abelianization but infinitely many 2-dimensional representations. – David Treumann Apr 30 '10 at 18:33
For a profinite group, you need at least that for all open normal subgroups N < G the abelianization of N is finite. Otherwise the character group of N is infinite, and for each 1-dimensional representation you get an induced representation of dimension [G:N] with two isomorphic if and only if they lie in the same orbit in the character group under the action of G/N. – Tyler Lawson Apr 30 '10 at 19:01

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.

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Brilliant. I'm not sure I ever would have guessed this, but it seems so clear in retrospect. – Ben Webster May 1 '10 at 0:47
Great! Quick question: "finitely generated" means "topologically finitely generated," right? – Qiaochu Yuan May 1 '10 at 3:37
@Qiaochu: That's right. – A Stasinski May 1 '10 at 13:56
This is a helpful reference to an out-of-the-way paper, shifting the focus from compact Lie groups to general compact groups. For the published version: MR1950883 (2004c:20069) 20F65 (20E18 20G20) Bass, Hyman (1-MI); Lubotzky, Alexander (IL-HEBR); Magid, Andy R. (1-OK); Mozes, Shahar (IL-HEBR), The proalgebraic completion of rigid groups. Proceedings of the Conference on Geometric and Combinatorial Group Theory, Part II (Haifa, 2000). Geom.Dedicata 95 (2002), 19–58. This seems to be the same as the preprint, but slightly corrected. – Jim Humphreys May 2 '10 at 14:03
.@AStasinski Brilliant +1 – Duchamp Gérard H. E. Apr 24 '15 at 12:13

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


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.

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Thanks! This is still good to know; I didn't know what the correct characterization was for Lie groups. – Qiaochu Yuan Apr 30 '10 at 17:59

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.)

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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

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