# Do compact groups acting irreducibly have finite subgroups which do the same?

Let $G$ be a closed subgroup of $U(n,{\bf C})$, not necessarily connected. Regard ${\bf C}^n$ as a complex $G$-module $M$.

Q. Suppose $M$ is irreducible as a $G$-module (equivalent, I think, to the condition that the linear span of $G$ is all of $M_n({\bf C})$). Does there always exist a finite subgroup $H < G$ such that $M$ is irreducible as an $H$-module?

Perhaps one should view the question as one about faithful irreducible representations of "abstract" compact Lie groups and their finite subgroups, but the more concrete picture is easier for me to think about.

It also seems possible to me that one could hammer the question by using structure theory for compact Lie groups in some way, and then looking down the lists to check everything works for A, B, C, D etc, but I'd prefer to have something that used the basic ideas that go into such structure theory rather than the final classification itself, since the original motivation for the question actually comes from an infinite-dimensional context.

(The original motivation for the question is probably not relevant here; but I include it in case it is of interest. The question is a generalization of one raised en passant in some work of A. Blanco, and if the answer is yes, it seems that improvements can be made to some results he's obtained concerning finite-dimensional Banach spaces $X$ for which $B(X)$ is "amenable with constant $1$". Moreover, a proof which was somehow "global" in nature might be adaptable to the study of $K(X)$ for certain well-behaved infinite-dimensional $X$.)

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I think this is not true. For a counterexample, consider $G = SU(2)$; any finite subgroup of $G$ is cyclic or dihedral up to finitely many exceptions, but $G$ has irreducible representations of arbitrarily large dimension. – ulrich Aug 28 '14 at 11:50
@ulrich thanks - it seems everyone arrived simultaneously at essentially the same answer with the same reasoning – Yemon Choi Aug 28 '14 at 12:14

No. Consider an irreducible representation of $\mathrm{SU}(2)$ of large dimension, say $n\ge n_0$. Then it cannot be irreducible for a finite subgroup, because these are either abelian or dihedral (with only 1 or 2-dimensional irreducible reps) or in a short finite list, with finitely many possible dimensions of irreducible reps for each. Maybe the best $n_0$ here is $n_0=7$ but I haven't checked.

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I should really have thought of this as I knew all the ingredients you mention but did not think how to put them together. Thanks! – Yemon Choi Aug 28 '14 at 12:22

The answer is already no for $G=U(2)$ acting on $Sym^6(\mathbb C^2)$.

Given a subgroup $\Gamma \leq U(n) \cong SU(n) \times U(1) / Z_n$, we can enlarge it by projecting to $SU(n)/Z_n$ and $U(1)/Z_n$, take the preimages in $SU(n),U(1)$, and multiply together. The $U(1)$ isn't going to help hold together an irrep so we've reduced to subgroups of $SU(n)$.

There aren't so many of these for $n=2$; the cyclic, binary dihedral, and the $E_{6,7,8}$ ones. Their largest irreps are 1, 2, 3, 4, 6-dimensional, i.e. the largest coefficient of a simple root in the expansion of the highest root of the $A,D,E$ root systems.

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D'oh! thanks. as is often the case I should really accept more than one anwer, but since I can't I'm accepting Yves's just because it's "closest to what I should have known had I been thinking more carefully" – Yemon Choi Aug 28 '14 at 12:16
Guralnick and Thiep arxiv.org/abs/0810.0853 show that for any finite group $G$ and representation $V$ of dimension $>1$, the representations $\mathrm{Sym}^k V$ are reducible for $k \geq 6$. In fact, already for $k \geq 4$ and $5$, and asking for Zar. closed subgroups of $GL(V)$ rather than finite, there are only a few cases where $\mathrm{Sym}^k V$ is irreducible, which they classify completely. – David Speyer Aug 28 '14 at 12:17
@DavidSpeyer: Actually, it's Tiep – Geoff Robinson Aug 28 '14 at 15:40