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Let $G$ be a compact group and let $V$ be it's continuous representation. It is well known that if $V$ is finite-dimensional, then there is an $G$-invariant inner product on $V$. I haven't found a corresponding statement in the literature for infinite-dimensional representation. What breaks down? Is there a simple counterexample?

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This question I asked a few months ago might be of help. Such an inner product does exist, and the proof is much the same as in the finite-dimensional case (depending on how you proved it, of course). –  Clive Newstead Jan 3 '13 at 23:07
$G$-invariant objects are often constructed by averaging over the group. You can average over a compact group by using its invariant measure (which is finite because the group is compact: the averaging becomes problematic when the measure of the group is infinite), so many statements about representations of finite groups carry over to compact groups, mutatis mutandis, with almost no additional effort. –  whuber Jan 3 '13 at 23:10
@CliveNewstead: Well, I am not so sure. One can write down the formula $(u,v)_{\mathrm{inv}} = \int_G (gu,gv)_{\mathrm{arb}} \mathrm{d}g$, but proving that it gives you a continuous inner product may be impossible. A lot of sources don't go into details of the proof even in the finite-dimensional case. Why wouldn't they stated a stronger version of the theorem then? –  Vít Tuček Jan 3 '13 at 23:16
@robot: because when you are teaching undergraduates about representation theory for the first time, dealing with finite groups is hard enough without having to digress into compact groups, especially if they haven't seen any topology or measure theory before. It is rarely pedagogically optimal to state the most general version of a theorem. –  Qiaochu Yuan Jan 3 '13 at 23:44
@robot: also, Maschke's theorem for compact groups is strictly speaking not a complete generalization of Maschke's theorem for finite groups; the former only holds over $\mathbb{R}$ or $\mathbb{C}$ because of the necessity of taking integrals but the latter holds over all fields of characteristic not dividing the order of $G$. –  Qiaochu Yuan Jan 3 '13 at 23:45
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1 Answer

up vote 1 down vote accepted

Theorem: Let $G$ be a compact (Hausdorff) group and let $H$ be a Hilbert space on which $G$ acts strongly continuously. Then

$$\langle u, v \rangle_G = \int_G \langle gu, gv \rangle \, dg$$

defines a $G$-invariant inner product on $H$.

Proof. By assumption $g \mapsto \langle gu, gv \rangle$ is a continuous function on $G$, so it is integrable and the above integral exists. From here the only nontrivial step is showing positive-definiteness (everything else follows from linearity of integration and invariance of Haar measure). So suppose $v \in H$ is a nonzero vector. Let $U$ be an open neighborhood of the origin so that $\langle gv, gv \rangle \ge \epsilon$ for all $g \in U$ and for some $\epsilon > 0$. Since $G$ is compact, the set of translates $gU$, which cover $G$, admits a finite subcover $g_1 U, ... g_n U$, from which it follows that $U$ has positive measure (at least $\frac{1}{n}$) and that

$$\langle v, v \rangle_G \ge \frac{\epsilon}{n}.$$

The conclusion follows.

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Thanks! If I remember correctly then representations on Hilbert spaces are automatically strongly continuous, are they not? I got stuck with the proof of continuity of $(u,v) \mapsto \langle u,v\rangle_G$. –  Vít Tuček Jan 4 '13 at 0:35
It was a trivial mistake on my side. I think I've got it correct now. Thanks again! –  Vít Tuček Jan 4 '13 at 0:44
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