While writing the chapters about compact groups in my notes, I had a few times the impression that it would be useful to use the fact that there is a basis of *conjugacy-invariant* neighborhoods of 1 in such a group. This thought would then fork in two subthreads, one in which I noticed that I didn’t really see how to prove this, and the second in which I realized I didn’t need it anyway.

This happened again during the last week-end, with the difference that I thought of trying to prove this property *using* representation theory. I failed at first, and finally tried to look it up online to make sure the property was actually true. Indeed, it is, and I found this book, which has three proofs (of a slightly more general statement). But I have to admit to having an inferiority complex with respect to general topology: any argument that goes on for more than a few lines without being transparent tends to make me uneasy and confuse me, and this is what happened with these arguments.

So I tried again to use representations, and indeed it works! Here’s the idea: the regular representation of on (with respect to Haar measure) is faithful, and gives a continuous injection

where the unitary group is given the strong operator topology (if is infinite, it is not continuous for the operator norm topology). Hence this map is a homeomorphism onto its image (we use here the compactness of ). What we gain from seeing in this way as a subgroup of the unitary group of a (typically) infinite-dimensional space, and with a “strange” topology to boot, is a concrete description of a basis of open neighborhoods of 1, which is open to further manipulations.

Indeed, from the definition of the strong topology on , any open neighborhood of 1 contains a finite intersection of sets of the type

where has norm 1 and . It is then easy to see, using unitarity, that the conjugacy-invariant subset

is equal to

where

is the orbit of under . But the strong continuity of implies that the orbit map is continuous for fixed , so that is compact in (as image of a compact set under a continuous map; here has the usual norm topology).

It is then a quick application of a standard compactness argument and splitting of epsilons to check that is a neighborhood of 1 contained in , and intersecting finitely many of these, we see that contains indeed a conjugacy invariant neighborhood of 1…

I really can’t say that it is simpler than the purely topological argument (mostly because one needs to know about Haar measure!), but I find this rather nice as an exercise. It illustrates how representation theory can be useful to study a general compact group, at a very basic level, and also shows that it may be useful to embed (or project, with the orbit map) a compact group into complicated-looking infinite-dimensional beasts… (Of course, if has a faithful finite-dimensional representation, this can be used instead of , but the purely topological argument becomes also much simpler.)

So, I think the pure topological neighborhood is pretty nice and doable. I don’t know if this is any better than what is in your book, but here it is…

Lemma: Let X be a topological space and let K be a compact topological space. Let U be an open subset of . For , write for the open subset of X such that . Set . Then V is open in X.

Proof: Let . We must show that V contains an open nbhd of x. Since , for every , we have . Using that U is open, and the defn of the product topology, there are open sets and in X and K, containing x and y, such that . Since K is compact, there are finitely many points such that . Then $\bigcap V(w_i)$ is the required nbhd of x in V. QED

Application: Take , your compact group. Let Omega be any nbhd of the identity, and let . So the lemma shows that is open. This is clear conjugacy invariant, contains the identity, and is contained in .

That is indeed nice and understandable! I didn’t think about introducing the product topology, but it really makes the argument clean. Thanks a lot!