Another exercise on compact groups

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 \rho of G on H=L^2(G) (with respect to Haar measure) is faithful, and gives a continuous injection
G\rightarrow U(H)
where the unitary group U(H) is given the strong operator topology (if G 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 G). What we gain from seeing G 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 U(H), any open neighborhood U of 1 contains a finite intersection of sets of the type
U_{f,\epsilon}=\{g\in G\,\mid\, \|\rho(g)f-f\|\text{\textless} \epsilon\}
where f\in H has norm 1 and \epsilon\text{\textgreater} 0. It is then easy to see, using unitarity, that the conjugacy-invariant subset
V_{f,\epsilon}=\bigcap_{x\in G}{x^{-1}U_{f,\epsilon}x}\subset U_{f,\epsilon}
is equal to
V_{f,\epsilon}=\{g\in G\,\mid\, \|\rho(g)\varphi-\varphi\|\text{\textless} \epsilon\text{ for all }\varphi\in A_f\}
where
A_f=\{\rho(x)f\,\mid\, x\in G\}\subset H
is the orbit of f under G. But the strong continuity of \rho implies that the orbit map g\mapsto \rho(g)f is continuous for fixed f, so that A_f is compact in H (as image of a compact set under a continuous map; here H has the usual norm topology).
It is then a quick application of a standard compactness argument and splitting of epsilons to check that V_{f,\epsilon} is a neighborhood of 1 contained in U_{f,\epsilon}, and intersecting finitely many of these, we see that U 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 G has a faithful finite-dimensional representation, this can be used instead of \rho, but the purely topological argument becomes also much simpler.)

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Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

2 thoughts on “Another exercise on compact groups”

  1. 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 X \times K. For y \in K, write U(y) for the open subset of X such that U \cap (X \times y) = U(y) \times y. Set V = \bigcap_{y \in K} U(y). Then V is open in X.

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

    Application: Take X=Y=G, your compact group. Let Omega be any nbhd of the identity, and let U = \{ (g,h) : h g h^{-1} \in \Omega \}. So the lemma shows that \bigcap_{h \in G} h^{-1} \Omega h \} is open. This is clear conjugacy invariant, contains the identity, and is contained in \Omega.

  2. 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!

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