# 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.)

### Kowalski

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

## 2 thoughts on “Another exercise on compact groups”

1. David Speyer says:

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!