0
$\begingroup$

Let $G$ be a compact Hausdorff topological group. Is it necessarily the case that every conjugacy class of $G$ is Borel? It's certainly true if $G$ is countable (in which case $G$ is actually finite and the result is trivial) or if $G$ is abelian (in which case each conjugacy class is a singleton so the result is again trivial since $G$ is Hausdorff). I haven't been able to come up with much beyond that; any thoughts would be appreciated!

$\endgroup$
1
$\begingroup$

In fact, conjugacy classes are compact (and thus closed, since $G$ is T2). The map $\rho:G\times G\times G\to G$, $\rho(x,y,z)=xzy^{-1}$ is continuous and $\operatorname{conj}(z)=\rho[\Delta\times \{z\}]$, where $\Delta=\{(x,x)\,:\, x\in G\}$ is the diagonal of $G\times G$. Since $G$ is T2, $\Delta$ is closed in $G\times G$ (and therefore compact).

$\endgroup$
  • $\begingroup$ Thanks! Could you also have said the following? Take some $z \in G$. The function $g \mapsto gzg^{-1} \colon G \to G$ is continuous because $G$ is a topological group. The conjugacy class of $z$ is the image of this function, hence compact. Then the Hausdorff condition only needs to be used once (to conclude that this compact set is closed); have I made an error? $\endgroup$ – diracdeltafunk Apr 12 at 22:08
  • $\begingroup$ @diracdeltafunk You are right. I was, in fact, considering that argument too (which proves that orbits are compact provided that $G$ is compact), but then I decided that it wasn't quite worth editing the answer. $\endgroup$ – Saucy O'Path Apr 12 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.