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!


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).

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  • $\begingroup$ 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 '19 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 '19 at 22:57

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