# Are conjugacy classes of compact groups Borel?

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).
• 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? – diracdeltafunk Apr 12 '19 at 22:08
• @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. – Saucy O'Path Apr 12 '19 at 22:57