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By Goto's theorem for any compact connected semisimple Lie group $G$ of dimension $n$, any element $x\in G$ is a commutator, namely $x=[y,z]$ for some $y, z\in G$. Another way to say it is that the commutator map $\pi:G\times G\rightarrow G$ is surjective. By Sard's Lemma it follows that typical element $w\in G$ is a regular value of $\pi$ and ${\pi}^{-1}(w)\subset G\times G$ is a smooth compact submanifold of dimension $n$.

Question: what is the homeomorphic type of this manifold for typical $w$?

Of course it is tempting to suspect that ${\pi}^{-1}(w)$ is homeomorphic to $G$ but somehow I have difficulty in checking it even in rather simple case of $G=SO(3)$...

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    $\begingroup$ It is natural to divide this preimage by the $G$-action (via conjugation); then it becomes the "relative character variety" of the once-punctured torus. Topology of such varieties was/is extensively studied, but I do not remember the answer off-hand. $\endgroup$ Commented Jun 2, 2020 at 16:00
  • $\begingroup$ @MoisheKohan , indeed this is how the question appeared, as I need to understand something about character varieties of one-punctured torus, fixing the holonomy element along the loop around the puncture. Could you provide me some standard references, if there are any , of course.. $\endgroup$ Commented Jun 2, 2020 at 16:04
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    $\begingroup$ Goldman wrote extensively on these; apparently, according to this paper, for $G=SU(2)$, the variety is homeomorphic to $S^2$, but no proof is given. $\endgroup$ Commented Jun 2, 2020 at 16:12
  • $\begingroup$ @MoisheKohan Thanks for the link. If indeed the factorized variety is homeomorphic to $S^2$ in this case then it would not be very surprizing if the preimage is actually homeomorphic to $S^3=SU(2)$ somehow Hopf-fibered over this $S^2$.. $\endgroup$ Commented Jun 2, 2020 at 16:18
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    $\begingroup$ Do we have a good reason to expect that generic fibers are connected? $\endgroup$
    – YCor
    Commented Jun 2, 2020 at 17:29

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