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Let $G$ be a (non-abelian) compact connected Lie group. Let $K\subseteq G^2$ be a set of pairs $(g_1, g_2)$ which $\textbf{do not}$ generate $G$ topologically.

For which Lie groups $G$ is it known that the $\textbf{closure}$ of $K$ in $G^2$ has zero measure? The idea of the proof and reference would be appreciated.

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    $\begingroup$ If $G$ is not semisimple it's clear that $K$ is dense. We can ask whether the converse holds. The most optimistic would be that for $G$ semisimple, $K$ is Zariski-closed. $\endgroup$
    – YCor
    Commented Sep 26, 2020 at 19:58
  • $\begingroup$ @YCor, seems that you are right. I found the reference, using your advise. Thanks. $\endgroup$
    – Dmitri Scheglov
    Commented Sep 27, 2020 at 0:16
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    $\begingroup$ ams.org/journals/proc/1999-127-11/S0002-9939-99-04959-X/… $\endgroup$
    – Dmitri Scheglov
    Commented Sep 27, 2020 at 0:16

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Looks like I found the reference. Thanks YCor!

https://www.ams.org/journals/proc/1999-127-11/S0002-9939-99-04959-X/S0002-9939-99-04959-X.pdf

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