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Edit: I'm specializing this question to the compact case I'll ask about the noncompact case as a new question.

Let $ G $ be a compact connected semisimple Lie group. Do there always exist two finite order elements of $ G $ which generate a dense subgroup?

Example: $$ \frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ and $$ \begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix} $$ generate a dense subgroup of $ SU_2 $.

This question is partially inspired by

Generating finite simple groups with $2$ elements

which shows that every finite simple group is 2-generated. Indeed even every finite quasisimple group is 2-generated Is every finite quasi-simple group generated by 2 elements? (however this fails for infinite simple groups: $ \mathrm{PSL}_2(\mathbb{Q}) $ is not 2-generated indeed not even finitely generated).

This is a cross-post of https://math.stackexchange.com/questions/4537024/dense-subgroups-generated-by-two-finite-order-elements an answer posted there points out that every compact semisimple Lie group can be topologically generated by 2 infinite order elements.

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    $\begingroup$ I think that there's a paper that proves this by some mechanism like showing the set of pairs that generate has positive Haar measure. I'll see if I can dig it up. $\endgroup$
    – LSpice
    Commented Sep 29, 2022 at 15:25
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    $\begingroup$ Theorem 2.1 in this J. Algebra paper of Breuillard-Gelander should help: indeed it is then enough to show that every (connected, finite center) semisimple Lie group is (topologically) generated by two circle subgroups. $\endgroup$
    – YCor
    Commented Sep 29, 2022 at 16:10
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    $\begingroup$ [Finite center should be assumed. Or, at least one has to exclude the case when $G$ surjects onto the universal cover of $\mathrm{SL}_2(\mathbf{R})$, which is torsion-free.] $\endgroup$
    – YCor
    Commented Sep 29, 2022 at 16:11
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    $\begingroup$ Good point I've added the assumption of finite center $\endgroup$
    – Ian Gershon Teixeira
    Commented Apr 10, 2023 at 14:46
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    $\begingroup$ You should add the assumption that $G$ is semisimple, otherwise the claim is clearly false since $G$ can be torsion-free, e.g. the subgroup $B< SL(2,{\mathbb R})$ of non-strictly upper triangular matrices with positive diagonal entries. $\endgroup$
    – Moishe Kohan
    Commented Apr 10, 2023 at 14:57

1 Answer 1

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You might want to look at MR0034766 (Kuranishi 1949, link). It deals with the connected semisimple case, which might be what you're after.

There's also a 1999 Proc AMS paper by M. Field, (freely accessible at AMS site — MR number for subscribers: MR1618662), which shows that the set of pairs generating a dense open subset of $G$ is non-empty Zariski open in $G \times G$ when $G$ is compact connected semisimple. This should lead to a proof that the elements can be taken to have finite order in this case, as the set of pairs of elements of finite order is dense in $G\times G$. Of course, in the non-compact case some extra condition on $G$ is required.

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    $\begingroup$ The MR link for this is erronious. The paper itself is here: projecteuclid.org/journals/kodai-mathematical-journal/volume-1/… The compact case is due to Auerbach. $\endgroup$
    – Dave Benson
    Commented Apr 12, 2023 at 19:21
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    $\begingroup$ What I mean is that it the "article" link is to the wrong paper. It links to a paper by Kawata. $\endgroup$
    – Dave Benson
    Commented Apr 12, 2023 at 19:27
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    $\begingroup$ @DaveBenson, there's a Contact Us link on the MathSciNet page, that I've used to good effect when dealing with bad MR links. I have just reported the error you mentioned, with credit to you. The correct DOI is 10.2996/kmj/1138833534. For your other reference, MR1618662, the DOI is 10.1090/S0002-9939-99-04959-X. $\endgroup$
    – LSpice
    Commented Apr 12, 2023 at 21:31
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    $\begingroup$ @IanGershonTeixeira Yes, exactly. This answers your question for compact connected semisimple, but not for noncompact connected semisimple with finite centre, where I don't know the answer. $\endgroup$
    – Dave Benson
    Commented Apr 13, 2023 at 17:05
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    $\begingroup$ @DaveBenson Thanks for pointing out the error. We will look into the erroneous article link. We get many of the links from the DOI repository, CrossRef. Lately, we seem to have had a higher number of bad links of this sort. We aren't sure if it's our matching algorithm or CrossRef's data. $\endgroup$
    – Edward Dunne
    Commented Apr 14, 2023 at 19:02

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