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Is there any reference that classifies all infinite subgroups of $SU(2)$?

I found this https://math.stackexchange.com/a/4398724/1079528 which claims that the only infinite maximal (closed) subgroup is $\langle U(1), \begin{pmatrix} 0 && -1\\ 1 && 0\end{pmatrix}\Big\rangle$. First, I can't find the reference for this claim. Second, I can't find anything about "non-closed" subgroups of $SU(2)$.

Robert Shore
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nick
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 18 '22 at 20:48
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    The infinite non-closed subgroups are very complicated; for example $SU(2)$ contains a copy of the free group on countably many generators. – Qiaochu Yuan Jul 18 '22 at 20:51
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    One can also consider, for example, $SU(2, R)$ where $R$ is any subring of $\mathbb{C}$; there are lots and lots of these because $\mathbb{C}$ is abstractly isomorphic to any algebraically closed field of characteristic $0$ of the same cardinality as it. Many of the corresponding subgroups will fail to be measurable, I think. – Qiaochu Yuan Jul 18 '22 at 20:55
  • I guess it would need to be a subring stable by the complex conjugation, but that still leaves a lot of possibilities. – Captain Lama Jul 18 '22 at 20:58
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    I would expect there are probably $2^{2^{\aleph_0}}$ different isomorphism types of such subgroups. – Eric Wofsey Jul 18 '22 at 22:13
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    You should read more of the linked question: There are also several maximal finite subgroups. As for a proof, one first classifies Lie subalgebras of $su(2)$. Could you do this? – Moishe Kohan Jul 19 '22 at 01:41
  • @EricWofsey nice question! (it could be asked on MO, I think) – YCor Jul 25 '22 at 06:44
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    https://arxiv.org/abs/1704.02106 mentions some interesting infinite, non closed subgroups of $ SU_2 $. In particular I believe that the examples listed there are related to $ S $-arithmetic groups. For example the group generated by $$ H=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} $$ and $$ T= \begin{bmatrix} 1 & 0 \ 0& \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \end{bmatrix} $$ Is exactly the $ R=\mathbb{Z}[1/\sqrt{2}] $ points of $ U_2(\mathbb{R}) $ (this is roughly an example of @QiaochuYuan comment). Related matrices should generate $ SU_2(\mathbb{Z}[1/\sqrt{2}]) $. – Ian Gershon Teixeira Sep 20 '22 at 19:04

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