For any simple Lie group $\mathrm{G}$ with a finite subgroup $\mathrm{H}$, does there exist a lattice $\Gamma$ with $\mathrm{H}\subseteq\Gamma\subset\mathrm{G}$?
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- 3$\begingroup$ This is true for $G=PSL(2,C)$ and $G=PU(2,1)$. link.springer.com/article/10.1007/s10711-006-9062-3 I don’t have a reference for the Kleinian case, but this is a well-known consequence of the orbifold theorem. $\endgroup$Ian Agol– Ian Agol2025-06-18 04:13:26 +00:00Commented Jun 18 at 4:13
- $\begingroup$ Just a remark: Simplicity condition is needed here: e.g. for $G=SE(2)$, subgroups of order $\ge 7$ do not embed in lattices. $\endgroup$Moishe Kohan– Moishe Kohan2025-06-21 20:30:32 +00:00Commented Jun 21 at 20:30
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1 Answer
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I can give a proof for $G=SL_2(\mathbb{R})$, but the general case is hard, I think, as in that area there are many unsolved problems like the Margulis Conjecture or the Lehmer conjecture.
Now any finite subgroup of $G=SL_2(\mathbb{R})$ lies in a maximal compact subgroup $K\cong SO(2)\cong\mathbb{R}/\mathbb{Z}$. So any such group is cyclic and any two groups of the same order are conjugate in $G$. To see that finite cyclic groups of any order appear in lattices in $G$, we make a Riemann surface $S$ as follows: take $\mathbb{P}^1(\mathbb{C})$, which is the 2-sphere and attach $n$ handles near the equator such that the whole thing has a rotational symmetry of order $n$. Then $S$ has the upper half plane for a universal covering, hence $S=\Gamma\backslash\mathbb{H}$ for a lattice $\Gamma$ in $PSL_2(\mathbb{R})$. Let $r$ denote the rotation of $S$ of order $n$. Then $r$ lifts to a symmetry of $\mathbb{H}$ and the group generated by this lift and $\Gamma$ is a lattice which contains a cyclic subgroup of order $n$. Now take the inverse image in $SL_2(\mathbb{R})$ and you get the wanted lattice.
