Given two matrices $A,B\in{\rm{SL}}_2(\Bbb{R})$, is there any criterion guaranteeing that the subgroup they generate is discrete? What if one puts restrictions on $A,B$ e.g. they are both elliptic? What could the topological closure of $\langle A,B\rangle$ be if it is not discrete?
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2$\begingroup$ Does this answer your question: mathoverflow.net/questions/109967/… ? $\endgroup$– Moishe KohanCommented Aug 13, 2020 at 23:51
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1$\begingroup$ Consider for instance the group generated by two upper triangular matrices with 1 one the diagonal, where the top right entry in the first matrix is 1 and in the second matrix is $\sqrt{2}$. The topological closure is isomorphic to $\mathbb{R}$. You can do the same trick be taking diagonal matrices with $e^1, e^{-1}$ and $e^{\sqrt{2}}, e^{-\sqrt{2}}$ and similarly with rotation matrices (the topological closure will be the circle). $\endgroup$– darklCommented Aug 13, 2020 at 23:55
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1$\begingroup$ @darkl: These are "elementary" subgroups and usually are excluded from the discussion of discreteness algorithms. In many ways, they are quite different from the nonelementary subgroups for which algorithms work. $\endgroup$– Moishe KohanCommented Aug 16, 2020 at 0:50
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1$\begingroup$ Here is a survey on this question: MR0974424 Gehring, F. W.; Martin, G. J. Iteration theory and inequalities for Kleinian groups. Bull. Amer. Math. Soc. (N.S.) 21 (1989), no. 1, 57–63. $\endgroup$– Alexandre EremenkoCommented Mar 27, 2021 at 11:22
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