Finding all rational (closed and connected) subgroups of $\text{SL}(2,\mathbb R)$ with respect to $\text{SL}(2,\mathbb Z)$

Question

A subgroup $H$ of $\text{SL}(2,\mathbb R)$ is called rational w.r.t. the lattice $\text{SL}(2,\mathbb Z)$ if $H \cap \text{SL}(2,\mathbb Z)$ is a lattice in $H$ (namely the quotient homogeneous space has a probability Haar measure).

I would like to find all rational, closed and connected subgroups of $\text{SL}(2,\mathbb R)$ with respect to $\text{SL}(2,\mathbb Z)$, but I don't know how hard this question is and where to start.

Answer 1

Yes, this can be listed in a reasonable way.

These are:

1. the trivial group $\{1\}$,
2. the whole group $\mathrm{SL}_2(\mathbf{R})$,
3. the $\mathrm{SL}_2(\mathbf{R})$-conjugates of the 1-dimensional compact group $\mathrm{SO}_2(\mathbf{R})$,
4. various non-compact 1-dimensional subgroups, namely: for each $n\ge 2$, the unit component $H'_n$ of the centralizer $H_n$ in $\mathrm{SL}_2(\mathbf{R})$ of the matrix $M_n=\begin{pmatrix}0 & -1\\ 1& n\end{pmatrix}$, and its $\mathrm{SL}_2(\mathbf{Q})$-conjugates. (Unit component is actually the unique subgroup of index $2$, except for $n=2$ in which case it has index $1$.)

These subgroups indeed satisfy the condition: indeed for conjugates of $H_n$ one gets the real points of a 1-dimensional $\mathbf{Q}$-anisotropic torus and hence integral matrices form a lattice therein, using Borel-HarishChandra (or a direct argument).

Conversely, let $L$ be a connected closed subgroup with the given condition. If its dimension is $0$ or $3$ we are done. If its dimension is $2$, then $L$ is conjugate to the group of upper triangular matrices with positive diagonal and det 1, which is not unimodular, and hence has no lattice, so this is excluded. Hence $L$ has dimension $1$. If $L$ is compact, it is a real conjugate of $\mathrm{SO}_2(\mathbf{R})$.

Now assume that $L$ is 1-dimensional and non-compact. It contains an integral matrix of determinant 1 and infinite order, hence of trace $\neq 0,\pm 1$; up to squaring we can suppose its trace is $n\ge 2$. Hence it is conjugate to $M_n$, and since then the centralizer of $M_n$ is 1-dimensional, we are done.

Remark: if we classify the subgroups $H_n$ (or equivalently $H'_n$) up to $\mathrm{SL}_2(\mathbf{Q})$-conjugacy, then we have to identify some values of $n$. Namely, $H_n$ and $H_m$ are $\mathrm{SL}_2(\mathbf{Q})$-conjugate iff $M_n,M_m$ have positive powers with the same trace. So for a "unique" classification, we need to discard $n\ge 3$ when there exist $m\ge 3$ and $k\ge 2$ such that $\mathrm{Tr}(M_m^k)=n$.

This discards the following set $X$ of values: 7, 14, 18, 23, 34, 47, 52, 62, 79, 98, 110, 119, 123, 142, 167, 194, 198, 223, 254, 287... (for instance, $7=\mathrm{Tr}(M_3^2)$, $14=\mathrm{Tr}(M_4^2)$, $18=\mathrm{Tr}(M_3^3)$...) these seem to coincide with the set of values $\ge 7$ of the sequence OEIS/A298878. The $\mathrm{SL}_2(\mathbf{Q})$-conjugacy classes of subgroups $H_n$ are thus indexed by $n\in\mathbf{N}_{\ge 2}\smallsetminus X$.

Remark 2 I'm not sure that "rational" is an adequate terminology: indeed it usually means that the subgroup is definable by equations with rational coefficients, and includes, for instance, the upper triangular group or the diagonal group. For rational connected subgroups $H\subset\mathrm{GL}_n$ (in this genuine sense), whether the integral matrices form a lattice is given by the Borel-Harish-Chandra criterion, namely iff $H$ has no nontrivial rational character (i.e. no nontrivial $\mathbf{Q}$-defined homomorphism into $\mathrm{GL}_1$).