Commensurator of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathrm{GL}_2^+(\mathbb{Q})$

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$I am trying to find the commensurator of $\SL_{2}(\mathbb{Z})$ on $\GL_{2}^+(\mathbb{R})$. So far I have been able to prove that $\GL_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it. Is there a way to prove the other inclusion, or to find (in case it is not $\GL_2^+(\mathbb{Q})$) the commensurator in another way?

By $\GL_2^+$ I mean invertible and positive determinant matrices.

Answer 1

It's not harder to compute the commensurator in $\mathrm{GL}_2(\mathbf{C})$. Since it contains the scalars, it is enough to compute the commensurator in $\mathrm{SL}_2(\mathbf{C})$.

If a matrix $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$ of determinant 1 commensurates $\mathrm{SL}_2(\mathbf{Z})$, then for some $n>0$, the matrices $A(I+nE_{12})A^{-1}$, $A(I+nE_{21})A^{-1}$, $A^{-1}(I+nE_{12})$, $A^{-1}(I+nE_{21})$ are all integral. Doing the computation, this implies that $a^2$, $b^2$, $c^2$, $d^2$, $ab$, $ac$, $cd$, $bd$ are all rational. One easily deduces that $A$ is a scalar multiple of a rational matrix.

Conversely, every scalar multiple of a rational matrix commensurates $\mathrm{SL}_2(\mathbf{Z})$.

For your precise question, the answer is then $\{\lambda A:\lambda\in\mathbf{R}_{>0},A\in\mathrm{SL}_2(\mathbf{Q})\}$