Orbits of action of $SL_2(\mathbb{Z})$ on matrices of fixed determinant

matricesnumber theory

The following question is an exercise from a lecture series on modular forms by M Ram Murty. Let $X_m$ be the set of matrices of determinant $m\in\mathbb{N}$. Then $SL_2(\mathbb{Z})$ acts on $X_m$ (say, on the left) by matrix multiplication. The exercise is to show that the number of orbits is finite and that the orbit representatives are given by
\begin{equation}T=\left\{
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}:d>0,ad=m, b\mod d
\right\}.
\end{equation}
I can show that these representatives are in disjoint orbits. But I'm not sure how to show that they are a complete set, i.e. I want to show that for any $M\in X_m$, there are matrices $A\in SL_2(\mathbb{Z})$ and $B\in T$ such that $AB=M$.

Best Answer

Let $$ X= \begin{pmatrix} x & u \\ y & v \end{pmatrix}, $$ where $x,y,u,v\in\mathbb{Z}$ and $\det(X)=m$.

  1. If $\gcd(x,y)=t$ and $x=tx'$, $y=ty'$, then $\gcd(x',y')=1$ and let's choose a matrix $A$ of the form

$$ A= \begin{pmatrix} x'' & y'' \\ y' & -x' \end{pmatrix}. $$

Prove that the $x'',y''$ can be replaced by integers so that $A\in SL_2(\mathbb{Z})$

  1. Check that

$$ X_1=AX= \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}, $$

where $a,b,d\in\mathbb{Z}$, $ad=m$.

  1. If $d<0$, then multiply $X_1$ by the matrix $-I$, where $I$ is a unit matrix. Consider further that $d>0$.

  2. Let $b=dq+r$, where $0\leq r<d$. Multiply $X_1$ by the matrix

$$ B= \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix} $$

Check that the matrix $BX_1$ has the form we want.

  1. It remains to prove that any two different matrices $X,Y$ of the specified kind lie in different orbits.
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