Stabilizers of an arbitrary element of $\mathbb Z^2$

Question

Let $G = SL(2,\mathbb Z)$ acting on $\mathbb Z^2$ by the product $Av$, I'm trying to define $\operatorname{Stab}(v)$ for all $v\in \mathbb Z^2$. $Av = v$, then $1$ is an eigenvalue associated with the eigenvector $v$.

For all $A\in \operatorname{Stab}(v)$, the characteristic polynomial is $(x-1)^2$, then the minimal polynomial $m_A$ is $(x-1)$ or $(x-1)^2$ for Hamilton–Cayley Theorem. If $m_A = (x-1)$, then $A = I_2$ and $I_2 \in \operatorname{Stab}(v)$. If $m_A = (x-1)^2$, then $A$ is not diagonalizable, then the Jordan normal form of A is \begin{equation} \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix} \end{equation} I don't know if this is a easy way, but I can't find the solution of this exercise.

Answer 1

You want to determine $A$'s entries $a,b,c,d\in\Bbb Z$ such that, given $(x,y)\in\Bbb Z^2$: \begin{cases} ax+by=x\\ cx+dy=y\\ ad-bc=1\\ \tag1 \end{cases} Therefore:

• for $v= \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}$, $\operatorname{Stab}(v)=\operatorname{SL}(2,\Bbb Z)$;
• for $v= \begin{pmatrix} x\ne 0 \\ 0 \\ \end{pmatrix}$, $\operatorname{Stab}(v)=\biggl\{ \begin{pmatrix} 1&b \\ 0&1 \\ \end{pmatrix}\mid b\in \Bbb Z \biggr\}$;
• for $v= \begin{pmatrix} 0 \\ y\ne 0 \\ \end{pmatrix}$, $\operatorname{Stab}(v)=\biggl\{ \begin{pmatrix} 1&0 \\ c&1 \\ \end{pmatrix}\mid c\in \Bbb Z \biggr\}$;
• if $x\mid y$, namely for $v= \begin{pmatrix} x \\ kx \\ \end{pmatrix}$, for some $k\in\Bbb Z$, $(1)$ yields $a+kb=1$, $c+kd=k$ and $d-bk=1$, and finally: \begin{alignat}{1} &bk=d-1\\ &a=2-d\\ &c=k(1-d) \end{alignat} So: $$\operatorname{Stab}(v)=\biggl\{ \begin{pmatrix} 2-d&\frac{d-1}{k} \\ -k(d-1)&d \\ \end{pmatrix}\mid d-1\in k\Bbb Z\biggr\}$$
• if $x\nmid y$, then $b=lx$ and $d=1+mx$ for some $l,m\in\Bbb Z$, whence $a=1-ly$ and $c=-my$. So: \begin{alignat}{1} \operatorname{Stab}(v) &= \biggl\{ \begin{pmatrix} 1-ly&lx \\ -my&1+mx \\ \end{pmatrix}\mid mx=ly, \text{ with }m,l\in\Bbb Z\biggr\} \\ &= \biggl\{ \begin{pmatrix} 1-mx&lx \\ -my&1+mx \\ \end{pmatrix}\mid mx=ly, \text{ with }m,l\in\Bbb Z\biggr\} \\ \end{alignat}