Let $A$ be a fixed matrix in $M_2\mathbb{Z}$ with determinant $n \neq 0$.

Question 1 How many ways can I write $A = XY$ for $X, Y \in M_2\mathbb{Z}$?

The answer to this question is pretty clearly infinite, since for any $\gamma \in GL_2\mathbb{Z}$ and any such pair $(X, Y)$, then $(X \gamma^{-1}, \gamma Y)$ is another such factorization. So let's get rid of that.

Question 2 Consider the set $S_A = \{(X, Y) \in (M_2\mathbb{Z})^2 \mid A = XY\}$. The group $G = GL_2\mathbb{Z}$ acts naturally on this set via $$ \gamma \cdot (X, Y) = (X \gamma^{-1}, \gamma Y) $$ What is the cardinality of $S_A/G$?

Unfortunately, this depends on our choice of $A$ (fixing $n$). If we compare $$ A_1 = \begin{pmatrix}4 & 0 \\ 0 & 1 \end{pmatrix} \qquad \qquad A_2 = \begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix} $$ then in the first case, we find $|S_{A_1}/G| = 3$ while in the second, $|S_{A_2}/G| = 4$. This is due to the fact that the second matrix is not primitive. We could exclude such matrices, or we could define an equivalence relation on factorizations via $$ (X, kY) \sim (kX, Y) $$ which is compatible with the action of $G$. For a fixed $A$, define $T_A = S_A/_\sim$.

The Real Question Does the cardinality of $T_A/G$ depend on $A$? If not, what is it?

By computation, it seems that this does not depend on $A$ and moreover, $$ |T_A/G| = \sum_{d \mid n} 1 = \sigma_0(n) $$

This seems like it should be a pretty obvious question about arithmetic/algebraic groups, but it's not really my area of expertise.

  • $\begingroup$ Notice that your second equivalence relation is just $(X,Y) \sim (k^{-1}X, kY)$, which is just the same as the action of scalar matrices, so you don't gain anything by quotienting by it. $\endgroup$ – Kevin Casto Mar 29 '16 at 22:34
  • $\begingroup$ Well... I think that depends on the definition of $GL_2Z$ being used. I've been thinking of this as $GL_2Z = \{A \in M_2Z \mid \det A = \pm 1\}$, in which case that is a different relation. But as David points out below, it's still the wrong relation, so the point is kind of moot. $\endgroup$ – Simon Rose Mar 30 '16 at 6:06

$\def\ZZ{\mathbb{Z}}$Your conjecture is right when you require $A$ to be primitive. The version where you set $(X,kY) \sim (kX,Y)$ doesn't work even in your chosen example.

$A$ gives a map $\ZZ^2 \to \ZZ^2$. Set $K = \ZZ^2/A \ZZ^2$; this is an abelian group of order $n$ generated by $2$ elements and the condition that $A$ is primitive implies that $G \cong \ZZ/n \ZZ$.

You want to know how many ways you can factor this as $\ZZ^2 \to L \to \ZZ^2$, where $L \cong \ZZ^2$ and we work up to isomorphisms on the middle factor. Such a factorization is uniquely determined by the subgroup $L/ZZ^2$ of $K$. So we are counting subgroups of $\ZZ/n \ZZ$, which there are $\sigma_0(n)$ of.

If we take $A = \left( \begin{smallmatrix} 2 & 0 \\ 0 & 2 \end{smallmatrix} \right)$, then $G \cong (\ZZ/2 \ZZ)^2$, with five subgroups. Representative factorizations are $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \quad \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$$ $$\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1& 1 \\ 1 & -1 \end{pmatrix}.$$ Your equivalence relation collapses these $5$ cases to $4$, not $3$.

It is clear that $\#(T_A/G)$ will only depend on the isomorphism type of the abelian group $\ZZ^2/A \ZZ^2$ (in other words, on the Smith normal form of $A$). If you really need it, I could work it out; I suspect you could too.

  • $\begingroup$ Ah, this seems exactly what I want, and I was clearly going about it the wrong way. Thanks! $\endgroup$ – Simon Rose Mar 29 '16 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.