13
$\begingroup$

I am porting this question across from StackExchange, since it has received no answers and perhaps is sufficiently deep to fit here.

I am considering the set of upper triangular matrices $$D_N=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}:a,b,d\in\mathbb{Z}, ad=N, \gcd(a,b,d)=1, 0\leq b<d\right\}.$$ Standard facts tell us that this is a set of representatives for $M_N$ under the action of $\operatorname{SL}_2(\mathbb{Z})$ from the left. ($M_N$ being the set of primitive integer matrices of determinant $N$.)

There is also an action of $\operatorname{SL}_2(\mathbb{Z})$ on $D_N$ from the right.

Namely, given $g\in D_N$ and $\gamma\in\operatorname{SL}_2(\mathbb{Z})$, the matrix $g\gamma$ is still a primitive integer matrix of determinant $N$, and hence takes the form $\gamma' h$ for some $\gamma'\in\operatorname{SL}_2(\mathbb{Z})$ and unique $h\in D_N$. So let $\gamma$ act on $D_N$ by sending each $g$ to the corresponding $h$.

The question is: is this action transitive? I thought this was a standard fact, and one seems to need it to prove that the $N$th modular polynomial $\Phi_N$ is irreducible (among other things).

However, I can't seem to find a reference for it, and after struggling for a time to prove it, I am increasingly pessimistic - it seems to rely on some properties of the "3-variable Euclidean algorithm" that I'm not certain I believe in.

(It would be enough to know, for instance, that one can write $$a\lambda m + bmn + dn\mu=1,$$ for some integers $\lambda, \mu,m,n$, but this doesn't seem particularly likely to hold in general - the Euclidean algorithm gets us close to this but not quite there.)

$\endgroup$
1
13
$\begingroup$

Yes, the action is transitive. It is easier not to choose a set of coset representatives for the left action, so the question is whether $SL_2(\mathbb{Z}) \times SL_2(\mathbb{Z})$ acts transitively on $M_N$ (under left and right action). We will want the following notations: Let $\tilde{M}_N$ be all integer matrices of determinant $N$; let $GL_2(\mathbb{Z})$ be the integer matrices with determinant $\pm 1$.

The double coset representatives for $GL_2(\mathbb{Z}) \times GL_2(\mathbb{Z})$ acting on $\tilde{M}_N$ are given by the matrices in Smith normal form: $\left( \begin{smallmatrix} a&0 \\ 0&b \end{smallmatrix} \right)$ with $a|b$, $ab=N$ and $a$, $b>0$. Such a matrix is only primitive if $a=1$, so the $GL_2(\mathbb{Z}) \times GL_2(\mathbb{Z})$ orbits on $M_N$ are represented by $\left( \begin{smallmatrix} N&0 \\ 0&1 \end{smallmatrix} \right)$.

We now must pass from $GL_2(\mathbb{Z})$ to $SL_2(\mathbb{Z})$. We have shown that any matrix in $M_N$ is of the form $\gamma_1 \left( \begin{smallmatrix} N&0 \\ 0&1 \end{smallmatrix} \right) \gamma_2$ with $\gamma_1$, $\gamma_2 \in GL_2(\mathbb{Z})$. Moreover, taking determinants, we must have $\det(\gamma_1) = \det(\gamma_2) = \pm 1$. If the sign is $+$, we are done. If it is $-$, replace $\gamma_1$ by $\gamma_1 \left( \begin{smallmatrix} 1&0 \\ 0&-1 \end{smallmatrix}\right)$ and $\gamma_2$ by $\left( \begin{smallmatrix} 1&0 \\ 0&-1 \end{smallmatrix}\right) \gamma_2$.

$\endgroup$
2
  • $\begingroup$ This is great, thankyou! I don't suppose you happen to have a reference for this fact as a whole? No worries if not, the argument you give is obviously fine. $\endgroup$ Jul 7 '17 at 15:55
  • $\begingroup$ I don't know a reference, but I wouldn't be surprised if some reader does. $\endgroup$ Jul 7 '17 at 21:05
7
$\begingroup$

Let me add another proof that highlights a connection with the Hecke congruence subgroup $\Gamma_0(N)$.

Let me abbreviate $\Gamma:=\operatorname{SL}_2(\mathbb{Z})$. We would like to know if the right action of $\Gamma$ on $\Gamma\backslash M_N$ is transitive. The size of the orbit of a coset $\Gamma g$ (where $g\in M_N$) equals the index of its stabilizer $(\Gamma:\Gamma\cap g^{-1}\Gamma g)$. Let us specify $g:=\left(\begin{smallmatrix} N&0 \\ 0&1 \end{smallmatrix}\right)\in M_N$. Then a quick calculation reveals that the stabilizer $\Gamma\cap g^{-1}\Gamma g$ is precisely $\Gamma_0(N)$, whence the orbit has size $(\Gamma:\Gamma_0(N))$. However, this index is known to be equal to the cardinality of $D_N$, so the action is transitive.

Added. The lastly mentioned fact has several proofs. One natural proof is to take a set of representatives of $\Gamma_0(N)\backslash\Gamma$ which has the same cardinality as $D_N$. Here is an explicit construction. For any decomposition $N=ad$, take a set of integers $c$ coprime with $a$ which represent all residue classes $b$ mod $d$ satisfying $\gcd(a,b,d)=1$; this is possible by an application of the Chinese remainder theorem. Extend the resulting pairs $(c,a)\in\mathbb{Z}^2$ to matrices $\left(\begin{smallmatrix} \ast &\ast \\ c&a \end{smallmatrix}\right)\in\Gamma$, then these matrices will represent $\Gamma_0(N)\backslash\Gamma$.

$\endgroup$
2
  • $\begingroup$ This is a nice different perspective on it, thank you. Is the final fact (the size of that index) obvious? I guess I can sort of vaguely see it. $\endgroup$ Jul 7 '17 at 17:42
  • $\begingroup$ @HadenSpence: See my added section. $\endgroup$
    – GH from MO
    Jul 7 '17 at 18:06

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.