10
$\begingroup$

One way of interpreting the question might be: Is the property of being congruence a topological property? Ie, is it detected at the level of Riemann surfaces $\mathcal{H}/\Gamma$?

My motivation is that I'm looking for an efficient algorithm to test if a given finite index subgroup of $SL_2(\mathbb{Z})$ is congruence.

The best algorithm I'm aware of is Hsu's algorithm, but this only tests if a finite index subgroup of $\text{PSL}_2(\mathbb{Z})$ is congruence.

Given a finite index $\Gamma\le\text{SL}_2(\mathbb{Z})$, the image of $\Gamma$ in $\text{PSL}_2(\mathbb{Z})$ is congruence if and only if $\pm\Gamma$ is congruence. Here, $\pm\Gamma$ is the subgroup generated by $\Gamma$ and $-I$.

Let $\ell$ be the "Wolhfart level" of $\Gamma$ - that is, $\ell$ is the least common multiple of the cusp widths of $\mathcal{H}/\Gamma$. Then, a classical theorem of Wolhfart/Klein says that $\pm\Gamma$ is congruence if and only if it contains $\Gamma(\ell)$, equivalently if it contains $\pm\Gamma(\ell)$ (this theorem is actually quoted incorrectly in many sources, where the $\pm$ in $\pm\Gamma$ is ignored!!)

The first question is: Is testing for congruence in $\text{PSL}_2(\mathbb{Z})$ enough? Ie, if $\pm\Gamma$ is congruence, then must $\Gamma$ be congruence?

Of course this is only relevant if $-I\notin\Gamma$, in which case $\pm\Gamma = \Gamma\times\{\pm I\}$, and $\pm\Gamma(\ell) = \Gamma(\ell)\times\{\pm I\}$. Intersecting $\Gamma$ with $\pm\Gamma(\ell)$, there are two cases:

A) If $\ell = 2$, then if $\pm\Gamma$ is congruence, it must contain $\Gamma(2) = \pm\Gamma(2)$, so $\Gamma\cap\Gamma(2)$ has index 2 in $\Gamma(2)$, with quotient an 2-torsion abelian group, but the maximal 2-torsion abelian quotient of $\Gamma(2)$ is $\Gamma(2)/\Gamma(4) = C_2\times C_2\times C_2$, hence $\Gamma$ contains $\Gamma(4)$ and is congruence.

In particular, this shows that the Sanov subgroup - the index 12 subgroup of $\text{SL}_2(\mathbb{Z})$ generated by $$\begin{bmatrix} 1&2\\0&1 \end{bmatrix} \quad\text{and}\quad\begin{bmatrix} 1&0\\2&1 \end{bmatrix}$$ is congruence, even though the cusp widths are all 2 and it doesn't contain $\Gamma(2)$.

B) If $\ell\ge 3$, then again $\Gamma\cap\pm\Gamma(\ell)$ has index 2 inside $\pm\Gamma(\ell)$, and certainly any index 2 subgroup of $\pm\Gamma(\ell)$ can appear, and so the question reduces to:

Reformulation of question: For any $\ell\ge 3$, is every index 2 subgroup of $\pm\Gamma(\ell)$ a congruence subgroup?

or equivalently: For any $\ell\ge 3$, is the composite $\Gamma(\ell)'\Gamma(\ell)^2$ congruence? (the first is the commutator subgroup, the second is the subgroup of squares)

I don't even know of a reasonable way to computationally test this, since without Wolhfart/Klein's result, one might a priori have to test non-containment of infinitely many $\Gamma(n)$'s to prove that some $\Gamma$ is noncongruence.

Of course, if the answer is negative, then naturally one might ask:

Is there an efficient algorithm to test whether a given finite index subgroup of $\text{SL}_2(\mathbb{Z})$ is congruence?

$\endgroup$
15
$\begingroup$
  • For the first question: it can happen that $\pm \Gamma$ is congruence but $\Gamma$ is not; there is a beautiful paper on this phenomenon, with lots of examples, by Kiming, Schütt and Verril here.

  • For the second question, the existence of an efficient algorithm: I set this once as a fourth-year research project to a Warwick undergrad, Tom Hamilton. Tom successfully generalised Hsu's algorithm to subgroups of SL2 instead of PSL2. You can read our paper on this here; the algorithm is implemented in recent versions of Sage.

In fact, in Sage there is a function which, given a finite-index subgroup of SL2Z containing -1, will enumerate all index 2 subgroups of that group not containing -1 (i.e. all liftings of $\Gamma / \{\pm 1\}$ to SL2Z). Using this, one can show (for instance) that there is an index 2 subgroup of $\pm \Gamma(7)$ which is non-congruence. It's generated by the following matrices:

(
[-1 -7]  [-48   7]  [209 -56]  [113 -35]  [ 55 -21]  [-120   49]
[ 0 -1], [ -7   1], [ 56 -15], [ 42 -13], [ 21  -8], [ -49   20],

[-15   7]  [ 239 -140]  [113 -70]  [-232  161]  [-181  133]  [-8  7]
[-28  13], [  70  -41], [ 21 -13], [ -49   34], [ -49   36], [-7  6],

[  76 -105]  [-169  238]  [ 43 -63]  [ 309 -490]  [ 134 -217]
[  21  -29], [ -49   69], [ 28 -41], [  70 -111], [  21  -34],

[-281  476]  [-230  399]  [-15  28]  [-97 231]  [ 218 -525]
[ -49   83], [ -49   85], [ -7  13], [-21  50], [  49 -118],

[ 279 -763]  [ 22 -63]  [-118  399]  [-29 112]  [ 139 -609]  [-36 175]
[  49 -134], [  7 -20], [ -21   71], [ -7  27], [  21  -92], [ -7  34],

[-43 252]
[ -7  41]
)
$\endgroup$

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.