Bounded denominators for modular forms

Question

I recently saw a conjecture that a modular form is a modular form for a congruence subgroup of $SL_2(Z)$ if and only if it has bounded denominators. Are both directions conjectures, or is one already known to be true?

Answer 1

As Ramsey pointed at, it is known that forms on congruence subgroups have bounded denominators. The other direction is still a conjecture, though some partial progress has been made. You might try looking at some of the papers by Ling Long ( http://orion.math.iastate.edu:80/linglong/ ). She has been interested in this problem for a while and has made some progress towards the conjecture. I'll briefly state one of her results (the others involve a bit more notation). All of her papers in this area are available on the arxiv.

Let $\Gamma$ be a finite index subgroup of $SL_2(\mathbb Z)$.

Unbounded denominators conjecture: Any meromorphic modular form on $\Gamma$ which is holomorphic on $\mathfrak h$ and has algebraic Fourier coefficients has bounded denominators if and only if it is a cusp form and $\Gamma$ is a congruence subgroup.

In their paper "Fourier Coefficients of Noncongruence Cusp Forms" Winnie Li and Ling Long prove the following theorem:

Theorem: Suppose the modular curve $X_\Gamma$ has a model defined over $\mathbb Q$ such that the cusp at $\infty$ is $\mathbb Q$-rational, $k\geq 2$ and $S_k(\Gamma)$ is one dimensional. Then a form in $S_k(\Gamma)$ with rational Fourier coefficients has bounded denominators if and only if it is a congruence modular form.

The other two papers by Long (joint with Chris Kurth) that I mentioned are available here and here. The idea of these papers is to prove the unbounded denominator property for certain classes of noncongruence subgroups.

Lastly, you may find this survey article a bit helpful.

Answer 2

This is a really old question but this is still the first one that shows up if you search for "unbounded denominators conjecture" so answering it here. This conjecture is now solved.

Here's the arXiv preprint from Frank Calegari, Vesselin Dimitrov, Yunqing Tang https://arxiv.org/abs/2109.09040 (Thanks @michael-albanese)

Here’s a quantamag post on it https://www.quantamagazine.org/long-sought-math-proof-unlocks-more-mysterious-modular-forms-20230309/

Answer 3

One direction is known. A modular form for a congruence subgroup has bounded denominators.

This is because such a form can be interpreted as a section of a natural line bundle on a moduli space of elliptic curves with "level structure." Equivalently, this form can be regarded as a rule that assigns to each relative elliptic curve over a ring $A$ with level structure an element in an a certain $A$-module. The $q$-expansion of such a form arises by considering the Tate elliptic curve and its level structures. Roughly speaking (i.e. ignoring the level structure and the bad fiber), these are defined over the ring $\mathbb{Z}[[q]]$. The result is that a modular form for a congruence subgroup that has rational coefficients moreover has a $q$-expansion in $\mathbb{Z}[[q]]\otimes\mathbb{Q}$, and hence has bounded denominators ($\mathbb{Z}[[q]]\otimes \mathbb{Q}$ being much smaller than $\mathbb{Q}[[q]]$).

Answer 4

The unbounded denominators conjecture was proven in the affirmative and is now a theorem.

https://arxiv.org/abs/2109.09040

https://www.quantamagazine.org/long-sought-math-proof-unlocks-more-mysterious-modular-forms-20230309/

Basically, the authors proved that there was an upper bound of possible noncongruence modular forms with bounded denominators, and if one existed, it would imply the existence of so many other such forms that it would exceed the upper bound.