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?



Bounded denominators for modular forms
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](http://arxiv.org/abs/1012.3062)" 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](http://arxiv.org/abs/math/0612100) and [here](http://arxiv.org/abs/0809.0020). The idea of these papers is to prove the unbounded denominator property for certain classes of noncongruence subgroups.


Lastly, you may find [this](http://arxiv.org/abs/0707.3315) survey article a bit helpful.


