I recently saw a conjecture that a modular form is a congruence modular form if and only if it has bounded denominators. I wonder if one direction or the other is already known to be true?

EDIT: For completeness, here is the answer I received at Math Overflow when I asked a while back.



I am familiar with this conjecture, specifically for cusp forms.

It is known that every cusp form of a congruence subgroup of $SL_2(\mathbb{Z})$ with algebraic Fourier coefficients has bounded denominators. This is because we can decompose the space of weight $k$ cuspforms of a congruence subgroup into eigenforms (newforms and pushups of newforms), which have algebraically integral Fourier coefficients. Then any form with algebraic Fourier coefficients is just a linear combination of these eigenforms over algebraic numbers, and so must have bounded denominators.

There are several examples of noncongruence cusp forms which are $n$th roots of congruence cusp forms, which means their expansions have arbitrarily large powers of $n$ in the denominator. But there is no general method yet for showing that every genuinely noncongruence cusp form has unbounded denominators.

  • $\begingroup$ Thanks for the help, once again. $\endgroup$ – Graphth Apr 17 '11 at 0:45

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