I have an elementary question on modular forms, but which I don't know how to solve.



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> a) Is there a congruence subgroup $\Gamma \leq \mathrm{SL}\_2(\Bbb Z)$, an integer $k \in \Bbb Z$ and a non-constant modular form $f \in M\_k(\Gamma)$ such that $f$ has only finitely many non-zero Fourier coefficients $a\_n(f)$ ?
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> b) What about $\{n \geq 0 \mid a\_n(f) \neq 0\}$ having zero density?
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One can easily have a non-zero modular form $f$ such that $a\_n(f) = 0$ for every odd integer $n$. For part a), I think the answer should be no : $f$ is just a trigonometric polynomial and I guess one can come up with some elementary argument, but I don't know exactly how. Part b) is maybe a more subtle question, I would be glad to have any information about it!


I already asked it [here](https://math.stackexchange.com/questions/4113252/), but got no comment nor any reply. Possibly related: [Modular forms with prime Fourier coefficients zero](https://mathoverflow.net/questions/25461/modular-forms-with-prime-fourier-coefficients-zero).



Modular forms with finitely many or very few non-zero Fourier coefficients
For simplicity, I will assume that $f$ is a cusp form (hence $k\geq 12$ and $k$ is even).


The answer to question (a) is negative. It was proved independently by [Rankin (1939)](https://doi.org/10.1017/S0305004100021095) and Selberg (1940) that the Dirichlet series
$$\sum\_n\frac{|a\_n(f)|^2}{n^s}$$
has a simple pole at $s=k$. Hence infinitely many coefficients $a\_n(f)$ are nonzero.


Regarding question (b), let me further assume that $f$ is a primitive Hecke eigenform. If $f$ is a CM form, then $a\_n(f)\neq 0$ implies that the prime factors of the square-free part of $n$ split in the quadratic number field associated with $f$, so the density of $\{n:a\_n(f)\neq 0\}$ is zero. If $f$ is not a CM form, then the density of $\{n:a\_n(f)\neq 0\}$ is positive, as proved by [Serre (1981)](http://www.numdam.org/item/?id=PMIHES_1981__54__123_0).


