There are several papers which compute zeroes of modular forms for genus 0 congruence subgroups, such as "Zeros of some level 2 Eisenstein series" by Garthwaite et al published in Proc AMS and work of Shigezumi and others in levels 3,5 and 7. However, there don't seem to be generalizations of this to higher genus subgroups.

I know a few examples of modular forms for higher genus subgroups where one can compute all the zeroes; for instance, the unique normalized cusp form of weight 1 and level $\Gamma_0(31)$ with character the Legendre character modulo 31 has simple zeroes at the two cusps and the two elliptic points because the valence formula forces them to be there. Similar ideas work for levels 17, 19, 21 and 39.

My question is this: is there a more general way to find the zeroes of modular forms in an explicit way for congruence subgroups?

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    $\begingroup$ Note that the zeros of weight k modular Hecke cusp forms become equidistributed as k→∞. See this talk of Sarnak: youtube.com/watch?v=EKuaQOHMgcE $\endgroup$
    – Marc Palm
    Commented Sep 27, 2013 at 14:49

3 Answers 3


If you are looking for examples of modular forms whose zeros can be described explicitly, then you probably want the zeros to be cusps or imaginary quadratic irrationals. In this case the Gross-Kohnen-Zagier theorem implicitly gives lots of examples, by describing the relations between Heegner points on modular elliptic curves. (Heegner points are closely related to imaginary quadratic numbers in the upper half plane.) Many examples of modular forms with zeros at imaginary quadratic irrationals can also be constructed explicitly as automorphic products.


Some vague thoughts, rather than an actual answer. The zeroes of a modular form $f$ of weight $k$ correspond (except at the cusps) to zeros of the modular function $F=f^{12k}/\Delta^k$ so the problem reduces to finding zeros of modular functions. In the level one case one can't avoid inverting the $j$-function, so let's suppose we can do this. This $F$ is integral over $\mathbb{C}[j]$. Multiplying together the conjugates of $F$ under the action of $\mathrm{SL}_2(\mathbb{Z})$ one gets a polynomial $g(j)$ in $j$. At a zero of $f$, $g(j)=0$ so that there are a finite number of possibilities for $j$. For each of these there will be at least one $z$ in the upper half-plane with $j(z)=f(z)=0$. But the $z$ you first think of with $j(z)=0$ needn't have $f(z)=0$. To find all such $z$ one can compute the images of $z$ under coset representatives for one's congruence subgroup and see at which $f$ vansihes (this might be tricky).

This all sounds like rather hard work if the index of one's congruence subgroup is large.

Added Finding the zeroes of a weight $12$ modular form for $\mathrm{SL}_2(\mathbb{Z})$ is equivalent to inverting the $j$-function numerically. This is essentially a period calculation. One writes down an elliptic curve over $\mathbb{C}$ with the given $j$ invariants, and then the corresponding points in the upper half-plane are the ratios of two periods (of the canonical differential on the curve) that generate the period lattice. I wonder if for more general congruence subgroups the problem reduces to something like a period computation?

  • $\begingroup$ You can reduce the large index problem a bit by replacing j with a principal modulus of a genus zero quotient that is larger than X(1). This requires the level to have a relatively small prime factor, though (at most 71). $\endgroup$
    – S. Carnahan
    Commented May 6, 2010 at 18:27

I don't have an answer, but I can give some context for the positions of zeros (which you may already know). The dimension formula for the space of modular forms of weight k is essentially (# of zeros) + 1 - restrictions. Here is the formula for even k, taken from Shimura (Introduction to the Arithmetic Theory of Automorphic Functions); I believe he has formulas for odd weight as well.

The space of modular forms of even weight k on a subgroup G of SL2(ℤ) with $I=[SL_2(\mathbb{Z}): \pm G]$, r2 order-2 elliptic points, and r3 order-3 elliptic points has dimension $d = Ik/12 + 1 -(k/4 -\lfloor k/4 \rfloor )r_2 - (k/3 - \lfloor k/3 \rfloor)r_3 - g$.

The r2 and r3 terms come from the zeros required by the valence formula. The -g can be seen as coming from conditions imposed by Abel's Theorem. What's left gives us the degrees of freedom in placing zeros and getting a unique modular form, up to a constant multiple.

In your example, specifying a cusp form of a certain weight leaves a 1-dimensional space, and we know where all the zeros are. For weight 12 forms on SL2(ℤ), we have genus 0 and don't need zeros at the elliptic points, so we have a 2-dimensional space. Specifying a zero at any point gives us a unique form: a zero at the cusp gives $\Delta$, a zero (of degree 3) at the order-3 point gives $E_4^3$, a zero (of degree 2) at the order-2 point gives $E_g^2$. A zero anywhere else comes from some linear combination, but I don't know how to relate the linear combination to the position of the zero.

It's more complicated in the higher genus case, because in a space with dimension d=n+1-g, you can place n zeros anywhere you want, but there are g additional zeros in the fundamental domain, somehow determined by the locations of the first n.


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