Number of zeroes of a modular form on $\Gamma_0(N)$ on fundamental domain

It is a well-known result that a modular form of weight $$k$$ on $${\rm SL}_2({\bf Z})$$ has $$k/12$$ zeros on any fundamental domain of the action on the upper half-plane. The proof is complex-analytic in nature. It can also be shown that any modular form of weight $$k$$ on $$\Gamma_0(4)$$ has $$k/2$$ zeroes on a fundamental domain of this action.

This certainly makes sense, since $$[{\rm SL}_2({\bf Z}) : \Gamma_0(4)] = 6$$ so the fundamental domain is six times bigger. Is it always true that the number of zeroes on a fundamental domain of $$\Gamma_0(N)\backslash {\cal H}$$ is $${k\over 12} [{\rm SL}_2({\bf Z}) : \Gamma_0(N)],$$ or is this just a coincidence? If this is true in general, is there a way of showing it that doesn't involve redoing the whole contour integration proof from scratch? I was thinking that all modular forms on $${\rm SL}_2({\bf Z})$$ are also modular forms on $$\Gamma_0(N)$$, so since the zeroes are counted fractionally on the boundary, we should just be able to multiply by the index to find that any $${\rm SL}_2({\bf Z})$$-modular form will have $$[{\rm SL}_2({\bf Z}): \Gamma_0(N)]$$ zeroes on $$\Gamma_0(N)\backslash {\cal H}$$. So if one can show that all modular forms of the same weight and level must have the same number of zeroes on a fundamental domain, then we will be done. But I wasn't able to show this (perhaps there is a an easy argument I'm missing?).

• I would suggest looking at the norm of the modular form for the map $X_0(N) \rightarrow X(1)$ (although there may be issues with elliptic points, to be carefully considered). Commented Dec 9, 2021 at 23:10
• Beyond low-level congruence subgroups, thinking in terms of Riemann-Roch (and Riemann-Hurwitz) is much more coherent, I think. As the genus goes up (as level goes up) the interval in which R-R is not decisive about divisors also increases... Commented Dec 9, 2021 at 23:10
• For a proof based on the norm map, as Mindlack suggests, take a look at Theorem 2.6.1 in these notes: mdave16.github.io/notes/Modular%20Forms%20-%20Marc%20Masdeu.pdf Commented Dec 10, 2021 at 11:18
• Thanks everyone for the references! I'm not familiar with the Riemann-Roch theorem and its friends, so I'll definitely have to look into these---the linked PDF looks like a great read. Commented Dec 10, 2021 at 13:27

• Take $$g\in M_k(\Gamma_0(n))$$.

It has $$d$$ zeros on $$X_0(n)$$.

Note that in contrary to meromorphic functions $$d$$ doesn't have to be an integer, for example $$E_4(z)$$ has a zero of order $$1/3$$ at $$SL_2(\Bbb{Z})e^{2i\pi/3}\in X_0(1)$$, this is because $$E_4(z)^3/\Delta(z)$$ is meromorphic so it has the same number of zeros and poles, and since $$\Delta$$ has only one simple zero, so does $$E_4(z)^3$$.

$$g^{12} \in M_{12k}(\Gamma_0(n))$$ has $$12 d$$ zeros on $$X_0(n)$$.

• Let $$f(z)=\Delta(i\pi)E_{12}(z)-E_{12}(i\pi)\Delta(z)\in M_{12}(\Gamma_0(1))$$.

I chose to put a zero at $$SL_2(\Bbb{Z}) i\pi$$ because the preimage of a neighborhood of it under the map $$X_0(n)\to X_0(1)$$ is obvious.

So $$f$$ has one zero on $$X_0(1)$$, it has $$[SL_2(\Bbb{Z}):\Gamma_0(n)]$$ zeros on $$X_0(n)$$ and $$f^k$$ has $$[SL_2(\Bbb{Z}):\Gamma_0(n)] k$$ zeros on $$X_0(n)$$.

• $$g^{12}/f^k$$ is meromorphic on $$X_0(n)$$ so it has the same number of zeros and poles on $$X_0(n)$$ ie. $$12 d = [SL_2(\Bbb{Z}):\Gamma_0(n)] k$$

• This is exactly what I'm looking for! I'm just wondering about a couple of details, since I'm very new to modular forms. Why exactly did we construct $f$ in this way (could any other modular form of weight 12 have worked)? How do we know that meromorphic quotient implies that the numerator and denominator have the same number of zeroes and poles? Thanks for the great answer and thanks in advance for the clarifications! Commented Dec 10, 2021 at 0:29
• Also, sorry but I just realised: is f is actually a modular form of weight $24$, since $E_{12}$ and $\Delta$ are both modular forms of weight $12$? Commented Dec 10, 2021 at 0:31
• Weigth 12. $X_0(n)$ is a compact Riemann surface that's why an $h$ meromorphic willl have the same number of zeros and poles. The proof is similar to the $k/12$ one, integrate something (namely $\frac{h'(z)}{h(z)}dz$) on the boundary of the fundamental domain and say that the edges cancel each other. Commented Dec 10, 2021 at 0:43