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?).

Solution 1:

• 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. $$12 d = [SL_2(\Bbb{Z}):\Gamma_0(n)] k$$