# Sturm bound - order at $\infty$ for $f \in M_k(\Gamma_0({N}))$

Let $$f \in M_k(\Gamma_0({N}))$$ and $$\text{ord}_{\infty}(f)>\vert \Gamma: \Gamma_0 (N) \vert \frac{k}{12}$$, then $$f=0$$

My idea is the following. Let $$L$$ be the representatives of the left cosets, than

$$g:=\prod_{\alpha \in L}{f\vert_{k\alpha^{-1}}} \; \in M_{nk}(SL2(\mathbb{Z}))$$ where $$n=\vert \Gamma: \Gamma_0 (N) \vert$$.

By the valence formula for modular forms in $$M_k(SL2(\mathbb{Z})))$$, $$f$$ vanishes if $$\text{ord}_{\infty}(g)> \vert \Gamma: \Gamma_0 (N) \vert \frac{k}{12}$$.

How do I go from here ?

Would appreciate any help.

For $$f\in M_k(G)$$, $$g = \prod_{\gamma\in G\backslash SL_2(\Bbb{Z})} f|_k \gamma$$
$$m= k |SL_2(\Bbb{Z})/G|$$
$$g^{12}$$ is a weight $$12 m$$ modular form for the full modular group, if it has a zero of order $$>m$$ at $$i\infty$$ then $$\Delta^{-m} g^{12}$$ is a weight $$0$$ modular form, thus it is constant, and since it vanishes at $$i\infty$$ it is zero, ie. $$g = 0$$ and $$f=0$$.
• The valence formula is saying that $g^{12}/\Delta^m$ is weight $0$ meromorphic so it has the same number of zeros and poles on the modular curve ie. $g^{12}$ has $m$ zeros on the modular curve (counting half the point $SL_2(\Bbb{Z})i$ and one-third the point $SL_2(\Bbb{Z})e^{2i\pi /3}$) Commented Dec 11, 2020 at 12:27