0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $f=0$.

$\endgroup$
1
  • $\begingroup$ 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 $SL_2(\Bbb{Z})e^{2i\pi /3}$) $\endgroup$
    – reuns
    Commented Dec 11, 2020 at 12:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .