Raising the level of a weakly holomorphic modular form

Question

If $f \in M^{!}_{k}(N)$ is a weakly holomorphic weight-$k$ modular form which is holomorphic at all cusps except $\infty$, and if $p \in \mathbb{N}$ is a prime which does not divide $N$, can I say that $f(pz) \in M^{!}_{k}(pN)$ is a weight-$k$ weakly holomorphic modular form of level $pN$ which also only has pole at $\infty$ ?

Answer 1

$f$ is a meromorphic modular form $\in M_k(\Gamma_0(N))$ with only one pole at the cusp $\Gamma_0(N).i\infty$

• Since $f(z) = (cz+d)^{-k}f(\gamma(z))$ for all $\gamma\in \Gamma_0(N)$ then $f(pz)= (cz+d)^{-k}f(p\gamma(z))$ for all $\gamma\in \Gamma_0(pN)$ (it works for $\Gamma_1(N)$ too)

  This follows from $\pmatrix{p & 0\\ 0&1}\pmatrix{a&b\\pNc&d}\pmatrix{1/p&0\\ 0&1}=\pmatrix{a&pb\\Nc&d}$

Let $p\nmid N$.

• That $f$ has only one pole at $\infty$ means that for $\gamma \in SL_2(\Bbb{Z})$, $\lim_{\Im(z)\to \infty} f(\gamma(z)))=i\infty$ iff $\gamma\in \Gamma_0(N)$ iff $\gamma.i\infty = \frac{a}{cN},\gcd(a,cN)=1$.

• Then let $\beta=\pmatrix{a&b\\ c&d}\in SL_2(\Bbb{Z}),\alpha=\pmatrix{a&B\\ c&D}\in GL_2(\Bbb{Q}),\det(\alpha)>0$. We find $\beta^{-1}\alpha=\pmatrix{1&r\\ 0&s},s>0$. Thus $$\lim_{\Im(z)\to \infty} f(\alpha.z)=\lim_{\Im(z)\to \infty} f(\beta.\beta^{-1}\alpha.z)=\lim_{\Im(z)\to \infty} f(\beta.z)$$ depends only on $\alpha.i\infty$.

• For $\delta \in SL_2(\Bbb{Z})$ we get that $f(p (\delta.i\infty))=\infty$ iff $\delta.i\infty= \frac{a}{pcN},\gcd(a,cN)=1$ iff "$\delta\in \Gamma_0(pN)$ or $p| a,p\nmid c$".

  In particular with $ \delta.i\infty= \frac1N,\delta=\pmatrix{1&0\\N& 1}$ then $f(pz)$ has a pole at the cusp $\Gamma_0(pN).\frac1N = \{ \frac{A/N+B}{cpN/N+D},AD-BcpN=1\}$ which is not the cusp $\Gamma_0(pN).i\infty$ because $\frac{A/N+B}{cpN/N+D}=\infty\implies D=-cp\implies p|AD-BcpN$.