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Is it a fact that if $f(z)$ is a modular form of weight $k$ for $\mathrm{SL}_2(\mathbb{Z})$ then $f(Nz)$ is a modular form of weight $k$ for $\Gamma_0(N)$?

I tried considering this by simply making a change of variables $Nz \mapsto \tau$ but I can't make my reasoning rigorous because I am unsure of which transformation law to check when finding the weight of $f(Nz)$. Is it $$ f(N\cdot \gamma(z)) = (cz+d)^kf(Nz) $$ or $$ f(\gamma(Nz)) = (c(Nz)+d)^kf(Nz) $$

If it's the second case I think my approach mentioned above would work but then I don't see the importance of emphasizing $\Gamma_0(N)$. In other words, what is the significance of this additional congruence relation we put on the bottom left entry $c$ of our matrix $\gamma$?

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Let $$f[\gamma]_k = f(\frac{az+b}{cz+d})\det(\gamma)^{k/2} (cz+d)^{-k}, \qquad \gamma = {\scriptstyle \begin{pmatrix} a & b \\ c & d \end{pmatrix}} \in GL_2(\mathbb{Q}) $$ So that $f[\gamma \gamma']_k = (f[\gamma]_k)[\gamma']_k$.

If $f[\gamma]_k = f$ for $\gamma \in \Gamma$ and $g = f[\alpha]_k$ then $g[\gamma]_k = g$ for $\gamma \in \alpha^{-1} \Gamma \alpha$

Here $\alpha = {\scriptstyle \begin{pmatrix} N & 0 \\ 0 & 1 \end{pmatrix}}$ so that $g(z) = f[\alpha]_k(z) = f(Nz)N^{k/2} $ and $f$ is a modular form of weight $k$ for $SL_2(\mathbb{Z})$ means $g$ is modular form of weight $k$ for $\Gamma_0(N) = \alpha^{-1} SL_2(\mathbb{Z})\alpha \cap SL_2(\mathbb{Z})$.

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  • $\begingroup$ In fact that holds when $f$ is a modular form for $\Gamma^0(N)$. $\endgroup$
    – ccorn
    Jul 14, 2017 at 15:43
  • $\begingroup$ @ccorn yes $\Gamma^0(N) = \alpha \Gamma_0(N) \alpha^{-1}$ $\endgroup$
    – reuns
    Jul 14, 2017 at 15:46
  • $\begingroup$ If you know this subject, I would like to understand how to find for which group $\sum_{n=1}^M a_n f(nz)$ and $\sum_{n=1}^M a_n f_n(z)$ are modular, where $f_n(z) = \frac{1}{n}\sum_{b=0}^{n-1} f(z+b/n)$, and also the converse theorems for showing an (entire) Dirichlet series with functional equation is the Mellin transform of a cusp form $\endgroup$
    – reuns
    Jul 14, 2017 at 15:54
  • $\begingroup$ It seems that $\alpha^{-1} SL_2(\mathbb{Z})\alpha = \mathrm{SL}_2(\mathbb{Z})$ is false since (2,1) entry is always divided by $N$. $\endgroup$
    – user682141
    Aug 29, 2022 at 14:06
  • $\begingroup$ @user682141 Yes obviously, did I write this? $\endgroup$
    – reuns
    Aug 29, 2022 at 14:12

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