If $f(z)$ is a modular form then $f(Nz)$ is in $M_k(\Gamma_0(N))$

Question

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 $\gamma$?

Answer 1

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 $\Gamma_0(N) = \alpha^{-1} SL_2(\mathbb{Z})\alpha \cap SL_2(\mathbb{Z})$.