TITLE: Why does it suffice to study modular forms for $\Gamma_0(N)$?
QUESTION [7 upvotes]: Every reference I've seen on modular forms seems to jump from the general definition of a modular form for congruence subgroups to studying modular forms just for $\Gamma_0(N)$.
Once upon a time, I saw something like
$$M_k(\Gamma(N)) \cong\bigoplus_{\chi\mod N} M_k(N^2,\chi)$$
where $M_k(\Gamma(N))$ is the vector space of all weight $k$ modular forms for $\Gamma(N)$, and on the right side $\chi$ ranges over all dirichlet characters mod $N$, and $M_k(N^2,\chi)$ is the vector space of weight $k$ modular forms $f$ satisfying
$$f(\gamma z) = \chi(d)(cz+d)^kf(z)$$
for all $\gamma = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\in \Gamma_0(N^2)$.
However, I don't see how this makes sense, since $\Gamma(N)$ is not a subgroup of $\Gamma_0(N^2)$.

REPLY [5 votes]: Conjugating $\Gamma(N)$ by $\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$
 $$\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix} \displaystyle\Gamma(N)\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & 1/N \end{bmatrix} 
\scriptstyle \ \ = \ \ \underbrace{\left\{\ \scriptstyle\begin{bmatrix} aN+1 & b \\ cN^2 & dN+1\end{bmatrix} \in SL_2(\mathbb{Z})\right\}}_{\displaystyle\tilde{\Gamma}_1(N^2)}\displaystyle \ \supset \Gamma_1(N^2)$$
Define the linear operator $T : M_k(\Gamma(N)) \to M_k(\tilde{\Gamma}_1(N^2)), \ \ T f(\tau) = f (N\tau)$,
and the usual linear operators for showing $M_k(\Gamma_1(N^2)) =\displaystyle \bigoplus_{\chi \bmod N^2}M_k(\Gamma_0(N^2),\chi)$ 

For $gcd(d,N^2)=1$, let $\langle d \rangle : M_k(\Gamma_1(N^2))  \to M_k(\Gamma_1(N^2)) $,  $\ \ \langle d \rangle g = g|_k\gamma, \quad \gamma \in \Gamma_0(N^2), \quad\gamma_d \equiv d \bmod N^2$ (which is well-defined, not depending on the chosen $\gamma$). Note that $\langle d d' \rangle = \langle d  \rangle\langle d' \rangle$
And for a $\chi \bmod N^2$ :
$$ \pi_\chi g= \frac{1}{\varphi(N^2)}\sum_{\begin{array}{l}d \bmod N^2\\gcd(d,N^2)  =1\end{array}} \overline{\chi(d) }\langle d \rangle g$$ 
$\pi_\chi$ is an orthogonal projection $M_k(\Gamma_1(N^2)) \to M_k(\Gamma_0(N^2),\chi)$, and $\displaystyle\sum_{\chi \bmod N^2} \pi_\chi g=\langle 1\rangle g= g$ and for any $\chi \ne \chi'$ : $\pi_\chi \pi_{\chi'} = 0$ 
Finally, $Tf = \langle dN+1\rangle T f$, so that $\langle dN+d'\rangle T f= \langle d'\rangle T f$ and hence $\pi_\chi T f = 0$ whenever $\chi$ isn't a character $\bmod N$. 

Thus $\sum_{\chi \bmod N} \pi_\chi Tf = Tf$, and together with $M_k(\Gamma_0(N^2),\chi) \subset M_k(\tilde{\Gamma}_1(N^2))$ for any $\chi \bmod N$, 
it means that $$M_k(\Gamma(N)) \simeq M_k(\tilde{\Gamma}_1(N^2)) =\bigoplus_{\chi \bmod N}M_k(\Gamma_0(N^2),\chi)$$