definition of cusp forms for a congruence group

Question

I'm reading the book A First Course in Modular Forms by Diamond and Shurman. On page 17, in the definition of cusp forms for a congruence group, they require the constant term to be zero in the Fourier expansion of $f[\alpha]_k$ for all $\alpha \in SL_2(\mathbb Z)$. My question is, why does $f[\alpha]_k$ have a Fourier expansion?

Answer 1

Let $\Gamma$ be a congruence subgroup of $SL(2,\mathbb{Z})$. We have $\Gamma(N)\subset \Gamma$ for some N where $\Gamma(N)$ is the Princpal congruence subgroup.

Now let $f\in M_k(\Gamma)$ be a (weak) modular form with respect to $\Gamma$. As $A=\begin{pmatrix} 1 & N\\ 0 & 1\end{pmatrix}\in \Gamma$ we have $f[A]_k=f$, that is, $f(z+n)=f(z) \ \ \ \forall z\in \mathbb{H}$. Then $f$ is periodic of period $N$, hence for a given $y\in \mathbb{R}^+$ we have a Fourier expansion for $f\mid_{\{z|Im(z)=y\}}$ of the form

$$ f(z)=\sum_{n\in \mathbb{N}} a_n e^{2\pi i n\frac{x}{N}}=\sum_{n\in \mathbb{N}} \hat{a}_n e^{2\pi i n\frac{z}{N}}$$

The last sum have an analytic expansion to all $\mathbb{H}$ with the exact same formula, this analytic function coincide with $f$ on a line and therefore coincide in $\mathbb{H}$. This shows that the fourier expansion of $f$ doesn't depend of $y$ and then is a global Fourier expansion for $f$.