I am trying to understand cusp forms, for a congruence subgroup. As an example, take $\Gamma_0(p)$, for $p$ a prime. As I understand there are two cusps, $0$ and $i\infty$, which are the $\Gamma_0(p)$ orbits of $i(\mathbb Q \cup \infty)$.
Checking a form $f$ for $\Gamma_0(p)$ vanishes at $\tau = i\infty$ is simple as you just check it vanishes for $q=0$. So let's assume that's true so that $f = \sum_{n=1}^\infty a_n q^n$.
To check $\tau = 0$, i.e. $q = 1$ means proving that $\sum_{n=1}^\infty a_n = 0$, which is often not trivial. So how does one usually approach showing this? I had the idea you could choose a different point in the same class, e.g. $\begin{pmatrix} 1 & p\\ 0 & 1 \end{pmatrix}$ takes you to $\tau = p$ but you still have to show the sum of the coefficients vanishes. Is there a trick to checking vanishing at cusps (and more generally checking holomorphicity at cusps when you have the $q$-expansion)? Also I noticed that $0$ is not in the upper half plane, so it seems there is an issue there, when it is touted as a cusp but I am assuming there is a point I am missing.
It depends on the amount of information you have on $f$.
For $f(z)=\sum_{n\ge 1} a_n e^{2i\pi nz} \in M_k(\Gamma_0(p))$ with integer coefficients bounded like $|a_n|\le C n^m$,
Then $$z^{-k} f(-1/(pz))=\sum_{n\ge 0} c_n e^{2i\pi nz} \in S_k(\Gamma_0(p)),\qquad c_0=\int_i^{i+1}z^{-k} f(-1/(pz)) dz\in \Bbb{Z}$$ Whence $c_0=0$ iff for some $N$ $$|\int_i^{i+1}z^{-k} \sum_{n=1}^N a_n e^{2i\pi(-1/(pz)) }dz|<1-\sum_{n=N+1}^\infty |\int_i^{i+1}z^{-k} e^{2i\pi(-1/(pz)) }dz|$$ In other words knowing the bound $Cn^m$ allows to check if it is a cusp form solely from the first few coefficients.
Certainly the bound $C n^m$ itself can be found from the first few coefficients, at this point I would look into https://arxiv.org/abs/1809.10908 and parigp and sage implementations of modular forms.