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Let $\psi$ be an odd Dirichlet character of $G_{\mathbb{Q}}$ with conductor equal to $N$ and $p \nmid N$ be a prime number. Assume that $\psi(Frob_p)=1$.
Denote by $E_{\psi,1} \in S_1(\Gamma_1(N))$ the weight one Eisenstein series associated to the characters $(\psi,1)$ and let $f \in S_1(\Gamma_1(N)\cap \Gamma_0(p))$ be a $p$-stablization of $E_{(\psi,1)}$ (i.e $f(z)=E_{(\psi,1)}(z)-E_{(\psi,1)}(pz))$. Since $\psi(p)=1$, the evaluation of $f$ at the cusp $\infty \in X^{rig}_1(N)$ corresponding to $\mathbb{G}_{m}$ is trivial (we can see $f$ as an overconvergent modular form on $X^{rig}_1(N)$ and $\infty$ is in the ordinary locus of $X_1^{rig}(N)$).

Can we compute the evaluation of $f$ on the cusps of $X(\Gamma_1(N) \cap \Gamma_0(p))$ which are in the $\Gamma_0(p)$-orbit of $\infty$?

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    $\begingroup$ I am confused by the question: the literal question asks a relatively elementary thing, but the proposed context is far more sophisticated. Perhaps it's possible that the modern-times inflation of sophistication has distracted the question/context away from something far more immediate? $\endgroup$
    – paul garrett
    Sep 9, 2016 at 0:19
  • $\begingroup$ For a classical modular form $f$, we say that $f$ is cuspidal if and only if the evaluation of $f$ at the cusps is trivial, but for cuspidal overconvergent modular form, the definition is different, we need only that the evaluation of $f$ at the orbit of $\Gamma_0(p)-\infty \in X(\Gamma_1(N) \cap \Gamma_0(p))$ is trivial. And my question is the following : can we say that $f$ cuspidal overconvergent ? $\endgroup$
    – Adel BETINA
    Sep 9, 2016 at 8:02

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