William Stein has a lovely book Modular Forms, A Computational Approach, and I'd like some clarification on one of the theorems from the section on Eisenstein series. Theorem 5.8 states that for characters $\chi$ and $\psi$ with conductors $L$ and $R$, respectively, that the Eisenstein series $E_{k,\chi,\psi}(z)$ is a modular form in $M_k(RL,\chi\psi)$ (I've elided a couple extra conditions for clarity).

Now, Theorem 5.10 says that $E_{k,\chi,\psi}(z) \in M_k(RL)$ are eigenforms for all Hecke operators. Is the intention of this that the form is only an eigenform when $\chi\psi=1$, or am I misunderstanding the notation? As far as I can tell, this is the only place in the entire book where this particular modular forms space notation $M_k(N)$ is used, so I don't know exactly what it refers to.


  • 2
    $\begingroup$ I think generally the notation is $M_k(N) := \bigoplus_{\chi \in (\mathbb{Z}/N \mathbb{Z})^* }M_k(N, \chi)$, so no you don't require $\chi \psi=1$. $\endgroup$ – ramanujan_dirac Jun 6 at 14:44
  • 1
    $\begingroup$ The Hecke operators are not the same for $M_k(\Gamma_0(N))$ and $M_k(\Gamma_0(N),\chi)$, to unify them you need to go in $M_k(\Gamma_1(N))$, each of them have their eigenforms $\endgroup$ – reuns Jun 6 at 16:20
  • $\begingroup$ Thanks @ramanujan_dirac and reuns. I suspected this was the case, but didn't want to rely on it without knowing for sure. $\endgroup$ – loofthri Jun 9 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.