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I have been reading some notes and books on modular forms and frequently meet the sententence "Let $f$ be a weight 2, level $N$ Hecke eigenform". But some of them have not make it clear whether $f \in S_2(\Gamma_0(N))$ or $f \in S_2(\Gamma_1(N))$. It makes me feel quite confused.

So my question is: is there any usual convention on that?

My attempts:

  1. Since we often define the Hecke operators $\langle d \rangle$ and $T_n$ (where $n$ is an integer and $d \in (\mathbb{Z}/N\mathbb{Z})^{\times}$) as linear operators on $S_2(\Gamma_1(N))$ (as in the book by Fred Diamond and Jerry Shurman) and Hecke eigenforms are defined as the "eigenvectors" of all operators in the sub-$\mathbb{C}$-algebra in $\mathrm{End}_{\mathbb{C}}(S_2(\Gamma_1(N)))$ generated by operators $\{\langle d \rangle, T_p \}_{d \in (\mathbb{Z}/N\mathbb{Z})^{\times}, p \, \text{prime}}$, the Hecke eigenforms are indeed cuspidal forms in $S_2(\Gamma_1(N))$.

  2. However, we see that in particular, a Hecke eigenform lies in all eigenspaces of diamond operators $$ S_2(N, \chi) := M_2(N, \chi) \cap S_2(\Gamma_1(N)) := \{ f \in M_2(\Gamma_1(N)) : \langle d \rangle f = \chi(d) f, \forall d \in (\mathbb{Z}/N\mathbb{Z})^{\times} \} \cap S_2(\Gamma_1(N)) , $$ where $\chi: (\mathbb{Z}/N\mathbb{Z})^{\times} \rightarrow \mathbb{C}$ is a Dirichlet character. In Exercise 4.4.3(a) of Fred Diamond and Jerry Shurman, we have shown that $$ M_2(N, \mathbf{1}_N) = M_2(\Gamma_0(N)). $$ Hence as a Hecke eigenform, $f$ ought to be an element in $S_2(N, \mathbf{1}_N) := M_2(N, \mathbf{1}_N) \cap S_2(\Gamma_1(N)) = S_2(\Gamma_0(N))$. Hence actually as a Hecke eigenform, $f$ also has level $\Gamma_0(N)$. So we do not need to distinguish the level "$\Gamma_0$" or "$\Gamma_1$".

So are my attempts correct?

All my knowledges on modular forms are self-teached, so I'm sorry if the question is too trivial or I had made some silly mistakes.

Thank you all for answering and commenting!

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  • $\begingroup$ It's also quite possible they mean $\Gamma(N)$. Or any other $\Gamma \supset\Gamma(N)$. I don't think there is a strong convention either way (although $\Gamma_1(N)$ is most likely). It's pretty common to hear someone define a modular form of level $N$ only for someone in the audience to ask them if they mean with character ($\Gamma_1$) or not ($\Gamma_0$). $\endgroup$
    – Mathmo123
    Commented Apr 23, 2021 at 14:26
  • $\begingroup$ Also, you really don't need to apologise about asking questions! For the record, none of your questions so far on this topic have been silly or trivial. $\endgroup$
    – Mathmo123
    Commented Apr 23, 2021 at 14:30
  • $\begingroup$ @Mathmo123 Thank you for your comments and especially for your encouragement! :) $\endgroup$
    – Hetong Xu
    Commented Apr 23, 2021 at 15:15

1 Answer 1

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Any $S_k(\Gamma_1(N))$ eigenform is in $S_k(\Gamma_0(N),\chi)$ for some $\chi\bmod N$.

This is because the projection map $\sum_{d\bmod N} \overline{\chi(d) }\langle d\rangle: S_k(\Gamma_1(N))\to S_k(\Gamma_0(N),\chi)$ commutes with the Hecke operators.

Whence some authors mean $f\in S_k(\Gamma_0(N))$, some others mean $f\in S_k(\Gamma_0(N),\chi)$, which doesn't make a big difference in most cases.

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  • $\begingroup$ Thank you for your answer! So it seems that the space with the nebentypus $\chi$, i.e. $S_k(\Gamma_0(N), \chi) := \{ f \in S_k(\Gamma_0(N)) : \langle d \rangle f = \chi(d) f, \forall d \in (\mathbb{Z}/N\mathbb{Z})^{\times} \}$ is a subspace of $S_k(\Gamma_0(N))$ and hence I can regard the character $\chi$ as some additional information for the eigenform $f$? Is my understanding correct? Besides, I'm actually reading things on modularity lifting, and hoping that the differences on $f \in S_k(\Gamma_0(N))$ and $f \in S_k(\Gamma_0(N), \chi)$ will not make a big difference on their Galois reps. $\endgroup$
    – Hetong Xu
    Commented Apr 23, 2021 at 12:51
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    $\begingroup$ If $\chi$ is not real then $f \in S_k(\Gamma_0(N), \chi)$ doesn't have rational coefficients. I'm a bit unsure for the quadratic character. $\endgroup$
    – reuns
    Commented Apr 23, 2021 at 12:53
  • $\begingroup$ Thank you! Yet sorry for that I'm not quite familiar with the meaning of quadratic twist of the underlying elliptic curve. And I'm a bit confused that if $S_k(\Gamma_0(N), \chi)$ is indeed a subspace of $S_k(\Gamma_0(N))$, where are the real differences between the two conventions (as you mentioned in the post). It seems that by writing $f \in S_k(\Gamma_0(N), \chi)$, we only want to adress the further information on $f$, namely the nebentypus $\chi$ of $f$? $\endgroup$
    – Hetong Xu
    Commented Apr 23, 2021 at 13:03
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    $\begingroup$ $$S_k(\Gamma_1(N))=\bigoplus_\chi S_k(\Gamma_0(N), \chi)$$ this follows from the projection map $\frac1{\varphi(N)}\sum_{d\bmod N} \overline{\chi(d) }\langle d\rangle$. For $f\in S_k(\Gamma_0(N),\phi)$ we have $\sum_n a_n(f)\chi(n)e^{2i\pi n z} \in S_k(\Gamma_0(N),\phi \chi^2)$ (or $\phi \overline{\chi}^2$ I should redo the calculation) so the $S_k(\Gamma_0(N),\phi \chi^2)$ are tightly related for different $\chi$, though the Galois representation will be a bit different. $\endgroup$
    – reuns
    Commented Apr 23, 2021 at 13:10
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    $\begingroup$ My quadratic twist comment was a mistake, if $\sum_{n\ge 1} a_n(f) e^{2i\pi nz}\in S_2(\Gamma_0(N))$ is an eigenform with rational coefficients then its L-function is that of an elliptic curve $y^2=x^3+ax+b$ and $\sum_{n\ge 1} (\frac{n}{N}) a_n(f) e^{2i\pi nz}\in S_2(\Gamma_0(N))$ is that of $\pm N y^2=x^3+ax+b$. The map $y^2=x^3+ax+b\to N y^2=x^3+ax+b$ has an easy interpretation in term of the Galois representation of the elliptic curve and its L-function. $\endgroup$
    – reuns
    Commented Apr 23, 2021 at 13:13

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