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I have a question reffering to a theorem by Weil, which gives sufficient conditions that a given L-series $$ L(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}$$ which is convergent somewhere comes from a modular form $f \in M_k(\Gamma_0(N), \chi)$ of weight $k$. This theorem is for example stated in the book Modular forms by T. Miyake on page 128 and the proof uses twists of the given $L$-function. Note that this is a generalization of Hecke's classical converse theorem (that is basically the case $N = 1$) which is much easier to prove since the full modular group is generated by the elements $S = \pmatrix{0 & -1 \\ 1 & 0}$ and $T = \pmatrix{1 & 1 \\ 0 & 1}$.

I was wondering if there are similar results for the case $M_k(\Gamma(N), \chi)$ with useful relaxations to the conditions of Weil.

Thank you!

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    $\begingroup$ $\Gamma(N)$ is conjugate to $\Gamma_1(N^2)$, and every form for $\Gamma_1(N^2)$ is a sum of forms for $\Gamma_0(N^2)$ with various nebentypus characters mod $N^2$, so it should be straightforward to obtain a variant of Weil's converse theorem for this situation. $\endgroup$
    – GH from MO
    Jun 4, 2018 at 12:07
  • $\begingroup$ Good point, thank you. Using this we obtain a map $\varphi : M_k(\Gamma(N)) \hookrightarrow \bigoplus_{\chi} M_k(\Gamma_0(N^2), \chi)$. Assume that $f(z) = \sum_{m=0}^\infty a(m)q^{m/N}$ and we would like to check whether $f \in M_k(\Gamma(N))$. But in order to apply a modified version of Weil's theorem we need a description of $\mathrm{im}(\varphi)$. Do I miss something? $\endgroup$
    – Johann Franke
    Jun 6, 2018 at 7:41
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    $\begingroup$ $\operatorname{im}(\varphi)$ is precisely the direct summands for which $\chi$ has conductor dividing $N$. $\endgroup$
    – David Loeffler
    Jun 6, 2018 at 12:31
  • $\begingroup$ All right, so we can write each modular form $f \in M_k(\Gamma(N))$ as a sum $f = \sum_{\chi} \alpha_\chi f_\chi$, where the $\chi$ have conductor dividing $N$ and the $f_\chi$ are elements of $M_k(\Gamma_0(N^2), \chi)$. The naive approach is to check all the $f_\chi$ being in $M_k(\Gamma_0(N^2), \chi)$ using Weil's result, but the decomposition on its own seems to be very espensive (even in theory), so this method is probably not the best way. $\endgroup$
    – Johann Franke
    Jun 6, 2018 at 17:06
  • $\begingroup$ All of this appears to be quite unnatural since actually it should be easier to follow that some function is a modular form for $\Gamma(N)$ than for $\Gamma_0(N)$, shouldn't it? $\endgroup$
    – Johann Franke
    Jun 6, 2018 at 17:10

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