3
$\begingroup$

Let $M_2(\Gamma_0(N))$ be the space of weight $2$ modular forms for $\Gamma_0(N)$ and let $f_1, \dots, f_r$ be a basis of normalized eigenforms for $M_2(\Gamma_0(N))$. Given a rational prime $p$, I'll say that the eigenforms $f_1, \dots, f_r$ are linearly independent mod $p$ if the mod $p$ reductions of the Fourier expansions of $f_1, \dots, f_r$ are $\mathbf{F}_{p^k}$-linearly independent in $\mathbf{F}_{p^k} [[ q ]]$.

My question is: what are sufficient conditions on $N$ and $p$ that ensure that $f_1, \dots, f_r$ are linearly independent mod $p$? For example, if $N$ is prime and $p$ divides the numerator of $(N-1)/12$, then $f_1, \dots, f_r$ are not linearly independent mod $p$, because there is an Eisenstein series which is congruent to a cuspform in $M_2(\Gamma_0(N))$. Are there sufficient conditions on $N$ and $p$ that ensure linear independence of the forms $f_1, \dots, f_r$?

Improve this question
$\endgroup$
7
  • $\begingroup$ Interesting - I didn't know about the "modular degree" condition you mentioned. In my case, I'm happy to assume that $N$ is prime. If $N$ is prime, then which primes $p$ do we have to exclude? Unfortunately, I don't think I can assume that one of the forms $f$ comes from an elliptic curve. $\endgroup$
    – Adithya Chakravarthy
    Commented Apr 16, 2023 at 13:31
  • $\begingroup$ I'm looking for a sufficient condition like: "if $N$ is prime and $p$ does not divide the numerator of $(N-1)/12$, then the eigenforms of $M_2(\Gamma_0(N))$ are linearly independent mod $p$". Mazur showed that if $p$ divides the numerator of $(N-1)/12$, then this false, so I'm wondering if once we exclude this case, we have linear independence mod $p$. $\endgroup$
    – Adithya Chakravarthy
    Commented Apr 16, 2023 at 14:51
  • $\begingroup$ I'm not entirely sure if I understand the question. The normalized eigenforms need not have coefficients in ring of algebraic integers that contains a prime ideal with norm $p$. Do you mean to allow the reductions to have Fourier expansions in $\mathbb{F}_{p^{k}}$? (For an explicit example, what about $N = 23$ and $p = 3$?) $\endgroup$
    – Jeremy Rouse
    Commented Apr 16, 2023 at 17:04
  • 1
    $\begingroup$ @user491858 I think you are not actually answering the question that is being asked; if there are three eigenforms $f, g, h$ in the space, and $f$ corresponds to an elliptic curve $E$ but $g, h$ do not, then the degree of the modular parametrisation of $E$ doesn't tell you anything about whether $g$ and $h$ satisfy a congruence. $\endgroup$
    – David Loeffler
    Commented Apr 17, 2023 at 18:23
  • 1
    $\begingroup$ @AdithyaChakravarthy The $\mathbb{Z}$-subalgebra of $End(M_2(\Gamma_0(N)))$ generated by the Hecke operators coprime to $N$ is a semisimple $\mathbb{Z}$-algebra, so its discriminant is a non-zero integer, well-defined up to sign, and the primes where congruences occur are precisely the primes which divide this integer. See Calegari and Stein, arxiv.org/pdf/math/0406243.pdf, for the prime-level case. But giving explicit bounds for the discriminant in terms of $N$ is probably very hard and messy. $\endgroup$
    – David Loeffler
    Commented Apr 17, 2023 at 18:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .