2
$\begingroup$

Using Hida theory, we can prove that there is a cusp form of weight 2 and level $\Gamma_0(11)$. Are there ways to prove that there is no cusp forms of weight 2 and level $\Gamma_0(p)$ where $p < 11$?

Edit. As suggested by Jakob (see comments), calculating the dimension of $M_2 (\Gamma_0(p))$ is one way to prove the fact. But I am interested in proving it using congruences like Hida theory.

Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ The space $M_2(\Gamma_0(p))$ just doesn't have very large dimension. See e.g. this and this (and I'm sure many other places, these are just what appeared on quick Googling). $\endgroup$
    – Jakob Streipel
    Commented Dec 12, 2023 at 16:09
  • 3
    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Commented Dec 12, 2023 at 16:21
  • 1
    $\begingroup$ Also, how does the Hida theory proof work for level 11? $\endgroup$
    – Kimball
    Commented Dec 12, 2023 at 19:28
  • 1
    $\begingroup$ The issue is that proof uses the existence of a cusp form to construct a family. It's translating one existence result into another. For non-existence you'd need something like, if there were a weight 2 cusp form of level 5, say, you'd get some other cusp form. And then you need to prove that doesn't exist (probably using something like dimension formulas). $\endgroup$
    – Kimball
    Commented Dec 14, 2023 at 0:43
  • 1
    $\begingroup$ @Offlaw The argument you give obviously works backwards as well, doesn't it? It establishes a bijection between cusp forms of weight 2 and level p, and p-ordinary cusp forms of weight (p+1) and level 1. If p < 11 there are none of the latter. $\endgroup$
    – David Loeffler
    Commented Dec 15, 2023 at 11:09

1 Answer 1

Reset to default
2
$\begingroup$

The dimension of $S_2(\Gamma_0(N))$ is the genus of the Riemann surface $\Gamma_0(N)\backslash\mathfrak{H}^*$. See Theorem 2.24 in Shimura: Introduction to the arithmetic theory of automorphic forms. The genus can be calculated via the Hurwitz formula applied to the branched cover $\Gamma_0(N)\backslash\mathfrak{H}^*\to\Gamma_0(1)\backslash\mathfrak{H}^*$. See Propositions 1.40 and 1.43 in the same book for explicit formulae.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Sorry! I forgot to mention that I'm aware of the result via dimension calculation. My motivation of this question to reach some kind of contradiction using congruences of modular forms. $\endgroup$
    – Offlaw
    Commented Dec 12, 2023 at 16:29
  • 1
    $\begingroup$ @Offlaw I see. It is a different question then. I leave my answer as non-experts might find it useful. $\endgroup$
    – GH from MO
    Commented Dec 12, 2023 at 16:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .