4

I want to find a basis of the space of modular forms of weight one $M_1(\Gamma_0(4), \chi)$, where $\chi$ is the character $\chi(d)=\Big(\frac{-1}{d} \Big)$, and

$$\Gamma_0(4)=\Big\lbrace \begin{pmatrix} a &b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z}) : \begin{pmatrix} a &b \\ c & d \end{pmatrix} \equiv \begin{pmatrix} * & * \\ 0 & * \end{pmatrix} \text{mod } N \Big\rbrace. $$

I would like to prove this as directly as possible. For example, I can not use the general formula for the dimension of $M_k(\Gamma)$ for a congruence subgroup $\Gamma \subseteq SL_2(\mathbb{Z})$*.

I know that $E_{1, \chi} = 1/4 \ + \ \sum_{n=1}^{\infty}\big(\sum_{d | n} \chi(d)\big)e(nz) \in M_1(\Gamma_0(4), \chi)$, where $e(nz)=\text{exp}(2\pi n z)$. I have also been able to show that $\text{dim} \ M_1(\Gamma_0(4), \chi) \leq 2$, but I do not know if the dimension is two or one. If it is two, I would like to find another element of the basis.

EDIT: Looking in the database LMFDB here I know that $M_1(\Gamma_0(4), \chi)$ has dimension 1, but I do not know how to prove this directly.

Thanks in advance.

Arriola
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  • How do you know that $,E_{1, \chi} \in M_1(\Gamma_0(4), \chi)$? – Somos Jun 06 '23 at 02:51
  • Have you tried to use Magma or SageMath yet? – Somos Jun 06 '23 at 03:04
  • @Somos Hi, i have tried using the database LMFDB but I don't know how to adjust the parameters correctly. I don't know how to use Magma or SageMath, but I will have a look :) – Arriola Jun 06 '23 at 06:41
  • @Somos According to LMFDB, it has dimension 1. Thus, no other element of the basis is needed. The problem is that I do not know how to show this. See: https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/?hst=SpaceDimensions&all_spaces=None&level=1-24&weight=1-12&dim=1-&char_label=4.b&char_order=2&search_type=SpaceDimensions – Arriola Jun 06 '23 at 15:37
  • The fact that you're working in weight $1$ makes this problem much harder. For higher weight, you can use the valence formula to compute the dimension, and then in the worst case fiddling around with enough Eisenstein series will get you there. But in weight $1$, there isn't even a known formula for the dimension in general. In the LMFDB, it was necessary to work rather hard to do this in general. – davidlowryduda Jun 06 '23 at 17:47
  • @davidlowryduda wow, I didn't know it was that hard, thanks for the info :). The ultimate goal of this was to find the formula for $r_2(n)=| \lbrace (x_1, x_2) \in \mathbb{Z}^2 : x_1^2+x_2^2=n \rbrace | $ by showing that $4E_{1, \chi}=\sum_{(x_1, x_2) \in \mathbb{Z}^2} e((x_1^2+x_2^2)z)$ – Arriola Jun 07 '23 at 09:31
  • Here is an alternate way to show your goal: explicitly compute the first several coefficients of $4 E_{1, \chi}$ and show that they agree. Then prove that there is a bound $X$ (often called the Sturm bound; see https://math.stackexchange.com/a/3944273/9754) such that if two forms agree on the first $X$ coefficients, then they are the same form. Here, it turns out that computing the first $2$ coefficients is provably enough. – davidlowryduda Jun 07 '23 at 14:49
  • @davidlowryduda Oh, thank you. I will have a look :) – Arriola Jun 07 '23 at 15:14

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