# Writing an integer as a sum of two squares: Why is this a modular form of weight 1?

I am trying to understand the following proof of the fact that the number of ways to write a positive integer $n$ as a sum of two squares (call it $s_n$) is given by $$s_n = 4\sum_{d|n} \chi_{-4}(d),$$ where $\chi_{-4}$ is the Dirichlet character defined by $$\chi_{-4}(m) = \begin{cases} 1, & m\equiv 1 \mod 4 \\ -1, &m \equiv 3 \mod 4\\ 0, &m \equiv 0,2 \mod 4.\end{cases}$$

The key observation is the following. Let

$$\theta(z) := \sum_{n \in \Bbb{Z}} q^{n^2} = 1+2q + 2q^2 +\ldots$$ be the Jacobi theta series where as usual $q = e^{2\pi i z}$. Then we have:

1. $s_n$ is the $n$-th coefficient of $\theta(z)^2$

2. $\theta(z)^2$ is a modular form of weight $1$ for the congruence subgroup $\Gamma_0(4) \subseteq \text{SL}_2(\Bbb{Z})$

3. The space $M_1(\Gamma_0(4))$ is one dimensional.

Of course, we are not done yet. We need to take advantage of 3 as follows. Defined the Eisenstein series $G_{1,\chi_{-4}}(z)$ (with respect to the Dirichlet character above) as

$$G_{1,\chi_{-4}}(z) = \frac{1}{4} + \sum_{n \geq 1}\left( \sum_{d |n} \chi_{-4}(d)\right) q^n .$$

My question is: I know that $G_{1, \chi_{-4}}$ satisfies the relation $$G_{1, \chi_{-4}}(\gamma\cdot z) = \chi_{-4}(a) (cz+d)^{-1} G_{1, \chi_{-4}}(z)$$ for a matrix $$\gamma = \left(\begin{array}{cc} a & b \\ c& d\end{array}\right) \in \Gamma_0(4).$$ I want to say from here that $G_{1,\chi_{-4}}(z)$ is actually a modular form of weight $1$ w.r.t. $\Gamma_0(4)$. The problem here is that the quantity $\chi_{-4}(a)$ is not always equal to $1$! How do we get around this?

• do you know the space $M_k(\Gamma_0(N),\chi)$ which is the set of holomorphic functions such that $f|_k\gamma(z) = \chi(d) f(z)$ for every $\gamma \in \Gamma$, with $f|_k\gamma(z) = (cz+d)^{-k} f(\gamma . z)$ Nov 3, 2016 at 7:32

By the appearance of $\chi_{-4}$ in the modular transformations law, the modular form $G_{1,\chi_{-4}}$ is not a modular form of weight 1 with respect to $\Gamma_{0}(4)$. But if we restrict the group to $$\Gamma_{1}(4)=\left\{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in SL_{2}(\mathbb{Z}):a\equiv d\equiv1\pmod{4},b\equiv c\equiv 0\pmod{4}\right\},$$ then we don't need care about $\chi_{-4}$ because $\chi_{-4}(a)=1$ for all $a\equiv 1\pmod{4}$. In other words, $G_{1,\chi_{-4}}$ is a modular form of weight 1 with respect to $\Gamma_{1}(4)$.
On the other hand, since $\Gamma_{1}(4)\subset\Gamma_{0}(4)$, $M_{1}(\Gamma_{0}(4))\subset M_{1}(\Gamma_{1}(4))$, where $M_{k}(\Gamma)$ denotes the space of modular forms of weight $k$ with respect to $\Gamma$. Thus, $\theta^{2}$ is can be considered as a modular form of weight 1 with respect to $\Gamma_{1}(4)$.
It is known that $\dim_{\mathbb{C}}(M_{1}(\Gamma_{1}(4)))=1$. So by working with $\Gamma_{1}(4)$ instead of $\Gamma_{0}(4)$ we can use the strategy described in the question to prove the desired formula.