Theta function squared is a weight $1$ modular form

Question

Let $$\vartheta(\tau) = \sum_{n\in\mathbb{Z}}e^{\pi in^2\tau}.$$ I know that $\vartheta$ satisfies the transfromation properties $$\vartheta(\tau + 2) = \vartheta(\tau), \quad \vartheta\left(-\frac{1}{\tau}\right) = \sqrt{\frac{\tau}{i}}\vartheta(\tau).$$ What I am interested in is the transformation properties of $$f(\tau) = \vartheta(2\tau)^2.$$ I have seen it mentioned in some notes by Kevin Buzzard and elsewhere that $f(\tau)$ is a weight $1$ modular form on the congruence subgroup $\Gamma_0(4)$ with Nebentypus $\chi$, the Dirichlet character coming from the Legendre symbol mod $4$. In other words, $$f\left(\frac{a\tau + b}{c\tau + d}\right) = \chi(d)(c\tau + d)f(\tau), \quad \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(4).$$ I cannot find a reference for this fact nor does it seem obvious from the transformation properties of $\vartheta(\tau)$. I know that given a modular form $g(\tau)$ of weight $k$ on $\Gamma_0(N)$ for some $N$, the function $g(2\tau)^2$ is modular of weight $2k$ on $\Gamma_0(2N)$. But the problem is that the subgroup on which $\vartheta(\tau)$ is modular (of weight $1/2$) is not one of the congruence subgroups $\Gamma_0(N)$. I would greatly appreciate any help tracking down a reference for this fact.

Answer 1

$g = \vartheta^2(2\tau)$ is invariant under $\Gamma_1(4)$ is easily checked: since $\Gamma_1(4)$ is generated by $$\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}\qquad \begin{pmatrix}1 & 0 \\ 4 & 1 \end{pmatrix}$$ Obviously $g$ is weight-$1$ invariant under the first one. For second one, $$\begin{aligned}g\left[\begin{pmatrix}1 & 0 \\ 4 & 1 \end{pmatrix}\right]_1 (\tau)&= (4\tau +1)^{-1}\vartheta^2(\frac{2\tau}{4\tau+1}) \\&= (4\tau +1)^{-1} \frac{i(4\tau+1)}{2\tau}\vartheta^2(-\frac{4\tau+1}{2\tau}) \\& = \frac{i}{2\tau}\vartheta^2(\frac{-1}{2\tau}) = \vartheta^2(2\tau) \end{aligned}$$

To check $g$ has nebentypus $\chi$, it suffices to show $$g\left[\begin{pmatrix}-1 & 0 \\ 4 & -1 \end{pmatrix}\right]_1(\tau)=-g(\tau)$$ I shall leave this checking to you.