The theta group $\Gamma$ is the subgroup of $SL(2;\mathbb{Z})$ generated by $T=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $S=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$ It is a subgroup of index $3$.
If $\vartheta(z) = 1 + \sum_{n=1}^{\infty} 2 q^{n^2}$, $q=e^{\pi i z}$, then $\vartheta^4$ is a modular form on $\Gamma$ of weight $2$ with respect to the multiplier system $\chi(T) = 1$ and $\chi(S) = -1$. In other words $$\vartheta^4(z+2) = \vartheta^4(z), \; \; \vartheta^4(-1/z) = -z^2 \vartheta^4(z).$$
I am wondering about the dimension of the space of such functions $M_{2,\chi}(\Gamma).$ The bound I know is $$\mathrm{dim} \, M_{2,\chi}(\Gamma) \le \frac{2 \cdot 2}{12} (SL(2;\mathbb{Z}) : \Gamma) + 1 = 2.$$ However I think the dimension is probably only $1$. Can anyone help me prove or disprove it?