I am asked to show $$(2\pi)^{-12}\Delta(\tau) = q \cdot \Big (\sum_{n\in \mathbb{Z}} (-1)^n \cdot q^{(3n^2+n)/2} \Big)^{24}$$ where $\Delta:\mathbb{H} \to \mathbb{C}$ is the modular discriminant, $q=e^{2\pi i \tau}$.
My approach: Call the RHS $\vartheta(\tau)$. I've shown $\vartheta$ converges normally on $\mathbb{H}$. Now, $\Delta$ is a (holomorphic) cusp form of weight $12$ on $SL_2(\mathbb{Z})$. Since the vector space of such cusp forms is one-dimensional, if we show $\vartheta$ is a cusp form then we know $\delta = \alpha \vartheta$ for some constant $\alpha$. Then examining the Fourier expansion one can show $\alpha = (2\pi)^{12}$.
To show $\vartheta \in S_k$ we must check for any $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$ that $\vartheta(\gamma \tau) = (c\tau+d)^{12}\vartheta(\tau)$, where $\gamma$ is the Mobius transform induced by the matrix. For this its enough to verify on the generators, which give transformations $\tau \to \tau + 1$, $\tau \to -1/\tau$. The first transformation is easy, but I can't figure out how to prove $\vartheta(-1/\tau) = \tau^{12} \vartheta(\tau)$.
I did a similar problem following the exact same procedure. It was to show $(2\pi)^{-12}\Delta(\tau) = 2^{-8}(\vartheta_1\vartheta_2\vartheta_3(\tau))^8$ for similar looking functions $\vartheta_j$. I was able to recognize these functions as theta functions and manipulate the theta transformation formula. In this problem, I'm failing to get theta transformation to work. Does anyone have any ideas? It should also be noted that we have been discussing vector valued theta functions recently on quadratic lattices, so maybe this is a necessary tool. Again, I can't find a way to apply these ideas though.
Another approach to the problem is using my previous exercise: note $\vartheta_1 = \sum_{n \in \mathbb{Z}}q^{n^2/2}$, $\vartheta_2 = \sum_{n \in \mathbb{Z}}(-1)^n \cdot q^{n^2/2}$, $\vartheta_3 = \sum_{n \in \mathbb{Z}}q^{(n+1/2)^2/2}$. Maybe one can quickly do the problem by the Cauchy product formula, but I'm bad at manipulating these messy sums. Thanks a lot.