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I was trying to verify modularity of the theta function $$\theta(z) = \sum_{t \in \mathbb{Z}}e^{2\pi it^{2}z}$$ for $\Gamma_{0}(4)$ directly. I know the factor of automorphy should be $$j(\gamma,z) = \left(\frac{c}{d}\right)\epsilon_{d}^{-1}(cz+d)^{\frac{1}{2}}$$ where $\left(\frac{c}{d}\right)$ is the extended Jacobi symbol, $\epsilon_{d}^{-1} = 1,i$ depending on if $d \equiv 1,-1 \pmod{4}$, and we take the principle branch of the square root. I've had two ideas, both ultimately unsucessful:

Attempt 1: It's known that $\Gamma_{0}(4)$ is generated by the matrices $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}$$ so if I can verify modularity for these three matrices I'm done. The first two cases are $$\theta(z+1) = \theta(z) \qquad \text{and} \qquad \theta(z) = \theta(z)$$ which are both obvious. The last case is $$\theta\left(\frac{z}{4z+1}\right) = (4z+1)^{1/2}\theta(z).$$ By the identity theorem it suffices to prove this on the line $z = iy$. Now I tried to use Poisson summation in the same way for the usual functional equation of the theta function (completing the square), but this doesn't seem to work because my change of variables in the fourier transform $$\int_{-\infty}^{\infty}e^{-2\pi t^{2}y}e^{-2\pi i nt}dt$$ needs to be something of the form $(4iy+1)^{1/2}t+\alpha \to t$ and this is not the same transformation that completes the square. As a side note, I'm also slightly worried I'm missing something with this attempt because the symbol is trivial in all three cases I have to check so it's as if I'm not using that data at all.

Method 2 We can decompose any Mobius transformation coming from a matrix in $SL_{2}(\mathbb{Z})$ as $$\frac{az+b}{cz+d} = \frac{a}{c}+\frac{\frac{-1}{c^{2}}}{z+\frac{d}{c}}$$ So if I had nice transformation laws for shifts, inversion, and scaling, I could piece everything together. From Poisson summation I have a nice transformation law for inversion which is well-known. There's also inariance for shifting, but only by integers and $\frac{a}{c}$ and $\frac{d}{c}$ certinaly don't have to be integers. I also don't think there is a nice transformation law for the theta function and scaling.

Maybe I need to work these ideas more, but is the automorphy of $\theta(z)$ coming from something deeper than Poisson summation?

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This is one of those cases where it's actually easier to verify a slightly stronger statement. Consider the subgroup $\Gamma$ of $\operatorname{SL}_2(\mathbb{R})$ generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & -\tfrac{1}{2} \\ 2 & 0 \end{pmatrix}$. This is not contained in $\operatorname{SL}_2(\mathbb{Z})$, evidently, but you can check that $\Gamma \cap \operatorname{SL}_2(\mathbb{Z}) = \Gamma_0(4)$ and $[\Gamma : \Gamma_0(4)]$ is finite. Then you can verify that $\theta$ is actually modular of weight $\tfrac{1}{2}$ for $\Gamma$, which amounts to verifying that it has the "correct" transformation law under $z \mapsto z + 1$ (easy) and $z \mapsto \tfrac{-1}{4z}$ (application of Poisson summation).

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    $\begingroup$ Ahhh I see your idea. One last thing, don't we need to expand our defintion of the symbols $\left(\frac{c}{d}\right)$ and $\epsilon_{d}^{-1}$ in this case because $d = 0$ for the second matrix? What should the extensions be? My guess is $\left(\frac{c}{0}\right) = 1$ and $\epsilon_{0}^{-1} = 1/\sqrt{i}$ (this should give the right transformation for $\theta^{2}$) but maybe this is wrong. $\endgroup$
    – modperspec
    Apr 12 at 2:14
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    $\begingroup$ More accurately, we need to extend the factor of automorphy to a cocycle on $\Gamma$. The extension won't necessarily have to be given by Jacobi symbols. $\endgroup$ Apr 12 at 5:15
  • $\begingroup$ Ah I see. Thank you! $\endgroup$
    – modperspec
    Apr 13 at 0:41

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