In number theory we learn that $\theta(z) = \sum q^{n^2}$ is a modular form with respect to $\Gamma = \Gamma_0(4)$. This boils down to two properties:
- $\theta(z)= \theta(z+1)$ this shift symmetry is pretty clear since perfect square are integers
- $\theta(- \frac{1}{4z}) = \sqrt{\frac{2z}{i}}\,\theta(z)$ this follows from Poisson summation, making it no less mysterious.
$$ \sum_{n \in \mathbb{Z}} e^{-\pi n^2 t}= \frac{1}{\sqrt{t}}\sum_{n \in \mathbb{Z}} e^{-\pi n^2 / t}$$
I am often drawn to speculate that $\pi\,n^2$ is the area of a circle. In any case the two transformations generate a group:
- $\langle z \mapsto z+1 ,z \mapsto -\frac{1}{4z}\rangle = \Gamma_0(4)$ it can also be written as a matrix group:
$$ \left\langle \left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array} \right), \left(\begin{array}{rc} 0 & 1 \\ -4 & 0\end{array} \right) \right\rangle$$
I also read that $\theta(z)$ is a "cusp" form meaning it vanishes at the cusps of $\mathbb{H}/\Gamma_0(4)$ but I have trouble since I don't know what this region looks like or why $\theta(z) = 0$ at these "corners".
Show the Dedekind $\eta(\tau)$ function is a modular form of weight $\frac{1}{2}$ for $\Gamma_0(6)$
$\Gamma_0(4)\backslash\mathbb{H}$ is a Riemann surface of genus zero with three cusps. The cusps can essentially be thought of as punctures, hence $\Gamma_0(4)\backslash\mathbb{H}$ looks like a sphere with three punctures.