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 $\Gamma_0(6)$

  • $\begingroup$ The definition of your $\theta$ misses a sum I think. $\endgroup$
    – user301452
    Commented Mar 18, 2016 at 16:58
  • $\begingroup$ math.stackexchange.com/questions/1288478/… $\endgroup$
    – cactus314
    Commented Mar 18, 2016 at 17:11
  • $\begingroup$ umm.. usually $\Gamma_0(4)$ is a subgroup of $\Gamma$ and $\begin{pmatrix}0 & 1 \\ -4 & 0 \end{pmatrix}$ has determinant $4$ and certainly can't belong in that group. Would you care to clarify your definitions ? $\endgroup$
    – mercio
    Commented Mar 18, 2016 at 17:57

1 Answer 1


$\Gamma_0(4)\backslash\mathbb{H}$ 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.


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