Find the $L^2$ norm of $\theta_\chi(z)^2$ in $\mathbb{H}/\Gamma_0(4)$

Question

I'd like to learn about $L^2$ norms on Hyperbolic space. My Automorphic forms textbook says this function is a "cusp form" so it's in $L^2(\mathbb{H}/\Gamma_0(4))$:

$$ \theta_\chi(z) = \sum_{n \in \mathbb{Z}} \chi(n) \, e^{-\pi n^2 \, z} $$

This is a cusp form, since $\chi(0) = 0$ for any Dirichlet character $\chi$. Can we do the integral over the fundamental region?

$$ L^2(\theta_\chi) = \int_{\mathbb{H}/\Gamma_0(4)} |\theta_\chi(z)|^2 \, \frac{dx^2 + dy^2 }{y^2} $$

Have I written the norm correctly? It's very likely that in some cases, my question is not well-posed and that can be discussed in the comments or I'll up-vote an answer.

---

It seems the original version of my question is problematic. At least $\theta_\chi(z)^2$ is a weight 1 (?) cusp form over $\Gamma_0(4)$ and the question $L^2$ makes sense.

$$ \theta_\chi(z)^2 = \sum_{(m,n) \in \mathbb{Z}^2} \chi(n) \, e^{-2\pi i (m^2+n^2 )\, z} $$

I may have to name a specific Dirichlet character. Perhaps the mod 4 character with $\chi(3)=-1$? And then we are integrating the absolute value squared over hyperbolic space:

$$| \theta_\chi(z)|^4 = \sum_{(m,n) \in \mathbb{Z}^4} \chi(n) \, e^{-2\pi i (a^2+b^2) -(c^2+d^2) )\, z} $$

We obtain a theta series associated to a quadratic form $q(z)=(a^2+b^2) -(c^2+d^2) $. The area form should be corrected as well

$$ L^2(\theta_\chi^2) = \int_{\mathbb{H}/\Gamma_0(4)} |\theta_\chi(z)|^4 \, \frac{dx \, dy}{y^2} $$

Is this question well stated?

Answer 1

A standard way to calculate norms of these types of objects is to unfold against an Eisenstein series. Let \[E(z,s) = \sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma_0(4q^2)} \Im(\gamma z)^s\] denote the usual Eisenstein series associated to the cusp at infinity of $\Gamma_0(4q^2) \backslash \mathbb{H}$. This has a pole at $s = 1$ with residue $1/\mathrm{vol}(\Gamma_0(4q^2) \backslash \mathbb{H})$ independently of $z$.

Suppose that $\chi$ is a primitive Dirichlet character modulo $q$ satisfying $\chi(-1) = (-1)^{\kappa}$. Define \[\theta_{\chi}(z) = \sum_{n = -\infty}^{\infty} n^{\kappa} \chi(n) e(n^2 z),\] where $e(z) = e^{2\pi i z}$. This satisfies \[\theta_{\chi}(\gamma z) = \chi(d) \overline{\varepsilon}_d \left(\frac{c}{d}\right) (cz + d)^{\kappa + 1/2} \theta_{\chi}(z)\] for all $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(4q^2)$. That is, this is a half-integral weight modular form. If $q > 1$, this vanishes at the cusp at infinity. In any case, we note that for $\Re(s) > 1$, the integrand of \[\int_{\Gamma_0(4q^2) \backslash \mathbb{H}} |\theta_{\chi}(z)|^2 E(z,s) \Im(z)^{\kappa + \frac{1}{2}} \, \frac{dx \, dy}{y^2}\] is equal to \[\sum_{\gamma \in \Gamma_{\infty} \backslash \Gamma_0(4q^2)} |\theta_{\chi}(\gamma z)|^2 \Im(\gamma z)^{s + \kappa + \frac{1}{2}}\] via the automorphy of $\theta_{\chi}(z)$. By unfolding the integral, we obtain \[\int_{\Gamma_{\infty} \backslash \mathbb{H}} |\theta_{\chi}(z)|^2 \Im(z)^{s + \kappa + \frac{1}{2}} \, \frac{dx \, dy}{y^2} = \int_{0}^{\infty} \int_{0}^{1} |\theta_{\chi}(z)|^2 y^{s + \kappa - \frac{1}{2}} \, dx \, \frac{dy}{y}.\] We replace $\theta_{\chi}(z)$ with its Fourier expansion and interchange the order of integration and summation, yielding \[\sum_{n_1 = -\infty}^{\infty} \sum_{n_2 = -\infty}^{\infty} (n_1 n_2)^{\kappa} \chi(n_1) \overline{\chi}(n_2) \int_{0}^{\infty} e^{-2\pi(n_1^2 + n_2^2) y} y^{s + \kappa - \frac{1}{2}} \, \frac{dy}{y} \int_{0}^{1} e((n_1^2 - n_2^2) x) \, dx.\] Only the diagonal terms $n_1 = \pm n_2$ survive the integration over $x$, yielding \[4 \sum_{\substack{n = 1 \\ (n,q) = 1}}^{\infty} n^{2\kappa} \int_{0}^{\infty} e^{-4\pi n^2 y} y^{s + \kappa - \frac{1}{2}} \, \frac{dy}{y}.\] We make the change of variables $y \mapsto y/(4\pi n^2)$, so that the integral becomes \[(4\pi n^2)^{-s - \kappa + \frac{1}{2}} \int_{0}^{\infty} e^{-y} y^{s + \kappa - \frac{1}{2}} \, \frac{dy}{y}.\] This integral is simply $\Gamma(s + \kappa - 1/2)$. So the original integral is equal to \[4 (4\pi)^{-s - \kappa + \frac{1}{2}} \Gamma\left(s + \kappa - \frac{1}{2}\right) \sum_{\substack{n = 1 \\ (n,q) = 1}}^{\infty} \frac{1}{n^{2s - 1}}.\] This Dirichlet series is equal to $\zeta(2s - 1) \prod_{p \mid q} (1 - p^{-(2s - 1)})$.

Now we take the residue at $s = 1$. The left-hand side is equal to \[\frac{1}{\mathrm{vol}\left(\Gamma_0(4q^2) \backslash \mathbb{H}\right)} \int_{\Gamma_0(4q^2) \backslash \mathbb{H}} |\theta_{\chi}(z)|^2 \Im(z)^{\kappa + \frac{1}{2}} \, \frac{dx \, dy}{y^2}.\] The right-hand side is equal to \[2 (4\pi)^{-\frac{1}{2} - \kappa} \Gamma\left(\frac{1}{2} + \kappa\right) \prod_{p \mid q} \left(1 - \frac{1}{p}\right).\]

Answer 2

The standard, but non-obvious, way to compute an $L^2$ norm of a modular form $f$ is as the residue at $s=1$ of the Rankin-Selberg $L$-function (or Dirichlet series) obtained as an integral $\int_{\Gamma\\\mathfrak H} E_s\cdot |f|^2$. This is because the residue at $s=1$ of that Eisenstein series is a constant (basically the multiplicative inverse of the natural volume of $\Gamma\backslash\mathfrak H$). For half-integral weight theta series, after the corrections indicated by @Peter Humphries are made, you can identify this Rankin-Selberg thing as an already-familiar Dirichlet series, whose reside at $1$ is known in other terms.

EDIT: "The" weight-zero Eisenstein series attached to the cusp $i\infty$ for $\Gamma_0(N)$ is $\sum_{c,d} y^2/|cz+d|^{2s}$ where $c,d$ are summed over relatively prime integers congruent to $0,1$ mod $N$, and multiplicatively modulo $\pm 1$ for $N=2$.