2025/06/22
Let $H(n)$ be the Hurwitz class number and $\Gamma(a,z)$ be the incomplete Gamma function. Zagier's famous harmonic Maass form is given by \[\mathcal{H}(z)=-\frac 1{12}+\sum_{n=1}^\infty H(n)q^n +\frac 1{4\sqrt \pi}\sum_{m=1}^\infty m\Gamma(-\tfrac 12,4\pi m^2 y)q^{-m^2}+\frac 1{8\pi \sqrt y}. \]
Let $\theta(z)=\sum_{n\in \Z}q^{n^2}$. It is known that $\mathcal{H}(z)$ is a weight $\frac 32$ harmonic Maass form of manageable growth on $(\Gamma_0(4),\nu_{\theta}^3)$, i.e. \[\mathcal{H}(\gamma z)=\nu_{\theta}(\gamma)^3 (cz+d)^{\frac 32} \mathcal{H}(z)\quad \text{for }\gamma\in \Gamma_0(4)\] where $\nu_{\theta}(\gamma)=\frac{\theta(\gamma z)}{(cz+d)^{\frac 12}\theta(z)}$. It is also known that $\xi_{\frac 32} \mathcal{H}=-\frac 1{16}\theta$.
By (Hirzebruch and Zagier, 1976), we have another expression of $\mathcal{H}$: \[\mathcal{F}(z):=-\frac 1{12}+\sum_{n=1}^{\infty} H(n)q^n\quad \text{and}\quad \mathcal{H}(z)-\mathcal{F}(z)=\frac{1+i}{16\pi} \psi(z)\] where \[\psi(z)=\int_{-\overline{\tau}}^{i\infty} \frac{\theta(v)}{(z+v)^{\frac 32}} dv. \] Here $\mathcal{F}(z)$ is the holomorphic part of $\mathcal{H}(z)$.
Although we know that $\theta(z)$ is a weight $\frac 12$ modular form on $(\Gamma_0(4),\nu_{\theta})$, there is a formula for the Fricke involution of $\theta(z)$. Let $W_{4}:=\lp\begin{smallmatrix} 0&-\frac 12\\2&0 \end{smallmatrix}\rp$. We have \[(\theta|_{\frac 12}W_4)(z)=(2z)^{-\frac 12} \theta \lp -\frac 1{4z} \rp=e(-\tfrac 18)\theta(z). \]
Therefore, $\theta$ is in fact living in $(\Gamma_0(4)^+,\nu_{\theta})$, where $\Gamma_0(4)^+=\langle \Gamma_0(4),W_4\rangle =\langle T,W_4\rangle$ (prove it!) and $\nu_{\theta}$ is extended by $\nu_{\theta}(W_4)=e(-\tfrac 18)$.
Look back to $\psi(z)$ above. It should has transformation formula similar to $\theta$ (also written in (Hirzebruch and Zagier, 1976)).
What is the transformation formula for $\mathcal{H}(z)$ under $W_4$?
Theorem 1. We have \[\lp \frac{2\tau}i \rp^{-\frac 32} \mathcal{H}\lp -\frac{1}{4\tau}\rp+\mathcal{H}(\tau)=-\frac 1{24} \theta(\tau)^3. \]
Proof. By the discussion under (Hirzebruch and Zagier, 1976, Theorem 2, Corollary), we can show that \[(-2i\tau)^{-\frac 32}\mathcal{F}(-\tfrac 1{4\tau} ) +\mathcal{F}(\tau)= -\frac 1{24}\theta(\tau)^3-\sqrt{\frac{\tau}{8i}} \int_{\R} e(\xi^2 \tau)\frac{1+e(2\xi\tau)}{1-e(2\xi\tau)} \xi d\xi\] (which comes actually from (Eichler, 1955)) and \[(-2i\tau)^{-\frac 32}\psi(-\tfrac 1{4\tau} ) +\psi(\tau)=\int_{0}^{i\infty} \frac{\theta(u)}{(\tau+u)^{\frac 32}} du.\] Then it suffices to prove that \begin{equation}\tag{$*$} \frac{1+i}{16\pi }\int_{0}^{i\infty} \frac{\theta(u)}{(\tau+u)^{\frac 32}} du =\sqrt{\frac{\tau}{8i}}\int_\R e(\xi^2 \tau) \frac{1+e(2\xi \tau)}{1-e(2\xi \tau)} \xi d\xi. \end{equation} To do this, we expand the left and right hand sides. On the left side, we use $\theta(u)=1+2\sum_{n=1}^\infty e(n^2 u)$ and get \[\int_0^{i\infty} (\tau+u)^{-\frac 32} du=2\tau^{-\frac 12},\quad \int_0^{i\infty} \frac{2e^{2\pi i n^2 u}}{(\tau+u)^{\frac 32}} du=2n\sqrt{2\pi} e(-\tfrac 18)e(-n^2 \tau)\Gamma(-\tfrac 12,-2\pi i n^2 \tau). \] On the right side, first notice that the integrand is an even function. When $\xi>0$ and $\im z>0$, we have $|e(2\xi z)| < 1 $ and we can expand the fraction as \[\frac{1+e(2\xi\tau)}{1-e(2\xi\tau)}=1+2\sum_{n=1}^\infty e(2n\xi \tau). \] Hence we get \[\int_{\R} e(\xi^2 \tau) \xi d\xi=2\int_0^\infty e(\xi^2\tau) \xi d\xi=\frac{i}{2\pi \tau}\] and \[4\int_{0}^\infty e^{2\pi i \xi^2 \tau + 4\pi i n \xi \tau} \xi d\xi=ne^{-2\pi i n^2 \tau}(-2\pi i \tau)^{-\frac 12}\Gamma(-\tfrac 12,-2\pi i n^2 \tau). \] The equation ($*$) is then proved by comparing the $n$-th terms and we have finished the proof of Theorem 1.