= Solution
Apply the <Poisson summation formula> to $h(x)=e^{2\pi i\tau x^2}$, with Fourier kernel $e^{-2\pi i x\xi}$. Its <complex Gaussian Fourier transform> is
$$
\widehat h(\xi)=(-2i\tau)^{-1/2}e^{-\pi i\xi^2/(2\tau)}.
$$
Choose the square root holomorphic on the half-plane and positive when $\tau$ is positive imaginary. Summing over integer $\xi$ gives
$$
\boxed{\theta\left(-\frac1{4\tau}\right)=\sqrt{\frac{2\tau}{i}}\,\theta(\tau).}
$$
Here the printed <theta series of integer squares> uses $q=e^{2\pi i\tau}$; the common theta-constant convention instead uses $e^{\pi iz}$.
To keep track of the shifted series, put $z=2\tau$ and define
$$
\Theta_3(z)=\sum_ne^{\pi izn^2},\quad\Theta_4(z)=\sum_n(-1)^ne^{\pi izn^2},\quad\Theta_2(z)=\sum_ne^{\pi iz(n+1/2)^2}.
$$
Thus $\phi(\tau)=\Theta_4(2\tau)$. The same Poisson calculation with a phase or shifted lattice gives $\Theta_4(-1/z)=\sqrt{-iz}\Theta_2(z)$ and $\Theta_2(-1/z)=\sqrt{-iz}\Theta_4(z)$. Translation gives $\Theta_4(z+2)=\Theta_4(z)$ and $\Theta_2(z-1)=e^{-\pi i/4}\Theta_2(z)$. These are <theta-constant inversion and translation laws>.
Let $U\tau=\tau/(2\tau+1)$ and put $w=-1/z-1$, so $2U\tau=-1/w$. Raising the preceding identities to the eighth power removes all square-root and phase ambiguities:
$$
\Theta_4(2U\tau)^8=w^4\Theta_2(w)^8=w^4z^4\Theta_4(z)^8=(2\tau+1)^4\phi(\tau)^8.
$$
Also $\phi(\tau+1)^8=\phi(\tau)^8$. The given generators, including their negatives, therefore establish weight four for $\phi^8$ on $\Gamma_0(2)$, and weight $4k$ for $\phi^{8k}$.
The defining series is holomorphic and its expansion at infinity has no negative powers. At the other <modular cusp>,
$$
(\phi^8|_4S)(\tau)=\frac1{16}\Theta_2(\tau/2)^8.
$$
Writing $q_0=e^{\pi i\tau}$ gives $\Theta_2(\tau/2)=2q_0^{1/8}\sum_{n\ge0}q_0^{n(n+1)/2}$, so the right side begins $16q_0$ and is a holomorphic power series in $q_0$. Taking its $k$th power proves <modular cusp> holomorphy for every $k>0$. Consequently
$$
\boxed{\phi^{8k}\in M_{4k}(\Gamma_0(2)).}
$$
The exponent in the shifted series is $n^2$, as printed in the PDF, not the corrupted exponent in the TeX aid.
Back to article page
