Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 23 / 2 / a / Solution

Apply the Poisson summation formula to $h(x)=e^{2\pi i\tau x^2}$, with Fourier kernel $e^{-2\pi ix\xi }$. Its complex Gaussian Fourier transform isChoose the square root holomorphic on the half-plane and positive when $\tau$ is positive imaginary. Summing over integer $\xi$ givesHere 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 defineThus $\varphi (\tau )={\Theta }_4(2\tau )$. 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:Also $\varphi (\tau +1{)}^8=\varphi (\tau {)}^8$. The given generators, including their negatives, therefore establish weight four for ${\varphi }^8$ on ${\Gamma }_0(2)$, and weight $4k$ for ${\varphi }^{8k}$.

The defining series is holomorphic and its expansion at infinity has no negative powers. At the other modular cusp,Writing $q_0=e^{\pi i\tau }$ gives ${\Theta }_2(\tau /2)=2q_0^{1/8}{\sum }_{n\ge 0}{q}_0^{n(n+1)/2}$, 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 $k>0$. ConsequentlyThe exponent in the shifted series is $n^2$, as printed in the PDF, not the corrupted exponent in the TeX aid.