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

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.

Use the three left-coset representatives $I,S,ST$ for ${\Gamma }_0(2)\backslash {\mathrm{SL}}_2(\mathbb{Z})$. If a nonzero weight-four cusp form $h$ existed, its coset norm of a modular formwould be a nonzero weight-twelve modular form for the full group: right multiplication permutes its factors. At infinity $h$ has order at least one in $q$. The other two factors correspond to the zero modular cusp of cusp width two and each have order at least $1/2$ in $q$. Thus $H$ has order at least two there.

The ratio $H/\Delta$ is weight zero, holomorphic on the half-plane because $\Delta$ has no zeros there, and holomorphic at the modular cusp with value zero. It descends to a holomorphic function on the compact full modular curve. Such a function is constant, hence zero, contradicting $H\ne 0$. ThereforeConstant terms at the two modular cusps define a linear map $M_4({\Gamma }_0(2))\to {\mathbb{C}}^2$ whose kernel is this modular cusp space. It is injective, so the dimension is at most two. The forms $E_4(\tau )$ and $E_4(2\tau )$ are holomorphic weight-four forms for this group. For the second, the same conjugation used in 1(c) proves transformation, and $E_4(2S\tau )|$ after the weight factor is $2^{-4}{E}_4(\tau /2)$, proving holomorphy at zero. Their constant terms are both one but their $q$ coefficients are respectively $240$ and zero, so they are independent. We obtainThis is the weight-four Eisenstein basis at level two.

The constant and first Fourier coefficients of ${\varphi }^8$ are $1$ and $-16$. In the weight-four Eisenstein basis at level two, these forceFor $n>0$, its $q^n$ coefficient is $-16{\sigma }_3(n)+256{\sigma }_3(n/2)$, where the second divisor sum is zero if $n$ is odd. The even divisors have cube sum $8{\sigma }_3(n/2)$, soOn the other hand, expanding the eighth power of the theta series of integer squares counts ordered integer eight-tuples of square sum $n$. Their sign is $(-1{)}^{{\sum }_j{m}_j}=(-1{)}^n$, since $m_j^2\equiv m_j(\mathrm{mod}2)$. If $r_8(n)$ denotes that count, thenSeparately $r_8(0)=1$, from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the eight-square representation formula.