Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 24 / 4

Use the Fourier transform convention $\hat{h}(\xi )={\int }_{\mathbb{R}}h(x)e^{-2\pi ix\xi }dx$. For a Schwartz function $h$, the Poisson summation formula isTo prove it, periodize $h$ as $P(x)={\sum }_{n\in \mathbb{Z}}h(x+n)$. Rapid decay of all derivatives makes this a smooth periodic function, with uniform termwise differentiability. Its $m$th Fourier coefficient iswhere absolute convergence justifies splitting the real-line integral into unit intervals. Integration by parts gives rapid decay of the coefficients of $P$, so its Fourier series converges absolutely and uniformly to $P$. Evaluate it at zero to obtain the formula.

For $\mathrm{Im}\tau >0$ and $z\in \mathbb{C}$, take the complex Gaussian function $h(x)=e^{\pi i\tau x^2+2\pi izx}$. It is a Schwartz function of the real variable $x$. Its Fourier transform isThe real Gaussian integral proves this first for $\tau =iy$, $y>0$, and real $z$. Normal convergence of the defining integrals makes both sides entire in $z$ and holomorphic in $\tau \in \mathbb{H}$, so the identity theorem extends it to the stated domain. Choose the square-root branch holomorphic for $-i\tau$ in the right half-plane and positive when $\tau =iy$.

Applying Poisson summation gives the transformation of the Jacobi theta function:The series on the right is ${\vartheta }_{00}(z/\tau ,-1/\tau )$. Rearranging therefore givesThe branch choice is part of the formula; at $\tau =i$ its square-root factor equals one.

Fix $\tau \in \mathbb{H}$ and let $\Lambda =\mathbb{Z}+\mathbb{Z}\tau$. We use the binary convention for the theta function with characteristics in the PDF. Its nonzero exponential prefactor does not change the zeros of the translated ${\vartheta }_{00}$. The four characteristic functions are entire in $z$, by normal convergence of their Gaussian series.

We first prove their zero information, rather than assume that the quotient has no poles. Pair the terms with indices $n$ and $-n-1$ to getThe logarithmic derivative $D(z)={\vartheta }_{00}^{\prime }(z)/{\vartheta }_{00}(z)$ is periodic under $z\mapsto z+1$ and satisfies $D(z+\tau )=D(z)-2\pi i$, by the spatial transformation formulas. Choose a fundamental parallelogram with zero-free boundary. Its vertical-side integrals cancel, and subtracting the top horizontal integral from the bottom one gives $2\pi i$. The argument principle therefore counts exactly one zero inside, with multiplicity. Thus the displayed zero is simple and is the unique one modulo $\Lambda$. The zeros of theta functions with characteristics are consequentlyAll are simple. In particular the three even theta constants are nonzero and ${\vartheta }_{11}^{\prime }(0)\ne 0$.

Let $R(z)$ be the product of these four functions divided by ${\vartheta }_{11}(2z)$. The denominator's simple zeros occur exactly at the four half-periods modulo $\Lambda$, and each is cancelled by the corresponding simple numerator zero. Hence $R$ extends to an entire function. Under $z\mapsto z+1$, the four numerator signs multiply to one, as do the two successive shifts of the denominator. Under $z\mapsto z+\tau$, the numerator has multiplier $q^{-2}{t}^{-4}$, where $t=e^{2\pi iz}$; the denominator, whose argument shifts by $2\tau$, has the same multiplier. Thus $R$ is an entire elliptic function with periods $1,\tau$. It is bounded on a fundamental parallelogram and hence on the plane, so Liouville's theorem makes it constant.

At $z=0$, the simple odd zero gives ${\vartheta }_{11}(z)/{\vartheta }_{11}(2z)\to 1/2$. Therefore the Jacobi theta duplication identity isAt a zero of the displayed denominator, this is interpreted by removable continuation. Equivalently, multiply through by the denominator to obtain an entire product identity valid for every $z$.