Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 / 4

For a full Euclidean lattice $\Lambda \subseteq {\mathbb{R}}^n$, its dual lattice isFor a Schwartz function, the Poisson summation formula for a Euclidean lattice statesTo prove it, periodize $f$ over $\Lambda$. The resulting function on ${\mathbb{R}}^n/\Lambda$ has Fourier coefficient $m(\Lambda {)}^{-1}\hat{f}(\mu )$ at $\mu \in {\Lambda }^{\vee }$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.

Apply part a in ${\mathbb{R}}^2$ toThe supplied identity ${\hat{f}}_k=(-i{)}^k{f}_k$ givesIn the notation of the weighted Gaussian theta sum of a complex lattice, this is

Put $L_{\tau }=y^{-1/2}(\mathbb{Z}\tau +\mathbb{Z})$, a covolume-one lattice, and defineTermwise Mellin transformation in the initial half-plane gives

Split the integral at $t=1$. The integral over $[1,\infty )$ is entire in $s$ because the theta sum decays exponentially. Apply the Poisson summation formula for a Euclidean lattice and the Fourier eigenfunction calculation from part b to the interval $(0,1]$, then substitute $t\mapsto 1/t$. This rewrites the small-time integral as another exponentially convergent integral over $[1,\infty )$ plus explicit elementary Mellin terms. Those terms are meromorphic, but for positive even $k$ their apparent poles are canceled by the zeros of $1/\Gamma (s+k)$. The displayed formula therefore continues holomorphically to every $s\in \mathbb{C}$, proving the analytic continuation of a weight-k real-analytic Eisenstein series.