Poisson summation formula for abelian groups

Remarks

Proof idea: let $F(x + H) = \int_H f(x + h)\, dh$. By Weil's formula $F$ is integrable on $G/H$, and its Fourier transform at $\chi \in H^\perp = \widehat{G/H}$ is $\widehat{f}(\chi)$. Since $F$ is continuous and $\widehat{F}$ is integrable, Fourier inversion holds at every point, and the point $0 + H$ gives the formula.

Example: $G = \R$ with Lebesgue measure, $H = \Z$ with counting measure, and $\chi_\xi(x) = e^{2\pi i \xi x}$. Then $G/H$ is the circle of length $1$, $H^\perp = \set{\chi_k : k \in \Z}$ carries counting measure, and the formula is the classical Poisson summation formula $\sum_{n} f(n) = \sum_{k} \widehat{f}(k)$. For $G = \Z/N\Z$ and $H$ the subgroup of multiples of $d$, where $d$ divides $N$, it is an identity between two finite sums.

Some hypotheses of this kind are needed. For a continuous integrable $f$ on $\R$ alone, the series $\sum_k \widehat{f}(k)$ need not converge, and then the right side has no meaning.

Going deeper

For a student

The abstract form explains where the classical formula comes from: periodising $f$ over $H$ gives a function on $G/H$, whose Fourier coefficients are the values of $\widehat{f}$ on $H^\perp$. With $G = \R^n$ and $H$ a lattice, $H^\perp$ is the dual lattice, which gives the theta function identities and the functional equation of the Riemann zeta function. With $G$ the adeles of a number field and $H$ the field itself, the same formula is the starting point of Tate's thesis.