Harmonic analysis on the adeles

Remarks

The basic facts of the theory are these. The map $y \mapsto (x \mapsto \psi_K(xy))$ is an isomorphism of topological groups from $\mathbb{A}_K$ onto its Pontryagin dual, so $\mathbb{A}_K$ is self-dual. Under this isomorphism the annihilator of the discrete subgroup $K$ is $K$ itself. For the self-dual measure, the compact quotient $\mathbb{A}_K / K$ has volume $1$. The adelic Poisson summation formula states that $\sum_{\gamma \in K} f(\gamma) = \sum_{\gamma \in K} \widehat{f}(\gamma)$ for every $f \in \mathcal{S}(\mathbb{A}_K)$.

Example: on $\mathbb{A}_{\Q}$, take $f_\infty(x) = e^{-\pi x^2}$ and $f_p = \mathbf{1}_{\Z_p}$ for every prime $p$. Then $f = \prod_v f_v$ lies in $\mathcal{S}(\mathbb{A}_{\Q})$, since every clause of item 2 holds. The Gaussian is its own Fourier transform for $\psi_\infty$, and $\mathbf{1}_{\Z_p}$ is its own Fourier transform for $\psi_p$ and the Haar measure with $\Z_p$ of volume $1$, so $\widehat{f} = f$. A rational number $r$ has $f(r) \ne 0$ only if $r \in \Z_p$ for all $p$, that is, $r \in \Z$. So Poisson summation gives the trivial identity $\sum_{n \in \Z} e^{-\pi n^2} = \sum_{n \in \Z} e^{-\pi n^2}$; with $f_\infty(x) = e^{-\pi t x^2}$ it gives the theta transformation formula.

Non-example: the product of $f_\infty(x) = e^{-\pi x^2}$ and $f_p = \mathbf{1}_{p\Z_p}$ for every prime $p$ is not in $\mathcal{S}(\mathbb{A}_{\Q})$. It fails the last clause of item 2, since $f_p \ne \mathbf{1}_{\Z_p}$ for infinitely many $p$. Indeed the set $\prod_p p\Z_p$ has empty interior in the finite adeles, so this function is not locally constant there.