Self-dual group

Example

A locally compact abelian group $G$ is self-dual if it is isomorphic, as a topological group, to its Pontryagin dual $\widehat{G}$.

1. The real line $\R$ is self-dual. For $\xi \in \R$ let $\chi_\xi(x) = e^{2\pi i \xi x}$. Each $\chi_\xi$ is a continuous character, and $\chi_{\xi + \eta} = \chi_\xi \chi_\eta$. Every continuous character of $\R$ has this form, and $\xi \mapsto \chi_\xi$ is a homeomorphism onto $\widehat{\R}$ with the compact-open topology.
2. The group $\Z/n\Z$ of integers modulo n is self-dual. For $k \in \Z/n\Z$ let $\chi_k(a) = e^{2\pi i k a / n}$. This is well defined, it is a character, and $k \mapsto \chi_k$ is an injective homomorphism. A character $\chi$ is determined by $\chi(1)$, which is an $n$-th root of unity, so all $n$ characters arise.
3. The p-adic numbers $\Q_p$ are self-dual. Here $\psi(x) = e^{2\pi i \set{x}_p}$, where $\set{x}_p$ is the $p$-adic fractional part, is a character, and $a \mapsto (x \mapsto \psi(ax))$ is an isomorphism $\Q_p \to \widehat{\Q_p}$.
4. Finite products of self-dual groups are self-dual, since the dual of a finite product is the product of the duals. So $\R^n$ is self-dual, and every finite abelian group is self-dual, being a product of cyclic groups.

Non-example: the circle group $\mathbb{T}$ is not self-dual. Its dual is $\Z$, via $n \mapsto (z \mapsto z^n)$, and $\Z$ is countable and discrete, while $\mathbb{T}$ is uncountable. Likewise $\Z$ is not self-dual.