# Pontryagin dual of the unit circle

Define the unit circle as $\frac{\mathbb{R}}{2\pi\mathbb{Z}}.$ I know the Pontryagin dual (looking at properties of the Fourier transform on locally compact Abelian groups) is $\mathbb{Z}$ but why?

Any notes or suggestions will be appreciated.

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The stupid answer is that $\operatorname{Hom}_{\text{cts}}(\mathbb Z, S^1) = \operatorname{Hom}_{\mathbb Z}(\mathbb Z,S^1) = S^1$, so we are done by duality.

Here, $S^1$ denotes the unit circle $S^1 \cong \frac{\mathbb R}{2\pi\mathbb Z} \cong \mathbb R/\mathbb Z$, a.k.a. $\mathbb T$.

The real answer is that all characters of $S^1$ are of the form $z \mapsto z^n$. The reason is explained in many places, for example here: https://mathoverflow.net/questions/89504/quick-computation-of-the-pontryagin-dual-group-of-torus

In particular, you can see there that it is enough to know that $\mathbb R$ is selfdual.