Why is that $\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}$?

Question

$\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}$, that is, every character of $\mathbb{R}/\mathbb{Z}$ is of the form $x\mapsto e(mx)$ for some integer $m$.

I was considering the dual of ${\mathbb{R}/\mathbb{Z}}$. What confused me is why we can assign an integer $m$ to the character $\chi.$

Thanks in advance! Any comment is appreciated!

Answer 1

We can identify $\mathbb R / \mathbb Z$ with the circle group $\mathbb T=\{z \in \mathbb C \mid |z|=1\}$, with the multiplication coming from $\mathbb C$. We map $x \to e^{2 \pi i x}$. Then $x+y \pmod 1 \mapsto e^{2 \pi i (x+y) \pmod {2\pi}}$, so this will be a group isomorphism.

Then the characters of $\mathbb T$ are the continuous group homomorphisms from $\mathbb T \to \mathbb T$. So, thinking about this geometrically, we start at $1$ and then wind our way around the circle an integer number of times. It must be an integer because group homomorphisms must map $1$ to $1$. Anticlockwise or clockwise will correspond to positive and negative integers. Clearly, each integer will give us a different character and we obtain all characters in this way.

So the set of characters is given by $$ \{x \mapsto e^{2 \pi imx} \mid m \in \mathbb Z \}, \text{ for } x \in \mathbb R/\mathbb Z. $$

Answer 2

$m$ is the size of the pre-image of 1. (And $m=0$ if the preimage is everything.)