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

$\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}$, 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!

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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 $$\{x \mapsto e^{2 \pi imx} \mid m \in \mathbb Z \}, \text{ for } x \in \mathbb R/\mathbb Z.$$

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Tom, it is a great answer! I have a further question, though. What are the things like if $m$ is not an integer, that is, the dual group of $\mathbb{R}^+$ or $\mathbb{R}^×$ with additive character $\chi_\alpha$ of the form $\chi_\alpha: x\mapsto\chi_\alpha(x)=e(x\alpha)$ for some complex number $\alpha$? It seems to me that we could not think it this geometrical way. –  Qiang Zhang May 23 '12 at 0:56

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

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@Phiar, thanks for the answer! I have one more question. Like my reponse to Tom, how can we explain if $m$ is not an integer, that is, the dual group of $\mathbb{R}^+$ or $\mathbb{R}^×$ with additive character $\chi_\alpha$ of the form $\chi_\alpha: x\mapsto\chi_\alpha(x)=e(x\alpha)$ for some complex number $\alpha$? –  Qiang Zhang May 23 '12 at 1:02