# Quotient of $R$ by $2πZ$

Let the additive group $2πZ$ act on $R$ on the right by $x · 2πn = x +2πn$, where $n$ is an integer. Show that the orbit space $R/2πZ$ is a smooth manifold.

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Can you think of what the space $\mathbb{R}/2\pi\mathbb{Z}$ looks like? [Hint: if $\theta \in \mathbb{R}/2\pi\mathbb{Z}$ then $\theta \sim \theta + 2n\pi$ for all $n \in \mathbb{Z}$. What does this remind you of?] – Clive Newstead Nov 1 '12 at 22:04
As a further hint, the orbit space is a compact, connected, 1-dimensional manifold, and there's only one of those. We'd like to see more of what work you have done so far! – Hew Wolff Nov 2 '12 at 0:26

To simplify notations, we prove that $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ is a smooth manifold. It can be proved similarly that $\mathbb{R}/2\pi\mathbb{Z}$ is also a smooth manifold.

Since $\mathbb{Z}$ is a closed subgroup of $\mathbb{R}$, $\mathbb{T}$ is a Hausdorff topological group. Let $p\colon \mathbb{R} \rightarrow \mathbb{T}$ be the canonical map. Let $V$ be an open subset of $\mathbb{R}$. It is easy to see that $p^{-1}p(V) = V + \mathbb{Z}$. Since $V + \mathbb{Z}$ is open, $p$ is an open map.

Let $U = \{x\in \mathbb{R} | |x| < 1/2\}$. Suppose $p(x) = p(y)$ for $x, y \in U$. Then $x - y \in \mathbb{Z}$. Since $|x - y| \le |x| + |y| < 1$, $x - y$ must be $0$. Hence $p|U$ is injective. Since $p$ is an open continuous map, $p|U$ is a homeomorphism onto $p(U)$.

Since $p$ is open, $p(U) + p(x)$ is an open neighborhood of $p(x)$ for $x \in \mathbb{R}$. Let $f_x\colon U \rightarrow p(U) + p(x)$ be the map defined by $f_x(t) = p(t) + p(x)$. Clearly $f_x$ is a homeomorphism. Hence $\mathbb{T}$ is a topological manifold.

Let $p(x), p(y)$ be any two points of $\mathbb{T}$. Suppose $p(z) \in (p(U) + p(x)) \cap (p(U) + p(y))$. There exist unique $s, t \in U$ such that

$p(z) = p(s) + p(x) = p(t) + p(y)$.

Note that $s$ and $t$ are local coordinates of $p(z)$ in $p(U) + p(x)$ and $p(U) + p(y)$ respectively. Since $s + x \equiv t + y$ mod $\mathbb{Z}$, $t = s + x - y + n$ for some $n \in \mathbb{Z}$. Since $t$ is a continuous function of $s$, $n$ is constant in a neighborhood of $s$. Thus $t$ is a smooth function of $s$. Hence $\mathbb{T}$ is a smooth manifold.

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