Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
    
Please help me,I need Answer –  Ciris Nov 1 '12 at 22:32
2  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.