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I'd like help for the problem:

Let the additive group $2\pi \mathbb{Z}$ act on $\mathbb{R}$ on the right by $x · 2\pi n = x+2\pi n$, where $n$ is an integer. Show that the orbit space $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $ is a smooth manifold.

I proved that $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $ is hausdorff and second countable, but I don't know how to find an atlas, I was thinking about $ \psi([x])=e^{ix}$ but I don't know how to show that this is a homeomorphism.

Jr.
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    Keep in mind the geometric interpretation that $\mathbb R/2\pi\mathbb Z$ is a circle. – Alex Becker Feb 26 '12 at 00:36
  • A continuous bijection between compact Hausdorff spaces is an homeo. – Mariano Suárez-Álvarez Feb 26 '12 at 00:50
  • @MarianoSuárez-Alvarez I'm under the impression that that's a rather hefty theorem though. – Alex Becker Feb 26 '12 at 00:57
  • @AlexBecker the idea is to conclude that,so suppose I don't know this. – Jr. Feb 26 '12 at 01:01
  • @MarianoSuárez-Alvarez How does one know that $\frac{\mathbb{R}}{2\pi \mathbb{Z}} is compact? – Jr. Feb 26 '12 at 01:03
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    It is the image of $[0,2\pi]$ under the projection map. – Mariano Suárez-Álvarez Feb 26 '12 at 01:06
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    Alex: not really. Since the function $f$ we start with is bijective, it has an inverse, so we want to show it is continuous: it is more or less obvious that it is enough to show that the image under $f$ of a closed set is closed. Closed sets in the domain are compact, the continuous image of a compact set is compact, and in the codomain compacts are closed. In any case, I'd expect this little piece of general topology to be known by anyone who knows the definition of a smooth manifold... His life will be hard if not! – Mariano Suárez-Álvarez Feb 26 '12 at 01:08
  • @AlexBecker I think the that the "theorem" is a consequence of the definition of homeomorphism – Jr. Feb 27 '12 at 02:09
  • @MarianoSuárez-Alvarez Sure! Thanks for the comments. – Jr. Feb 27 '12 at 02:12

1 Answers1

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This is easier to do in general than in the special case, so I'll just generalize. Most of the claims below require proof, by the way!

Suppose $M$ is a smooth manifold and that $G$ is a group which acts on $M$ smoothly and properly discontinuously. Let $N=M/G$ be the quotient topological space and $\pi:M\to N$ the quotient map. Since the action is properly discontinuous, $N$ is Hausdorff and the map $\pi$ is open —in particular, the second-countability of $M$ implies that of $N$.

Let us now construct an atlas.

Let $p\in M$. Proper-discontinuity implies there is an open neighborhood $U\subseteq M$ of $p$ such that $U\cap gU=\emptyset$ for all $g\in G\setminus\{1\}$. The restriction $\pi|_U:U\to N$ is an homeomorphism onto its image, which is open in $N$. By replacing $U$ by a smaller open neighborhood of $p$, we can suppose that there is a chart $\phi:V\to\mathbb R^n$ in the (maximal) atlas of $M$ with $U\subseteq V$. We define $\psi_p=\phi\circ(\pi|_U)^{-1}:\pi(U)\to\mathbb R^n$.

Now the set $\mathcal A=\{\psi_p:p\in M\}$ is an atlas on $N$.

Mariano Suárez-Álvarez
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  • Why not take the most natural idea: restrict the quotient map $f:\mathbb R \rightarrow S^1$ to any open Intervall $I$ of length $<1$. It is then clear that $f|_I$ is injective and continous. Now show that it is open and that the coordinate changes $f_J^{-1} f_I$ are smooth. – Blah Feb 26 '12 at 07:30
  • @Blah, Dear Blah, that is precisely what I did above, modulo the observation that the standard atlas on $\mathbb R$ contains the identity function $\mathrm{id}:\mathbb R\to\mathbb R$. – Mariano Suárez-Álvarez Feb 26 '12 at 08:57
  • Mariano, sorry for my duplicate: Of course you are right, I just wanted to give a hint that the atlas can be constructed really elementary. – Blah Feb 26 '12 at 10:54
  • sorry but, what is "acts on $M$ smoothly and properly discontinuously",and why this is the case of the action above? – Jr. Feb 27 '12 at 02:19
  • That the action be smooth meansthat each element acts by a diffeo of $M$. The definition of "properly discontinuously" you will find in Wikipedia. – Mariano Suárez-Álvarez Feb 27 '12 at 02:29
  • How to prove that $\psi$ is $C^{\infty}$? – Jr. Feb 27 '12 at 02:31
  • You cannot prove that $\psi$ is $C^\infty$: it does not make sense! I am using the maps $\psi$ to construct the atlas... What you probably want is to ask is «how can I prove that the transition functions of the atlas I constructed are smooth?» – Mariano Suárez-Álvarez Feb 27 '12 at 02:34
  • @MarianoSuárez-Alvarez No,I mean look $\psi$ a function between two spaces. – Jr. Feb 27 '12 at 02:38
  • I don't understand your clarification :/ – Mariano Suárez-Álvarez Feb 27 '12 at 02:39
  • $\frac{\mathbb{R}}{2 \pi \mathbb{Z}}$ is a smooth manifold so is $\mathbb{S}^1$, it does make sense talk about differentiability of a function between manifolds. – Jr. Feb 27 '12 at 02:42