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I am sorry because it is probably a trivial question. I consider a homeomorphism of the circle that preserves orientation and that has a rational rotation number p/q (it is an irreducible fraction). I assume that ALL the orbits are periodic. Then we know that all the orbits have the same period q. I want to know if such a homeomorphism is conjugate to the rotation of angle p/q. Is that true ? If yes, can you tell me how to construct such a conjugacy ? I tried but I failed... If it is not true, can you give me a counterexample ? Thanks in advance.

GH from MO
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Maxence
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  • This questions seems borderline for this site, but let me give you hint : What happens when you look at the interval $[x,x'[$ where $x$ is any point and $x'$ is the closest point forward that lies in the orbit of $x$, and then at iterates of $x+\varepsilon$ for small positive $\varepsilon$? – Benoît Kloeckner May 28 '13 at 18:08
  • Thank you for your hint. It is the first time I ask a question on this site, so can you explain what it seems borderline ? Thanks in advance? – Maxence May 29 '13 at 11:15
  • @Maxence: You could consider accepting the answer if it was helpful to you. The question was borderline because it is about a fairly classical fact that you should have found in introductory books on the topic… – Loïc Teyssier May 29 '13 at 16:35
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    @BenoîtKloeckner I don't agree with the "borderline" label. Nobody is assumed to know everything on every subject, even on its own subject. I appreciate when specialists give their input even for some "naive" questions because sometime (depending on the qualities of the specialist) it can be enlightening. – Patrick I-Z Dec 08 '13 at 10:45
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    @PatrickI-Z: I find this question borderline not because everybody should know the answer, but because it seemed to me that any mathematician could find the answer, or at least explain what he or she tried in the question. And borderline does not mean clearly out of scope. – Benoît Kloeckner Dec 08 '13 at 20:30
  • @BenoîtKloeckner If people can teach me and prevent me to go open a book, I'm thankful. You see my point. OK then, I think we understand each other. – Patrick I-Z Dec 08 '13 at 20:52
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    This exchange reminds me of a classic Andre Weil anecdote. A mathematician addresses Weil: "Can I ask you a stupid question?" - "You just did", Weil snaps back. – Victor Protsak Dec 08 '13 at 21:04

1 Answers1

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The answer is "yes". Consider $\mathbb S^1$ as the quotient $\mathbb R/\mathbb Z$. Your homeomorphism $f$ lifts to a homeomorphism $\phi : \mathbb R \to \mathbb R$ such that $\phi(x+1)=\phi(x)+1$. Form the map $h:=\frac{1}{q} \sum _{n=1} ^q (\phi^{\circ n}-pn)$, where $\phi ^{\circ n}$ is the composition $n$ times of $\phi$ with itself. By construction $h\circ \phi = h+\frac{p}{q} $ and $h(x+1)=1+h(x)$, so that $h$ factors as a homeomorphism of the circle conjugating $f$ to the rotation. By the way this approach wors in $\mathbb R^n$ too.

Loïc Teyssier
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