---
tag: 5N5SDY3K53
kind: statement
role: theorem
name: Poincaré classification of circle homeomorphisms
msc: ["37E10", "37E45"]
uses: [DP4WWQBJJF, 0QT00PB5G2, T75CZ5H47N, AEXJNRBASV, HNXXAPQKG3, 2Q1J7RZ4TK, 0BXNDZFQME, 0BSJSX7ME2]
about: [DP4WWQBJJF]
sources:
  - url: https://en.wikipedia.org/wiki/Rotation_number
---

## Statement

Let $f$ be an orientation-preserving [[homeomorphism|0QT00PB5G2]] of the circle $S^1 = \R / \Z$, with [[rotation number|DP4WWQBJJF]] $\rho = \rho(f)$.

1. Let $\rho = p/q$ with $p, q$ coprime integers and $q \ge 1$. Then $f$ has a [[periodic point|T75CZ5H47N]], and every periodic point of $f$ has least period $q$. The omega-limit set of every point is a periodic orbit.
2. Let $\rho$ be irrational. Then $f$ has no periodic point. There is a continuous surjective map $h \colon S^1 \to S^1$ of degree one, weakly preserving the cyclic order, with $h \circ f = R_\rho \circ h$. Here $R_\rho(x) = x + \rho$.
3. If $\rho$ is irrational and $f$ has a [[dense|0BSJSX7ME2]] orbit, then $h$ is a homeomorphism, so $f$ is [[topologically conjugate|HNXXAPQKG3]] to the rotation $R_\rho$.

## Remarks

The converse parts also hold: $f$ has a periodic point if and only if $\rho$ is rational.
For part 1, one shows that $\rho(F) = p/q$ for a lift $F$ forces the equation $F^q(x) = x + p$ to have a solution, which gives a periodic point.
Since $f$ preserves the cyclic order, a periodic orbit of $f$ is arranged on the circle like an orbit of the rotation by $p/q$.
This forces its least period to be $q$.
Between two consecutive fixed points of $f^q$ the map $f^q$ moves every point in the same direction.
So every orbit of $f^q$ is monotone and converges to a fixed point of $f^q$.
For part 2, $h$ is built by sending the orbit of a point $x$ to the orbit of $0$ under $R_\rho$.
This works because $f$ orders an orbit on the circle in the same way as $R_\rho$ does, and $h$ then extends by monotonicity.
When $\rho$ is irrational and $f$ has no dense orbit, the omega-limit set of every point is the same [[Cantor set|0BXNDZFQME]], and $f$ is only [[semiconjugate|AEXJNRBASV]] to $R_\rho$.
Example: for the rotation $R_{2/5}$ every point has period $5$, and for an irrational $\alpha$ the map $R_\alpha$ is an [[irrational rotation|2Q1J7RZ4TK]] with $h = \id$.

## Explanations

### L1 human

A homeomorphism of the circle moves points around while keeping their order.
Poincaré showed that the long-run average turn, the rotation number, tells what kind of motion is possible.
If it is a fraction $p/q$, there are points that come back after exactly $q$ steps, and all other points drift towards such cycles.
If it is irrational, nothing ever comes back exactly, and the map looks like a rigid irrational rotation, possibly after collapsing some intervals.

### L2 human

The theorem reduces the topological study of circle homeomorphisms to one number plus a little extra structure.
In the irrational case the remaining question is whether the semiconjugacy $h$ is a conjugacy, that is, whether $f$ has wandering intervals.
Denjoy's theorem answers this for smooth enough diffeomorphisms: a $C^2$ diffeomorphism with irrational rotation number is conjugate to the rotation.
In the rational case the periodic orbits all have the same period and the same combinatorics as the rotation by $p/q$.
This is the starting point of the study of Arnold tongues.