Let $f : \mathbb S^1 \to \mathbb S^1$ be an orientation-preserving homeomorphism. The classical definition of the rotation number is the following: we lift $f$ to a homeomorphism $F : \mathbb R \to \mathbb R$ and we define the rotation number to be $$\rho(f) =\lim_{n\to \infty} \frac{F^n(x)}{n}$$ (for a fixed point $x \in \mathbb S^1$).

Apparently an equivalent definition is the following: $$ \rho(f)= \lim_{n\to \infty} \frac{1}{n} \textrm{#}\Big \{ 0\leq i \leq n: \ f^i(x) \in [z ,f(z)) \Big\} $$ where $ x \in \mathbb{S} ^1$ and $\textrm{#}X$ denotes number elements of a set $X$.

Can you help me to prove that?

  • $\begingroup$ I took the liberty of rewriting the question (which I like a lot). I hope you don't mind. $\endgroup$ – PseudoNeo Jul 15 '15 at 14:50
  • 2
    $\begingroup$ What is $z$ in the second definition? $\endgroup$ – user226970 Jan 18 '16 at 1:58
  • $\begingroup$ @user226970 Any point of the circle; part of the claim is that the result is the same. $\endgroup$ – user147263 Jan 19 '16 at 1:11

To make sure I got the terminology correctly,

  • I assume the lifting is given by the standard projection $x\mapsto x \bmod 1$, so $F(x)\bmod 1=f(x\bmod 1)$. For brevity, I will denote $x\bmod 1$ by $\overline{x}$.
  • I assume orientation-preserving means the lifting $F$ can be chosen to be an increasing function.

I write $\rho(f)$ for the first definition and $\hat{\rho}(f)$ for the second.

Intuitively, $\rho(f)$ is the average rotation speed of the orbits of $f$. The criterion $f^i(x)\in [z,f(z))$ is an indicator that the orbit of $x$ has turned one more time around the circle at step $i$, so $\hat{\rho}(f)$ is simply the average number of turns per iteration of $f$, which is another way to measure the rotation speed.

Let us try to make this intuition precise. By subtracting a suitable integer if necessary, we can assume that $z-1<F(z)<z+1$ for every $z\in\mathbb{R}$. If $f$ has a fixed point, then every orbit is asymptotic to a fixed point of $f$, and both definitions give $0$. So, without loss of generality, we may also assume that $z<F(z)<z+1$ for every $z\in\mathbb{R}$.

Let $x,z\in\mathbb{R}$ be arbitrary. For $n\in\mathbb{N}$, define \begin{align} r_n &:= \sup\{r\in\mathbb{Z}: r+z\leq F^n(x)\} \\ &= \lfloor F^n(x) - z\rfloor \;. \end{align} Intuitively, $r_n$ is more or less the number of times the orbit of $x$ has passed $z$.

Observation. $\lim_{n\to\infty} {\displaystyle\frac{r_n}{n}} \bmod 1 = \rho(f)$.

The relationship between $r_n$ and the second definition is a consequence of the lemma below. I divide the lemma into three claims.

Claim I. $r_i\leq r_{i+1}\leq r_i+1$.

Argument. By definition, $r_i+z\leq F^i(x)<r_i+1+z$. By the monotonicity of $F$, we get \begin{align} r_i+F(z) &\leq F^{i+1}(x) < r_i+1+z + \left(F(z)-z)\right) \;. \end{align} The left-hand inequality, together with $z<F(z)$ gives $r_i\leq r_{i+1}$. The right-hand inequality, along with $F(z)-z<1$ implies $r_{i+1}\leq r_i + 1$.

Claim II. $r_{i+1}=r_i+1$ if and only if $F^{i+1}(x)\in [\,r_{i+1}+z,r_{i+1}+F(z)\,)$.

Argument. We have $r_{i+1}=r_i+1$ if and only if \begin{align} F^i(x) &< r_{i+1} + z \leq F^{i+1}(x) \;. \end{align} Applying $F$ and using its monotonicity, the left-hand inequality is equivalent to \begin{align} F^{i+1}(x) &< r_{i+1} + F(z) \;. \end{align} Therefore, $r_{i+1}=r_i+1$ if and only if \begin{align} r_{i+1} + z &\leq F^{i+1}(x) < r_{i+1} + F(z) \;. \end{align}

Claim III. $F^i(x)\in[\,m+z,m+F(z)\,)$ implies $m=r_i$.

Argument. The assumption $m+z\leq F^i(x)$ implies $m\leq r_i$. On the other hand, $F^i(x)<m+F(z)$ and $F(z)<z+1$ give $F^i(x)<m+1+z$, which implies $r_i\leq m$.

Combining the above claims, we have

Lemma. $r_i\leq r_{i+1}\leq r_i+1$, and the equality $r_{i+1}=r_i+1$ happens if and only if $f(\overline{x}) \in [\,\overline{z}, f(\overline{z})\,)$.

The above lemma immediately gives $\lim_{n\to\infty} {\displaystyle\frac{r_n}{n}} = \hat{\rho}(f)$, concluding the proof.


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