Circle rotation number invariant under topological semi-conjugacy.

For a circle homeomorphism $f: S^1 \rightarrow S^1$ we can define the the rotation number $$\rho(f) = \lim_{n \rightarrow \infty} \frac{1}{n}(F^n(x) - x) \mod 1,$$ for a lift $F:\mathbb{R} \rightarrow \mathbb{R}$ of $f$ and a point $x \in \mathbb{R}$. $\rho(f)$ exists always and is independent of $x$ and $F$. In this wikipedia article it says that the rotation number is invariant under topological semi-conjugacy, i.e. for two homeomorphisms $f,g$ of the circle the rotation numbers $\rho(f) = \rho(g)$ are equal if there is a continuous, surjective and monotone map $h:S^1 \rightarrow S^1$ with $$f \circ h = h \circ g.$$ I proof why this is true for homeomorphisms $h$ can be found here in chapter 11. I don't understand why this is still true if $h$ is no homeomorphism.

Thanks

The following argument is an extension of the one for a homeomorphism $h$ in [Hasselblatt & Katok. A First Course in Dynamics. Cambridge University Press, 2003, pg. 128].

We will show (for the right choice of $G$), $\frac{|F^n(x) - G^n(x)|}{n} \to 0$ as $n \to \infty$ for all $x$, where $F$ and $G$ are lifts of $f$ and $g$ respectively, i.e. $\pi F = f \pi$, where $\pi: \mathbb{R} \to \mathbb{S}^1$ is the projection operator.

Let $H$ be a lift of $h$. Then $\pi F H = f \pi H = f h \pi= h g \pi = h \pi G = \pi H G$. This implies $F \circ H = H \circ G + k$ for some $k \in \mathbb{Z}$. Another lift $G'$ of $g$ may be chosen so that $F \circ H = H \circ G'$. In particular, let $G'(x) = G(x) + k$. Since $H$ is a degree-one lift, $H(x + k) = H(x) + k$ for any integer $k$. Then $H \circ G'(x) = H (G(x) + k) = H(G(x)) + k = F \circ H(x)$.

For the rest of the argument, assume the lift $G$ is chosen so that $F \circ H = H \circ G$.

By induction $F^n\circ H = H \circ G^n$: this holds true for $n = 1$ and assuming $F^{n-1} \circ H = H \circ G^{n-1}$ then

\begin{aligned} F^n\circ H &= F^{n-1}\circ F \circ H\\ &= F^{n-1} \circ H \circ G\\ &= H \circ G^{n-1} \circ G\\ &= H \circ G^n \end{aligned}.

This implies $G$ may be chosen so that $F^n\circ H = H \circ G^n$, where $H$ is any lift of $h$.

Suppose $H$ is the lift such that $H(0) \in [0,1)$. Then for all $x \in [0,1)$, since $0 < H(0) \leq H(x) < H(1) = H(0) + 1 < 2$, we have $-1 \leq H(x) -x \leq 2$ for all $x \in [0,1)$. Since $H$ is a degree-one lift, $H(x) - x$ is periodic with period $1$, so this implies for all $x \in \mathbb{R}$, $|H(x) - x| < 2$.

Thus, $|H\circ G^n(x) - G^n(x)| < 2$.

Using the fact $F^n(x + 1) = F^n(x) + 1$ for all $x$, we have $F^n(x + 2) = F^n(x) +2$, and $F^n(x-2) = F^n(x) - 2$. Suppose $| y - x | < 2$, so $x - 2 < y < x + 2$. Since $F^n$ is strictly increasing, $F^n(x-2) < F^n(y) < F^n(x + 2)$, hence $F(x) - 2 < F^n(y) < F^n(x) + 2$. Therefore, if $|y-x| < 2$, we have $|F^n(y) - F^n(x)| < 2$.

Since $|x - H(x)| < 2$, we have $|F^n(x) - F^n\circ H(x)| < 2$.

Therefore,

\begin{aligned} |F^n(x) - G^n(x)| & = |F^n(x) - F^n\circ H(x) + F^n\circ H(x) - G^n(x)|\\ &= |F^n(x) - F^n\circ H(x) + H\circ G^n(x) - G^n(x)|\\ &\leq |F^n(x) - F^n\circ H(x)| + |H\circ G^n(x) - G^n(x)|\\ &< 2 + 2 \end{aligned}

Hence, $\frac{|F^n(x) - G^n(x)|}{n} < \frac{4}{n} \to 0$ as $n \to \infty$.

This implies $\rho(f) = \rho(g)$.

• Nice answer, but I have a litter puzzle, why $F$ is strictly increasing? – Hu xiyu Dec 4 '17 at 7:01
• How do you guarantee that $H$ is a degree-one map? Since $h$ is not bijective this is not true anymore. – ghreis Apr 2 '18 at 16:11