Rotation number and semiconjugacy.

Question

Here is the question I am thinking about:

Suppose $f,g : S^1 \to S^1$ are two homeomorphisms so that, for some homeomorphism $h: S^1 \to S^1,$ we have $$h \circ f \circ h^{-1} = g$$ That is, $f$ is conjugate to $g$ in the group of circle homeomorphisms.

I want to show that $f$ and $g$ have the same rotation number.

some thoughts:

I believe that my solution should be divided into two cases:

If the rotation number of $f$ denoted by $\tau (f)$ is rational or irrational. In case of rational number $p/q,$ I know that any periodic point has period $q.$

I am also guessing that this post Rotation number invariant under topological semi-conjugacy. might be helpful to me but I do not know how.

Any help will be greatly appreciated!

Answer 1

Let $\pi:\mathbb{R}\rightarrow S^1$ be the covering map; $F,\ H$ be the corresponding lifts (of $f,\ h$) $\mathbb{R}\rightarrow\mathbb{R}$. We also additionally enforce $F(0),H(0)\in[0,1]$. W.l.o.g. we take $F$ to be increasing (i.e. $f$ is orientation-preserving).

Lemma 1: $H^{-1}:\mathbb{R}\rightarrow\mathbb{R}$ is actually the lift of $h^{-1}:S^1\rightarrow S^1$.
Proof: $h^{-1}\circ\pi=h^{-1}\circ\pi\circ H\circ H^{-1} = h^{-1}\circ h\circ \pi\circ H^{-1}=\pi\circ H^{-1}$. $\blacksquare$

We then have, $G:=H\circ F\circ H^{-1}$ is a valid lift of $g$ with $G(0)\in[0,1]$.

Lemma 2: $\forall x\in\mathbb{R}$ we have
$|H(x)-x|\leq 1$
$|H^{-1}(x)-x|\leq 1$
Proof: $h^{-1}$ and $h$ are homeomorphisms, hence degree 1 maps. The claim follows. $\blacksquare$

We need another uniform bound estimate as follows:

Lemma 2: $\forall x,y\in[0,1]$ we have
$|F^n(x)-F^n(y)|\leq 2\ \forall n\in\mathbb{N}$
Proof: We take $x<y$ w.l.o.g. If $F(0)\notin(x,y)$, $\pi$ is a length-preserving homeomorphism from $F^n([x,y])$ to a subset of $S^1$, so, $|F^n(x)-F^n(y)|\leq 1\ \forall n\in\mathbb{N}$
Otherwise, put $z:= F(0)$, $\pi$ is a length-preserving homeomorphism from $F^n([x,z])$ and $F^n([z,y])$ to subsets of $S^1$. Adding the two lengths and using the triangle-inequality, we get $|F^n(x)-F^n(y)|\leq 2\ \forall n\in\mathbb{N}$ $\blacksquare$

Now, to show that the rotation numbers are equal (from definition), all we have to show is, for appropriate $x\in\mathbb{R}$ ($x=H(0)$ works); $$\lim_{n\rightarrow\infty}{\frac{|F^n(x)-x|}{n}} = \lim_{n\rightarrow\infty}{\frac{|G^n(x)-x|}{n}}=\lim_{n\rightarrow\infty}{\frac{|H\circ F^n\circ H^{-1}(x)-x|}{n}}$$ We simply do the following: $$||F^n(x)-x|-|H\circ F^n\circ H^{-1}(x)-x||\leq |F^n(x)-H\circ F^n\circ H^{-1}(x)|$$ and re-write RHS as follows: $$|F^n(x)-H\circ F^n\circ H^{-1}(x)|= |F^n(x)-F^n\circ H^{-1}(x)+F^n\circ H^{-1}(x)-H\circ F^n\circ H^{-1}(x)|$$ Now, clearly, we can bound $$|F^n\circ H^{-1}(x)-H\circ F^n\circ H^{-1}(x)|\leq 1 \ \text{[Lemma 2]}$$ $$|F^n(x)-F^n\circ H^{-1}(x)|\leq 2\ \text{[Lemma 2 and 3]}$$ And, by triangle inequality $$|F^n(x)-H\circ F^n\circ H^{-1}(x)|\leq 3$$ and, combining these inequalities, we have $$\lim_{n\rightarrow\infty}{\left\lvert\frac{|F^n(x)-x|}{n}-\frac{|H\circ F^n\circ H^{-1}(x)-x|}{n}\right\rvert}\leq \lim_{n\rightarrow\infty}{\frac{3}{n}}=0$$

Note that we needed $h$ only to satisfy Lemma 1 and 2; and we can, even in this proof, relax the conditions a bit. We can also try to relax Lemma 1 with a weaker requirement, if $h$ is only surjective.