Rotation number of inverse maps on the circle.

Question

I'm still a bit lost in my studies of rotation numbers. Any help is much appreciated!

Let's say we have a homeomorphism $F: \mathbb{R} \rightarrow \mathbb{R}$ which is a lift of a homeomorphism $f:S^1 \rightarrow S^1$ of the circle. The homeo $f$ is assumed to be orientation preserving, i.e. $F(x+1) = F(x) +1$ for all $x \in \mathbb{R}$.

The rotation number $$ \rho(F,x) = \lim_{n \rightarrow \infty} \frac{F^n(x) - x}{n} $$ exists for every $x \in \mathbb{R}$ and is constant, i.e. $\rho(F,x) = \rho(F,y)$ for all $x,y \in \mathbb{R}$. Let $F^{-1}$ be the inverse of $F$. I know that $\rho(F^{-1},x)$ also exists for every $x \in \mathbb{R}$. What I want to show now is that $$ \rho(F,x) + \rho(F^{-1},x) = 0 \quad\text{for all } x \in \mathbb{R}. $$ Somehow I am still stuck. What I managed so far was to calculate $$ \frac{F^n(x) - x}{n} = -\frac{x - F^n(x)}{n} = -\frac{F^{-n} \circ F^n(x) - F^n(x)}{n}. $$ This almost looks like a solution to me since if I can show that if the right hand side $\frac{F^{-n} \circ F^n(x) - F^n(x)}{n}$ converges to $\rho(F^{-1},x)$ for $n \rightarrow \infty$, I am done. I know already that this term is convergent because the left hand side is convergent. I also now that for every fixed $k \in \mathbb{N}$ the term $$ \frac{F^{-n} \circ F^k(x) - F^k(x)}{n} \quad\text{converges to } \rho(F^{-1}) \text{ for } n \rightarrow \infty. $$

Meh, I'm lost. Sorry if this is a stupid question.

Answer 1

Intuitively, $F$ is a homeomorphism so $F^{-1}$ should exist and it should "rotate" in the opposite direction of $F$.

Can you possibly show $\rho(G \circ F,x) = \rho(F,x) + \rho(G, F(x)) $ ?

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Maybe easier, let $y_n = F^{-n}(x)$ since $F$ is a homeomorphism there is only one point like this:

$$ \frac{ F^{-n}(x) - x} {n} = \frac{ y_n - F^n(y_n)}{n} \to \rho(F, y_n)$$

Unfortunately, you don't know much about except possibly: $y_n \approx x - n \rho(F^{-1}, x)$.

Maybe you can get uniform convergence to $\rho(F, \cdot)$ since $S^1$ is compact?

Answer 2

Lemma: If $f,g:S^1\to S^1$ are two orientation-preserving homeomorphisms and $f\circ g=g\circ f$, then $\rho(f\circ g)=\rho(f)+ \rho(g) (\mod 1)$ (This is an exercise from Barreira & Valls' Dynamical Systems: an Introduction, p. 84).

Proof of Lemma: Since $f,g$ are orientation-preserving, there are lifts $F,G:\mathbb{R}\to\mathbb{R}:F \uparrow,G\uparrow$ of $f,g$, respectively. Since $f,g$ commute, so do $F,G$: Let $\pi:\mathbb{R}\to S^1$ be the projection. Then $F\circ G=\pi^{-1}\circ f\circ \pi \circ \pi^{-1} \circ g\circ \pi=\pi^{-1} \circ f\circ g \circ \pi=\pi^{-1}\circ g\circ f\circ \pi=G\circ F$.

$F\circ G$ is a lift of $f\circ g$ and it is increasing. Hence $\rho(F\circ G)$ is defined. Then $\rho(F\circ G)=\lim_{n\to\infty}\dfrac{(F\circ G)^n(x)-x}{n}= \lim_{n\to\infty}\left(\dfrac{F^n(G^n(x))-G^n(x)}{n}+\dfrac{G^n(x)-x}{n}\right)= \rho(F)+\rho(G)$ since the used limits exist independently of $x\in\mathbb{R}$. Also note that in the second equality we use the fact that $F,G$ commute.

Finally since by definition $\rho(f)=\pi(\rho(F))$, modding the acquired equality by $1$, we are done (Admittedly we won't be using the complete form of the lemma, but I figured I should record it completely for future reference.).

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Turning back to your question, we have an orientation-preserving homeomorphism $f:S^1\to S^1$. Let $F$ be an increasing lift of $f$. Then we have $f\circ f^{-1}=1_{S^1}=f^{-1}\circ f$ and $1_{\mathbb{R}}$ is a lift of $1_{S^1}$. As $f$ and its inverse commute, we may employ the above lemma. Noting that $\rho(1_{\mathbb{R}})=0$, as (again) the limit in the definition of the rotation number is independent of $x$, the result follows.