# A question about the rotation number of homeomorphisms of the circle

Let $f: S^1 \rightarrow S^1$ be an orientation-preserving homeomorphism of the circle and let $F: \mathbb{R} \rightarrow \mathbb{R}$ be any lift of $f$. Usually one defines the rotation number $\rho(x,F)$ of $F$ for a point $x \in \mathbb{R}$ as the limit $$\rho(x,F) = \lim_{n \mapsto \infty} \frac{F^n(x) - x}{n}.$$ This limit exists always and is independent of the point $x$. Hence we can just write $\rho(F)$. It makes sense to me that the rotation number respects the composition of homeomorphisms, i.e. $$\rho(G \circ F) = \rho(G) + \rho(F),$$ but I seem to be unable to prove it. Is it true at all? Any hints on how to prove it?

• Even if $G$ does not commute with $F$? – Tomás Jun 18 '14 at 13:08

It is not true.

Let the universal covering map $\mathbb{R} \to S^1$ be $t \mapsto e^{2 \pi i t}$.

Let $f : S^1 \to S^1$ fix $1+0i$, and let $f$ act on $S^1 - \{1+0i\}$ by moving each point a little bit in the counterclockwise direction. Lift $f$ to $F$ so that it fixes each integer $n$ and displaces every other point a little bit to the right.

Let $g : S^1 \to S^1$ fix $-1+0i$ and let $g$ act on $S^1 - \{-1+0i\}$ by moving each point a little bit in the counterclockwise direction. Lift $g$ to $G$ so that it fixes each half integer $n+\frac{1}{2}$ and displaces every other point a little bit to the right.

Then $F \circ G$ displaces each point to the right by an amount bounded away from zero and so $\rho(F \circ G) > 0$. But $\rho(F) + \rho(G) = 0 + 0 = 0$.

• Thank you, your example is very helpful indeed! Any idea if it is true if $F$ and $G$ commute? – wabu wabu rabu Jun 18 '14 at 13:34
• See the answer of Tomas. – Lee Mosher Jun 18 '14 at 14:28

If $F$ and $G$ commutes then, the result is true. Indeed, note that

$$\frac{(G\circ F)^n(x)-x}{n}=\frac{G^n\circ F^n(x)-F^n(x)}{n}+\frac{F^n(x)-x}{n}$$

• okay, I see that the first term on the right hand side converges. But why is $\lim_{n \mapsto \infty} \frac{G^n \circ F^n (x) - F^n(x)}{n} = \rho(G)$? – wabu wabu rabu Jun 23 '14 at 8:47

We define the mean rotation vector of F is an element of $R^{2}$ by : $$\rho_{\mu}(F) = \int \rho(F,x) d\mu$$ This $\rho_{\mu}(F)$ in $R^{1}$ is equal to $\rho(F)$ ; thus: $$\rho_{\mu}(GoF)=\int \rho(GoF,x) d\mu = \int \lim_{n} {(GoF)^{n}(x)-x \over n} d\mu$$ $$=\int \lim_{n} {G^{n}(F^{n}(x))-x \over n} d\mu =\int \lim_{n} {G^{n}(F^{n}(x))-F^{n}(x) + F^{n}(x) - x \over n} d\mu$$ $$=\int \lim_{n} {G^{n}(F^{n}(x))-F^{n}(x)\over n}d\mu + \int \lim_{n}{F^{n}(x) - x \over n} d\mu$$ $$=\int \rho(G,x) d\mu + \int \rho(F,x) d\mu$$ $$=\rho_{\mu}(G) + \rho_{\mu}(F)$$ Thus in $R^{1}$ , we have: $$\rho(GoF)=\rho(G)+\rho(F).$$