0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Even if $G$ does not commute with $F$? $\endgroup$ – Tomás Jun 18 '14 at 13:08
2
$\begingroup$

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 $\rho(F) + \rho(G) = 0 + 0 = 0$.

$\endgroup$
  • $\begingroup$ Thank you, your example is very helpful indeed! Any idea if it is true if $F$ and $G$ commute? $\endgroup$ – wabu wabu rabu Jun 18 '14 at 13:34
  • $\begingroup$ See the answer of Tomas. $\endgroup$ – Lee Mosher Jun 18 '14 at 14:28
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ okay, I see that the first term on the right hand side converges. But why is $\lim_{n \mapsto \infty} \frac{G^n \circ F^n (x) - F^n(x)}{n} = \rho(G)$? $\endgroup$ – wabu wabu rabu Jun 23 '14 at 8:47
0
$\begingroup$

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: $$\rho(GoF)=\rho(G)+\rho(F).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.