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This question concerns a previous question, Rotation number of inverse maps on the circle. in which all the terminology and notation used below is defined.

The question is given the rotation number $\rho (f)$, calculate the rotation number $\rho (f^{-1})$. All proofs I have found of this statement calculate $\rho(F \circ F^{-1})$ (for $F, F^{-1}$ lifts of $f, f^{-1}$) and end up with something like:

$\rho(F\circ G)=\lim_{n\to\infty}\dfrac{(F\circ G)^n(x)-x}{n}= \lim_{n\to\infty}\left(\dfrac{F^n(G^n(x))-G^n(x)}{n}+\dfrac{G^n(x)-x}{n}\right)=\rho(F)+\rho(G)$

For $F, G$ lifts of commuting circle homeomorphisms $f, g$, and the proof ends by letting $G = F^{-1}$. But in the definition of rotation number for $f$, i.e. $\rho(F) = \lim_{n\to\infty}\dfrac{F^n(x)-x}{n} = \lim_{n\to\infty}\dfrac{F^n(x)}{n}$, if we let $x = F^{-n}(y)$ for some $y$ as above we get that the rotation number is $\rho(F) = \lim_{n\to\infty}\dfrac{y}{n} = 0$ for any $f$ (note that the rotation number is always defined and has the same value for any $x$ in the domain of $F$).

So my question is are we allowed to use a function $G^n(x)$ in place of a variable $x$ in the definition of rotation number as the proofs commonly do, or are these proofs incorrect?

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You are pretty confused. They are used as such in the definition. When you substitute $x = F^{n}(y)$, it changes to $y - F^{-n}y$. Since $f$ and $g$ commutes, their lifts too. Only that property is being exploited here.

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  • $\begingroup$ I agree I'm confused, but what about when you use the equivalent definition of the rotation number without the $-x$ term in the numerator? The $F^n(x)$ and $x = F^{-n}(y)$ cancel each other out to get $y / n$ as $n \to \infty$ surely? The reason I ask is my course uses the definition $\rho(F) = \lim_{n\to\infty}\dfrac{F^n(x)}{n}$, without a $-x$ in the numerator. $\endgroup$ – D. P Apr 28 '16 at 15:36
  • $\begingroup$ @D.P Let $f$ be the anticlockwise rotation of $\Bbb{S}^1$ with angle $2\pi \theta$. Then what do you think the rotation number is ? $\endgroup$ – Mambo Apr 28 '16 at 19:25
  • $\begingroup$ $f(x) = x + 2\pi\theta \mod 1$, so lift $F(x) = x + 2\pi\theta$ and $F^n(x) = x+ 2n\pi\theta$ giving $\rho(F) = 2\pi\theta$, so $\rho(f) = 2\pi\theta \mod 1$. $\endgroup$ – D. P Apr 28 '16 at 19:35
  • $\begingroup$ If $f(z) = e^{2\pi i \theta }z$, then $F(x) = x + \theta$ will be lift and $F^n(x) = x + n\theta$. Then $\rho(f) = \theta \mod 1$. How is this possible by your definition? $\endgroup$ – Mambo Apr 28 '16 at 20:01
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    $\begingroup$ Ignore everything I have written on this comment section, I have finally realised that I am being dumb and the counterexample is wrong because $y = F^n(x)$ so $\dfrac{y}{n} \to \rho(F)$. $\endgroup$ – D. P May 1 '16 at 17:14

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