Let $f, g: \mathbb{S}^1 \rightarrow \mathbb{S}^1$ be orientation-preserving homeomorphisms. Show that if $f \circ g = g \circ f$, then $\rho(f \circ g) = \rho(f) + \rho(g)$ mod $1$.

My attempt:

Let $F \circ G$ be the lift of $f \circ g$. Then $\rho(F \circ G) = \lim_{n \to \infty} \frac{(F\circ G)^n(x) - x}{n} = \lim_{n \to \infty} \frac{(F^n\circ G^n)(x) - x}{n} = \lim_{n \to \infty} \frac{(F^n(G^n(x)) - x}{n} = \lim_{n \to \infty} \frac{(F^n(G^n(x)) - G^n(x) + G^n(x) - x}{n} = \lim_{n \to \infty} \frac{(F^n(G^n(x)) - G^n(x)}{n} + \lim_{n \to \infty} \frac{G^n(x) - x}{n} = \rho(F) + \rho(G)$

and by the definition of rotation number of lifts, the desired equality holds.

  • $\begingroup$ Btw, the only part I'm not sure about in my attempt is the first limit in the penultimate step. My argument is that the rotation number does not depend on the selected point, so even if we change it in every step of limit (I mean that it is different for every $n$), it should still converge to $\rho(F)$ $\endgroup$ – Pan Miroslav Oct 16 '17 at 18:30

What you conjecture is indeed true, but you need to prove it. Hint: use lifts. Namely, show that $$ \left|\frac{F^n(x)-x}{n}-\frac{F^n(y)-y}{n}\right|\le\frac{2k}n $$ whenever $|x-y|\le k$ for some integer $k$. What you want follows readily from this property.

  • $\begingroup$ I think this is true: $\left|\frac{F^n(x) - x}{n} - \frac{F^n(y) - y}{n}\right| = \left|\frac{F^n(x) - x}{n} + \frac{y - F^n(y)}{n}\right| \leq \left|\frac{F^n(x) - x}{n}\right| + \left| \frac{y - F^n(y)}{n}\right| \leq \frac{1}{n} + \frac{1}{n} = \frac{2}{n}$ no matter how close $x, y$ are (thanks to $f$ being homeomorphism). So even stronger property holds, but I don't know how does it help me. $\endgroup$ – Pan Miroslav Oct 17 '17 at 13:19
  • $\begingroup$ If it were like you say all rotations numbers would be zero, right? Do use what I suggested. $\endgroup$ – John B Oct 17 '17 at 13:25
  • $\begingroup$ Sorry, I misread something in my notes, it actually says this: if $f$ is homeomorphism with the degree $1$ (in other words, orientation-preserving homeomorphism - OPH) (so $f^n$ is OPH too), $F$ is a lift of $f$ (homeomorphism too) then for any $x, y$ we have that the distance between $F(x) - x, F(y) - y$ is lower than $1$ and the same holds for $F^n(x) - x, F^n(y) - y$. Then the given absolute value is actually lower or equal than $\frac{1}{n}$. $\endgroup$ – Pan Miroslav Oct 17 '17 at 13:39
  • $\begingroup$ If you say so. :D $\endgroup$ – John B Oct 17 '17 at 19:43
  • $\begingroup$ I will try to ask my professor tomorrow. Anyway, I accepted your answer, it definitely helped me a lot! $\endgroup$ – Pan Miroslav Oct 17 '17 at 20:24

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