Hi I'm trying to probe for an exercise that given $f,g \in Hom_+(S^1)$ such that f and g commute i.e: $f \circ g = g \circ f$ that the rotation number $\rho (f \circ g) = \rho(f) + \rho(g)$ where the rotation number is defined as:
$\rho(F)=lim_{n\rightarrow \infty} \frac{F^n(x)-x}{n}$ and $\rho(f)=\rho(F) \bmod 1$
Here F is a lifting of f meaning that the projection $\pi(F(x)) = f(\pi(x))$ and $\pi(x)=x \bmod 1$.
While trying to probe this I proved that $F^n \circ G^n$ is a lift of $f \circ g$ and that $F^n \circ G^n - G^n \circ F^n=k \in Z$
I also proved that given $ \| x - y \| < k$ then $\| F^n(x) - F^n(y) \| < k$.
So trying to do the final step I reach
$\rho(F \circ G)=lim_{n\rightarrow \infty} \frac{F^n(G^n(x))-x}{n}=lim_{n\rightarrow \infty} \frac{F^n(G^n(x))-G^n(x)}{n}+\frac{G^n(x)-x}{n}=\rho(G)+lim_{n\rightarrow \infty} \frac{F^n(G^n(x))-G^n(x)}{n}$
so I suppose I need to probe $lim_{n\rightarrow \infty} \frac{F^n(G^n(x))-G^n(x)}{n}=\rho(F)$
I tried to reach the result by both sides, the bound didn't help me much as to probe it both x and y are fixed, so it means that I need to attach it to a fixed point and I endup getting n again:
Suppose that $\| G(x) - x\| < k$ then $\| F^n(G^n(x))-F^n(x) \| \leq \| F^n(G^n(x))-F^n(G^{n-1}(x)) \| + ... + \| F^n(G(x))-x \|=nk$
so by taking the limit $\| F^n(G^n(x))-F^n(x) \| \rightarrow s \leq k$ which is probably far from being 0.
On the other path using the way that the liftings commute I did get:
$\rho(F \circ G)=lim_{n\rightarrow \infty} \frac{G^n(F^n(x))-x}{n}=lim_{n\rightarrow \infty} \frac{G^n(F^n(x))-F^n(x)}{n}+\frac{F^n(x)-x}{n}=\rho(F)+lim_{n\rightarrow \infty} \frac{G^n(F^n(x))-F^n(x)}{n}=\rho(G)+lim_{n\rightarrow \infty} \frac{F^n(G^n(x))-G^n(x)}{n}$
the last part given that $F^n \circ G^n - G^n \circ F^n=k \in Z$ constant so it converges to 0.
I'm stuck, I do not know how to move forward on any of those paths.