Question

I know the definition of k -quasi-symmetric maps $f$ on $R$,it is

there exists $k>0$ such that $\frac{1}{k}\leq\frac{f(x+t)-f(x)}{f(x)-f(x-t)} \leq k \forall x,t\in R.$

So I just want to double check the definition of the same for circle, since I was not able to find a specific definition :

Can I say $h: S^1\to S^1$ is k -quasisymmtric, if any lift $\tilde{h}: R\to R$ of $h$ is k -quasisymmtric according to the definition of a k-q.s. map$:R\to R$. This dfinition does not dpend on which lift I choose.

Answer 1

You are right: the exact same definition is true for the circle. There is no need to lift your map. So, a homeomorphism $h:S^{1} \rightarrow S^{1}$ is k-quasisymmetric if, for any two intervals $I_{1}$, $I_{2}$ with a common endpoint and having same length ($\vert I_{1}\vert=\vert I_{2} \vert$), we have the following: $$ \frac{\vert h(I_{1}) \vert}{\vert h(I_{2}) \vert} \leq k $$

You can find such a definition for example on the online Encyclopedia of Math. (http://eom.springer.de)