Analytic diffeomorphisms of the circle from complex domains

Let $\gamma \subset \mathbb C$ be a simple closed analytic curve and let $\Delta$ be the closure of the disk it bounds. The Riemann mapping theorem gives two biholomorphisms: $$\phi : (D^2,S^1) \to (\Delta,\gamma)$$ and $$\psi : (\mathbb{CP}^1 - \text{Int}\,\Delta,\gamma) \to (D^2,S^1)\,,$$ where $\text{Int}$ means interior and $D^2 \subset \mathbb C$ is the closed unit disk. We have the induced maps $$\alpha:= \phi|_{S^1}\,,\quad \beta:=\psi|_\gamma\,.$$ Let $$\Gamma(\gamma):=\overline \beta \circ \alpha : S^1 \to S^1$$ where the bar denotes complex conjugation. This $\Gamma(\gamma)$ is an orientation-preserving analytic diffeomorphism of $S^1$. Moreover, $\alpha$ and $\beta$ are well-defined up to precomposition and postcomposition with elements of the biholomorphism group $G$ of $(D^2,S^1)$, restricted to $S^1$, respectively, which implies that $\Gamma(\gamma)$ is also well-defined up to pre- and postcomposition with elements of $G$, since conjugation can be moved inside an automorphism of $D^2$ by changing the automorphism. Therefore the double coset of $\Gamma(\gamma)$ in the analytic diffeomorphism group of $S^1$ by $G$ is well-defined.

For instance if $\gamma = S^1$, this is the double coset of the identity. Probably this can also be computed for ellipses in terms of elliptic functions.

Question: What can be said about this double coset in general? Can the curve $\gamma$ be reconstructed from it? What are its dynamical properties?

I have to admit that I'm asking this out of sheer curiosity, although this does tie with a line of research I tried not so long ago.

• MR0902292 Kirillov, A. A. Kähler structure on the K-orbits of a group of diffeomorphisms of the circle. – Alexandre Eremenko Jan 12 '15 at 23:57

This is the so-called conformal welding problem. One can ask the same question for any Jordan curve $\gamma$ (non necessarily analytic). With this domain of definition, your map $\Gamma$ is well-known to be neither injective nor surjective. There are even orientation-preserving homeomorphisms of the circle analytic everywhere except at one point that are not the welding homeomorphism of any Jordan curve. However, the image of your map $\Gamma$ contains all quasisymmetric orientation-preserving homeomorphisms of the circle, this is sometimes referred to as the fundamental theorem of conformal welding and it was first proved by Pfluger in 1960. A simpler proof was later given by Lehto and Virtanen.
In general, it is difficult to explicitely reconstruct the curve $\gamma$ from the welding homeomorphism. In some cases though, the associated curve $\gamma$ has a special form. For instance, if the homeomorphism is the $n$-th root of a Blaschke product of degree $n$, then the corresponding curve $\gamma$ is a proper polynomial lemniscate of the same degree. Conversely, the welding homeomorphism of a proper polynomial lemniscate of degree $n$ is a $n$-th root of a Blaschke product. This was proved by Ebenfelt, Khavinson and Shapiro in the paper "Two-dimensional shapes and lemniscates", arXiv:1003.4567. See also arXiv:1406.3545 for a simpler proof and a generalization to rational lemniscates.
EDIT Perhaps I should add more details to what I mean exactly by the fact that the map $\Gamma$ is in general not injective. It is easy to see that if $T$ is a Möbius transformation, then $\gamma$ and $T(\gamma)$ have the same welding homeomorphism. The map $\Gamma$ is not injective even modulo Möbius transformations. The easiest way to see this is to consider a curve $\gamma$ of positive area and use the measurable Riemann mapping theorem to obtain an infinite-dimensional family of homeomorphisms of the sphere conformal outside $\gamma$. If $f$ is any such map, then it is easy to see that $\gamma$ and $f(\gamma)$ give rise to same welding homeomorphism. However, a dimension argument shows that the image of $\gamma$ under such a map $f$ cannot be always Möbius-equivalent to $\gamma$.
A sufficient condition for the uniqueness of the curve $\gamma$ from its welding homeomorphism is if $\gamma$ is conformally removable, i.e. if every homeomorphism of the sphere conformal outside $\gamma$ is a Möbius transformations.