# 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

## 1 Answer

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.

Furthermore, there are numerical methods to compute the curve from its welding homeomorphism. Probably the most efficient one is Marshall's Zipper algorithm, see his paper "Conformal Welding for Finitely Connected Regions". There is also Sharon and Mumford's paper "2d-shape analysis using conformal mapping".

If you're interested in conformal welding, a good survey is the one by Hamilton MR1966191 (2005e:30012) Hamilton, D. H.(1-MD) Conformal welding. Handbook of complex analysis: geometric function theory, Vol. 1, 137–146, North-Holland, Amsterdam, 2002. 30C35

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.