Finding a homeomorphism guaranteed by Schoenflies Theorem

Question

Assume I have a Jordan curve $C \subset \mathbb{R}^2$. Then by Schoenflies Theorem there exists a homeomorphism $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(C)$ is the unit circle. Is there a simple way to go about finding one given a parametrized curve in the plane?

Answer 1

There are constructive versions of the Schoenflies theorem, but of course you have to be more precise about your input data.

For example, if you give a piecewise-linear simple closed curve in the plane, then there's a constructive version of the Riemann Mapping Theorem that applies. I believe the proof is fairly old. I'd look in Stephenson's Circle Packings book: http://www.math.utk.edu/~kens/

If your curve is smooth, then there's a less finitary but still fairly explicit way to construct the diffeomorphism. The idea would be to use the Gage-Hamilton curve straightening differential equation. This will turn your smooth curve into the round circle. The isotopy extension theorem would then provide your diffeomorphism.

Of course, there's Alexander's proof of the theorem. It's yet less explicit but it works quite well. The idea is that your curve in the plane divides the plane into an "inside" and "outside". A general linear function $f : \mathbb R^2 \to \mathbb R$ is a Morse function on the "inside" region. If all the critical points are at different "heights" this Morse function divides the "inside" into combinatorial discs, glued together along arcs in their boundary. So there's an inductive step where you have to prove that two discs glued together along arcs in their boundary is a disc. The problem with this is that you've got to use this inductive step many times, so you see the actual map you get may be quite complicated.