# Contractibility of the space of Jordan curves

Is the space of Jordan curves in $$\textbf{R}^2$$ contractible? In other words, is there a canonical or continuous way to deform each Jordan curve to the unit circle $$\textbf{S}^1$$.

If the curves are smooth one can do this via curve shortening flow/rescaling. I think this approach even works for rectifiable curves, by a paper of Lauer. But I do not know a reference for the general topological case. In the smooth case, is there a way to do this without using flows?

By the space of Jordan curves here I mean one-to-one continuous maps from $$\textbf{S}^1$$ to $$\textbf{R}^2$$, modulo homeomorphisms of $$\textbf{S}^1$$. So two Jordan curves are close provided that they admit parameterizations which are pointwise close.

• Is there a solution in the space of simple polygons? I wasn't able to find anything conclusive online for simultaneous deformation of polygons. Jun 18 at 15:54
• @ Mikhail Katz: Curve shortening flow works for polygonal curves as well, and there are also other methods like the "Carpenter Rule Problem" which convexifies polygons in a canonical way. Jun 18 at 16:02
• Thanks! Do you have some references for curve-shortening flow for simple polygons? The texts I saw on line tend to be conjectural. Jun 18 at 16:04
• @ Mikhail Katz: I don't know the best reference for CSF for polygons, but Carpenter's rule problem has been well studied in the paper of Connelly, Demaine and Rote. Jun 18 at 16:12

It seems that the answer is "yes".

Let us identify $$\mathbb{R}^2$$ with $$\mathbb{S}^2\setminus\{n\}$$. Each Jordan curve $$\gamma$$ bounds a disc containing $$n$$. This disc admits a conformal parametrization by the unit disc $$\mathbb{D}$$ such that its center goes to $$n$$. This parametrization is unique up to rotation of $$\mathbb{D}$$. In particular, the image $$\gamma_r$$ of the circle of radius $$r$$ in $$\mathbb{D}$$ is completely determined by $$\gamma$$. Note that there is a homotopy from that sends $$\gamma$$ to $$\gamma_{1/2}$$. The continuity at $$t=1$$ follows from Thm 15 (VIII, §81) in "Automorphic Functions" by Lester R Ford.

So the question is reduced to the smooth case, and you know it already.

• Is the conformal parameterization always continuous up to the boundary? Jun 16 at 21:36
• @John Pardon: Yes, by Caratheodory's extension theorem. Jun 16 at 21:41
• Anton, Is there a solution in the space of simple polygons? I wasn't able to find anything conclusive online for simultaneous deformation of polygons. Jun 18 at 16:00
• @MikhailKatz good question. I do not see optimal formulation so far --- maybe one should use some kind of polyhedral homotopy (similar to def of tame knots). Jun 22 at 20:46

Anton's construction depends continuously on the curve but does not seem quite canonical, in the sense that smoothings of two congruent curves may not be congruent. We can ensure that this will be the case by refining the argument as follows.

Given a Jordan curve $$J$$ in $$\textbf{R}^2$$, let $$C$$ be the smallest circle which encloses it, and $$C_\epsilon$$ be the outer parallel circle at some fixed distance $$\epsilon>0$$. Then the region between $$J$$ and $$C_\epsilon$$ forms a topological annulus $$A$$. There exists a unique annulus $$A'$$ of the form $$r\leq |z|\leq 1$$ which is conformal to $$A$$ (e.g. see Thm 2.7.3 in Conformal Maps and Geometry by D. Beliaev). Let $$f\colon A'\to A$$ be a conformal map. Restricting $$f$$ to circles $$|z|=t$$ yields a homotopy between $$J$$ and $$C_\epsilon$$ which depends only on the congruence class of $$J$$ (the choice of $$f$$ does not matter because the only conformal maps $$A'\to A'$$ are rotations, or an inversion).

Addendum (8/8/2023): Igor Belegradek and I just finished a paper where we show that the space of Jordan domains in any surface of constant curvature admits a strong deformation retraction onto the space of round disks, and this deformation is equivariant under isometries of the surface. The proof is based on a point selection theorem from the interior of domains which allows us to apply Riemann's mapping theorem, or more precisely Caratheodory's kernel theorem, to a family of domains, as discussed in a related post.

• Do you have a reference (or an argument) for the fact that this procedure depends continuously on the curve? I could not extract this easily from Beliaev's notes. Jun 20 at 8:47
• @Pierre PC: Continuous dependence on prescribed boundary data follows from Caratheodory's kernel theorem, which gives conditions for local uniform convergence of conformal maps. See "Boundary Behavior of Conformal maps" by Pommerenke; specifically, Thm. 2.11 which Pommerenke attributes to Rado. This takes care of continuous dependence on a single boundary component. The case of multiple boundary components should be very similar, since the arguments are local. Jun 20 at 12:41