## Why are there no wild arcs in the plane?

On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can extend your Jordan arc to a Jordan curve. That is, as long as you can extend an embedding of an interval to an embedding of a circle. However, I have to admit I don't readily see how to do this, and there are examples which show this is a subtle question. E.g. you can take an arc which spirals infinitely into a point. So my question is how one can show an embedding of an arc into $\mathbb R^2$ extends to an embedding of a circle into $\mathbb R^2$, or, failing that, if someone knows some other proof that all planar Jordan arcs are standard.

Edit: I want to highlight Bill Thurston's elegant answer (in a comment) to the question of whether you can extend an arc to a circle, even though I accepted his other answer using Caratheodory's theorem. Namely, assume your arc runs from $0$ to $\infty$ in $\mathbb C\cup{\infty}$. Then take the pre-image under the double branched cover $z\mapsto z^2$. The original arc can be identified with one of its two preimages, while the other preimage fills it in to a circle. Then one can apply the Schönflies theorem.

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One way to show all arcs are tame is to apply the Riemann mapping theorem to the complement of the arc on $S^2$. Caratheodory proved that whenever the complement of a simply-connected domain is locally connected, then the Riemann mapping extends to a continuous map of the disk to the plane. One half of the disk parametrizes the arc.

Another (related) method is to take the stereographic projection of the arc to a sphere in $\mathbb E^3$, and form its convex hull. If the interior of the ball is interpreted as the projective model for hyperbolic 3-space, then the boundary of the convex hull intersect the interior of the ball is a developable surface, i.e. its path metric (that is, distance between points is the minimum arc length of a path connecting them) is a complete metric locally isometric to the path-metric of the hyperbolic plane. Since it's simply-connected, it's globally isometric to the hyperbolic plane. For reasons parallel to Caratheodory's the map extends continuously to a map of the closed disk (2-dimensional projective disk model of the hyperbolic plane, together with its circle at infinity) to the closed ball.

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Is there a more direct relation between these two proofs, other than that they end in a similar way? E.g., is there a correspondence or a relation between the conformal structure on the Jordan domain at infinity, and the hyperbolic structure on the pleated surface? – Greg Kuperberg Mar 22 2011 at 7:46
Dennis Sullivan found a simple way to construct a map from the boundary of the convex hull of a simply-connected domain to the Poincaré metric on a simply-connected domain that has a bounded bi-Lipschitz constant. David Epstein and Al Marden subsequently gave a very detailed and explicit construction, with an explicit constant. The natural conjecture (or at least a conjecture that is attributed to me) was that the best quasiconformal constant should be 2, but this was disproved: cf. Epstein, Marden and Markovic, Annals of Math 159 (2004) pp 305--336. – Bill Thurston Mar 24 2011 at 1:29

By stereographic projection we can assume that the given Jordan arc lies on a sphere, and that its two ends are at opposite poles, $N$ and $S$. Now project the arc onto a cylinder that touches the sphere at the equator, so that $N$ and $S$ go to infinity on opposite ends of the cylinder.

Unrolling the cylinder, we now have our arc on a strip of the plane, with the ends of the arc at opposite ends of the strip. Now we can inverse stereographically project the arc back onto the sphere, so that both ends go to the point of projection. We now have a closed Jordan curve on the sphere, and we can apply the Schoenflies theorem.

Edit. Ryan Budney has pointed out the flaw in this argument, so I withdraw it as it stands. Thanks for the amendment, Ryan.

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How do you know the arc has ends? – Ryan Budney Mar 8 2011 at 4:14
Based on the stackexchange question, I also assumed the arc is a compact one. – Richard Kent Mar 8 2011 at 4:19
@Richard: in case your response is aimed at me, what I mean is in Stillwell's argument "our arc on a strip in the plane" is an arc in the universal cover of a twice punctured sphere -- this cover is homeomorphic to $\mathbb R^2$, so showing the arc has endpoints on the boundary is a non-trivial problem (it's easy to construct ones that don't, even if they come from embedded intervals in $S^2$ via this construction). – Ryan Budney Mar 8 2011 at 4:24
Oh, I think I see how one can tweak Stillwell's argument to make it work -- the lifted arc is in $(-1,1) \times \mathbb R$ which you could compactify to $[-1,1] \times [-\infty,\infty]$. Crush $\{-1\}\times [-\infty,\infty]$ and $\{1\}\times [-\infty,\infty]$ to points and identify the quotient space with a compact disc -- the arc has ends in this quotient space. – Ryan Budney Mar 8 2011 at 4:33
Actually, if the arc goes from $0$ to $\infty$ in $\hat \mathbb C$, then just take the preimage under $z \mapsto z^2$: it's a circle. It's nice, but it reduces the problem to another problem that ab initio is just as hard, just better known. – Bill Thurston Mar 8 2011 at 4:46