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How do you prove to any two simply-connected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice.

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Such a proof would certainly need the Jordan curve theorem (at least for smooth simple plane curves), and you might not call that elementary. –  John Pardon May 26 '11 at 13:12

2 Answers 2

Here's how one proof goes. I'm omitting some details, at least for now.

If $U$ is a simply-connected domain in the plane $\mathbb C$, then define the function $r: U \to \mathbb R_+$ to be the maximum radius of a (round) open disk in the standard Euclidean metric whose interior is contained in $U$. If $r(z) = \infty$ for any $z$, then $U$ is the entire plane, so there is nothing to prove. Otherwise, define a new Riemannian metric using the conformal factor $1/r$, that is, $ds' = 1/r\ ds$ where $ds$ is arc length in the standard metric.

Note that if $U$ is the upper half-plane, the metric coincides with the hyperbolic metric. If $U = \mathbb C \setminus \{0\}$, then all complex linear automorphisms of $\mathbb C$ preserve the metric, and the metric is that of a cylinder that is parametrized isometrically by $\log(z) / \left \langle 2 \pi i \right \rangle$. If $U$ is a round disk, then $ds'$ is a smooth negatively-curved metric except at the center of the disk, where there is a non-smooth point, but no cone angle: if the disk has radius 1, then a circle about the center of radius $\epsilon$ has length $2 \pi \epsilon / (1-\epsilon)$

In general, although the metric $ds'$ need not be smooth, it always has non-positive curvature. Intuitively: the $1/r$ factor means it takes infinite arc length to reach the boundary, and shortest $ds'$ geodesics try to thread their way into any bays and inlets of $U$ keeping far from the shoreline, since the speed limit is drastically reduced near the shore. In particular, there is a unique $ds'$ geodesic between any two points in $U$, and geodesics have the unique continuation property, they are determined by the tangent vector at the beginning and the length.

To parametrize $U$ by $\mathbb R^2$, choose any point $z_0$ in $U$. The tangent space to $U$ at $z_0$ parametrizes $U$, by $V$ goes to the geodesic through $V$ whose length is the length of $V$.

Sorry for leaving off details. Somewhere else on MO I believe I posted an alternate way to do this, using the convex hull of $S^2 \setminus U'$, where $U'$ is the stereographic image of $U$ on a sphere; in the projective hyperbolic metric, this boundary of the convex hull always is isometric to the hyperbolic plane, from which a proof is easy.

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Since you ask for an elementary argument, let's assume that the domains have polygonal boundary. Then, the solution to the "carpenter's rule problem" (by Connelly-Demaine-Rote and Streinu) gives an actual algorithm for transfroming the domains into convex domains, for which any number of arguments work. (the carpenter's rule problem is described here:


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