5
$\begingroup$

Is there a way to prove the Riemann mapping theorem using the theory of harmonic maps (in the sense of "Harmonic Mappings of Riemannian Manifolds" by Eells and Sampson)?

$\endgroup$
  • 3
    $\begingroup$ Riemann's own (incomplete) proof did exactly that: en.wikipedia.org/wiki/Riemann_mapping_theorem#A_sketch_proof $\endgroup$ – Christian Remling Sep 15 '16 at 16:48
  • $\begingroup$ @Christian Remling. Thanks. Apparently Riemann was assuming smooth boundary, a hypothesis that I would like to avoid. Nonetheless it would be interesting to have a reference to a proof of the smooth version based on Riemann's original approach. $\endgroup$ – Chris Judge Sep 15 '16 at 17:35
  • 1
    $\begingroup$ Note also that in the case of a domain bounded by a continuous simple curve of finite length, the existence of a Riemann mapping continuous up to the boundary is a special case of Tibor Rado's solution of the Plateau problem via minimization of the Dirichlet integral over parametrizations. $\endgroup$ – Pietro Majer Sep 15 '16 at 21:36
9
$\begingroup$

Yes, there is a classical proof of the Riemann mapping theorem using harmonic maps and the Dirichlet problem. Riemann's original assumption of boundary smoothness can be removed using Perron's method and a simple argument due to Osgood.

For the detailed proof, see this note by Greene and Kim.

$\endgroup$
  • $\begingroup$ I did a search on "harmonic map" in the paper of Green and Kim and got no hits. By harmonic map I'm really thinking of, for example, Eeels and Sampson. However, this article by Greene and Kim is interesting. $\endgroup$ – Chris Judge Sep 16 '16 at 17:20
  • 2
    $\begingroup$ @ChrisJudge: Harmonic maps into $\mathbb{C}$ are just complex valued harmonic functions, since the metric $\mathrm{d}z\mathrm{d}\bar{z}$ is flat. That in the paper of Greene and Kim the mapping is found as a minimizer of the Dirichlet energy also should be a big hint that it is a harmonic map. $\endgroup$ – Willie Wong Sep 16 '16 at 18:28
  • $\begingroup$ @Willie Wong: There doesn't seem to be a target Riemannian metric on the disk being used. Riemann's trick reduces the target to 1-D and I was looking for a solution with a 2-D target. There's something unsatisfactory about depending on harmonic conjugate. $\endgroup$ – Chris Judge Sep 16 '16 at 21:15
  • $\begingroup$ @ChrisJudge I would be very surprised if you find a proof more "harmonic" than this. Of course, I leave it to you to decide whether it is what you're looking for or not. $\endgroup$ – Malik Younsi Sep 17 '16 at 3:53
  • $\begingroup$ @ChrisJudge: Okay, I see what you mean. In that case I would suggest editing your original post to stress this point. $\endgroup$ – Willie Wong Sep 19 '16 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.