Talk:Riemann mapping theorem
|WikiProject Mathematics||(Rated Start-class, Mid-importance)|
Clarification for fractal boundaries
The article states that the theorem holds for domains with fractal (non-differentiable) boundaries. An appeal is made to the Dirichlet principle, but its then hinted that the Dirichlet principle is suspect. Thus, the proof for the general case of a fractal boundary seems uncertain ... can the status be clarified? linas 03:28, 26 August 2006 (UTC)
- It seems like Riemann's proof used the Dirichlet principle, which as far as I can tell is valid in only a subset of the cases which the theorem is actually valid (Hilbert claimed I think that it was valid for the case Riemann was considering). The later proofs seem to have used a different method which doesn't have this frailty.--22.214.171.124 (talk) 18:55, 29 October 2010 (UTC)
Why is this theorem impressive?
The title of this section doesn't seem very encyclopedic to me; in mathematics it is pretty common that people say things like this, but it is not the business of an encyclopedia to say what is "impressive" and what isn't. (And speaking just for myself, I don't like it when a written reference tries to sell me on the beauty of something by making explicit aesthetic claims about it. The math can speak for itself.) Everything in the section seems very relevant to the entry and the appreciation of the Riemann mapping theorem, but perhaps it should be under a more neutral title. 126.96.36.199 (talk) 07:10, 25 March 2009 (UTC)
I completely disagree. I saw the section title, instantly wanted to know what the reasons would be, and learned some very interesting things that I had not known. With a more neutral section title I would probably have missed this. —Preceding unsigned comment added by 188.8.131.52 (talk) 16:05, 18 May 2010 (UTC)
Doubly connected domain
Maybe this isn't the best place to ask this, but the phrase "doubly connected domain" is used here without comment, it's not a phrase I'm familiar with, and I can't find it defined properly anywhere. I found one paper by Osserman and Schiffer that defines a "double connected minimal surface" in R^2 to be the image of an annulus under a conformal mapping. Is this the most general definition for "doubly connected"? 184.108.40.206 (talk) 04:12, 11 June 2009 (UTC)
- In the complex plane, a doubly connected set is one that is topologically equivalent to an annulus. That is my understanding of it anyway.--220.127.116.11 (talk) 19:02, 29 October 2010 (UTC)
How many Riemann mappings are there?
I have heard that, the set of Riemann mappings between a certain shape and the unit disc has a dimension of 3 (i.e. every mapping in this set can be defined by 3 parameters). But I don't have references. Can you please add some information about this? --Erel Segal (talk) 18:24, 27 June 2013 (UTC)
- Suppose that U is a non-empty simply connected open subset of C (as in the hypotheses of the Riemann mapping theorem). Let's say that we have one Riemann mapping f : U → D. How do we find others? One obvious technique is to post-compose f with an automorphism of D (a biholomorphism D → D). This determines another Riemann mapping of U. So there are at least as many Riemann mappings as there are biholomorphisms of the disk.
- Suppose that we had a second Riemann mapping g. Then by composing f and g−1 we get a biholomorphism of the disk. Therefore there are no more Riemann mappings than there are biholomorphisms of the disk.
- This reduces the problem to understanding biholomorphisms of the disk. This can be done using Möbius transformations and the Schwarz lemma. Suppose that we pick three points ζ1, ζ2, ζ3 on the unit circle and three other points ξ1, ξ2, ξ3 on the unit circle. As long as the ζ's and ξ's share the same betweenness relations, there is a unique Möbius transformation that sends ζi to ξi for all i. So there is at least a three parameter family of biholomorphisms of the disk. Now you need to ask whether there are any biholomorphisms that do not come from Möbius transformations. The Schwarz lemma tells you that the answer is no: After composing with a suitable Möbius transformation, the biholomorphism sends zero to zero; then, by conjugating by an appropriate Möbius transformation, the derivative at zero must have unit length; but then the Schwarz lemma says that the function is multiplication by a constant. Ozob (talk) 04:09, 28 June 2013 (UTC)