Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In dimension 2 we know by the Riemann mapping theorem that any simply connected domain ( $\neq \mathbb{R}^{2}$) can be mapped bijectively to the unit disk with a function that preserves angles between curves, ie is conformal.

I have read the claim that conformal maps in higher dimensions are pretty boring but does anyone know a proof or even a intuitive argument that conformal maps in higher dimensions are trivial?

share|improve this question

4 Answers 4

up vote 22 down vote accepted

I think you're looking for Liouville's theorem. This theorem states that for $n >2$, if $V_1,V_2 \subset \mathbb{R}^n$ are open subsets and $f : V_1 \rightarrow V_2$ is a smooth conformal map, then $f$ is the restriction of a higher-dimensional analogue of a Mobius transformation.

By the way, observe that there are no assumptions on the topology of the $V_i$ -- they don't have to be simply-connected, etc.


EDIT : I'm updating this ancient answer to link to a blog post by Danny Calegari which contains a sketch of a beautifully geometric argument for Liouville's theorem.

share|improve this answer
    
Andy, this is a precise answer! I just wonder what is the underlying intuitive argument... Can we prove this geometrically, or we need to use analysis? –  Dmitri Dec 29 '09 at 21:37
    
Thank you, yes that is the claim I have seen before, though the name was not mentioned and I do not think any smoothness condition was mentioned. Unfortunately the claim does not seem to have as simple a proof as I hoped. –  Johan Dec 29 '09 at 21:38
    
I believe the smoothness condition can be relaxed. Wikipedia claims that the result holds for maps which are only weakly differentiable, but the only proof I've read required a certain amount of differentiability -- you might want to check the references it gives. I recall that M. Berger's book "Geometry" contains a proof which is reasonably geometric, though since the assumptions of the theorem are infinitesimal there is some analysis needed. –  Andy Putman Dec 29 '09 at 21:52

Because the Riemann mapping theorem does not hold in higher dimensions. While there are all sorts of conformal mappings in dimension 2, for higher dimensions Liouville's theorem restricts all possible conformal mappings to the ones that are compositions of similarities, translations, orthogonal transformations and inversions. In generality there are contractible spaces not homeomorphic (therefore not conformal) to $\mathbb{R}^n$ such as Whitehead continuum

As for the proof of Liouville's theorem, maybe this article is of interest where you can see a sketch of Nevanlinna's original proof and a proof by nonstandard analysis.

share|improve this answer

There is a proof of Liouville's theorem by Charles Frances with less computations than most others, and carrying some intuition. However it is restricted to real-analytic transforms, and it is written in French. It has been published in "L'enseignement mathématique", and is available there: http://www.math.u-psud.fr/~frances/liouville2.pdf

share|improve this answer

I think this is a good reference for it. Iwaniec, Tadeusz; Martin, Gaven Geometric function theory and non-linear analysis. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2001. xvi+552.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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