20
$\begingroup$

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?

$\endgroup$

4 Answers 4

32
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ Commented Dec 29, 2009 at 21:37
  • $\begingroup$ 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. $\endgroup$
    – Johan
    Commented Dec 29, 2009 at 21:38
  • $\begingroup$ 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. $\endgroup$ Commented Dec 29, 2009 at 21:52
  • $\begingroup$ You forgot to add that the open subsets are connected, otherwise the LT fails! As for weakly differentiable, yes, the class $W^{1,n}$ suffices. Any book on quasiconformal maps will have a proof that 1-quasiconformal mappings of open connected subsets are restrictions of Moebius maps. Even Mostow in his book on strong rigidity proves the result (which is critical for his theorem). $\endgroup$ Commented Jul 18, 2022 at 19:29
  • $\begingroup$ @MoisheKohan: Good point about needing the open sets to be connected. As far as weakly differentiable being enough, it's definitely not the case that every book on quasiconformal maps proves Liouville's theorem or the smoothness of $1$-quasiconformal maps in the generality needed for it, for instance neither Hubbard's book nor Ahlfor's book do this since they entirely focus on the case of two real dimensions. $\endgroup$ Commented Jul 18, 2022 at 19:42
16
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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: https://irma.math.unistra.fr/~frances/publiouville4.pdf

$\endgroup$
7
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .