Conformal maps in higher dimensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:33:40Z http://mathoverflow.net/feeds/question/10066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10066/conformal-maps-in-higher-dimensions Conformal maps in higher dimensions Johan 2009-12-29T21:09:44Z 2012-09-28T11:39:47Z <p>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. </p> <p>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? </p> http://mathoverflow.net/questions/10066/conformal-maps-in-higher-dimensions/10068#10068 Answer by Andy Putman for Conformal maps in higher dimensions Andy Putman 2009-12-29T21:19:25Z 2009-12-29T21:24:38Z <p>I think you're looking for <a href="http://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)" rel="nofollow">Liouville's theorem</a>. 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.</p> <p>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.</p> http://mathoverflow.net/questions/10066/conformal-maps-in-higher-dimensions/10069#10069 Answer by Gjergji Zaimi for Conformal maps in higher dimensions Gjergji Zaimi 2009-12-29T21:38:28Z 2009-12-29T21:38:28Z <p>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 <a href="http://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)" rel="nofollow">Liouville's theorem</a> 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 <a href="http://en.wikipedia.org/wiki/Whitehead_continuum" rel="nofollow">Whitehead continuum</a></p> <p>As for the proof of Liouville's theorem, maybe <a href="http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.pjm/1102983461&amp;page=record" rel="nofollow"> this article</a> is of interest where you can see a sketch of Nevanlinna's original proof and a proof by nonstandard analysis.</p> http://mathoverflow.net/questions/10066/conformal-maps-in-higher-dimensions/74421#74421 Answer by Benoît Kloeckner for Conformal maps in higher dimensions Benoît Kloeckner 2011-09-03T06:57:09Z 2011-09-03T06:57:09Z <p>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: <a href="http://www.math.u-psud.fr/~frances/liouville2.pdf" rel="nofollow">http://www.math.u-psud.fr/~frances/liouville2.pdf</a></p> http://mathoverflow.net/questions/10066/conformal-maps-in-higher-dimensions/108333#108333 Answer by Changyu Guo for Conformal maps in higher dimensions Changyu Guo 2012-09-28T11:39:47Z 2012-09-28T11:39:47Z <p>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.</p>