Dear colleagues:
Is there a conformal mapping that transforms non-spherical surfaces to spherical ones? or what is the methodology to construct that mapping?
if there is such mapping it's better to be convertible one.
Best Regards.
Dear Khaled Mohamed,
Suppose S is a compact metric surface, then for each point p, there
exits a local coordinate chart (U,f ), such that p belongs U and the
local coordinates are isothermal ( and metric conformal in local coordinates ) . In particular, it is true for surfaces in Euclidean 3-dimensional space.
See https://en.wikipedia.org/wiki/Beltrami_equation,
Ahlfors, Lars V. (1966), Lectures on quasiconformal mappings, Van Nostrand
So using Beltrami equation, we can local coordinates in which metric is conformal.
best,
MM
Have a look at this paper https://smartech.gatech.edu/bitstream/handle/1853/32156/1999_IEEE_EMBS_001.pdf
Geometrically,
a conformal mapping maps an infinitesimal circle on the source surface to a circle on the target surface.
A quasi-conformal mapping maps an infinitesimal circle to an ellipse.
See for example
W. Zeng , L.M. Lui , F. Luo, J.S. Liu, T.F. Chan, S.T. Yau, X.F. Gu,Computing Quasiconformal Maps on Riemann
surfaces using Discrete Curvature Flow,
arXiv:1005.4648v2 [math.NA] 1 Dec 2010
Wei Zeng, Feng Luo, Shing-Tung Yau, David Gu, Surface Quasi-Conformal Mapping by Solving Beltrami Equations
http://www3.cs.stonybrook.edu/~manifold/pdf/2009/IMA_2009_talk.pdf
. In this work, the authors develop an effective numerical algorithm
to compute the quasi-conformal mapping on general Riemann surfaces of any genus.
best,
MM
Does it imply that any mapping of 2D (or higher) to 1D is non-conformal ? (As there is no angles in 1D)..Thanks
As far as I understand it implies that any mapping of 2D (or higher) to 1D is non-conformal. Furthermore there is
Invariance of domain is a theorem which due to L. E. J. Brouwer in topology about homeomorphic subsets of Euclidean space Rn.
An important consequence of the domain invariance theorem is that Rn cannot be homeomorphic to Rm if m is different then n. Indeed, no non-empty open subset of Rn can be homeomorphic to any open subset of Rm in this case.
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