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.