On Wed, Mar 2, 2011 at 5:22 PM, David Makin <makinmagic@tiscali.co.uk> wrote:

HI all,

This is a question relating to the elusive "holy grail" of a perfect Mandelbrot Set in 3D.

As I understand it the problems with respect to getting "whipped cream" in non-linear pure-3D fractals is related to the facts that there are no complete algebraic fields in R3 and that (by Louiville's (conformality) theorem) conformal mappings are restricted to mobius transforms only.

With respect to Louiville's theorem:

http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28conformal_mappings%29

The proof shows that any Rn (n>2) mapping/transform that is not a mobius transform will not be conformal.

My question (due to my own ignorance/lack of understanding) is:

Does Louiville's theorem also discount the possibility of an Rn non-mobius mapping/transform being conformal in Rx space (x<n) if only Rx of the Rn space is considered ?

For example is it not possible that if we have Rn (n>3) then a non-Mobius mapping/transform in Rn could be such that all non-conformality is restricted to a particular Rx of Rn where x>=1 and x<n-2 ?

I'm not sure exactly what you're asking. If the question is whether there are maps from Rm to Rn (the restriction of the original Rn ->Rn map to Rm that is conformal, the answer is yes (consider the higher-dimensional analogue of rolling a plane around a cylinder), but if we map the image back to R^m, then the composition, by Liouvile's theorem, is a Moebius transformation. Nor can you get any "extra" conformal maps by starting with some other m-dimensional manifold embedded in R^n. The wikipedia article says: "Similar rigidity results (in the smooth case) hold on any conformal manifold. The group of conformal isometries of an n-dimensional conformal Riemannian manifold always has dimension that cannot exceed that of the full conformal group SO(n+1,1). " So in dimension > 2, there are never any "extra" conformal maps besides the "obvious" ones. Andy