# 3D conformal mappings

Are there analogues to conformal mapping in 3 dimensions?

I have a specific example I am trying to solve.. Laplace's equation in 3D with slightly complicated rectilinear boundaries. (Think of solving a harmonic function over a 3D boundary which is a cube but with a sub-cube "bitten" out of one corner.)

Laplace's equation is still valid under conformal transformations, so for example in 2D I could take a square domain with a subsquare bitten out of a corner, and apply an inverse tranformation like some of these and solve the equation in a simple square domain.

Are there similar conformal-like transformations in 3D? Perhaps they wouldn't be called conformal maps, but maybe something exists which would work similarly for my 3D Laplace equations.

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Well, there are conformal map in higher dimensions (defined by saying that they preserve angles). However, they are quite a bit less flexible than in 2d; see my answer to the following question : mathoverflow.net/questions/10066/… –  Andy Putman Jan 16 '11 at 4:54
Can I requestion "complicated" rather than "complex" in the second paragraph? The word "complex" has all sorts of technical meaning. –  Theo Johnson-Freyd Jan 16 '11 at 8:12
Theo, you're right.. done! –  MathGeek Jan 16 '11 at 8:46
Andy, yes, those affine transforms aren't useful for this. So the question stands.. are there other transforms that DO work in 3D for the Laplace equation? Or is that really EQUIVALENT to a conformal map, meaning that only affine transforms have that characteristic in 3D? –  MathGeek Jan 16 '11 at 8:48
@MathGeek: note that higher-dimensional conformal maps are not all affine, they are slightly more general than that (restriction of Möbius transform on the sphere), but this does not help for your original question. –  Benoît Kloeckner Sep 3 '11 at 6:53

Definition: given two open subsets $U,V\subset \mathbb R^3$, a smooth map $f:U\to V$ is called weak conformal if, at every $x\in U$, the three eigenvalues of $P_x:\mathbb R^3\to \mathbb R^3$ are in geometric progression. Here, the positive operator $P_x$ is the one coming from the polar decomposition of the tangent map $T_xf:T_x \mathbb R^3 = \mathbb R^3\to T_{f(x)} \mathbb R^3 = \mathbb R^3$.