On Thu, 26 Jun 2003 17:17:58 +0200, "David Turner"

>1. I read that it is fairly widely believed, but not proved, that [the

>surfaces of] all convex polyhedra can be unfolded to form a planar surface.

>I think this is called an isometric embedding in R^2?

>2. At any rate: It is well established that a sphere cannot be similarly

>embedded in R^2.

>3. A sphere is supposed to be the limiting case of a convex regular

>polyhedron as the number of faces gets larger.

>So: why is it that every c.r.p. is embeddable, but the limit is not?

You are mixing apples and oranges, as the expression goes. Shall we

look at a few facts?

If we are discussing topology, there is no difference between a sphere

and the surface of a convex polyhedron. In order to flatten either one

you must first disrupt its topology with at least a puncture, if not

some cuts. But in topology we do not speak of isometric embedding.

We must be talking about some kind of geometry, but what kind? If it's

differential geometry, there's no problem applying it to a sphere; but

how do we treat a polyhedron as a differentiable surface? For example,

we know how to compute "curvature" on a sphere, and it's one of the

things that tells us a sphere is *intrinsically* not flat (whereas a

cylinder *is* intrinsically flat). But how should we compute curvature

on a polyhedron?

Actually, there is a technology that says all the curvature of the

polyhedron is concentrated at the vertices, though this is not a kind

of theory one often encounters. In such a formalism you could take a

limit as the number of vertices increases, and find that curvature

distributes across the surface in the limit. But in such a formalism

you could not flatten a corner, because that would change curvature.

<http://geom.math.uiuc.edu/docs/education/institute91/handouts/node21....>

<http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node29.html>