On Oct 11, 2007, at 2:20 PM, Dan Asimov wrote:
The Nash embedding theorem -- as improved by Kuiper (cf. http://en.wikipedia.org/wiki/Nash_embedding_theorem ) -- implies that S^2 can be C^1 isometrically embedded into an arbitrarily small ball of R^3. The wikipedia is very handy! I wasn't acquainted with the terminology "short map" for a non-distance-increasing map. How common is that? Anyway, it sounds like a good name for the concept.
QUESTION: Suppose we consider only isometric embeddings h_1 that arise from a continuous family of isometric embeddings h_t: S^2 -> R^3, 0 <= t <= 1, where h_0 is the inclusion mapping. Does there still exist an isometry h_1 of S^2 into an arbitrarily small ball in R^3? If not, how small can the image of such an isometry be? (It's not immediately obvious that the image can even have a smaller diameter than the original sphere, but it can.)
--Dan
The Nash-Kuiper embedding theorem can be done in a parametrized way, i.e. any C^1 continuous family of short C^1 embeddings parametrized say by a cell complex, i.e. P -> {C^1 short embeddings S^2 -> R^3}, can be deformed through short maps to a family of isometric embeddings, by the same proof. I.e. there would be a homotopy P x [0,1] -> {C^1 short embeddings S^2 -> R} that ends up in {C^1 isometric embeddings S^2 -> R}. So the answer is yes. On the other hand, concrete finite-complexity constructions seem harder. You can iteratively dimple a surface, by moving a plane through it and reflecting the part that sticks through the plane; this stays embedded for a while. Are you thinking about doing this from multiple directions, and seeing how far you can go? I wonder if there's a version of this that makes it into a good puzzle.One could ask for isometric embeddings obtained by dividing the sphere into finitely many pieces which are transformed rigidly. This seems tricky. Bill