Can't someone CONSTRUCT at least one example of a disc teabag of positive volume. Below is the simple example of an origami solution for the square teabag. I trust it's understandable. Fold on the dotted lines, but the 4 corner squares are folded on the diagonals so they get squenched into triangles. In other words, this is not really an embedding or even an immersion because it's not injective. The mapping two-to one on the corner squares. __1/4___________1/2____________1/4___ |* * * * | | * * * * 1/4 | * * * * | |* * * * * * * * * * * * * * *| | * * | | * * | | * * 1/2 | * * | | * * | | * * | | * * | |* * * * * * * * * * * * * * * | |" * * * * 1/4 | * * * * | |* _____ *__________________*_______*| Maybe some of the other origami solutions are honest embeddings or maybe it's not possible (?). david At 10:04 PM 10/11/2007, you wrote:
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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
David Gale Professor Emeritus Department of Mathematics University of California, Berkeley