Re: [math-fun] Re: Teabag Problem
Bill AT. writers: << [I wrote] << 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.)
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.
Very interesting! << 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? . . .
Yes, exactly. Iterated reflections of this kind through a moving plane are the only concrete isometries of S^2 into R^3 that I thought of. --Dan
Some more thoughts on isometrically crumpling a sphere into a small wad. Imagine taking a very thin copper shell, and tapping it repeatedly and very lightly in a random spot with a ball-peen hammer. Mathematically, repeatedly choose a random ray through the origin, find the orthogonal plane that just grazes the surface, move it inward by some small amount epsilon, and reflect what is outside into the inside. Question: in the limit, as epsilon -> 0, do the spheres remain embedded until they are an arbitrarily small size, with probability -
1? If not, what is the probabilistic limit until they stop being embedded?
Heuristically, the normal vectors should evolve by small changes in random directions, so a typical vector should evolve more or less like Brownian motion. Unfortunately the size of the disk for an epsilon-size dent of the unit sphere is sqrt(epsilon), so the Brownian motion wouldn't seem to scale to 0, but to real Brownian motion on the surface of the sphere. It looks to me like a race between the decrease of radius (which needs to get to 1) and the change of angle (which shouldn't get further than pi/2), so it could seem to go either way. This picture isn't quite accurate, because the shapes of the dents will be irregular and probably hard to analyze as soon as the sphere is dented. This might be a fun thing to see simulated. This also suggests that a more carefully constructed random process might work more readily: try to design a sequence of taps that balance the changes in angle of the surface. This reminds me of the question of arranging charged particles on the sphere to minimize energy, as in some of Neil Sloane's computer experiments a while ago. If we were doing this in hyperbolic space on a horosphere, we could use an equilateral triangle grid, divided as a tripartite graph, and cycle between tapping dents around the 3 sets of vertices. Bill On Oct 12, 2007, at 11:32 AM, Dan Asimov wrote:
Bill writes:
<< [I wrote]
<< 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.)
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.
Very interesting!
<< 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? . . .
Yes, exactly. Iterated reflections of this kind through a moving plane are the only concrete isometries of S^2 into R^3 that I thought of.
--Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Bill Thurston -
Dan Asimov