Re: [math-fun] Re: Teabag Problem
Bill T. writes: << . . . [T]here's a general principle, I think due to Nash (who proved a C^1 isometric embedding theorem) that was motivational at least implicitly if not explicit in some of Gromov's early work on partial differential inequalities, that any sub-isometric smooth embedding of a surface in space can be approximated by an isometric embedding. If you accept this principle, it's very easy to enclose a positive volume by gluing together two C^1 smooth disks which contract distances, then approximating by an isometric embedding. . . . . . .
Let S^2 denote the unit sphere in R^3. 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. 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
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
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
Okay: I'll give a word description of a construction for the two-disk teabag. It would be fun to make pictures, but II'm in a hurry and I couldn't send the attachment anyway. Start with a unit disk. Arrange n=5 or more equally-spaced radii on the edges of a shallow pyramid, and shape the n sectors of disk between them into sectors of cones that go upward. Now the edge of the disk is the union of n circular arcs of radii somewhat less than 1 (depending how steep or shallow the pyramid). For each of the n circular arcs, construct a plane i through its endpoints and bisecting the angle between the plane of the arc and the base of the pyramid. This plane should be tipped nearly vertical, but tipped slightly toward the apex of the pyramid. Reflect the distal portion of the cone in this plane. Now we have an isometric embedding of a disk into a half-space, with boundary consisting of n circular arcs on the plane bounding the half- space. Double to get the teabag enclosing positive volume. In the limiting case where the pyramid is flat, the seam is obtained by inscribing a regular n-gon in the circle and reflecting the circular arcs in the sides of the polygon. That's why n must be 5 or more: for a triangle, these arcs intersect, and for a square, they are tangent, so they would intersect as soon as the pyramid isn't flat. You can make a similar origami type embedded construction for the square. Make concave-upward creases on the diagonal lines, and concave-downward creases on the altitudes. Add extra vertices on the altitudes at some fixed distance from the center, with new concave- downward creases from these vertices to the two adjacent corners, and reverse the direction of the fold on the distal portions of the altitudes. This should fold into a figure in upper half-space whose base looks like a 4-pointed star in the plane. (But I haven't actually done it) Bill On Oct 12, 2007, at 6:30 AM, David Gale wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 10/12/07, Bill Thurston <wpthurston@mac.com> wrote:
... You can make a similar origami type embedded construction for the square. Make concave-upward creases on the diagonal lines, and concave-downward creases on the altitudes. Add extra vertices on the altitudes at some fixed distance from the center, with new concave- downward creases from these vertices to the two adjacent corners, and reverse the direction of the fold on the distal portions of the altitudes. This should fold into a figure in upper half-space whose base looks like a 4-pointed star in the plane. (But I haven't actually done it)
The optimum position for this "extra" vertex is 0.21628 units approx from the edge of the flat bag; it lies at that height vertically above a point 0.18452 inboard from the edge of the square hull of edge 0.92941 occupied by the inflated bag on its horizontal plane of symmetry. In this configuration the (maximised) volume equals 0.1656982335 --- not bad at all for a straight-off-the-cuff construction, when compared with the best attempt of 0.2055 claimed by Andrew Kepert! WFL
On 10/14/07, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/12/07, Bill Thurston <wpthurston@mac.com> wrote:
... You can make a similar origami type embedded construction for the square. Make concave-upward creases on the diagonal lines, and concave-downward creases on the altitudes. Add extra vertices on the altitudes at some fixed distance from the center, with new concave- downward creases from these vertices to the two adjacent corners, and reverse the direction of the fold on the distal portions of the altitudes. This should fold into a figure in upper half-space whose base looks like a 4-pointed star in the plane. (But I haven't actually done it)
The optimum position for this "extra" vertex is 0.21628 units approx from the edge of the flat bag; it lies at that height vertically above a point 0.18452 inboard from the edge of the square hull of edge 0.92941 occupied by the inflated bag on its horizontal plane of symmetry.
In this configuration the (maximised) volume equals 0.1656982335 --- not bad at all for a straight-off-the-cuff construction, when compared with the best attempt of 0.2055 claimed by Andrew Kepert! WFL
In practice, 0.21628 is sufficiently close to tan(\pi/8)/2 that the "extra" creases can be placed nearly optimally via the traditional origami trick of (twice) halving the angle at the corner of the flat square. All of which --- to a (former) outdoor enthusiast --- prompts the unexpected observation that, using just 5 poles, 8 pegs, 13 reinforcement patches and a 5-metre or so square of mylar, it is possible to fashion a pretty serviceable tent. Entry and exit might prove problematic, of course ... Details, my dear chap, details! WFL
On 10/12/07, Bill Thurston <wpthurston@mac.com> wrote:
... You can make a similar origami type embedded construction for the square. Make concave-upward creases on the diagonal lines, and concave-downward creases on the altitudes. Add extra vertices on the altitudes at some fixed distance from the center, with new concave- downward creases from these vertices to the two adjacent corners, and reverse the direction of the fold on the distal portions of the altitudes. This should fold into a figure in upper half-space whose base looks like a 4-pointed star in the plane. (But I haven't actually done it)
I've been attempting to add more creases to the construction above, so as to increase the (maximum attainable) volume --- so far without success. Is it the case that an isometric embedding is always approximable by polyhedral isometric embeddings? WFL
participants (5)
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Bill Thurston -
Bill Thurston -
Dan Asimov -
David Gale -
Fred lunnon