Yesterday we watched the video Outside In. Is it possible to turn a torus inside-out without breaking or creasing it? R.
On Wed, 31 Mar 2004, Richard Guy wrote:
Is it possible to turn a torus inside-out without breaking or creasing it? R.
Yes, i.e., the standard embedding of a torus in R^3 is regularly homotopic to the (relatively) inside-out embedding, just as this is the case for the sphere. The idea is, we all know from Martin Gardner that a torus T minus a disk can be turned inside out. So while keeping a small disk D of T fixed, we use that method to turn the rest of the torus inside-out. Now the only part of the torus that isn't inside-out is that small disk. By imagining D to be essentially a sphere S (while keeping a subdisk of S fixed), we can use the eversion of the sphere S to turn D inside-out as well, which completes turning the torus inside out. Dan Asimov
made it into today's New York Times! --Dan
Here's the URL http://www.nytimes.com/2004/04/03/arts/03MATH.html I attended this event and took quite a few photos and some low res video. Not much is on my web site yet (url below, scroll down a bit) but I did put up a video of tyler macready flying the walkalong glider (it's mentioned in the article). Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org ----- Original Message ----- From: "Dan Asimov" <asimov@msri.org> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Saturday, April 03, 2004 10:02 AM Subject: [math-fun] 2004 Gathering for Gardner
made it into today's New York Times!
--Dan
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On Wed, Mar 31, 2004 at 11:21:49PM -0800, Dan Asimov wrote:
On Wed, 31 Mar 2004, Richard Guy wrote:
Is it possible to turn a torus inside-out without breaking or creasing it? R.
Yes, i.e., the standard embedding of a torus in R^3 is regularly homotopic to the (relatively) inside-out embedding, just as this is the case for the sphere. ...
So are all the embeddings of a torus regularly homotopic? There are a number of "obvious" embeddings differing by elements of GL(2,Z); I can see how to get a number of elements: * The regular homotopy you mention gives the element ( 0 1 ) ( 1 0 ) of GL(2,Z); * By thinking of the torus as the neighborhood of an unknot and passing it through itself, you change the longitude by 2*the meridian, giving the element of ( 1 2 ) ( 0 1 ) of GL(2,Z). * A simple rotation in R^3 gives the element ( -1 0 ) ( 0 -1 ) These elements do not generate all of GL(2,Z), however. For instance, no element which is equivalent modulo 2 to ( 1 1 ) ( 0 1 ) is in the subgroup generated by these elements. Can the whole group be generated? ---------- In these embeddings of the torus in S^3, both complementary regions are solid tori. In any embedding in S^3, one of the complementary regions will be a solid torus and the other will be the complement of a knot; by a regular homotopy of the torus embedding, you can unknot the knotted side, turn any embedding into one of the ones above. The trick as Dan mentioned in his message lets you treat embeddings in R^3 as though they were in S^3. Peace, Dylan
As your reasoning suggests, there are 4 regular homotopy classes of embeddings (or immersions) of a torus in R^3. This is a consequence of the fact that there are two regular homotopy classes of immersions of immersions of an annulus---one that looks like a cylinder, the other like a stack of figure eights. You can look at an annulus going the short way around, and an annulus going the long way around --- each one has two independent configurations. The immersions of an annulus are distinguished because they represent the two elements of the fundamental group of SO(3). (This is the same thing that says you can hold a full cup of coffee in one hand, and without letting go or spilling, turn it around twice and get back in the same position. However, if you turn it around just once, your arm will be in a different, probably awkward, position.) For a g-holed torus, there are 2^(2g) different regular homotopy classes. --Bill
On Wed, Mar 31, 2004 at 11:21:49PM -0800, Dan Asimov wrote:
On Wed, 31 Mar 2004, Richard Guy wrote:
Is it possible to turn a torus inside-out without breaking or creasing it? R.
Yes, i.e., the standard embedding of a torus in R^3 is regularly homotopic to the (relatively) inside-out embedding, just as this is the case for the sphere. ...
So are all the embeddings of a torus regularly homotopic? There are a number of "obvious" embeddings differing by elements of GL(2,Z); I can see how to get a number of elements:
* The regular homotopy you mention gives the element
( 0 1 ) ( 1 0 )
of GL(2,Z);
* By thinking of the torus as the neighborhood of an unknot and passing it through itself, you change the longitude by 2*the meridian, giving the element of
( 1 2 ) ( 0 1 )
of GL(2,Z).
* A simple rotation in R^3 gives the element
( -1 0 ) ( 0 -1 )
These elements do not generate all of GL(2,Z), however. For instance, no element which is equivalent modulo 2 to
( 1 1 ) ( 0 1 )
is in the subgroup generated by these elements. Can the whole group be generated?
----------
In these embeddings of the torus in S^3, both complementary regions are solid tori. In any embedding in S^3, one of the complementary regions will be a solid torus and the other will be the complement of a knot; by a regular homotopy of the torus embedding, you can unknot the knotted side, turn any embedding into one of the ones above. The trick as Dan mentioned in his message lets you treat embeddings in R^3 as though they were in S^3.
Peace, Dylan
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On Sun, Apr 04, 2004 at 07:28:02AM -0400, William P. Thurston wrote:
As your reasoning suggests, there are 4 regular homotopy classes of embeddings (or immersions) of a torus in R^3.
Are you sure? After thinking about it for a while, I think there should be 4 regular homotopy classes of immersions, but only 3 of them are represented by embeddings. The regular homotopy classes of immersions are distinguished by taking the standard meridian and longitude, and seeing whether the band obtained by taking a regular neighborhood in the surfaces is twisted by an even or odd number of times. In the standard embedding of the torus, both the meridian (the (1,0) curve) and longitude (0,1) are untwisted, while the diagonal (1,1) is twisted. In general, curves (p,q) where p and q are both odd (and relatively prime, to give a simple curve on the torus) are twisted an odd number of times. But two such curves cannot form a basis for the homology of the torus. I believe another way to say this is that of the 4 different spin structures on the torus, 3 of them are even (and bound 3-manifolds) while one of them is odd. This also resolves another point I was having difficulty reconciling; namely, the subgroup of SL(2,Z) I found before, generated by the matrices ( 1 2 ) (-1 0 ) ( 0 1 ) ( 0 1 ) ( 0 -1 ) ( 1 0 ) seemed to be of index 3 in SL(2,Z). Namely, it seems to be the inverse image of a subgroup of order 2 in SL(2,Z_2) =~ S_3. (I haven't checked that these matrices generate that subgroup.) Peace, Dylan
Yes, you're right --- three of the regular homotopy classes are represented by embeddings, while the fourth is only represented by an immersion. The fourth class is represented by a surface that is swept out by a figure eight going in a figure-eight-shaped trajectory. This particular regular homotopy class counts as "0" since the map of tangent planes is homotopic to a constant map. BTW, this immersion was the basis for the method of proving the immersion theorem that inspired "Outside In". The situation is similar for higher genus surfaces --- if I'm not mistaken, all but one of the 2^(2g) regular homotopy classes is represented by embeddings, but the 0 class is only representable by immersions. --Bill On Apr 8, 2004, at 9:56 PM, Dylan Thurston wrote:
On Sun, Apr 04, 2004 at 07:28:02AM -0400, William P. Thurston wrote:
As your reasoning suggests, there are 4 regular homotopy classes of embeddings (or immersions) of a torus in R^3.
Are you sure? After thinking about it for a while, I think there should be 4 regular homotopy classes of immersions, but only 3 of them are represented by embeddings. The regular homotopy classes of immersions are distinguished by taking the standard meridian and longitude, and seeing whether the band obtained by taking a regular neighborhood in the surfaces is twisted by an even or odd number of times. In the standard embedding of the torus, both the meridian (the (1,0) curve) and longitude (0,1) are untwisted, while the diagonal (1,1) is twisted. In general, curves (p,q) where p and q are both odd (and relatively prime, to give a simple curve on the torus) are twisted an odd number of times. But two such curves cannot form a basis for the homology of the torus.
I believe another way to say this is that of the 4 different spin structures on the torus, 3 of them are even (and bound 3-manifolds) while one of them is odd.
This also resolves another point I was having difficulty reconciling; namely, the subgroup of SL(2,Z) I found before, generated by the matrices ( 1 2 ) (-1 0 ) ( 0 1 ) ( 0 1 ) ( 0 -1 ) ( 1 0 ) seemed to be of index 3 in SL(2,Z). Namely, it seems to be the inverse image of a subgroup of order 2 in SL(2,Z_2) =~ S_3. (I haven't checked that these matrices generate that subgroup.)
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participants (6)
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Dan Asimov -
dpt＠lotus.bostoncoop.net -
Richard Guy -
Thane Plambeck -
William P. Thurston -
William P. Thurston