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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