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