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.)
Peace, Dylan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun