What is the intuitive explanation for the homology of S^3\knot?

[pp31]

Hey guys,

There was a definition of linking number that used the fact that H_1(S3\knot, Z) = Z. But to do that I was trying to compute the homology of S^3\knot and had no idea how to do it. Any help would be appreciated.

Thanks

 

[fzero]

A simple closed curve is a 1-1 mapping of S^1 into our space. The complement of such a curve in a simply-connected space obviously has first homology group equal to \mathbb{Z}. The linking number is the homology class of the 2nd curve.

 

[lavinia]

Hey guys,

There was a definition of linking number that used the fact that H_1(S3\knot, Z) = Z. But to do that I was trying to compute the homology of S^3\knot and had no idea how to do it. Any help would be appreciated.

Thanks

Try using a Mayer-Vietoris sequence by cutting out a solid tube around the curve. The 3 sphere minus the tube is a strong deformation retract of the 3 sphere minus the curve and so has the same homology.

Intuitively think of the close curve as a wire carrying a steady current. the current will generate a static magnetic field at any point in space. Very near the curve, the magnetic field will look like the field generated by a straight wire. This field is planar and circlulates around the wire in a circle. This circle is the generator of the homology group. You know that it is not null homotopic in 3 space because work done by the magnetic field on a magnetic particle that flows around the circle in one complete circuit will not be zero. In fact, if you look at the expression for the magnetic field given in the Law of Biot and Savart you will see that the magnetic field is approximately tangent to the circle.