most recent 30 from http://mathoverflow.net 2013-05-20T12:09:50Z http://mathoverflow.net/feeds/question/42557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42557/computing-the-hopf-invariant-without-integration-or-homology-as-in-milnor-of-t Harry Gindi 2010-10-17T23:36:59Z 2010-10-18T00:13:49Z

<p>In exercise 15 of Milnor's <em>Topology from a Differentiable Viewpoint</em>, one is asked to compute the Hopf invariant of the Hopf map. The way one is supposed to do this is to compute the linking number of two of the fibres, but Milnor doesn't define the linking number in terms of an integral. He says to compute it as the degree of the map $\frac{x-y}{||x-y||}$ from the product of two compact oriented boundaryless manifolds embedded in $\mathbf{R}^{k+1}$ to the sphere of dimension $k$ where the sum of the dimension of the manifolds is $k$.</p> <p>I'm aware of other ways to compute the Hopf invariant by using deRham cohomology (see Bott and Tu, for instance), but I'm curious how one is actually supposed to do it by hand. Is there a particularly concrete way to compute the linking number without using this other machinery? Most of the other exercises in the book have cute little solutions, but is that true of this problem?</p> <p>(Not homework!!)</p>

http://mathoverflow.net/questions/42557/computing-the-hopf-invariant-without-integration-or-homology-as-in-milnor-of-t/42559#42559 Dylan Thurston 2010-10-17T23:52:38Z 2010-10-17T23:52:38Z

<p>If you have the Hopf link embedded in some standard way in $\mathbb{R}^3$, you can see the linking number as given by the degree of a map $S^1 \times S^1 \to S^2$ in a number of ways. For instance, the pre-image of the north pole in $S^2$ consists of pairs of points stacked vertically above each other, i.e., crossings between the two components in the knot diagram given by projection to the $xy$ plane. (Crossings will correspond to preimages of the north pole or south pole, depending on your conventions.) For the standard diagram for the Hopf link, there's only one crossing that counts. The hard part from this point of view is getting the orientation right (is the Hopf invariant $-1$ or $+1$?), but that can be done with care and attention.</p>