Computing the hopf invariant (without integration or homology, as in Milnor) of the hopf map

In exercise 15 of Milnor's Topology from a Differentiable Viewpoint, 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 $k$.

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?

(Not homework!!)

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The degree of a map $f : M \to N$ provided $M$ and $N$ are compact, orientable and of the same dimension is given by an integral. Think about $\int_M f^* \omega$, provided $\int_N \omega = 1$. – Ryan Budney Oct 17 2010 at 23:45
Sure, but integration is not covered in the book, and all of the other exercises only use material covered in the book. – Harry Gindi Oct 17 2010 at 23:59
This is not research level. I voted to close. – Andy Putman Oct 18 2010 at 0:35
People have asked problems from Atiyah-MacDonald here before, and this is certainly more research-level than those. – Harry Gindi Oct 18 2010 at 1:08
Atiyah-MacDonald and Milnor's "Topology from the Differentiable Viewpoint" are at similar levels (1st year grad), though Milnor is maybe a little easier. However, AM contains a couple of exercises that are notoriously difficult (even for experts) and thus are borderline appropriate. Milnor does not, and what you asked is absolutely standard 1st year graduate topology. – Andy Putman Oct 18 2010 at 3:26

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.

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what's a standard way to embed the hopf link in $\mathbf{R}^3$? – Harry Gindi Oct 18 2010 at 0:06
Two circles in orthogonal planes, the center of one circle being a point of the other. – Ryan Budney Oct 18 2010 at 0:08