So the idea is to find a continuous map f : S3 --> tangent vectors on S2, where the tangent vectors are "spinorial".

In the Hopf fibration, S3 is the set of points (U,W) in C2 with |U|2 + |W|2 = 1, and p(U,W) = U/W on the Riemann sphere.

If U = U1 + iU2, W = W1 + iW2 then

p(U,W) = (2(U1W1 + U2W2), 2(U2W1 - U1W2), U12 + U22 - W12 - W22).

He says in the book that the existence of this map f implies that the Hopf fibration is a fiber bundle that does not have a cross-section. That's because if you had a cross-section of the Hopf fibration on S3, i.e. a map g: S2 --> S3 such that g composed with the projection map p: S3 --> S2 is the identity on S2, then there would be a continuous vector field V on S2, where V(x) = fg(x) for x in S2. And this would be a contradiction, since you can't "comb the hair on a sphere". (your head is not a sphere since your neck is not being combed, gruesome thought)

So, f has to interpret a point (U,W) on S3 as a tangent vector to a point on S2. You might hope it'd go to a tangent vector to p(U,W). The only trouble is, the tangent vector he gives in the hint,

U1∂/∂U2 - U2∂/∂U1 + W1∂/∂W2 - W2∂/∂W1,

is the exact kernel of the natural map from the tangent space of S3 to the tangent space of S2! When you have a map between manifolds, there is an associated map between the tangent spaces of the manifolds that is given by multiplying a tangent vector by the Jacobian matrix of the map. It's the best linear approximation of the map at a given point.

So, what you can do instead is to see that

p(−W1, W2, U1, −U2) = −p(U1, U2, W1, W2), the antipodal point on S2;

and the vector U1∂/∂U2 - U2∂/∂U1 + W1∂/∂W2 - W2∂/∂W1 is orthogonal to (−W1, W2, U1, −U2), so it's a tangent vector at that point too! And it gets sent to the tangent space at the antipodal point on S2 by the natural map between tangent spaces, and it's not in the kernel of the natural map at that point!

So, set f(−W1, W2, U1, −U2) = (one half of) natural map (U1∂/∂U2 - U2∂/∂U1 + W1∂/∂W2 - W2∂/∂W1).

This does all that's wanted. f(U,W) is a tangent vector to p(U,W) that's of unit length for any (U,W).

Since p(U,W) = p(-U,-W),  f(U,W) must be f(-U,-W), which it is.

And as you go around a fiber on S3 - a fiber containing (U,W) is the circle (Ue,We) - the tangent vector f(U,W) rotates 2 full circles around p(U,W). Which is what is wanted, one gathers, that is the spinoriality ...

I originally solved it, just by looking at what f(U,W) has to be for (U,W) in p−1(0,0,1) and p−1(0,0,−1), which was no harder if less ingenious.

Laura