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
(Ueiθ,Weiθ) - 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