deant
Joined: 12 Jul 2010, 07:44 Posts: 154

Exercise [15.08]
"Interpreted as a real bundle" just means we're free to use (u,v,x,y) real coordinates, rather than just using (w,z) complex ones throughout.
Given that, what the exercise seems to be asking is to find a 11 continous map between tangent vectors on S2 and points in B'C.
In my solution to exercise [15.05], equations (3) and a modified (6) (halve the RHS) provide a continuous 11 map between B' and unit vectors of S2.
These equations assume that (u,v,x,y) is a point on S3. If we remove that assumption, and allow other points (ku,kv,kx,ky) where k is real (i.e. we consider the entire ray from the origin through a point on S3, we find that (3) maps the entire ray to the same point on S2 (except the origin: but that's why the actual B'C manifold needs a 'blown up' origin; this isn't represented in the equations). (6) On the other hand retains the vector direction, but vector length varies as k^2.
Thus these two equations already do in fact already provide a 11 map between B'C and tangent vectors on S2: To adjust the length of the tangent vector, just travel along the ray (ku,kv,kx,ky). If we'd like vector lengths on S2 to scale linearly with k, we can divide the RHS of (6) by 2*sqrt(u^2+v^2+x^2+y^2), ...(rather than just dividing it by 2 as discussed earlier). It doesn't affect the 11 nature of the map either way, though.
Note that it doesn't matter if we pick +k or k, since these points are identified (considered to be the same point) on the manifold B'C anyway.
