A Universe of Sorts

§ Concrete calculation of hopf fibration

Take any point (a, b, c, d) ∈ S3.
This determines a quaternion a + bi + cj + dk.
Let the rotation determined by a quaternion q be written as R_q.
Recall that this determines a rotation of 3 space (b, c, d) and angle $2 cos^{-1}(a)$ .
Let $P_0$ be any point on S2, say (1, 0, 0).
Then, the hopf fibration h : S3 → S2 sends p ∈ S3 → R_p(P_0) ∈ R3.
Now, see that there are many points in S3 which have image (1, 0, 0).
For example, one can check that all points (cos t, sin t, 0, 0).
We can show that for any point p ∈ S2, the inverse image h^{-1}(p) ∈ S3 will be a unit circle in S^3.
The full fibration will therefore be S1 → S3 -h→ S2, since each of the fibers of h is a circle ( S1).

For a given point on p ∈ S2, we can render $h^{-1}(p) \subseteq S^3$ as a set of points in R3 via stereographic projection, which furthermore preserves circles (sends circles to circles).
The map sends a point $(w, x, y, z)$ to the point $(x/(1-w), y/(1-w), z/(1-w))$ . The only singularity is at $w=1$ which we conveniently ignore.
See that this map sends circles to circles. Suppose we have the locus of points (w, x, y, z) where w^2 + x^2 + y^2 + z^2 = 1.
Then the length of the sterographic projection vector is (x^2 + y^2 + z^2)/(1-w)^2, which is 1-w^2/(1-w)^2 ?? SOMETHING IS WRONG!
Recall the intuition for the sterographic projection, wrt S2 and R3.
For a point p ∈ S2, draw a line l, from the north pole (0, 0, 1)to the point p. The projection is the point of intersection of l with the xy plane.
Same thing a dimension up.