• 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).

§ Rendering

• 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.

§ References

• An Elementary Introduction to the Hopf Fibration, by David W Lyons.