## § 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`

).

#### § 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.