Proving Linkedness of Hopf Fibers

Question

So I've been working on understanding the Hopf fibration in terms of quaternions for the past few months, following along with the investigations in David Lyon's paper, "An Elementary Introduction to the Hopf Fibration". I've done a good job getting through almost the entirety of the paper, until I get to the last investigation: Investigation (M).

The Hopf fibration is a mapping from $S^3 \mapsto S^2$ such that the preimage of a point on $S^2$ is a circle (fiber) in $S^3$. The previous investigation, which I already did successfully, shows that these fibers, when stereographically projected, are linked with the unit circle in the $yz$-plane. This is simple enough. Here is investigation (M):

To show that any two projected fiber circles $C$ and $D$ are linked, we exhibit a continuous one-to-one map $\psi: \mathbb{R}^3 → \mathbb{R}^3$ that takes $C$ to the unit circle in the $y, z$-plane, and takes $D$ to some other projected fiber circle $E$. Since $E$ is linked with the unit circle in the $y, z$-plane, $C$ and $D$ must also be linked. See Figure 9. (Students who have never studied topology may accept the intuitively reasonable statement that linked-ness of circles cannot be altered by a continuous bijective map. Students with experience in topology may enjoy trying to prove this.) Here is how to construct the map $\psi$. Let $P$ be any point on the circle $C$, and let $r = s^{−1}(P)$. Define $f: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ by $f(x) = kr^{−1}x$ (quaternion multiplication). The map $\psi$ is the composition $s \circ f \circ s^{−1}$.

Just for some background:

$s$ is the stereographic projection map given by $$(w,x,y,z) \mapsto \left(\frac{x}{1-w}, \frac{y}{1-w}, \frac{z}{1-w} \right)$$ where $(w,x,y,z)$ is the vector in $\mathbb{R}^4$ associated with the quaternion given by the Hopf map. A long investigation previous to this (Investigation (J)) explains that these quaternions are $h^{-1}(P) = \{r_1 e^{it}\}_{0\leq t \leq 2\pi}$ where $P$ is any point on $S^2$. This is not the same P as the one on the fiber in Investigation M.

So this is how I'm thinking about the problem:

I understand that if we can construct a continuous map that takes 2 fibers and sends one to the unit circle in the $yz$-plane, then we have shown that since all fibers are linked with the unit circle, and continuous maps don't break linked-ness, then the fibers must be linked with each other. The part I can't get is actually proving this.

Instead of creating a fiber, I just took a specific quaternion corresponding with a particular fiber and defined that as $r = s^{-1}(P)$. If you apply $s(f(r))$, you end up with $k$. Then I picked another point on that fiber and used the same $r^{-1}$ to calculate $f$, but the algebra ended up really messy and I wasn't getting anywhere. I know what should follow is a unit circle, but I don't think I'm approaching the problem correctly, because it seems overly tedious.

Does anyone have any thoughts about how to prove this?

Thanks!

Answer 1

I know the paper and in fact, the author made a mistake in the homeomorphism $\psi$ that is supposed to take two fibers to the unit circle and some other fiber. Also, I feel it is somewhat overcomplicated.

Construct the map from scratch. First, we demand that for fibers $C, D$ the fiber $C$ is sent to the fiber over $(-1, 0, 0)$, which is the unit circle under stereographic projection. Guess that $\psi$ is of the form $\psi(x):=\rho x$ where $\rho$ is some unit Quaternion since this simple map will take great circles to great circles (not necessarily fibers though, that needs to be proved separately!).

Calculate $$ h(\psi(C)) = (\rho C) i \overline{(\rho C)} = \rho C i \overline{C} \overline{\rho} = \rho h(C) \rho^{-1} $$ and recognize the Quaternion rotation (in the paper, this was another investigation).

To satisfy our demands, $\rho$ must therefore rotate $\mathbb R^3$ such that $h(C)$ is sent to $(-1, 0, 0)$. Now, use the construction of the Hopf-fibration as presented in the paper. By it, we know for any $r\in C$ we have $rh(C)\bar r = (1, 0, 0)$. Now we simply rotate by $k$ afterwards to get to $(-1, 0, 0)$ as desired. Together this yields $\rho = kr$. This is the mistake in the paper.

This yields $h(\psi(C)) = k(rh(C)\bar r)\bar k = (-1, 0, 0)$ and you can satisfy yourself, that $h(\psi(D))$ is a well-defined point so that $\psi$ also sent $D$ to another fiber.