Proving that the Hopf Fibration is a fiber bundle

algebraic-topologyfiber-bundlesfibrationgeneral-topologyhopf-fibration

I'm having trouble understanding a proof that the Hopf Fibration is a fiber bundle. Here is the paper that I'm working through: hopf fibration

Here is the paragraph from the paper that I am most confused about:

"Now consider a covering of $S^2$, consisting of $S^2 − [0 : 1]$ and $S^2 − [1 : 0]$. Referring to Figure 2.1, and noting that [0 : 1] corresponds to the point $0 ∈ \mathbb{C} \cup \{\infty\}$, i.e. $\textbf{0} ∈ S^2$, and [1 : 0] corresponds to the point ∞ ∈ $\mathbb{C} \cup \{\infty\}$, we see immediately that $S^2 − [0 : 1]$ is really $S^2 − \{\textbf{0}\}$, and $S^2 − [1 : 0]$ is equivalent to $\mathbb{R}^2$. These two sets $S^2 − \{\textbf{0}\}$ and $\mathbb{R}^2$ therefore form an open covering of $S^2$."

I think I understand all of the basic notation used here but I am confused about the following things:

  1. What even is $S^2 – \{\textbf{0}\}$? Figure 2.1 has $S^2$ sitting on top of the projective plane, so I'm assuming that $S^2 – \{\textbf{0}\}$ is taken to be this same sphere with the point at the origin removed? But then isn't this just homeomorphic to $\mathbb{R^2}$?

  2. What does the author mean by "is really" and "is equivalent to"? Homeomorphic? I guess I'm just generally confused what the author means by $S^2 – [0:1]$ and $S^2 – [1:0]$ since $[0:1]$ and $[1:0]$ are elements are elments of $\mathbb{C}\mathbb{P}^1$. (If this is just a notational shortcut to describe $S^2 − \{\textbf{0}\}$ and $\mathbb{R}^2$ why even bring it up in the first place?)

  3. Why is it true that $S^2 – \{\textbf{0}\}$ and $\mathbb{R}^2$ must form an open covering? I'm literally confused about the justification here…. Does it have something to do with the notation I'm don't understand in point 2?

I have a feeling I'm missing something to help me put all of the pieces together here. If anyone has any ideas about how to understand this argument or even another way to show that the hopf fibration is a bundle I would be very grateful. Thank you!

Best Answer

The Hopf fibration is a map $S^3 \to S^2$. The author, however, defines it as the map $$p : S^3 \to \mathbb C P^1, p(z_1,z_2) = [z_1:z_2] .$$ This is a very elegant definition, but to obtain the desired map $S^3 \to S^2$ we have to identify $\mathbb C P^1$ with $S^2$. This is done in Proposition 2.2 where $\mathbb C P^1$ is identified with the one-point-compactification $\hat{\mathbb C} = \mathbb C \cup \{\infty\}$ of $\mathbb C$. An explicit homeomorphism is given by $$\phi : \mathbb C P^1 \to \hat{\mathbb C}, \phi([z_1:z_2]) = \begin{cases} z_1/z_2 & z_2 \ne 0 \\ \infty & z_2 = 0 \end{cases}$$

$\hat{\mathbb C}$ is well-known to be homeomorphic to $S^2$ (use the stereographic projection $\chi_N : S^2 \setminus \{N\} \to \mathbb R^2 = \mathbb C$, where $N$ is the north pole of $S^2$). Let $\psi : \hat{\mathbb C} \to S^2$ denote the corresponding homeomorphism.

Now the author makes a double abuse of notation:

  1. He writes the points of $S^2$ in the form $[z_1:z_2] \in \mathbb C P^1$. This explains what $S^2 − [0 : 1]$ and $S^2 − [1 : 0]$ are. The correct notation would be $S^2 \setminus \{\psi(\phi([0:1]))\}$ and $S^2 \setminus \{\psi(\phi([1:0]))\}$. Tracking the homeomorphisms $\phi$ and $\psi$ we see that $\psi(\phi([0:1]))$ is the south pole of $S^2$ and $\psi(\phi([1:0]))$ the north pole.

  2. He also writes the points of $S^2$ in the form $z \in \hat{\mathbb C}$. This explains that he means $$S^2 - [0:1] = \hat{\mathbb C} - \{0\} \quad, \quad S^2 - [1:0] = \hat{\mathbb C} - \{\infty\} = \mathbb C = \mathbb R^2 .$$ But in my opinion writing $\hat{\mathbb C} - \{0\} = S^2 − \{0\}$ is really unfortunate (not to say nonsense).

Anyway, that $S^2 - \{\textbf{0}\}$ and $\mathbb{R}^2$ form an open cover of $S^2$ is obvious: These two sets are obtained by removing from $S^2$ two distinct points.

For additional information concerning the Hopf map and $\mathbb C P^1$ you can look at Prove that Hopf maps on $S^3, S^2$ and $\mathbb{C} \mathbb{P}^1$ are smooth submersions and Problem 2.9 in Lee's Smooth Manifolds: finding a map between complex projective spaces that makes the diagram commutative.

You will find in particular explicit descriptions of the Hopf map $S^3 \to S^2$ with the "true" space $S^2$ as its range.

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