The fundamental group of the space $Y=S^3-(p^{-1}(\alpha_1)\cup ... \cup p^{-1}(\alpha_n))$, where $p :S^3 \to \Bbb CP^1$ is the quotient map

Question

Consider $S^3$ as a subspace of $\Bbb C^2$, let $q : \Bbb C^2 \to \Bbb CP^1$ be the canonical quotient map, and let $p=q|_{S^3}$. (I heard that $p$ is a Hopf fibration, but I don't know what it is.) It is easy to see that $p^{-1}(\alpha)$ is homeomorphic to $S^1$, for each $\alpha \in \Bbb CP^1$.

Now, take finitely many points $\alpha_1, ..., \alpha_n \in \Bbb CP^1$, and conisder the subspace $Y=S^3-(p^{-1}(\alpha_1)\cup ... \cup p^{-1}(\alpha_n))$ of $S^3$. I want to compute $\pi_1(Y).$

1. Is $Y$ path-connected? This is not obvious to me.

2. How do I have to compute $\pi_1(Y)$? I only know some basic theory of fundamental groups such as van Kampen theorem and covering spaces, so I need an elementary approach. Any hints?

Answer 1

The following is not a fully rigorous proof, but being elementary it might help you.

Suppose at first $\,n=1$.

Seeing $S^3$ as embedded in $\mathbb{C}^2$, you can verify that the following map

$\begin{align} f:\,&S^1 \times\mathring{D^2}\rightarrow S^3\setminus p^{-1}(x_1)\\ &(e^{i\theta}, re^{i\phi})\mapsto\:(\sqrt{1-r^2}e^{i\theta}, re^{i(\phi+\theta)}) \end{align}$

is a homeomorphism. Notice that $re^{i\phi}$ is the polar form for the elements in $\mathring{D^2}$, the open unit disc. Also observe that to construct this specific map I put $x_1=[0:1]\in \mathbb{C}P^1$, but there is no loss of generality since you can rotate $S^3$ in the first place.

Next, how can we make additional ring-shaped holes in $S^3$? Just puncture in $\mathring{D^2}$!

Specifically, choose $n-1$ distinct points $r_2e^{i\phi_2},\ldots, r_{n}e^{i\phi_{n}}$ in $\mathring{D^2}$ and restrict $f$ to $S^1\times\bigl(\mathring{D^2}\setminus\{r_2e^{i\phi_2},\ldots, r_{n}e^{i\phi_{n}}\}\bigr)$. Then there correspond $n-1$ distinct points $x_2,\ldots,x_n$ in $\mathbb{C}P^1$ in the same way as $x_1$ correspond to the boundary of $\mathring{D^2}$ in $\mathbb{C}$.

Now you must see how to compute the desired fundamental group, since each component of the space

$S^1\times\bigl(\mathring{D^2}\setminus\{r_2e^{i\phi_2},\ldots, r_{n}e^{i\phi_{n}}\}\bigr)$

has a familiar homotopy type.

Answer 2

A broad but not complete hint

Once you realize that the map $p$ is $(z_1, z_2) \mapsto z_1/z_2$, it gets a little easier. It's clear that $p^{-1}(\alpha)$, for any $\alpha$ is just a curve on $S^3$.

In particular, if you think of $S^3$ as being $\Bbb R^3$ with an extra point at infinity, you can pick your $\alpha$s so that the point at infinity is not in the preimage of any of them, so what you've got is a bunch of curves in 3-space.

What do they look like? Well, let's that the case where the ratio is $0$ (i.e., $\alpha = 1$). Then we're looking at points $(z_1, z_2) \in S^3$ where $z_1 = 0$, i.e., we're looking at points $(0 + 0i, c + di)$ where $c^2 + d^2 = 1$. In $\Bbb R^3$, that looks just like a nice round circle.

It turns out that if you change $\alpha$ away from $0$ a tiny it, say, to $0.1$, you can again solve the equations, and you find that the preimage is more or less another circle...but it's linked with the first one. In fact any two preimages are linked exactly once.

Now the problem itself suggest that the choice of the $\alpha_i$s doesn't matter much, so you might as well pick them to suit yourself, and if you do (e.g., if you pick them to lie equally spaced around a unit circle in $\Bbb C$), the preimages will all lie on a single torus in $\Bbb R^3$. In fact, each one will be a $(1,1)$ torus knot: as you walk around the torus in one direction, the curve will "walk" around it once in the other direction as well. That's a little vague. Instead, suppose you regard the torus as a square with its edges identified pairwise as usual. The the preimages of the $\alpha_i$ will each be a 45-degree diagonal line in the square, equally spaced.

Proving all this requires writing down a bunch of formulas for things like stereographic projection from the sphere to 3-space, etc., and is kind of a pain in the neck.

But now at least I've got you started: the thing you want to compute is the fundamental group of 3-space with $n$ $(1,1)$-torus-knots removed.

Here: I found a link to a nice picture of a bunch of $(1,1)$ torus knots all on the same torus: https://collegemathteaching.files.wordpress.com/2015/07/torusknot.jpg

So you want to compute the fundamental group of $\Bbb R^3$ minus $n$ of those colored curves. (I'd suggest starting with $n = 1$ and $n = 2$, of course.)