I am trying to understand why the preimages of two points under the Hopf fibration are linked.
I thought that two circles in $\mathbb{C}^n$ are linked iff one circle intersects the convex hull of the other.
$$p: S^3\to\mathbb{C}P^1,\quad p(z_1,z_2)=[z_1,z_2].$$
Suppose that $z_1\neq 0$. Then the image is defined only by the ratio $z_2/z_1$. Suppose that we have $v, u\in\mathbb{C}P^1$, $v_2/v_1=a, u_2/u_1=b.$ Then
$$p^{-1}(v)=\{(z_1,z_2)\in\mathbb{C}^2\ |\ z_2=az_1, |z_1|^2+|z_2|^2=1 \},$$
$$p^{-1}(u)=\{(z_1,z_2)\in\mathbb{C}^2\ |\ z_2=bz_1, |z_1|^2+|z_2|^2=1 \}.$$
But it seems that the convex hulls of these circles intersect only at one point (0,0), so they don't seem to be linked.
What's wrong with my reasoning? And how can I show that any two fibers are linked?
Best Answer
You can also do this with the fundamental group. Because if $L$ is two linked circles, then $\pi_1(S^3\setminus L)$ is $\mathbb{Z}\times \mathbb{Z}$, a free abelian group; if $L$ is two unlinked circles, then $\pi_1(S^3\setminus L)$ is $\mathbb{Z}\ast \mathbb{Z}$, a free group.
If we remove two points from $S^2$, we now have the fibration $$ S^1 \rightarrow S^3\setminus L\rightarrow S^2\setminus \{x,y\} $$ where that last space is homotopic to a circle. Thus we have the exact sequence of fundamental groups $$ \{1\}\rightarrow \pi_1(S^1)\rightarrow \pi_1(S^3\setminus L)\rightarrow \pi_1(S^2\setminus\{x,y\})\rightarrow\{1\} $$
And that means we can't get the free group, so $L$ consists of linked circles.