4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I don't think you can link circles in $\mathbb{C}^2$: there is too much room. They are only linked in $S^3$. There you can check that any disc with boundary on on circle will intersect the other circle in one point. $\endgroup$ – Thomas Rot Aug 23 '18 at 13:27
  • $\begingroup$ @Thomas Rot. Thank you, this clears up my confusion. $\endgroup$ – Serg Aug 23 '18 at 13:42
  • $\begingroup$ The "convex hull" criterion that you quoted is for linking in $\mathbb R^3$. It won't work in $\mathbb C^2\cong\mathbb R^4$. It could be made to work in the sphere minus one point $S^3-\{p\}\subseteq\mathbb R^4$, if you used the notion of convexity that results from identifying sphere-minus-point with $\mathbb R^3$, say via stereographic projection. (One could also define a suitable notion of convexity on the sphere, using short arcs of great circles instead of line segments, because no line segments lie on the sphere.) $\endgroup$ – Andreas Blass Aug 23 '18 at 22:01
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.