I am a condensed matter physicist, and want to understand the Hopf link from analytic point of view. My question is as follows.

We have two sets of equations, and each set of equations describes a circle in certain space, for example in $\mathbb{R}^3$. In that case, how can we tell from the equations if the two circles are linked, and whether the link is a Hopf link? I want the conditions from analytic geometry point of view.

Could anybody point out some references for this problem?

  • 4
    $\begingroup$ en.wikipedia.org/wiki/… . Is this what you were looking for, perhaps? $\endgroup$ – Marco Golla Apr 9 '15 at 21:29
  • 2
    $\begingroup$ @MarcoGolla The linking number cannot distinguish the Hopf link among links with linking number $\pm 1$. $\endgroup$ – Jim Conant Apr 10 '15 at 2:33
  • $\begingroup$ @DelDon: perhaps you meant to ask a slightly different question? Linking numbers is a very coarse way of looking at things -- it fails to distinguish most links. If would be like being a condensed matter physicist and the only physical quantity you understood was viscosity. It gets you somewhere, but not very far, at least, not without a lot of work. $\endgroup$ – Ryan Budney Apr 10 '15 at 4:57
  • $\begingroup$ @JimConant Yes, sure; this was only a partial answer (specifically to "how can we tell if the two circles are linked"). $\endgroup$ – Marco Golla Apr 10 '15 at 6:36

It seems there are two different questions here: 1) how can we tell from the equations if the two circles are linked? and 2) how can we tell if the link is the Hopf link?

To answer the first question, a two component link is unlinked in $R^3$ (or $S^3$) if and only if there is a embedded $S^2$ that separates the two components. (Similar arguments and definitions will apply to links with more than two components. However, I will just consider two component links here.)

If the two components have non-zero linking number, this provides an obstruction to the two components being unlinked. However, this is not a necessary condition. There are several two component links where both components have linking number 0 with the other component. The Whitehead link is such an example (image from wikicommons).

The Whitehead link, image from wikicommons

There are many ways to see that this link cannot be unlinked that involve either computing a hyperbolic structure on the complement of the link, an obstruction to the link complement containing a sphere that does not bound a 3-ball, or using arguments with the fundamental group $G$ of the complement to show that $G$ does not surject the free group on two letters. A good reference for the first argument is either: Thurston's notes or Thurston's article Three dimensional manifolds, Kleinian groups and hyperbolic geometry and good reference for the second argument would be Burde and Zieschang's book "Knots". A final method would be to use double branched covers.

A double branched cover of a knot or link in $S^3$ is defined as follows. Take an embedded tubular neighborhood around each component of the link. Remove each tubular neighborhood and call the new space $M$. (This space is a link exterior and it's interior is homeomorphic to a link complement.) Take the double cover of $M$ corresponding to the kernel of the map $f=g(h(\pi_1(M)))$ where $h: \pi_1(M) \to Z \times Z$ given by abelianization and $g: Z \times Z \to Z/2Z$ given by $g(a,b)= a+b \mod 2.$ Then take the double cover of the tubular neighborhoods removed at the first step and reattach them to this cover of $M$. A more detailed construction of this is given in Rolfsen's "Knots and Link" (Chapters 5C and 10C).

While double branched covers are not a complete invariant of knots and links, they do completely distinguish a class of knots and links, two bridge links by work of Hodgson and Rubinstein. It is relevant to this discussion because the Hopf link is a two bridge link and it's double branched covering is $RP^3$, the unique manifold with $S^3$ as a double cover. To wrap up the discussion, a link its split if and only if its double branched cover contains an embedded $S^2$ that does not bound a 3-ball.


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.