9
$\begingroup$

I recently got into Lickorish's paper Prime knots and tangles and a question, which I didn't have the first time I read it, naturally emerged.

The Thurston-Perelman Geometrization Theorem asserts that given a compact, closed, orientable three-manifold $M$, either:

1) $M$ contains an essentrial sphere, or

2) $M$ contains an incompressible torus, or

3) $M$ is Seifert fibered, or

4) $M$ is hyperbolic.

Given a link $L \subset S^3$ we can form a closed three-manifold $M_L$ by considering the double-branched cover of $S^3$ over $L$.

QUESTION. How do conditions 1-4, regarding the manifold $M_L$, translate to the topology of the link $L$?

Following Lickorish's paper it is easy to find some sufficient conditions based on the possibility to recognize certain tangle configurations in a diagram of $L$. What about "if and only if" conditions?

$\endgroup$
  • $\begingroup$ In particular I would like to know a sufficient condition for hyperbolicity... $\endgroup$ – Antonio Alfieri Dec 16 '16 at 21:04
  • 4
    $\begingroup$ Your question can be recast in terms of orbifolds. In particular, make the link into a $\pi$-orbifold. Then the analogue of the geometrization theorem is the orbifold theorem. For background on this and a statement, see ams.org/mathscinet-getitem?mr=1778789 $\endgroup$ – Ian Agol Dec 17 '16 at 8:20
  • $\begingroup$ Thank you. I think you already gave this reference to me, I think this time I'm gonna read this! $\endgroup$ – Antonio Alfieri Jan 11 '17 at 19:58
  • $\begingroup$ Another (much earlier) reference for this material is the Bonahon and Siebenmann paper: www-bcf.usc.edu/~fbonahon/Research/Preprints/BonSieb.pdf $\endgroup$ – Ryan Budney Jan 12 '17 at 3:50
7
$\begingroup$

As Ian Agol mentioned in his comment, the OP's question can be thought of in terms of the Orbifold Theorem. There are two (contemporaneous and) independent proofs of the Orbifold Theorem:

Daryl Cooper, Craig D. Hodgson, and Steven P. Kerckhoff, MR 1778789 Three-dimensional orbifolds and cone-manifolds, ISBN: 4-931469-05-1. (The relevant background chapter is available here https://projecteuclid.org/download/pdf_1/euclid.msjm/1389985818)

Michel Boileau, Bernhard Leeb, and Joan Porti, MR 2178962 Geometrization of 3-dimensional orbifolds, Ann. of Math. (2) 162 (2005), no. 1, 195--290.

The background section of either paper seems to answer the OP's question. However, here is my attempt to present the necessary details:

Consider the warm up case, if the double branched cover $M_L$ of a link (embedded in $S^3$) $L$ is geometric (in the sense of admitting one the eight Thurston geometries), then $M$ double covers a geometric orbifold $Q$ which has underlying space $S^3$ and the singular locus has cone angle $\pi$ and is isotopic to $L$ (as an embedded link in $S^3$). This covers cases 3) and 4) of the OP's question. Good background for Case 3) is the work of Montesinos, especially his book Classical Tessellations and Three-Manifolds.

Otherwise, there is an obstruction to geometrization. For closed, orientable 3-manifolds, the obstructions to geometrization are embedded $S^2$ which do not bound $B^3$'s and incompressible $T^2$. As a result of the two papers cited above, the obstructions to (closed) orbifold geometrization are more or less analogous.

First, we need to discuss bad 2-orbifolds of $Q$, i.e. 2-orbifolds which have underlying space $S^2$ transversely intersect the singular locus $Q$ in exactly one point. This is equivalent to such a 2-orbifold not having a manifold cover (see for example Chapter 13 of Thurston's notes).

For the rest of the argument we will restrict to the case of an orbifold $Q$ with underlying space $S^3$ and having a singular locus a link $L$ labeled only by cone angle $\pi$.

EDIT: the paragraph below originally forgot to consider split links. Thanks to the OP for pointing this out.

1) In this context $Q$ is reducible if it contains a) an embedded bad sub-2-orbifold, b) an embedded 2-sphere which does not bound a 3-ball, or c) an embedded 2-fold orientable quotient of 2-sphere which does not bound the quotient obtained by rotating a 3-ball along an unknotted arc. This perspective more naturally generalizes to arguments of the above two papers. In the case of the question, we see that $Q$ will contain an embedded 2-sphere which does not bound a 3-ball if and only if $L$ is split and $Q$ will contain an embedded 2-fold orientable quotient of 2-sphere as in case c) if and only if $L$ is not prime. Furthermore, if $L$ is split or non-prime link, then each piece of a prime decomposition of the link $L$ will correspond to a prime decomposition of the orbifold $Q$ and a prime decomposition of $M_L$. (The general case involves consideration of more 2-orbifold quotients of $S^2$.) So the orbifold version of reducible involves both contains an essential 2-sphere quotient and bad 2-orbifolds.

2) The final case to consider is if $Q$ is irreducible but not geometric. In this case, $Q$ is the orbifold equivalent of being toroidal. The second reference uses the term toric for this case, but we will follow the first reference an say that an orbifold $Q$ is orbifold-atoroidal. Before the precise definition, notice that a torus $T^2$ has a unique orientable 2-fold quotient $S^2(2,2,2,2)$ (a 2-sphere with four cone points of order 2, sometimes called a pillow case). A rational tangle in a 3-ball can be obtained by taking two embedded unknotted arcs in 3-ball and then twisting along their boundaries. Assuming $Q$ irreducible, an embedded $T^2$ is incompressible if it does not bound solid torus to one side. In an irreducible 3-orbifold, an embedded $S^2(2,2,2,2)$ is incompressible if it does not bound a rational tangle to one side. If $Q$ contains an incompressible $T^2$ or $S^2(2,2,2,2)$, then we say $Q$ is orbifold-toroidal and otherwise $Q$ is orbifold-atoroidal. (The more general definition of orbifold-atoroidal involves considering more general quotients of $T^2$.) For the specific instance above, $Q$ is orbifold-toroidal if $L$ is a satellite link or if $L$ contains an essential Conway sphere, which is really just a concise restatement of the above paragraph. The decomposition of $L$ along satellite tori and essential Conway spheres leads to a JSJ-decomposition of $M_L$.

$\endgroup$
  • $\begingroup$ Thank you for your answer. I can't understand one thing: is condition 1) equivalent to say that the link $L$ is non-split and prime? $\endgroup$ – Antonio Alfieri Jan 11 '17 at 19:59
  • $\begingroup$ Yes, you are correct. I didn't include a discussion split links. I will edit that section to correct for this omission. $\endgroup$ – Neil Hoffman Jan 11 '17 at 21:04
  • $\begingroup$ Let me summarise what happens in the case of knots. According to Thurston a knot is either a satellite of a non-trivial knot, a torus knot or an hyperbolic knot. So if a knot is not a satellite (it is prime) and it is either a torus knot or an hyperbolic knot. In the first case its double cover $M_L$ is a Seifert manifold, while in the second one if there are no essential Conway spheres I can conclude that $M_L$ is hyperbolic. Am I right? $\endgroup$ – Antonio Alfieri Jan 11 '17 at 23:10
  • $\begingroup$ No, the double branched cover is a closed manifold making the problem more subtle. There are hyperbolic knots and links with non-hyperbolic double branched covers. For example: a two bridge (knot or) link has double branched cover a lens space; and a Montesinos (knot or) link has double branched cover a Seifert fibered space. Except for (2,n) torus links a two bridge link complement is hyperbolic. For Montesinos links of length 3, there are no essential Conway spheres while only a small subset are not hyperbolic. $\endgroup$ – Neil Hoffman Jan 11 '17 at 23:22
  • $\begingroup$ There are also links have double branched cover a Solv manifold, I believe these are generically hyperbolic as well. $\endgroup$ – Neil Hoffman Jan 11 '17 at 23:22

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.