4
$\begingroup$

It's well known that any oriented closed 3-manifold (topological or smooth) can be obtained by surgerizing along a (framed oriented) link $L$ in the 3-sphere $S^{3}$. Even better, Kirby found a complete set of relations that classify all 3-manifold in terms of links. In short, we have the map

$$\{\mbox{framed oriented link in }S^3\}/\{\mbox{kirby move}\} \xrightarrow{\sim} \{3\mbox{-manifold (oriented and closed)}\}.$$

A natural question arise.

  1. Given two (oriented and framed) link presentations, is there an algorithm (that halts in finite time) determines if they give rise to the same $3$-manifold?

Or even better, one may ask

  1. Given a link presentation $l$, is there any sort of normal form $[l]$ for it (in the sense that $l$ and $l'$ are related by Kirby moves if and only if $[l] = [l']$?

Related

  • A program KLO lets you play with link presentations under Kirby moves. But it doesn't seem to determine if two link presentations produce the same $3$-manifold.

  • The most popular programs snappea and regina present a $3$-manifold in terms of triangulation. It does not seem that they provide a way to determine from their Kirby diagram.

Improve this question
$\endgroup$
2
  • $\begingroup$ For your first question, the answer is yes because there is an algorithm (that uses geometrization) to tell if two closed 3-manifolds are homeomorphic. $\endgroup$
    – user101010
    Commented Sep 21, 2021 at 14:55
  • $\begingroup$ Regarding your final comment, if you have a Kirby (surgery) diagram, you can create a filled triangulation in SnapPea, and then convert that to a semi-simplicial triangulation afterwards. So yes it can be done, but it is a 2-step process. Or perhaps you are referring to having a "dictionary" of Kirby diagrams, i.e. a census of 3-manifolds indexed by Kirby diagrams. This is certainly a "doable thing" with the current technology. $\endgroup$
    – Ryan Budney
    Commented Sep 21, 2021 at 17:09

1 Answer 1

Reset to default
3
$\begingroup$

  1. It is a folklore result that geometrisation solves the homeomorphism problem for (compact, connected, oriented) three-manifolds. Kuperberg discusses this, and improves the running time to elementary recursive. So, use Snappy to convert the given Kirby diagrams to triangulations and appeal to the above.

  2. Sure. First place a reasonable complexity on diagrams. Say, count the number of crossings, add the bit-size of the framings, and break ties using some lexicographic complexity on encodings of graphs (as Burton does using isoSigs of triangulations). Now generate all diagrams up to the complexity of the given framed link. Use (1) to decide homeomorphism and take the smallest. This algorithm to compute normal forms is basically as fast as your solution to the homeomorphism problem (times the number of Kirby diagrams up to the given complexity, of course!).

And to repeat a remark from above - snappy can produce a triangulation from a framed link.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I was reading Matveev's book "Algorithmic Topology Classification of 3-Manifolds". In section 6.1, it seems to suggest that we know how to distinguish Haken 3-manifolds, but not in general yet. Luckily, all (but the trivial knot) knots in $S^{3}$ have their complement being Haken, so we can do that. The discussion extends to those of links by a relative distinguishing theorem (also discussed in 6.1). $\endgroup$
    – Student
    Commented Sep 25, 2021 at 20:11
  • $\begingroup$ You are correct that, before Perelman, the homeomorphism problem for three-manifolds was only known in the Haken case (and several other special cases). However, with geometrisation in hand, the general problem now has a solution. Again, see the paper of Kuperberg: arxiv.org/abs/1508.06720 $\endgroup$
    – Sam Nead
    Commented Sep 26, 2021 at 6:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .