Given a closed orientable 3-manifold $M^3$ and two knots $K_1$ and $K_2$ in $M$, is there an algorithm to decide if $K_1$ and $K_2$ are isotopic? Is there an algorithm to decide if there is a homeomorphism $f: M \to M$ with $f(K_1) = K_2$?

I would expect these questions to be involved (at least as difficult as the recognition problem for 3-manifolds) but I thought that the answer is possibly a folklore consequence of geometrization. The special case of the first question for $K_1$ unknotted follows from a result of Hass and Lagarias (see Theorem 1.2). The case where $M = S^3$ is a result of Waldhausen using normal surface theory.

As an aside, I don't know anything about the corresponding questions where instead of considering knots we consider closed embedded surfaces $F_1$ and $F_2$. Even in the case where $M = S^3$ (so only the isotopy question is relevant) the genus of $F_1$ and $F_2$ is greater than one, this is a total mystery to me. By looking at tori bounding regular neighborhoods of knots, this is strictly more difficult than the corresponding questions for knots. Are there results in this direction?

  • 1
    $\begingroup$ There's a good amount of work on the isotopy problem for surfaces. Much of the work is for special classes of surfaces where the solution is very clean. But the isotopy problem for surfaces is very close to a lot of fairly subtle problems, like the isotopy problem for embedded graphs, or the isotopy problem for 3-dimensional submanifolds. So usually in the literature the work on these problems in phrased in a somewhat specialized language. $\endgroup$ Apr 7 at 18:01
  • $\begingroup$ @RyanBudney Do you by any chance know if the surface isotopy problem is known in $S^3$? Is there maybe an analog of the Gordon-Luecke result for surfaces (it is certainly true for genus < 2)? $\endgroup$
    – user101010
    Apr 8 at 9:00
  • $\begingroup$ There won't be anything quite like Gordon-Luecke for closed surfaces of genus $g \geq 2$. With closed surfaces you have the issue that often neither side is a handlebody. To determine if two surfaces are isotopic you look at the decomposition they give of $S^3$, $S^3 = V_1 \cup V_2$ for the first surface, $S^3 = W_1 \cup W_2$ for the second. Call the $V_i$ manifolds the "sides" of the surface. You then check to see if there are diffeomorphisms $V_i \simeq W_j$ and if that diffeomorphism extends over $V_j \simeq W_i$. $\endgroup$ Apr 8 at 18:01
  • $\begingroup$ Only fairly recently has the isotopy problem for Heegaard surfaces in non-Haken manifolds been resolved. mathscinet.ams.org/mathscinet-getitem?mr=3865653 To me this seems like the hardest case of the isotopy problem, but I guess that no one has worked out an isotopy algorithm in general. $\endgroup$
    – Ian Agol
    May 2 at 16:26

Regarding the second question. This reduces to the homeomorphism problem for three manifolds, which is known due to geometrisation. Here is a sketch. Fix $M$ as well as the knots $K$ and $L$. Let $X_K$ and $X_L$ be the knot complements. We mark the boundary of each with the meridional slope of $K$ and $L$, respectively. We ask our solution to the homeomorphism problem if $X_K$ and $X_L$ are homeomorphic. If the answer is "no" then we are done. If the answer is "yes" we ask for all homeomorphisms from $X_K$ to $X_L$. For each of these we restrict to the boundary and see if the resulting map takes the meridian of $K$ to that of $L$. If any do, the answer is "yes". If none do, the answer is "no".

[Edit - as pointed out in the comments, there may be infinitely many homoeomorphisms between $X_K$ and $X_L$. This happens, for example, when $K$ and $L$ are both unknots in the three-sphere. In any case, the self-homeomorphisms of $X_K$ restrict to give a virtually cyclic group of homeomorphisms of $\partial X_K$. When this group is infinite, the cyclic subgroup is generated by a Dehn twist about the slope that dies in homology.]

Regarding the first question. If $M = S^3$ then again this reduces to the homeomorphism problem for three-manifolds, by appealing to Property P (due to Gordon-Luecke). Some fiddling about is needed to deal with mirror images.

I don't know the status of the first question for other manifolds off the top of my head. I'll report back if anything occurs to me. :)

  • $\begingroup$ The isotopy problem is the diffeomorphism of pairs problem for pairs $(M,K)$ where you add the additional problem of deciding if the diffeomorphism of $M$ is isotopic to the identity. Well, generally it's weaker than that problem but this certainly "covers" the problem. I do not believe there is an algorithm for this problem in the literature but it would seem this is a problem that you could build an algorithmic solution to. Certainly in practice this is a reasonable-enough problem that does not take long to solve. $\endgroup$ Apr 7 at 17:45
  • $\begingroup$ Thanks for the answer :-) Is there a place in the literature where 3-manifold recognition is written down for compact orientable manifolds with torus boundary? Also how do I go through the potentially infinite set of homeomorphisms between $X_K$ and $X_L$? I imagine generically this set is finite, but to turn what you suggest into an algorithm... $\endgroup$
    – user101010
    Apr 8 at 9:04
  • $\begingroup$ @user101010 - Greg Kuperberg has an article on the folklore result that the homeomorphism problem follows from geometrisation. Here is a link to the arXiv version: arxiv.org/abs/1508.06720 - he also gives bounds beyond "the problem is decidable". :) $\endgroup$
    – Sam Nead
    Apr 8 at 12:34
  • $\begingroup$ @SamNead Thanks - I linked that article in the question actually. Unfortunately, it is mentioned there that he only carries things out in the closed case and leaves the case with boundary for future work. $\endgroup$
    – user101010
    Apr 8 at 14:35
  • $\begingroup$ Ah, I didn't click through! Thank you for pointing that out. Hmm. Looking at GK's paper again, I believe that his discussion goes through without changes for three-manifolds where all boundary components are tori. $\endgroup$
    – Sam Nead
    Apr 8 at 15:21

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.