The Mazur swindle can be used to show that any knot with an inverse under connected sum must be isotopic to the unknot. The isotopy, as I understand it, involves instantiating a pair of the knot and its inverse, canceling the first two knots, and repeating the process ad infinitum until there are no knots left. While this is perfectly fine using the given definitions, it seems to fit under the definition of a supertask, which seems like a bit of a cheat. The overall question is, how inherent is a supertask to this proof?

To make this more concrete, let us instead define the equivalence of knots by saying that they are related by a finite sequence of Reidemeister moves.

1) Can the proof by swindle be done under this definition?

2) Since the swindle involves a wild knot, does Reidemeister’s theorem apply?

3) If so, how does Reidemeister’s theorem reduce the supertask above to a finite task?


Here's a description of the Mazur swindle as I understand it (see https://math.stackexchange.com/a/3348047/172988):

You take the $x$-axis in $\mathbb{R}^3$ and tie into it $K_1$ in the intervals $[2n,2n+1]$ and $K_2$ into $[2n+1,2n+2]$ for $n\geq 0$ an integer. Then, since we assume $K_1\mathbin{\#} K_2$ is isotopic to the unknot, we can take intervals $I_n=[2n,2n+2]$ and $J_n=[2n+1,2n+3]$ and individually isotope $I_n$ or $J_n$ back to the $x$-axis for a fixed $n$. Then, the idea is that you can simultaneously do all the $I_n$ isotopies to get the $x$-axis itself; similarly you can simultaneously do all the $J_n$ isotopies to get just $K_1$ tied into $[0,1]$. One can then conclude $K_1$ is isotopic to the unknot. I am only sketching things out here, and there are various topological details to check. Also, this infinite knotted object is not a knot; perhaps one might call it a "longer knot." The key is that everything can be done with piecewise linear topology, which makes the usual 3-manifold arguments work out.

This version of the argument constructs an actual isotopy of $\mathbb{R}^3$, but it can be turned into diagrammatic manipulations by Reidemeister moves, in a certain sense. The only complication is that infinitely many non-conflicting Reidemeister moves are performed in parallel. But, it will complete after only finitely many such parallel moves.

I would expect it cannot in general require only finitely many Reidemeister moves. At least, the Reidemeister moves suggested above must be infinite in number since there are steps going from a diagram with infinitely many crossings to one with only finitely many. I doubt it's possible to use only finitely many Reidemeister moves except if $K_1$ is already known to be the unknot.

1) While the proof requires infinitely many Reidemeister moves to be performed, it's a bit different from a supertask because it's possible to do all of them in finite time by doing them in parallel. The diagram is split into countably many regions, at at each moment in time at most one Reidemeister move is performed in any given region.

2) Reidemeister's theorem does not apply since the object being isotoped is not a knot. However, all Reidemeister's theorem says is that if $D_1$ and $D_2$ are respectively diagrams for knots $K_1$ and $K_2$, then if $K_1$ is isotopic to $K_2$ there is a sequence of Reidemeister moves that transform $D_1$ into $D_2$. The swindle (when done diagrammatically) only needs the converse, that Reidemeister moves correspond to isotopies, which is true even in this setting. Furthermore, we don't need any theorems about existence of diagrams for wild knots since we provide the diagram ourselves.

3) Just to make what I said a bit clearer: Reidemeister's theorem is only used for converting isotopies into sequences of moves, but we can do the swindle by producing the program to perform infinitely many Reidemeister moves in finite time ourselves.

By the way, Mazur's swindle isn't necessary to prove that only the unknot has an inverse under connect sum. The reason is that Seifert genus satisfies $$g(K_1\mathbin{\#}K_2) = g(K_1) + g(K_2)$$ and one can prove rather easily that $g(K)=0$ iff $K$ is the unknot.

There are some possible examples of "supertasks" in topology that rely on the way certain limits converge. One is in things like mapping telescopes for CW complexes (see Hatcher's book on algebraic topology). The weak topology lets you push problems away into higher dimensions, so to speak.

You might be able to consider Alexander's trick as a sort of supertask, but this is a stretch.

  • $\begingroup$ Thank you. I believe this contains the answers to my question, but I disagree that what you say isn’t a supertask. Doing two Reidemeister moves in parallel isn’t a Reidemeister move; it’s two. As I read what you’ve written, it is not possible to prove that a knot with an inverse is equivalent to the unknot using the swindle, when the definition of equivalence is “related by a finite sequence of Reidemeister moves”. So, I would say that the swindle really is a swindle.... But, as you say, other proofs exist, so it doesn’t really change anything. $\endgroup$
    – user60338
    Jun 4 '20 at 2:21
  • $\begingroup$ @user60338 I missed the part where you defined knot equivalence through Reidemeister moves. Be aware that this is a somewhat arbitrary restriction because Reidemeister's theorem is for tame knots; but I don't want to discourage you from studying the ramifications here. From my point of view, the fundamental definition for knot equivalence is whether there is an ambient isotopy carrying the first knot to the second. There truly is an isotopy $[0,1]\times \mathbb{R}^3\to \mathbb{R}^3$ that carries the first "longer knot" to the second. $\endgroup$ Jun 4 '20 at 3:23
  • $\begingroup$ @user60338 I think the generalized Reidemeister theorem for these "longer knots" would be that every isotopy corresponds to finite sequences of parallel Reidemeister moves. It just seems different from the usual sorts of supertasks because the causal chain is only finitely long for any given part of the transformed knot. $\endgroup$ Jun 4 '20 at 3:30
  • $\begingroup$ @user60338 When I said "Reidemeister's theorem is for tame knots," I misspoke. I meant it's for tame $S^1$ knots. The "longer knots" are tame, but they aren't $S^1$ knots since they are a certain kind of embedded $\mathbb{R}$. $\endgroup$ Jun 4 '20 at 4:03
  • $\begingroup$ Sure, it's clear that there's a perfectly clear isotopy involved. But no one unties knots using an isotopy. The Reidemeister theorem (in my mind) tells us that isotopy for nice knots is equivalent to one way to thinking about the usual way people untie knots. The fact that this equivalence no longer follows through the swindle was the point of the question: from the point of view of Reidemeister moves, the swindle only proves that you can perform the unknot as a supertask. $\endgroup$
    – user60338
    Jun 4 '20 at 17:06

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.