4
$\begingroup$

It is well known that any knot diagram can be unknotted by a sequence of crossing changes (i.e., changing an overcrossing with an undercrossing or vice versa) and of Reidemeister moves. More precisely, one can first perform a certain number of crossing exchanges to modify the given knot diagram into a diagram representing the unknot (the minimal number of such changes is known as the unknotting number of the knot diagram), and then one can perform a sequence of Reidemeister moves to actually unknot the diagram. Yet, during this process the number of crossings in not necessarily decreasing. A classical example is the knot diagram at the bottom of page 4 here, where the knot diagram actually represents an unknot, and so one can effectively unknot it by a sequence of Reidemeister moves, but the sequence of Reidemeister moves needed to unknot will increase the number of crossings at some point. Yet, if one can use both Reidemeister moves and crossing changes, then it is easy to see how to build a sequence of unknotting moves always decreasing (or better, never increasing) the number of crossings.

So my question is: it is always true that given a knot diagram $\Gamma$ there exists a sequence of Reidemeister moves and crossing changes that transforms $\Gamma$ in the standard diagram for the unknot in such a way that the number of crossings is decreasing (or better, non-increasing) along the process?

The reason for this question is, apart from knowing it for the sake of itself, an attempt to rigorously understand a line in Link polynomials and a graphical calculus by Louis Kauffman and Pierre Vogel, where they say "Since any 4-valent planar graph can be undone by a series of moves of the type -shadow of a Reidemeister move" (page 78). Namely, although this line is pretty clear at an intuitive level, my feeling is that in order to make their argument completely rigorous one should know that number of crossing decreasing sequences (possibly involving crossing changes) always exist.

$\endgroup$

1 Answer 1

6
$\begingroup$

Edit: This answer has been edited to correct a mistake graciously pointed out by Ian Agol in the comments below.

The answer to you question is yes. It is possible to take a sequence of crossing changes and Reidemeister moves to unknot a diagram in a non-decreasing manner.

Here is an algorithm to accomplish this. For a fixed (finite crossing) diagram D of a knot K (with at least one crossing), pick a crossing, number it 0 and choose one of the two arcs in the under crossing. Label that arc a0. Follow a0 to the next crossing label that 1 and the next arc a1. Continue the process until either you get back to crossing 0 via the over crossing or you get back to a crossing already numbered. In the second case, start at that crossing and label it 0 choose the under crossing that is part of the path and label that a0 and start the process over.

Added case (selecting an 'innermost loop'):

This path is a loop which bounds a disk in the plane. Shade in this plane and check to see if there are arcs of the diagram contained in the disk which begin and end one the boundary loop. If such an arc exists, check the arc for loops, and repeat the process until you have found an `innermost loop', i.e. a loop that 1) bounds a disk which contains no loops and 2) such that no subpath of the loop is itself a loop. For each loop there is a well defined crossing to begin and end the loop determined by the diagram. Call that crossing 0.

Thus we can assume that we have a path that begins and ends at crossing 0 which is innermost. Make crossing changes so that this path only contains over crossings (except at crossing 0). Then use Reidemeister II and III moves to consolidate this path to a small neighborhood of crossing 0 such that no crossing of the diagram are in this neighborhood. The innermost property assures that Reidemeister II moves will decrease the number of crossings between the loop and the rest of the altered diagram. Finally, use a Reidemeister I to remove crossing 0.

If crossings remain, apply the method above until there are no more crossings.

Of course, it well known that the crossing changes are required to run such an algorithm. For example, there is a 10 crossing diagram of the unknot aptly dubbed "the Culprit", Figure 2 of:

Allison Henrich and Louis H. Kauffman, MR 3193721 Unknotting unknots, Amer. Math. Monthly 121 (2014), no. 5, 379--390.

which cannot be simplified in a non-decreasing fashion using only Reidemeister moves.

$\endgroup$
6
  • 2
    $\begingroup$ I don't think that your description is quite right. I think that you need to assume that the loop is ``innermost", in order to be able to apply only Reidemeister II and III moves. In this picture, I don't think it is possible to simplify using only II and III: dl.dropboxusercontent.com/u/8592391/Loop.jpeg If the loop is innermost, then one can show that all arcs crossing through are simple, and then one can show that using II and III moves, one can isotope these arcs one at a time off of the loop, at which point one can apply the Type I move. $\endgroup$
    – Ian Agol
    Aug 8, 2016 at 18:52
  • $\begingroup$ @IanAgol Thanks for pointing this out. You are right. The example you give can not be simplified monotonically using just Reidemeister II and III moves (with Reidemeister II causing the problem). You are also right that the notion of "innermost" should fix the problem. $\endgroup$ Aug 9, 2016 at 15:56
  • $\begingroup$ Okay, that's better, although there's still some justification needed to see that type II and III suffice .... $\endgroup$
    – Ian Agol
    Aug 9, 2016 at 16:26
  • $\begingroup$ To see that type II and III moves suffice to simplify an innermost loop, take an innermost bigon inside of this loop. Then one can show that all arcs cross from one side of this bigon to the other, and intersect each other at most once. Now one can use type III moves to "move" triangles to the boundary at which point the number of intersections between arcs in this bigon decreases after another type III move. Once there are no intersections between arcs left, type III moves can be made to empty the bigon, then a type II move applied to decrease the crossing number. $\endgroup$
    – Ian Agol
    Aug 10, 2016 at 0:13
  • $\begingroup$ A different way to see this is that any innermost loop is either empty, or contains a bigon. As a note on the history of this technique, Steinitz, in his proof to the famous theorem about convex polyhedra, proved that any inclusionwise-minimal bigon can be reduced using type III moves. $\endgroup$ Jan 24, 2017 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.