# Finite application of one of Reidemeister moves on a knot diagram

It is known that given a knot diagram we can transform it into a trivial unknot diagram by a series of Reidemeister moves. The key word is "series".

Can we transform any knot diagram using a finite number of exclusively one of Reidemeister moves into a trivial unknot? (given that we can deform its links in any way)

The immediate answer that comes up in my mind is "no", but I have no idea how to correctly prove it. Twist (type 1 Reidemeister move) and poke (type 2 Reidemeister move) moves, for example, decrease the number of crossings in a knot diagram, so a "greedy" way of thinking about this problem is that only first two Reidemeister moves can actually achieve it in a finite number of moves (guaranteed we can use unknot operations), and the third one can't.

It would be great if someone could provide some resources on further reading or point me into the right direction.

• Are "unknot operations" crossing changes? If so, the answer is yes. You perform crossing changes so that the knot is in "climbing position", i.e. it goes out of the page while the crossings occur, and then there is one rapid descent with no crossings. Then it's just a finite number of Reidemeister moves that decrease the number of crossings. Mar 19 at 17:55
• @RyanBudney Sorry, but it seems that I phrased it rather poorly (English is not my first language). By saying "unknot operations" I meant deformation of links, or just untangling of a knot. Mar 19 at 18:35
• I'm not certain how you are "counting" Reidemeister moves. You talk about "only first two", which sounds like you are counting. Are you asking if you can unknot applying only a totality of two Reidemeister moves? Or are you using the numbering system for Reidemeister moves, i.e. the "type 1" move vs. the "type 2" move? Mar 19 at 19:02
• @RyanBudney I again apologize for my wording. I am using the numbering system for Reidemeister moves. Mar 19 at 20:00

Here is an elementary proof that there exist unknot diagrams that can't be simplified using only R1, only R2, or only R3.

Let $$D$$ be a diagram of the unknot with an even number of crossings. Replace an arc locally with the following tangle: which can be done with three R1 and one R2 moves. Call the resulting diagram $$D'$$.

$$D'$$ cannot be turned into the trivial unknot with only R3 moves because those don't change the number of crossings. It cannot be turned into the trivial unknot with only R2 moves because those don't change the parity of the number of crossings. It cannot be turned into the trivial unknot with only R1 moves because reductions using R1 moves require a local property at the crossing that two adjacent arcs at the crossing are directly connected with an arc with no crossings, so the four upper crossings in this tangle do not admit R1 moves.

If you start with an unknot diagram with an odd number of crossings, you can do something similar by introducing a tangle without the kink at the bottom.

Moves that reduce crossings are not enough, because this paper

Kauffman, Louis H., and Sofia Lambropoulou. "Hard unknots and collapsing tangles." Introductory lectures on knot theory, Ser. Knots Everything 46 (2012): 187-247. arXiv link

established that

"This paper gives infinitely many examples of unknot diagrams that are hard, in the sense that the diagrams need to be made more complicated by Reidemeister moves before they can be simplified." "View Figure 4 for an unknotting sequence for the Culprit. Notice that we undo it by swinging the arc that passes underneath most of the diagram outward, and that in this process the number of crossings in the intermediate diagrams increases. In the diagrams of Figure 4 the largest increase is to a diagram of 12 crossings. This is the best possible result for this diagram."

• I can add that the Conjecture in this paper (on page 40) about the set of minimal hard unknot diagrams was resolved in arxiv version link (pages 10-11) Apr 2 at 20:09