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Consider the following move on diagrams. I dimly recall hearing or reading that a sequence of such moves is sufficient to unknot any knot but I don't recall where I saw this. The strands in the diagram can be oriented arbitrarily. If anyone know a reference or proof I'd be grateful.

By the way, it is clear that this move is not sufficient for links since it preserves linking number modulo 2.

enter image description here

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Let $K$ be a knot. I've not worked out the details yet. I think a way to go about is to show that if $\beta$ is an $n$-braid with the closure of the braid $\bar{\beta}=K$. Then we can apply the relations $\sigma_i^2=\sigma_i^{-2}$, $i=1,2\ldots, (n-1)$ a finite number of times to $\beta$ to get $\beta\simeq(\sigma_1\sigma_2,\ldots,\sigma_{n-1})$. Now, this new braid closes to give the unknot, and applying these relations is equivalent to applying the moves in your diagram, so every knot can be transformed in to the unknot. This shouldn't be too difficult but you'd have to work out the details. – Dan Rust Apr 28 '12 at 22:01
@DanielR: Good idea, but you are only using the version of the move where both strands are oriented in the same direction. I suspect you also need to use the version where they are oriented in opposite directions. – Grumpy Parsnip Apr 30 '12 at 12:33
You're right Jim and it does seem harder than I first thought. It's at least obvious for closed 2-braids. Perhaps one approach might be to show that you can always reduce the braid index of a knot using the move in your diagram. – Dan Rust Apr 30 '12 at 23:38
I also dimly remember the same thing. But a google search turned nothing up. – Daniel Moskovich May 1 '12 at 3:45

The way that I would approach this problem would be using the machinery of claspers. Below, I use clasper language freely because I know that you are familiar with it. Your move implies that clasper edges behave like combinatorial objects- I can delete twists in them, and pass clasper edges through one another.

Begin by untying the knot using clasper surgery (for example using Y-claspers only, by unknotting using delta-moves, as in Murakami-Nakanishi/ Matveev). I don't need to remember twisting and linkage of edges thanks to your move- only the position of the leaves, and the combinatorial structure of the clasper (uni-trivalent graphs which end on the leaves) matters.

Next, I notice that the result of a Ck-move, if it happens inside a small ball with one unknotted line segment and no other clasper leaves inside, is ambient isotopic to a line segment whatever the combinatorial ordering of the leaves I think (draw it! The picture unravels "from the left". An illustration is Diagram 32 of [Edit: This is true for some orderings and not others, so more work is needed at this step] I also notice that I can pass one leaf through another "at the cost" of introducing a clasper-move with one more trivalent vertex, and that I can perform a "topological IHX" move inside a clasper to reduce it to "comb form", in which it represents a Ck move. This is enough- I choose a small ball, choose a clasper C, and pull all leaves of C inside the small ball. IHX so it becomes a Ck-move (maybe with leaves arranged in a strange order), and cancel it. I get left with a diagram with one fewer clasper (although the remaining claspers may be more complicated). Induction finishes. [Edit: It isn't clear that this process "converges"- see comments.]

This is one thing that clasper machinery is really well suited for, I think- it's the right language to discuss unknotting moves. Choose a clasper decomposition of the knot or link (replacing it by the unknot, with some tangled web of claspers inside it), identify moves on claspers induced by moves on knots, and show that they suffice to untangle the web, by pulling leaves into standard positions. To my taste, this leads to the nicest proof of "delta moves unknot".

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I like the spirit of this proof, as I am a big fan of claspers, but I'm worried that it doesn't "converge." When you pull claspers into a small ball, you create error terms, as you mentioned, but these error terms themselves will have to be pulled into a ball, and this may proceed ad infinitum. I think your proof shows that my stated move is sufficient modulo arbitrarily high order clasper moves. – Grumpy Parsnip May 1 '12 at 10:52
I see another problem. You can only ignore twisting of bands modulo 2... – Grumpy Parsnip May 1 '12 at 17:40
I found out that this is an open problem, see my answer above, but I awarded you the bounty since I didn't want it to go to waste, and you put some thought into the problem. – Grumpy Parsnip May 1 '12 at 23:48
You're right. It isn't trivial that this "converges", although it seems reasonable to try to prove "convergence"... I'm confused- I think I can ignore all twisting of edges, not just modulo 2. Clasper surgery on a basic clasper with a half-twisted edge is the LHS of your move, if clasper surgery on the untwisted edge is the right- no? – Daniel Moskovich May 2 '12 at 0:13
No I think, clasper surgery on a +1 twisted edge clasper is the left side and surgery on a -1 twisted edge clasper is the right side (or the reverse depending on convention). – Grumpy Parsnip May 2 '12 at 0:18
up vote 3 down vote accepted

At someone's suggestion I emailed Jozef Przytycki, who kindly sent me the following reply:

It is still an open problem (proven for knots of 12 or less crossings). I call this 4-move conjecture (Nakanishi 4-move conjecture). For links of two component, conjecture is that the target is the trivial link or a Hopf link. The possible counterexample is a planar 2-cable of trefoil (so 12 crossings). For 3 or more components nothing like this is possible (even if link is link homotopic to trivial one).

Look for example at:

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Is "region crossing change" the sort of move you're looking for? Here's a web page with a link to a news article (in Japanese) and a article preprint (in English):

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The operation in the diagram is a special case of a region crossing change (on a specific class of regions with 2 crossing points on the boundary). So, on its own, this doesn't prove that the operation in the diagram is an unkotting operation. – Dan Rust May 1 '12 at 0:39
A region crossing change is a move on a specific fixed diagram (no Reidemeister moves allowed), and this is a move on a knot (modulo Reidemeister moves). So it looks unrelated to me (but I really like Shimizu's result). – Daniel Moskovich May 2 '12 at 13:29

I think it can't be true as stated, since there are knot diagrams on which no such move can be made at all. See, e.g., $8_{18}$ at There is no place where you have two consecutive crossings involving the same two strands.

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I was careful to say any knot can be transformed to the unknot by such a move, not any knot diagram. You might have to perform additional Reidemeister moves. – Grumpy Parsnip May 1 '12 at 1:13
This doesn't remain true under reidemeister moves. I was under the impression that the operation in the image was allowed in addition to reidemeister moves. – Dan Rust May 1 '12 at 1:13
@DanielR: yes.${}$ – Grumpy Parsnip May 1 '12 at 1:14
Fair enough. I'd like to see an unknotting of $8_{18}$, then. – Gerry Myerson May 1 '12 at 1:30
There's aa recent paper of Duzhin and Shkolnikov in which they prove that no diagram of 8<sub>18</sub> has a diagram all of whose crossings are paired like this (conjecture of Przytycki), albiet with "opposite orientation". This is just a casual aside- the Duzhin-Shkolnikov result has no direct bearing on the OP's question, as far as I know. – Daniel Moskovich May 1 '12 at 3:51

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