Consider the move R2d which applies to tangle diagrams, described in the Figure 1.

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Question: is it possible to achieve the move R2d by a sequence of splices and loop adding or erasing, i.e. is there any sequence of moves SPLICE 1, SPLICE 2, LOOP 1, LOOP 2 (described in Figure 2.) which transforms the LHS tangle diagram into the RHS tangle diagram from the Figure 1?

My guess is that it is not possible, but I am looking for a proof.

The moves SPLICE and LOOP are the following:

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2 Answers 2


This rather an idea which may turn out to be wrong. I suppose that you want to prove that given the diagram on the left with endpoints marked you cannot go the the diagram on the right with the same marking. (By marking I mean that there are no implicit identifications introduced by some symmetry of the rest of the tangle.)

Define the spliced form of a diagram with four endpoints to be the diagram obtained by elimination of all crossings by SPLICE1 or SPLICE2 and then erasing all loops. It seems to be definable. Then I think you can prove that it is invariant under all of your operations, but it is different for the diagrams concerned.

  • 1
    $\begingroup$ You can also show that these diagrams admit different good 2-colorings. Call a 2-coloring of lines and endpoints good if the endpoints have the same color as the lines they belong to, while crossings are colored in the only way compatible with splice1/2. $\endgroup$ Apr 24, 2013 at 6:40
  • $\begingroup$ (Crossings are not colored, the lines at crossings are colored) $\endgroup$ Apr 24, 2013 at 6:45
  • $\begingroup$ Thanks, I'll try. But is not obvious that the spliced form is invariant. This would mean that the order of applications of splices 1/2 and loop 1/2. Notice that all moves go in both directions and one may obtain complex diagrams from simple ones. In case anybody interested, what is not clear at all to me is what happens if we work not with tangle diagrams, which are globally planar graphs, but with "locally planar" tangle diagrams, i.e. we allow the possibility that two edges of the diagram cross without counting this a real crossing. $\endgroup$ Apr 24, 2013 at 8:01
  • $\begingroup$ Correction: "This would mean that the order of applications of splices 1/2 and loop 1/2" ... is commutative. $\endgroup$ Apr 24, 2013 at 8:03

Peter Kravchuk seems to be right. I shall write a more detailed version of his answer.

To any diagram A associate it's reduced diagram R(A), which is obtained by the following procedure: first use SPLICE 1/2 from left to right for all crossings, then use LOOP 1/2 from right to left in order to eliminate all loops which are present at this stage. Notice that:

  1. The order of application of the SPLICE 1/2 moves does not matter, because they are applied only once per crossing. There is a finite number of splices, equal to the number of crossings. Define the bag of splices SPLICE(A) to be the set of SPLICE 1 applied.
  2. Same is true for the order of eliminations of loops by LOOP 1/2. There is a finite number of loop eliminations, because the number of loops (at this stage) cannot be bigger than the number of edges of the initial diagram. Define the bag of loops LOOP(A) to be the set of all loops which are present after all splices are done.
  3. All this is true for locally planar tangle diagrams, i.e. if we allow that the 4-valent graph with decorations which represents the tangle diagram to be only locally planar (in this case we can safely ignore the accidental edge-crossings which are not nodes of the graph).

Let us now check that the reduced diagram does not change if one of the 4 moves is applied to the initial diagram.

Apply a SPLICE 1/2 move to the initial diagram A, from left to right, and get B. Then SPLICE(B) is what is left in the bag SPLICE(A) after taking out the respective splice. Also LOOP(B) = LOOP(A) because of the definition of bags of loops. Therefore R(A) = R(B).

Apply a SPLICE 1/2 from right to left to A and get B. Then R(A) = R(B) by the same proof, with A, B switching places.

Apply a LOOP1/2 from left to right to A and get B. The new loop introduced in the diagram does not participate to any crossing (therefore SPLICE(A) = SPLICE(B)), so we find it in the bag of loops of B, which is made by all the elements of LOOP(A) and this new loop. Therefore R(A) = R(B).

Same goes for LOOP1/2 applied from right to left.

Finally, remark that the reduced diagram of the LHS in Figure 1 is different than the reduced diagram of the RHS in Figure 1, therefore the move R2d cannot be achieved with a sequence of splices and loops addition/elimination.

Bonus 1: same is true in the class of locally planar tangle diagrams.

Bonus 2: With the notations for the oriented Reidemeister moves from the article Minimal generating sets of Reidemeister moves by Michael Polyak, (only that I put "R" where he puts "$\Omega$"), we notice that from the list of 16 oriented Reidemeister moves, only 4 of them cannot be obtained by a sequence of SPLICE and LOOP (we just proved that!), namely R2c, R2d, R3a, R3h. This gives:

The set of 12 oriented Reidemeister moves made by all moves excepting R2c, R2d, R3a, R3h is NOT generating.


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