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Assume you have a 4-valent graph (i.e., a knot universe, i.e. a collection of self-intersecting curves). Your allowed moves are the equivalents of Reidemeister 1, 2, 3, just with 4-nodes instead of over/underpasses. (I don't think you need a pic :-) Is the following logic OK? Effectively, your graph is a knot where overpasses can be turned into underpasses ad lib. Since this is an unknotting move, and the rule set above is equivalent to the "real" Reidemeister plus some crossing flips, the pseudo-Reidemeister moves can turn any graph into disconnected loops.

(If yes, I think I have a complete no-brainer proof for the invariance of the Dubrovnik 2-variable polynomial, but I bet you find loads of brains lacking in the fine print :-) BTW, is there a no-brainer proof for the fact that crossing flip is an unknotting move? (Or maybe the graph version is even simpler to prove.) Obviously, you as knot theory profis don't actually need another proof for the Dubrovnik, since one exists... but I prefer one my no-brain can understand :-)

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A no-brainer proof that crossing flips unknot knots, is imagining walking along your knot. The first time you see a crossing change it (if necessary) so that you walk along the top strand. Once you have traversed the knot it will be unknotted. As to your actual question, I am unsure what you mean. Are you thinking of working with a planar 4-valent graph, and allowing planar projections of the Reidemeister moves? –  Jana Archibald Sep 30 '11 at 13:12
    
I think the argument in your first paragraph is correct. –  Kevin Walker Sep 30 '11 at 15:26

2 Answers 2

The comments here seem to be answers and the answers comments, so I do not know if you feel that your question has been answered or not! At the risk of beating a dead horse, let me make sure. I am assuming you mean a quadravalent graph in the plane, so there are no ordinary crossings. Following the algorithm of the first commenter (traverse the graph, and mark the first arc you go over in each crossing as nominally the "over" strand), you can identify the graph with an unknot with an overlapping projection. The sequence of Reidemeister moves that turns that unknot into the standard projection of the unknot give you a sequence of your Rmoves that does the same thing.

In answer to the first answer, I believe the 4-valent version of the Reidemeister moves means draw the ordinary Reidemeister moves and the replace each crossing with a 4-valent vertex. Anyway, that was the assumption of my answer!

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+ for comments are answers and answers are comments! :-) –  Joseph O'Rourke Oct 8 '11 at 14:32

This is a comment requesting clarification, not an answer. At least to me, it is not self-evident what is the 4-valent equivalent of the three Reidemeister moves. For example, Reidemeister I seems fundamentally 3-valent. If you allow the move shown below, which is akin to Reidemeister II, and if your collection of curves includes only proper crossings, as illustrated, then indeed the graph can be converted to a collection of disjoint (but often nested) loops: in the example, four loops nested inside a fifth.
       Reidemeister 4
Perhaps you could clarify your intent?

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(sorry for the late answer - chess tournament :-) Here are my rules for expanding a 4-node graph. imgur.com/XQVuI Some needed comments on it: 1. Since you can gauge-transform the 4-valent node by adding equal amounts of infinity- and 0-tangle, one variable will be superfluous anyway. 2. You'll see the rules are NOT completely analogous to Reidemeister 1-3, but the difference is "debris" and won't count for the "any graph can be simplified" argument. <continued> –  Hauke Reddmann Oct 4 '11 at 9:36
    
3. After solving, the two variables (e.g. writhe normalization and unlink value) of the Dubrovnik polynomial remain. 4. The last equation shows how to expand a knot into 4-nodes. The approach is a somewhat trickier version of Kuperbergs G2 invariant, only with 4-nodes instead of 3-nodes. The straightforward version (expand a triangle into lower terms) only leads to a special Dubrovnik (the same as in his Spider paper). –  Hauke Reddmann Oct 4 '11 at 9:41

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