show/hide this revision's text 3 Fixed image.

If you take the diagram of the Reidemeister 3 move and "shortcircuit" two ends, you get (click http://imgur.com/kRvZa if Imgur hotlink doesn't work):
          Weak R3
I have circumstantial evidence that this weaker version is actually equivalent to R3. (Only in a computational sense! My hypothesis: If A and B are two diagrams of the same knot, while it might not be actually possible to go from A to B by applying weak R3 moves (+R2+R1, of course), the assumption that weak R3 holds forces invariant(A)=invariant(B) for any Lie group derived invariant. Likewise, in Kauffmans abstract tensor framework, just assume weak R3, solve and get the rest of the Yang-Baxter equation for free.)
Thus: Is there work on "alternative moves"? Can you construct a counterexample? (I.e. an pseudo-invariant which is constant under weak R3+R2+R1, but not under R3? The example must have a skein equation, though.)
show/hide this revision's text 2 Total overhaul

Another thing I "know" since 30 years but only recently I found a clue:
Assume a R matrix (graphic: %) can be decomposed into "projectors" (graphic: >=< ).(See the conditions here: http://mathoverflow.net/questions/65332/matrix-decomposition-the-other-way). It seems to me that this is always possible if it's

If you take the R matrix diagram of a Lie group, but may work even if not (i.e. just a random Yang-Baxter solution).In any case, you can apply the usual trick - if you can pull a string over all nodes >= ,i.e. Reidemeister 3 for nodes holds, Reidemeister 3 for crossings follows. Of course yourprojectors must obey the usual 6j rules either. Now I found something fascinating:The diagrams for Biedenharn-Elliott move and Reidemeister 3 for nodes have five open ends.I played with "quadratic" graphs with only four open shortcircuit" two ends, you get (click http://imgur.com/kRvZa if Imgur hotlink doesn't work):
I have circumstantial evidence that this weaker version is actually equivalent to R3.(Only in a computational sense! My hypothesis: If A and the results B are exactly two diagrams of the same - the consistency rules following from the diagrams are exactlyBiedenharn-Elliott and Reidemeister 3 for nodes again.
After "undecomposing" everything againknot, the following weakened Reidemeister 3 versionresults: (hope the image comes through - http://imgur.com/kRvZa)
The above musings are while it might not even a handwaving proof, so I won't dare be actually possible to state "For Lie groups, Reidemeister 3 follows go from A to B by applying weak Reidemeister 3". StillR3 moves (+R2+R1, of course), it would be epic if one could replace the annoying 3+3-tangle condition with 2+2assumption that weak R3 holds forces invariant(A)=invariant(B) for any Lie group derived invariant. And when I solve Likewise, in Kauffmans abstract tensor equationsframework, I know not a single counterexample either. So my questionjust assume weak R3, solve and get the rest of the Yang-Baxter equation for free.)
Thus: Do Is there work on "alternative moves"? Can you know construct a counterexample- any construction where ? (I.e. an pseudo-invariant which is constant under weak Reidemeister 3 holds R3+R2+R1, but Reidemeister 3 is violatednot under R3? The example must have a skein equation, though.)
show/hide this revision's text 1

Can Reidemeister 3 be weakened?

Another thing I "know" since 30 years but only recently I found a clue:
Assume a R matrix (graphic: %) can be decomposed into "projectors" (graphic: >=< ). (See the conditions here: http://mathoverflow.net/questions/65332/matrix-decomposition-the-other-way). It seems to me that this is always possible if it's the R matrix of a Lie group, but may work even if not (i.e. just a random Yang-Baxter solution). In any case, you can apply the usual trick - if you can pull a string over all nodes >= , i.e. Reidemeister 3 for nodes holds, Reidemeister 3 for crossings follows. Of course your projectors must obey the usual 6j rules either. Now I found something fascinating: The diagrams for Biedenharn-Elliott and Reidemeister 3 for nodes have five open ends. I played with "quadratic" graphs with only four open ends, and the results are exactly the same - the consistency rules following from the diagrams are exactly Biedenharn-Elliott and Reidemeister 3 for nodes again.
After "undecomposing" everything again, the following weakened Reidemeister 3 version results: (hope the image comes through - http://imgur.com/kRvZa)
The above musings are not even a handwaving proof, so I won't dare to state "For Lie groups, Reidemeister 3 follows from weak Reidemeister 3". Still, it would be epic if one could replace the annoying 3+3-tangle condition with 2+2. And when I solve abstract tensor equations, I know not a single counterexample either. So my question: Do you know a counterexample - any construction where weak Reidemeister 3 holds but Reidemeister 3 is violated?