# Around the Montesino-Nakanishi 3-move

I have a few questions around the 3-move. I know it's NOT an unknotting move (but who needs knots with 20+ crossings anyway :-) by the recent proof of Przytycki.
1. In another paper about the third skein module, he conjectures about a knot polynomial based on a cubic skein equation. Is this polynomial still conjectural (at last so I understood the paper)? (I'm not up-to-date with the literature, naturally.)
1a) Regardless whether yes or no, is this polynomial "meant" for directed links?
2. Instead of 3-move, consider the skein equation
$S^2+a1*S^1+a2*S^0+a3*S^{-1}+a0*H=0$
(where S is the S matrix and H the Temperley-Lieb generator/the 0 tangle). This is more or less equivalent to the 3-move (e.g. I always "knew" the Borromean double would spell trouble, which also was the timewise first counterexample for Montesino-Nakanishi). I say "more or less" because if you eliminate H you are already in the fourth skein module. Also this is for undirected links only. But I only want to know if the unknotting power is the same, i.e. deleting clasps unknots exactly the same knots as the 3-move.

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Removing a clasp involves to crossings, a $3$-move involves three crossings. They aren't the same. –  Charlie Frohman Oct 10 '11 at 12:18
Of course not! But consider e.g. this: In the cubic skein paper mentioned above, there is a skein computation showing how to flip all crossings of two adjacent triangles with the 3-move. About the same equation lies around in my collected trash, only I used the clasp equation above and the deduction was less elegant. The formulae relating the parameters are uncannily similar too. Thus I think, even if the moves are not the same, the unknotting powers are equal. (I have more "debris".) –  Hauke Reddmann Oct 11 '11 at 9:08