please note that this question deals with undirected knots/links!

The most generic cubic skein relation for a knot polynome would be


where $w^3$ is one positive writhe unit. The form is fairly obvious from some self-consistency demands. Now since 20 years or so I try to prove that this already IS a knot polynome but I can't even prove that it is defined for all links, let alone that this relation is self-consistent.

Lately, I used Kauffman's abstract tensor approach to classify "all" S matrix solutions (when you have a state model, the proof is in the computing). I found also some S matrices included having a cubic skein relation, some maybe yet unknown.

The above relation would generalize a) Kauffman AND Dubrovnik polynome, b) the product Jones(x)Jones(y), c) the Kuperberg G2 spider and d) of course all solutions I mentioned above.*

Now sordidly I'm a complete amateur, and if yesterday a paper appeared proving my hypothesis, I might not even recognize THAT it does, let alone find it in the literature.

So, this is my question: Do you know of additional knot polynomes with cubic skein relations, possibly falling under d)? Or maybe even a compendium of all known polynomes? One even an amateur can understand? (E.g. Reshitikhin/Turaev definitely goes over my head.)

If I even write a paper on my S matrix work, of course I'd like to at least identify the solutions already known.

Hauke Reddmann

*: $w=z^5,u-z^{14}+z^2-z^{-6},v=z^6-z^{-2}+z^{-14}$ is an example. It also pops up in a possible generalization of the B2 spider as I very recently found.

  • 2
    $\begingroup$ I can't see the diagram. $\endgroup$ Commented Dec 20, 2010 at 18:20
  • $\begingroup$ @OP: I hope you don't mind; I fixed the link to the diagram. $\endgroup$ Commented Dec 20, 2010 at 19:28
  • $\begingroup$ Which now seems to be down again. $\endgroup$ Commented Dec 20, 2010 at 19:38
  • $\begingroup$ Do you really want 2 equals signs there? The square of a crossing is invertible, it's typically not 0. $\endgroup$ Commented Dec 20, 2010 at 22:33
  • $\begingroup$ I still can't see the diagram $\endgroup$ Commented Dec 21, 2010 at 0:20

1 Answer 1


First off I think there's at least one other knot polynomial satisfying these skein relations: the Reshetikhin-Turaev invariant coming from the 133-dimensional representation of E7. Unfortunately for you, I don't think anyone's ever given an elementary description of that knot polynomial. On the other hand, it would be nice (and, in my opinion, publishable) to see a purely elementary description of this RT invariant, so if you find a knot polynomial which you can prove exists by elemenatary means you should feel free to contact me and I'll let you know if it's E7.

I may have missed something (I only did a quick heuristic search), but I suspect that this E7 example is the only other RT invariant which satisfies this sort of skein relation (other than the ones that you already listed). [Update: there's also the spin representation of Spin(12).] I expect that there are no known knot polynomials satisfying this skein relation which don't come from RT... I'd have to do some more checking to be totally sure...

On the other hand, if I understand everything correctly, according to the introduction of this fascinating (though somewhat mysterious) paper your cubic skein relation is not enough to define its value on all links. They claim that this result is proved here, but I'm a little confused as Dabkowski and Przytycki result seems to me to be slightly weaker. That is, it seems to me that they're only proving that the most natural way you might prove that you can evaluate all links using this relation doesn't work. However, I might be missing something here. At any rate, I would not be very optimistic about that skein relation being enough to reduce everything.

  • $\begingroup$ Thanks for the answer and especially the links - only a new overview article on knot polynomials (with some actual values - I may dream :-) would be more helpful here. Hauke P.S. Apologies to everyone - I should choose my image provider more carefully... $\endgroup$ Commented Dec 22, 2010 at 16:10

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