Cubic skein relations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T11:26:24Z http://mathoverflow.net/feeds/question/49981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49981/cubic-skein-relations Cubic skein relations Hauke Reddmann 2010-12-20T17:33:54Z 2011-06-25T01:26:04Z <p>Hi,</p> <p>please note that this question deals with <em>undirected</em> knots/links!</p> <p>The most generic cubic skein relation for a knot polynome would be <img src="http://pic-hoster.net/upload/22041/SKEIN.gif" alt="S^2=wvS+w/S+w^2(u-v)I-u*inf"></p> <p>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.</p> <p>Lately, I used Kauffmans 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.</p> <p>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.*</p> <p>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 <em>find</em> it in the literature. </p> <p>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.)</p> <p>If I even write a paper on my S matrix work, of course I'd like to at least identify the solutions already known.</p> <p>Hauke Reddmann</p> <ul> <li>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.</li> </ul> http://mathoverflow.net/questions/49981/cubic-skein-relations/50010#50010 Answer by Noah Snyder for Cubic skein relations Noah Snyder 2010-12-20T23:15:21Z 2010-12-21T17:27:56Z <p>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. </p> <p>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). [<strong>Update</strong>: 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...</p> <p>On the other hand, if I understand everything correctly, according to the introduction of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.ajm/1098301002" rel="nofollow">this fascinating (though somewhat mysterious) paper</a> your cubic skein relation is <em>not</em> enough to define its value on all links. They claim that this result is proved <a href="http://arxiv.org/pdf/math/0205040v2" rel="nofollow">here</a>, 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 <em>the most natural way you might prove that you can evaluate all links using this relation</em> 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.</p>