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Has anybody seen seen quantum knot invariants associated to (E6, 27) or (E7, 56) worked out in the literature? Even for just simple knots like the trefoil or figure-8?

I suspect these haven't been worked out, but if anybody knows of a reference containing these or even just discussing them, please let me know.

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It is certainly possible to evaluate these for links which can be obtained as the closure of a braid with at most three strings. –  Bruce Westbury Jan 22 '12 at 22:17
    
I'm pretty sure that the quantum groups Mathematica package which is part of the Knot Atlas package (katlas.org/wiki/Main_Page) can do this. It'll probably be pretty slow. You can ask Scott Morrison if you want more info. –  Noah Snyder Jan 22 '12 at 22:17
    
I spoke to Scott a couple of days ago. His package cannot compute quantum invariants for F4 or En due to the complexity of some intermediate computations. He noted that one could avoid these issues by, e.g. writing down the quantum positive roots in terms of the PBW bases... working this out is [currently] a bit beyond my primitive background in QA. However, if anybody knows of a good source for such things, that would be equally useful. –  Ross Elliot Jan 22 '12 at 22:38
    
Ok, my bad. I'd used it for some more basic quantum En calculations before, but I guess they were all a lot simpler than what you need. –  Noah Snyder Jan 23 '12 at 0:15
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My recollection from conversations with Scott is that the main obstacle is that arithmetic with rational functions is slow. For the 3-strand braid group you're already talking about composing matrices that are roughly 20,000 by 20,000. –  Noah Snyder Jan 23 '12 at 16:59
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2 Answers 2

$(E7, 56)$ belongs to a series for which I computed some skein relations in my phd thesis (see p55-56 of http://web.univ-ubs.fr/lmam/patureau/articles/these.ps.gz with $Y=X^{19}$). Unfortunately the set of skein relations is not complete but you can use it to compute the quantum invariants of small knots.

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Dear Bertrand & Ross - could you check the following? I neither speak French nor Math :-) but I computed the same skein relations with magic and trickery (so my results are unproven, of course). I get 11+6+2*8+2*3=39 basis 3+3 tangles (which I here already split into the D6h symmetry classes) and under the E7 family polynome 10+5+2*7+2*3=35 are independent. The linear rest should be your skein relations. So, are there exactly 4 (including symmetry-transformed!) of them, i.e. your Figs. 2/3? Since they don't look symmetric under rotation, it's hard to check for me.

(BTW, for 2+2 tangles, I "know" an even generalized result (of your Fig. 1) for 20 years :-)

Maybe going to 4+4 tangles could solve the problem of having a complete reducing skein set (or at least give an unproven solution) but since that might need hundreds of basis tangles you a) either have to crowdsource the calculation or b) do it with birdtracks or c) use math (I'm out then :-)

EDIT: Now that I'm back at my PC with access to the literature, 4+4 doesn't look too promising either. Kuperbergs G2 paper says there are 455 crossingless freeways with 8 endpoints. Uck. (And under E7 instead, a lot of them are inaccessible.) So if one could do the following: 1. Prove that: IF you can reduce 1-gons and 2-gons and two adjacent 3-gons (works for all the E7 family) AND an adjacent 3- and 4-gon pair AND two 3-gons touching on a corner (five crossings; i.e. simpler diagrams now span the 8-endpoint vector space...or so I hope!) THEN you could reduce any link diagram, and: 2. the compatibility rules of this reducing set are finite (I don't have much hope for that either!)... the whole E7 family polynomial proof would reduce to skein diagram manipulation (of gargantuan size, of course). Yup, it's probably better to use math instead :-)

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