# Quantum knot invariants dummyfied

This construction should define an invariant for colored tangled trivalent graphs.

1. Choose a quantum group G. (Without loss of geniality, G=A1(q) :-)
2. Color the edges with representations of G.
3. The representation 1 is "invisible"!
4. Assign to each trivalent node Y a Clebsch. (Yes, you told me, for general G these are not necessarily known.)
5. Throw in some spin and phase factors for elegance. They are not really required, but e.g. if the down leg of the Y is colored with 1, graphically (remember 3.!) you have a cap. And it's nice to have cap=cup, and since cap*cup=$\delta$ this requires some phase fumbling.
6. As formula: If the Y is colored with the representations J1,J2,J3 and the tensor indices of Y are m1,m2,m3 then
$Y=Clebsch{[j1j2m1m2|j3m3]}(-1)^{(j1+j2-j3)/2}({[2j1+1]*[2j2+1]/[2j3+1]})^{1/4}$
7. I can't give a formula for the R matrices in terms of the Y tensors yet, because I would need the R-symbols (see e.g. math-qa1004.5456v2), i.e. the eigenvalues of the R matrices. (Side Question: Are these generally known for given G? For A1, at least, they are easy.)
8. In any case, the only :-) thing effectively to prove in this approach would be an overfiendish identity as an integral over 15 Legendre polynomials. Assuming you know the Clebsches and the R-symbols for G, of course. This is not completely wishful thinking - most 3j/6j identities translate to interesting graph identities.(Essentially the construction which I asked about here Matrix decomposition the other way can be used to patch R matrices from Y tensors.)

OK, here comes the real question. Assuming that 8 follows from some magic property of Clebsches, and my sketchy construction would be made mathematically exact, would I have done something new? Or is this just equivalent to a "dummyfication" of Reshitikhine-Turaev/Turaev-Viro? (Invent. Math. 92, 527-553 (1988) and so on)
Or to specialize to A1 again, is the colored Jones polynomial generalizable as an invariant of tangled trivalent graphs?

• A corollary to Noah's answer below is that when you say "tangled trivalent graph" you should mean among other things that your tangled graph is framed or fat or something along those lines. Now, this is what most people mean, so I have no objection to you responding "yes, of course I mean that". But it does deserve saying. – Theo Johnson-Freyd Aug 16 '11 at 19:55
• Indeed. The point here is that quantum knot invariants typically don't satisfy the Reidemeister 1 move. If you look at a single object at a time you can change your normalizations to make it invariant under R1, but for trivalent graphs you're looking at multiple representations and there's no way to get invariance under R1 for all the representations. – Noah Snyder Aug 16 '11 at 21:15
• I like the idea of "without loss of geniality." I understand the questioner is not a native speaker, but still it is a pleasant was to do math. – Scott Carter Aug 17 '11 at 1:21
• @Scott - you mean I fell for a classical "false friend"? Indeed. I should have known. Nevermind, it was a pun anyway :-) @Noah - Ah, I knew I had the RT paper lying around somewhere. I reread it now, but even with a crash course in Lie algebra I still barely understand what they do but not how. (And yes, the devil sits in the details as always.) – Hauke Reddmann Aug 17 '11 at 13:17