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

- Choose a quantum group G. (Without loss of geniality, G=A1(q) :-)
- Color the edges with representations of G.
- The representation 1 is "invisible"!
- Assign to each trivalent node Y a Clebsch. (Yes, you told me, for general G these are not necessarily known.)
- 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.
- As formula: If the Y is colored with the representations J1,J2,J3
and the tensor indices of Y are m1,m2,m3 then

$Y=Clebsch{[j1j2m1m2|j3m3]}(-1)^{(j1+j2-j3)/2}({[2j1+1]*[2j2+1]/[2j3+1]})^{1/4}$ - 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.)
- 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?

whatthey do but nothow. (And yes, the devil sits in the details as always.) $\endgroup$ – Hauke Reddmann Aug 17 '11 at 13:17