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?