Background: I'm a quantum chemist and even once wrote a proggie for 3j and 6j symbols. Imagine my gaping mouth when I read the paper (Reshetikhin/Turaev, I think) who let them pop up in my fave, knot theory. My ultimate goal would be to rewrite my zoo of S matrices with 3j symbols etc. so that checking the invariance under Reidemeister moves would just need a few applications of Biedenharn-Elliot and whatsnot. (AND I don't have to solve 1000 nonlinear equations in 20 variables to get my S in the first place :-)
Namagiri is throwing epiphanies at me lately in torrents: meanwhile I can rewrite most of my zoo into simpler trivalent vertexes (looking like this: =< ) - and maybe the matrix elements of those tensors are just 3j symbols? One day later I tripped over q-alg/9706029v2 and if you look closely, the intertwiners ARE 3j symbols, look like trivalent vertices and fulfil same relations as those in the Kuperberg G2 paper!
But now I'm caught between two contradictory statements. My instinct says that you just draw quantum brackets around any integer in the standard sum formula: http://en.wikipedia.org/wiki/Table_of_Clebsch-Gordan_coefficients and voila, Quantum Clebsch. But in the paper above, a psi for a representation replaces the usual J, and also the Wiki implies Quantum Clebsches depend on the quantum group. (Which makes more sense anyway, because trivalent-vertex-for-G2 != trivalent-vertex-for-A2, for example).
Question: Can you confirm that Quantum Clebsches depend on the quantum group? If yes, why does the intuitive approach (replace all integers by quantum integers) fail here? Or maybe it fails generally, except for a special quantum group, which is? (And does, by chance, Scott Morrisons QuantumKnot package contain a function for computing Quantum Clebsches for which the formula is known?)