OK, I have to ask a dumb question again: Where do Lie groups enter in the construction of the Reshitikhin-Turaev invariant? The parts of the proof I understand are that 6j symbols take care of themselves. So why not define 6j symbols axiomatically (Biedenharn-Elliott plus symmetry plus function value at a 0 argument should suffice)? You also need an integer-valued triangle function f(a,b,c) giving the number of "irrep" c in the tensor product of a and b for a start (to construct an infinite version of a fusion category, or whatchamacallit), so {abc|def}=0 if f(a,b,c)=0. Not even talking of multiplicity hell. But these only look like "technical" difficulties to me, but not impossible. Of course, with a Lie group you get the 6j symbols "for free" (eh, semisimple? Otherwise I don't see why you can't use e.g. Vogels general Lie group).

In fact, the last months I constructed general 6j symbols for the E7 series, just for fun, here is one:

{JJA|VVV}=-I*Sqrt[Q20]*Sqrt[Q11]*Sqrt[Q10]*Sqrt[Q02]*Q11/Q32/Q30/Q43/Q22

(V defining, J adjoint, A antisymmetric, Qxy means QuantumInteger[x*m/2+y], where m is the parameter in Westburys "Magic" paper. Insert m=-2/3 to get the SU2 6j. If not, there is a typo...). Unfortunately, with higher irreps involved the 6j symbols lost such "pretty" form, I landed in phase choice hell and stopped.

At the moment I construct 6j symbols without resorting to Lie groups at all, just with diagrams. (I'm not even past the Clebsch-Gordan series of V*V, since the technical difficulties are enormous.) So again, is there any reason why the Reshitikhin-Turaev construction with 6j symbols, but without Lie groups, shouldn't work? A non-constructive existence proof would completely suffice, and I could lay myself to rest finally after 20 years :-)