Are Quantum Clebsch-Gordan coeffients quantum group dependent? 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?)
 A: In chemistry the only Lie group that comes up is SU(2).  Naively you'd think the relevant group would be SO(3), which is the group of symmetries of space and so acts on the wave functions.  But actually because electrons are Fermions you probably also care about SU(2) (which is the double cover of SO(3)).
Thus, traditional "Clebsch-Gordon" coefficients involve the group SU(2).  You can generalize this in two independent directions:


*

*Replacing SU(2) with other Lie groups.  (This is classical mathematics, but not so relevant to chemistry, unless you somehow ended up doing chemistry in higher dimensions.)

*Quantizing.  (This is much newer mathematics involving quantum groups.)
I think you're getting confused because you hadn't heard of generalization 1 before.
(Warning: "replace all the integers with quantum integers" will cause some errors in general.  For example, the dimension of the fundamental representation of SO(n) is n, but the quantum dimension of the fundamental rep of quantum SO(n) is either $[n-1]_q+1$ or $[n-1]_{q^2}+1$ depending on parity.  In neither case is it [n].)
