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A formula for (SU2) quantum 6j symbols exists. A formula expressing ordinary (q=1) 9j symbols in terms of 6j symbols is long known. Unfortunately, combining both (I tried it myself) got tricky - the associated graph K3,3 is nonplanar, at least one knot-type crossing is needed and first of all, this ruins the symmetry.

Can I find the quantum analogon of the standard sum over the product of three 6j symbols in the literature (or can someone post it here)?

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Regarding the symmetry (and with almost complete lack of understanding regarding the specific question): As long as you have a choice like which knot crossing you use, there is still hope for a symmetry if you use formulas that combine all such choices. Perhaps the easiest (but perhaps to easy) way to do something like this would be to make all choices and compute an average. – Johannes Hahn Jun 11 '13 at 12:14
Turaev writes that the standard definition of 9j symbols in terms of 6j symbols carries over directly to the quantum case (unlike the 3j symbols, which need separate consideration). This doesn't work? V.G. Turaev, Quantum Invariants of Knots and 3-Manifolds, page 343. – Carlo Beenakker Jun 11 '13 at 12:58
@ Carlo: This would mean that the standard definition already is "normalized" with respect to the annoying crossing. (I "pulled it through" in my own computations; if it's not needed at all - the better! :-) My lib has the book, I can look up the details. THX!) – Hauke Reddmann Jun 12 '13 at 10:57

to save you a trip to the library, here's the relevant paragraph from V.G. Turaev, Quantum Invariants of Knots and 3-Manifolds, page 343.

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