Can the usual definition of a Lie algebra via commutators be simply adapted to quantum Lie algebras? Graphically you have the IHX scheme, with the X being a virtual crossing (so to say). Does it suffice to replace it with a real (knot-like) crossing and adapt the structure constants and coefficients before IHX a bit?
EDIT: $[x,y]_q=xy-q*yx$ and replacing all brackets by this q-bracket is a valid definition, but it does not answer my question, since in tensor form, yx is xy times a virtual crossing, so to say, and virtual and real crossings don't mix very well. I do have a conjecture for the formula, though. It is probably correct at q=1 (I guess it's eq. 4.36 of "Birdtracks", modulo a funny weight factor which is due to my choice of gauge, or that I use undirected graphs), and this graphic should make it clear (the blue J is the adjoint irrep, everything else can be any irrep):
The formula I'm seeking would then follow by applying it to a trivalent node made from two defining and one adjoint irrep. I will subsequently verify (or falsify :-) it by applying it to some simple examples like SU2(q) but that doesn't replace a proof...

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    $\begingroup$ Based on the use of the terms "crossing" and "virtual crossing" this appears to be a question about knot invariants. Certainly, it makes no sense as a question about "representation theory" (of what?), or even about Lie algebras in the traditional algebraic realm. $\endgroup$ Nov 27 '16 at 16:54

Third Option - Black: Any irrep - Blue: The adjoint irrep

This looks like the correct linear combination which at least partially answers my problem:
a) In the classic limit, the P term crosscancels (as long as lim P isn't infinite), and if lim Q=1/2 the IHX equation results. (It remains to express the unknown scalars P and Q as 6j salad.)
b) In the cubic skein case (E7 series), if the black line is the defining irrep, the lower pics sketch how to eliminate the adjoint. The result is identical to Przytyckis 6-tangle equation for cubic skeins, for a certain choice of P and Q.
The obvious generalization is asked in my next question on MO. :-)


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