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
(knotlike) crossing and adapt the structure constants and coefficients
before IHX a bit?
EDIT: $[x,y]_q=xyq*yx$ and replacing all brackets by this qbracket 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):
http://imgur.com/sz8Nylh
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...

3$\begingroup$ 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. $\endgroup$– Victor ProtsakNov 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 6tangle equation for cubic skeins, for a certain choice of P and Q.
The obvious generalization is asked in my next question on MO. :)