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Jun 24, 2012 at 18:10 comment added MTS I know this is nitpicky, however: $U_q(\mathfrak{g})$ is not a quasitriangular Hopf algebra in the true sense since the R-matrix does not live in the algebraic tensor product $U_q(\mathfrak{g}) \otimes U_q(\mathfrak{g})$. Of course the R-matrix acts in tensor products of finite-dimensional representations, and this gives the braiding, etc. But I think it pays to be careful about how one refers these things.
Mar 12, 2012 at 14:40 comment added Daniel Moskovich You might also be interested in Dancso's paper, which avoids the use of associators in the construction of a universal invariant (it's written for KTG's, but they replace q-tangles, and seem like a better formalism), and is quite elegant: arxiv.org/abs/0811.4615
Nov 12, 2011 at 4:29 vote accept John Pardon
Oct 18, 2011 at 22:36 answer added Adrien timeline score: 10
Oct 18, 2011 at 22:16 comment added Jim Conant This seems highly unlikely to me. My approach to rule it out would be to look at simple knot and link diagrams to constrain what the chord-diagram valued R-matrix could possibly be.
Oct 18, 2011 at 22:02 history asked John Pardon CC BY-SA 3.0