Notes for AKT-170113/0:50:48

Roland At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting [math]\displaystyle{ R = 1 + hr + \frac{1}{2!}h^2r^2 }[/math] and working modulo [math]\displaystyle{ h^3 }[/math]. I put the [math]\displaystyle{ h^2 }[/math] term to make the inverse [math]\displaystyle{ R^{-1} }[/math] be identical but with negative [math]\displaystyle{ h }[/math], the factorial is just a hint of more to come. I thought it was fun to have an example of this in [math]\displaystyle{ U(sl_2) }[/math] where you can check that [math]\displaystyle{ r_{12} = E_1F_2 + \frac{1}{4} H_1H_2 }[/math] is a solution to CYBE.