Notes for AKT-170113/0:50:48

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