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 $r_{ij} = E_iF_j + \frac{1}{4} H_iH_j$ is a solution to CYBE. If I got it right the positive Reidemeister 1 curl yields the value $1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8})$ bad news, we need the element $S$ to fix it.