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 $h^3$. Notice I added the $h^2$ term to make the inverse $R^{-1}$ be identical but with negative $h$, the factorial is just for fun. I wanted to test this idea 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 get an invariant in this case. Taking $\tilde{r}_{ij} = r_{ij}+r_{ji}$ we may do a little better in that now the curl yields $1+ h(EF+FE+\frac{1}{2}H^2)+\mathcal{O}(h^2)$ indicating our ambiguity may now be a central element (the Casimir at order h).