Notes for AKT-170113/0:50:48

Roland At 38:12 Dror mentions a solution to CYBE already gives a knot invariant by setting $R=1+hr+{\frac {1}{2!}}h^{2}r^{2}$ 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 $r_{ij}=E_{i}F_{j}+{\frac {1}{4}}H_{i}H_{j}$ is a solution to CYBE. If I got it right the positive Reidemeister 1 curl yields the value $1+h(EF+{\frac {1}{4}}H^{2})+{\frac {1}{2}}h^{2}(2E^{2}F^{2}+EH^{2}F+EFH+{\frac {H^{4}}{8}})$ 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).