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]. Notice I added 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 for fun. I wanted to test this idea in [math]\displaystyle{ U(sl_2) }[/math] where you can check that [math]\displaystyle{ r_{ij} = E_iF_j + \frac{1}{4} H_iH_j }[/math] is a solution to CYBE. If I got it right the positive Reidemeister 1 curl yields the value [math]\displaystyle{ 1+ h(EF+\frac{1}{4}H^2)+\frac{1}{2}h^2(2E^2F^2 + EH^2F+EFH+\frac{H^4}{8}) }[/math] bad news, we need the element [math]\displaystyle{ S }[/math] to get an invariant in this case. Taking [math]\displaystyle{ \tilde{r}_{ij} = r_{ij}+r_{ji} }[/math] we may do a little better in that now the curl yields [math]\displaystyle{ 1+ h(EF+FE+\frac{1}{2}H^2)+\mathcal{O}(h^2) }[/math] indicating our ambiguity may now be a central element (the Casimir at order h).