1. Problem
Let $(H,R)$ be a quasitriangular, finite-dimensional Hopf algebra with antipode $S$ and coproduct $\Delta$. Using Sweedler notation define the Drinfeld element $u$ as $u:= S(R_{(2)})R_{(1)}$.
A text I am reading proposes (without proof) that the Drinfeld element is a group-like element up to a correction term, i.e. that the following identity holds:
$$\Delta (u)=\overline R \cdot \overline R_{21} \cdot (u \otimes u).$$
Here $\overline R = R^{-1}$ and $\overline R_{21}= \tau_{H,H}(R)^{-1} $ with $\tau_{H,H}$ the twist map on $H$.
I am unable to prove this identity.
2. My thoughts so far
Another way to think about it:
The above identity is equivalent to saying that $Q \cdot \Delta (u)= u \otimes u$ (or equivalently $\Delta (u) \cdot Q = u \otimes u$) with $Q:=R_{21}\cdot R_{12}$ the monodromy element.
I tried using the fact that $S^2(h)=uhu^{-1}$ for all $h \in H$ (i.e. writing $u$ as $u:= S^{-1}(S^2(R_{(2)}))R_{(1)}$, the antipode is invertible because we are finite-dimensional), and then using that $S$ is an antialgebra homorphism and that the coproduct is an algebra homomorphism. Further, I have tried using the defining properties of a quasitriangular Hopf algebra (the second and third here relate the coproduct to the universal $R$-matrix) — with no success.
