# Coproduct of the Drinfeld element?

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.

• Do you know about tensor categories? Have a look at Kassel's book "Quantum Groups". The element $u$ has an interpretation in the category of $H$-modules, and then proving the coproduct is just a matter of deforming certain string diagrams. Commented Oct 21, 2020 at 12:16
• @Jo Mo: Thanks a lot! I know about tensor categories, and about Kassel's book. I haven't read it yet, though. I will, if I find the time. Commented Oct 29, 2020 at 13:49