TITLE: Drinfeld't map, centre of quantum group, representation category of quantum group
QUESTION [9 upvotes]: My question is about the Drinfeld't map between $Rep(U_q(\mathfrak{g}))$ and $Z(U_q(\mathfrak{g}))$. I have heard the reference 1989 paper by Drinfeld't "Almost cocommutative Hopf algebras" - but this paper I cannot find in English. 
Define the Drinfeld't map as follows. Given a representation $\pi_V: U_q(\mathfrak{g}) \rightarrow End(V)$, $(1 \otimes \pi_V)(R^{21}R) \in U_q(\mathfrak{g}) \otimes End(V)$, so let $c_V = tr((1 \otimes \pi_V)(R^{21}R)) \in U_q(\mathfrak{g})$. It is easy to see that the map $c$ is compatible with the additive structure on $Rep(U_q(\mathfrak{g}))$; is it also compatible with the multiplicative (tensor) structure on that category (my calculations say otherwise, but I may have messed up somewhere)? Then the claim is that $c_V \in Z(U_q(\mathfrak{g}))$ and these elements generate the center under suitable hypothesis. Why is this? 
If there is an ambiguity: by $U_q(\mathfrak{g})$, I mean the quantum group defined in Jantzen's "Lectures on Quantum Groups", Chapter 4. 
Edit: Cf. the comment below, feel free to make any assumptions that make the problem easier, e.g. characteristic $0$ and $q$ not a root of unity.

REPLY [2 votes]: The fact that these elements are in the center is quite direct, just using the basic property of $R$ and of the trace, and is true in any quasitriangular Hopf algebra.  You really need to know what you want the quantum group for, because it is certainly possible to define the QG so that this is all the center, or so that it is not!