In `Hidden Quantum Groups Inside Kac-Moody Algebra', by Alekseev, Faddeev, and Semenov-Tian-Shansky, a relationship between **quantum groups** and **affine Kac-Moody algebras** is shown for the WZW model.

Towards this end, the authors show that a certain **exchange algebra** (that is implied by the quantum group relations) implies the **algebra between a current and local operator** given in equation (45), i.e.,
$$[J_1(x), u_2(y)] = 2~\textrm{ln}(q)Pu_2(y)δ(x-y), $$
where the subscripts '1' and '2' indicate the relevant tensor factor the operators are valued in.
They claim this is a straightforward computation, but I cannot arrive at their result precisely, and am wondering if they have made an error.

My derivation is as follows.

They start from the exchange algebra (equation (33)) $$ u_1(x)u_2(y)=u_2(y)u_1(x)R(x-y), $$ where $R(x-y)$ is given in (42) as $$ R(x-y)=F_{12}q^{Psign(x-y)}F^{-1}_{21}, $$ and essentially encodes two

**R-matrices**that satisfy the quantum group relations. Here $P$ is the**permutation operator**, $F_{12}$ is an invertible matrix in $\mathbb{C}\otimes \mathbb{C}$, and $F_{21}=PF_{12}P$.Playing around with the algebra (33), it is easy to arrive at (44), which is equivalent to \begin{equation}\tag{1} u'_1(x+\epsilon)u_1(x)^{-1}u_2(y+\epsilon)=u'_1(x+\epsilon)u_2(y+\epsilon)R(x-y-\epsilon)^{-1}u_1(x)^{-1}. \end{equation} I am also able to derive the expansion of $R(x-y-\epsilon)^{-1}$, i.e.,

$$R(x-y-\epsilon)^{-1}=R(x-y)^{-1}+2 \epsilon \textrm{ ln}(q) F_{21}P F_{12}^{-1}\delta(x-y)+O(\epsilon^2),$$ where I have used $ R(x-y)^{-1}=F_{21}q^{-Psign(x-y)}F^{-1}_{12} $, $sign (0)=0$ and $\frac{d}{dx}sign(x)=2\delta(x)$. This is equivalent to the expression below (44) since $F_{21}P F_{12}^{-1}=P=F_{12}P F_{21}^{-1}$.Next I expand (1) in $\epsilon$, using equation (36), which is equivalent to $$ u'_1(x+\epsilon)u_1(x)^{-1}=A_1+\epsilon J_1(x) + O(\epsilon^2) $$ (where $A=a I_1$ for some constant $a$, and $J(x)$ is the WZW current), and $$ u_2(y+\epsilon)=u_2(y)+\epsilon u'_2(y)+O(\epsilon^2). $$ I also use $$ u_2(y+\epsilon)u'_1(x+\epsilon)=u'_1(x+\epsilon)u_2(y+\epsilon)R(x-y)^{-1}+u_1(x+\epsilon)u_2(y+\epsilon) \frac{d}{dx}R(x-y)^{-1},$$ (where $\frac{d}{dx}R(x-y)^{-1}=-2\textrm{ln}(q)P\delta(x-y)$) which follows from the exchange algebra relation.

Doing so, I arrive at $$ \begin{aligned} A_1 u_2(y)+\epsilon J_1(x)u_2(y) +\epsilon A_1 u'_2(y)=&u_2(y)A_1+\epsilon u_2'(y)A_1 + \epsilon u_2(y)J_1(x) \\ & +2\epsilon \textrm{ln}(q) Pu_2(y)\delta(x-y) \\& + 2\epsilon \textrm{ln}(q) Pu_2(y) A_1 \delta(x-y) + O(\epsilon^2), \end{aligned} $$ (assuming that $u_1(x+\epsilon)u_1(x)^{-1}=\epsilon I_1 + O(\epsilon^2)$, which is not provided, but seems natural and brings me closest to their answer) which implies (by collecting order $\epsilon$ terms) that $$ [J_1(x), u_2(y)] = 2(1+a)~\textrm{ln}(q)Pu_2(y)δ(x-y). $$ This is different from equation (45), as there is a factor of $(1+a)$ on the RHS.

Have I made a mistake, or is there an error in the paper?

**Note**: Based on მამუკა ჯიბლაძე's comments, I tried using $u'_1(x+\epsilon)u_1(x)^{-1}=\frac{A_1}{\epsilon}+J_1(x) + O(\epsilon)$ and $u_1(x+\epsilon)u_1(x)^{-1}=I_1 + O(\epsilon)$, but the problem still persists.