Affine Kac-Moody algebra from quantum group exchange algebra 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.
 A: $\let\eps\varepsilon$
I believe there is an error in the expression for the derivative of $R^{-1}$, it should be
$$
\frac{d}{dx}R(x-y)^{-1}=-2\textrm{ln}(q)P\delta(x-y)R(x-y)^{-1}
$$
Accordingly, one must have
$$
R(x-y-\epsilon)^{-1}=(1+2\eps\ln(q)\delta(x-y)P)R(x-y)^{-1}+O(\epsilon^2).
$$
Substituting this into your
$$
u'_1(x+\eps)u_1(x)^{-1}u_2(y+\eps)=u'_1(x+\eps)u_2(y+\eps)R(x-y-\eps)^{-1}u_1(x)^{-1}.
$$
gives
\begin{align*}
u'_1(x+\eps)u_1(x)^{-1}u_2(y+\eps)=&
u'_1(x+\eps)u_2(y+\eps)R(x-y)^{-1}u_1(x)^{-1}\\
+&2\eps\ln(q)\delta(x-y)u'_1(x+\eps)u_2(y+\eps)PR(x-y)^{-1}u_1(x)^{-1}
\end{align*}
Also your expression at the end of 3. should be
\begin{multline*}
u'_1(x+\eps)u_2(y+\eps)R(x-y)^{-1}\\
=u_2(y+\eps)u'_1(x+\eps)+2\ln(q)\delta(x-y)Pu_1(x+\eps)u_2(y+\eps)R(x-y)^{-1}.
\end{multline*}
Substituting it gives
\begin{align*}
u'_1(x+\eps)u_1(x)^{-1}u_2(y+\eps&)=
u_2(y+\eps)u'_1(x+\eps)u_1(x)^{-1}\\
&+2\ln(q)\delta(x-y)Pu_1(x+\eps)u_2(y+\eps)R(x-y)^{-1}u_1(x)^{-1}\\
&+2\eps\ln(q)\delta(x-y)u'_1(x+\eps)u_2(y+\eps)PR(x-y)^{-1}u_1(x)^{-1}.
\end{align*}
Then, presuming the correct version of (36) is
$$
u'(x+\eps)u(x)^{-1}=\frac1\eps A+J(x)+O(\eps),
$$
we get
$$
[J_1(x),u_2(y)]=2\ln(q)\delta(x-y)Pu_1(x)u_2(y)R(x-y)^{-1}u_1(x)^{-1}+O(\eps),
$$
which gives the needed equality.
