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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.

  1. 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 $F_{21}=PF_{12}P$.

  2. 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 $F_{21}P F_{12}^{-1}=P=F_{12}P F_{21}^{-1}$.

  3. 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.

  4. 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.

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    $\begingroup$ Cross-post at physics.stackexchange.com/questions/584708/… $\endgroup$
    – Mtheorist
    Oct 8, 2020 at 6:57
  • $\begingroup$ Re "slightly different" [from the expression below (44)]: it is the same. Since $F_{21}=PF_{12}P$, both $F_{12}PF_{21}^{-1}$ and $F_{21}PF_{12}^{-1}$ are equal to just $P$. $\endgroup$
    – მამუკა ჯიბლაძე
    Oct 8, 2020 at 7:56
  • $\begingroup$ მამუკა ჯიბლაძე Thanks, I will edit the question to reflect this. $\endgroup$
    – Mtheorist
    Oct 8, 2020 at 8:06
  • $\begingroup$ I think there is something wrong with your expression for $u_1(x+\epsilon)u_1(x)^{-1}$. It implies that this quantity is $\epsilon$-close to zero, while actually it must be $\epsilon$-close to the identity. I believe it should be $I_1+O(\epsilon)$, no? $\endgroup$
    – მამუკა ჯიბლაძე
    Oct 8, 2020 at 12:38
  • $\begingroup$ I thought so too, but when I look at equation (36), it should go to $u’(x)u(x)^{-1}=J(x)$ as $\epsilon$ goes to zero, but this is not the case either. I am rather confused about this. $\endgroup$
    – Mtheorist
    Oct 8, 2020 at 12:42

1 Answer 1

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$\let\eps\varepsilon$

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.

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  • $\begingroup$ Thank you very much, it is a very nice proof. I am a little confused by the first line though, when I take the derivative of $R(x-y)^{-1}$, I get $F_{21} (- 2\textrm{ln}(q) P \delta(x-y)) q^{-Psign(x-y)}F^{-1}_{12}$. I can't seem to change this into the form you get, unless $F$ is trivial. $\endgroup$
    – Mtheorist
    Oct 13, 2020 at 9:01
  • $\begingroup$ @Mtheorist I thought that $F_{12}$ does not depend on $x$, while $\frac d{dx}$ of $q^{-P\mathrm{sign}(x-y)}=e^{-P\mathrm{sign}(x-y)\ln(q)}$ is $\frac d{dx}(-P\mathrm{sign}(x-y)\ln(q))e^{-P\mathrm{sign}(x-y)\ln(q)}$, is not it so? $\endgroup$
    – მამუკა ჯიბლაძე
    Oct 13, 2020 at 9:51
  • $\begingroup$ You are right. But $F_21 (-2\textrm{ln}(q)P\delta(x-y))q^{-Psign(x-y)}$ only equals $(-2\textrm{ln}(q)P\delta(x-y))q^{-Psign(x-y)} F_{12}^{-1}$ if we note that $sign(0)=0$. $\endgroup$
    – Mtheorist
    Oct 13, 2020 at 9:57
  • $\begingroup$ Another point I am confused about is your version of my expression at the end of 3., you get a factor of $Pu_1(x+\epsilon)u_2(y+\epsilon)R(x-y)^{-1}$. Should it not be $u_1(x+\epsilon)u_2(y+\epsilon)PR(x-y)^{-1}$? $\endgroup$
    – Mtheorist
    Oct 13, 2020 at 9:59
  • $\begingroup$ You are right, this $P$ interferes with what I tried to say. In fact there are two things that are not clear to me too. First, what exactly is $R^+(x-y)$: they have $R^+$ and $R(x-y)$ but I could not find their definition for $R^+(x-y)$. Second, what exactly is $\frac d{dx}q^{-P\operatorname{sign}(x-y)}$, is it $-2\ln(q)\delta(x-y)Pq^{-P\operatorname{sign}(x-y)}$ or $-2\ln(q)\delta(x-y)q^{-P\operatorname{sign}(x-y)}P$ or are these two things actually the same? Or is it still something different? $\endgroup$
    – მამუკა ჯიბლაძე
    Oct 13, 2020 at 12:04

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