I have been using Lorenz Eberhardt's 2019 ESI lecture notes on WZW model. Below Equation 3.5 on Page 8, it is written that the current algebra, which forms a Kac-Moody Algebra, is the main organizing principle for WZW models. However, we can similarly write the current algebra for any 2d CFT as a Kac-Moody algebra(explained below), for example in Equation 2.60, Section 2.6, Page 37 of [1]. Is it then possible to make any 2d CFT into a WZW model by designing a Verma module which is a representation of the Kac-Moody algebra for the given CFT (For example this is done for free bosons in Section 15.6 of the Yellow book) ?

[1] Introduction to Conformal Field Theory: With Applications to String Theory by Plauschinn and Ralph Blumenhagen

Argument why conserved current forms a current algebra -

Consider quasi-primary chiral fields, $\phi_i(z)$ and $\phi_j(z)$, with mode expansion given by $\phi_i(z) = \sum_{m} \phi_{(i)m}z^{-m-h_i}$ and similarly for $\phi_{j}$. Then using contour integral representation of the modes and OPE of two chiral quasi-primary fields(expressed in terms of other quasi-primary fields)

$$ \left[\phi_{(i) m}, \phi_{(j) n}\right]=\sum_{k} C_{i j}^{k} p_{i j k}(m, n) \phi_{(k) m+n}+d_{i j} \delta_{m,-n}\left(\begin{array}{c} m+h_{i}-1 \\ 2 h_{i}-1 \end{array}\right) $$ where \begin{aligned} &p_{i j k}(m, n)=\sum_{r, s \in Z_{0}^{+}} C_{r, s}^{i j k} \cdot\left(\begin{array}{c} -m+h_{i}-1 \\ r \end{array}\right) \cdot\left(\begin{array}{c} -n+h_{j}-1 \\ s \end{array}\right) \\ &r+s=h_{l}+h_{j}-h_{k}-1 \\ &C_{r, s}^{i j k}=(-1)^{r} \frac{\left(2 h_{k}-1\right) !}{\left(h_{i}+h_{j}+h_{k}-2\right) !} \prod_{t=0}^{s-1}\left(2 h_{i}-2-r-t\right) \prod_{u=0}^{r-1}\left(2 h_{j}-2-s-u\right) \end{aligned} Now, let us assume that our theory has N - chiral quasi-primary fields of conformal dimension 1(which [1] calls current). It can be shown using the above relation that $$\left[j_{(i) m}, j_{(j) n}\right]=\sum_{k} C_{i j}^{k} p_{111}(m, n) j_{(k) m+n}+d_{i j} m \delta_{m,-n}$$ which after rotating the fields and rescaling them becomes into $$ \left[j^{(i)}_{m}, j^{(j)}_{n}\right] = \sum_{l} f^{ijl} j^{l}_{m+n} + k m \delta^{ij} \delta_{m,-n} $$ where j with superscripts are the new fields. This is the affine Kac-Moody algebra structure.

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    $\begingroup$ "We can similarly write the current algebra for any 2d CFT as a Kac-Moody algebra": this is wrong, a 2d CFT needs not have any current algebra. $\endgroup$ Commented Sep 14, 2021 at 7:08
  • $\begingroup$ @SylvainRibault Blumenhagen's CFT on Page 37 shows that if we assume the existence of N quasi-primary fields of conformal dimension (1,0), then the Laurent modes of these fields can be arranged into a Kac-Moody algebra and isn't this the current algebra of the theory ? So instead of 2d CFT, if I say that 2d CFTs which have at least 1 primary(or quasi-primary) field of conformal dimension (1,0), then we have a current algebra which is Kac-Moody. Can such a CFT be made into a WZW model ? $\endgroup$
    – alpha
    Commented Sep 14, 2021 at 13:15
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    $\begingroup$ Having fields of dimensions $(1, 0)$ does not imply a Kac-Moody algebra. For example, in the $O(n)$ model you have a bunch of primary fields $J^a$ of dimension $(1, 0)$, but they are not conserved currents i.e. $\bar\partial J^a\neq 0$. $\endgroup$ Commented Sep 14, 2021 at 13:32
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    $\begingroup$ I do not see why conserved currents would necessarily form a Kac-Moody algebra. If there is an argument to that effect in ref. [1], maybe you could paraphrase it in your question. $\endgroup$ Commented Sep 14, 2021 at 15:33
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    $\begingroup$ Thank you for the argument. It looks quite sound and I do not see a counter-example, so I am happy to accept that conserved currents of dimension (1, 0) have to form a Kac-Moody algebra. (Even if they are only quasi-primary.) $\endgroup$ Commented Sep 15, 2021 at 7:40

1 Answer 1


Accepting the argument of [1] that conserved currents of dimension $(1,0)$ form a Kac-Moody algebra, the remaining question is whether any CFT with Kac-Moody symmetry (left and right) is a WZW model. The answer is no: https://en.wikipedia.org/wiki/Wess%E2%80%93Zumino%E2%80%93Witten_model#Other_theories_based_on_affine_Lie_algebras


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