I came across this question while reading Chapter 15 of Conformal Field Theory by Di Francesco.
So the action of the Weiss-Zumino-Witten(WZW) model is as follows:
$S = \frac{1}{4a^2}\int d^2x \rm{Tr}'(\partial^{\mu}g^{-1}\partial_\mu g) + k\Gamma$$$S = \frac{1}{4a^2}\int d^2x {\rm Tr}'(\partial^{\mu}g^{-1}\partial_\mu g) + k\Gamma$$
where g$g$ is in some representation of Lie Group G$G$ and $\Gamma$ is the Weiss-Zumino-Witten term. The action is invariant under the transformation
$g \rightarrow \Omega_1 g \Omega_2$$$g \rightarrow \Omega_1 g \Omega_2$$
where $\Omega_{1,2}$ belongs to G$G$. So it has symmetry $G\times G$. But, it later pointed out that due to the holomorphic factorization of the g$g$ field, this symmetry is actually to $G(z)\times G(\bar{z})$.
My question is, this enlargement from a global symmetry to a local symmetry is reminicent of gauge symmetry. But I think things are here are a bit different because here the symmetry is not a redundancy. Is there a way to make this contrast/comparison between symmetry in WZW model to gauge symmetry more precise? Is it because some special properties of gauge transformations in 2d?
