# Vanishing of field strength in gauged WZW model

Consider gWZW action $S_{gWZW}(g,A)=S_{WZW}(g)+S_{gauge}(g,A)$, where $S_{WZW}$ is usual WZW action being sum of sigma model and WZ terms for field $g$ taking values in group $G$ and $S_{gauge}=\int d^2\sigma\text{Tr}(A_+\partial _-gg^{-1}-A_-g^{-1}\partial _+g-g^{-1}A_+gA_-+A_+A_-)$ is gauge action which depends on field $g$ and gauge fields $A_\pm$, the later take values in subalgebra of group $G$ algebra which we want to gauge.

According to equations of motion following from gWZW action field strength $F_{+-}=\partial _+A_--\partial _-A_+ +[A_+,A_-]$ vanishes on-shell. To show it one shall project equation of motion for field $g$ on gauge subalgebra and substitute constraints for gauge fields $A_\pm$.

My question is: are we allowed to use $F_{+-}=0$ equation of motion while studying symmetries of action - that is are we allowed to substitute $F_{+-}=0$ in variation of action justifying it by claim that fields $A_\pm$ are non-dynamical?

I have natural doubts that this is allowed because as described above equation $F_{+-}=0$ is obtained not just with the help of equations following from variation of action by non-dynamical $A_\pm$ fields but also using $g$ field equations. This is topical issue e.g. when one deals with sort of supersymmetric extension of gWZW model (some non-ordinary or potentially-deformed kinds of them - because there's no necessity to set $F_{+-}=0$ in usual supersymmetric gWZW model). If so to obtain $F_{+-}=0$ equation one also needs to use fermionic fields equations of motion.

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Your intuition is right: you are not allowed to assume the equations of motion if you're determining the symmetries of the action. Quantum mechanically, the theory may be defined as a path integral over all configurations - including (and especially) those that don't satisfy the classical equations of motion - and all of them contribute to the transition amplitudes.

So if the symmetry didn't exist for the configurations that don't obey the equations of motion, the calculated amplitudes wouldn't obey the symmetry, either.

While you're not allowed to assume the equations of motion, you do allow variations of the Lagrangian that are total derivatives - so that the action which is the integral is still preserved assuming a good behavior at infinity. In some contexts, this "symmetry up to total derivatives" plays a similar role to using the equations of motion.

Was there some particular symmetry you wondered about here? The essence of your question is general and has no special relationship with the WZNW models.

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