Is it true that in abelian Chern-Simons theory diffeomorphisms differ from ordinary gauge transformations trivially?

Question

In Henneaux's Lectures on the Antifield BRST Formalism for Gauge Theories, it is claimed in Exercise 1 that diffeomorphisms $\delta_\xi A_\mu=\xi^\rho\partial_\rho A_\mu+\partial_\mu\xi^\rho A_\rho$ differ from ordinary gauge transformations $\delta_\Lambda A_\mu=\partial_\mu\Lambda$ by a trivial gauge transformation $\delta_\mu A=\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\frac{\delta S}{\delta A_\nu(y)}$, for some $\mu_{\mu\nu}(x,y)=-\mu_{\nu\mu}(y,x)$. In here $S=\int\text{d}x\epsilon^{\mu\nu\rho}F_{\mu\nu}A_\rho$ is the action for Abelian Chern-Simons theory.

I tried to proof this. The equations of motion are $\frac{\delta S}{\delta A_\mu(x)}=\epsilon^{\mu\alpha\beta}\partial_\alpha A_\beta(x)$. Therefore, what we want to proof is that there is a $\Lambda$ and $\mu_{\mu\nu}(x,y)$ such that $$\delta_\xi A_\mu-\delta_\Lambda A_\mu=\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(y).$$ Choosing $\mu_{\mu\nu}(x,y)=\epsilon_{\mu\nu\alpha}\xi^\alpha(x)\delta(x-y)$ yields $$\int\text{d}^Dy\,\mu_{\mu\nu}(x,y)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(y)=\epsilon_{\mu\nu\gamma}\xi^\gamma(x)\epsilon^{\nu\alpha\beta}\partial_\alpha A_\beta(x)=\xi^\alpha\partial_\alpha A_\mu+\partial_\mu \xi^\beta A_\beta-\partial_\mu (\xi^\beta A_\beta)=\delta_\xi A_\mu-\delta_{\xi^\nu A_\nu}A_\mu.$$ Thus, this choice almost works. However, the parameter $\Lambda$ shouldn't depend on $A$. Is there another choice of $\mu$ and $\Lambda$ so that $\Lambda$ doesn't depend on $A$?

Investigating the problem further , I noticed the following. If indeed there is such a $\mu$ and $\Lambda$, then, whenever the equations of motion are satisfied we have that $\delta_\xi A_\mu-\delta_\Lambda A_\mu=0$. The equations of motion imply that $A$ is pure gauge, i.e. $A_\mu=\partial_\mu \Omega$ for some $\Omega$. Then $$\delta_\xi A_\mu-\delta_\Lambda A_\mu=\xi^\rho\partial_\rho\partial_\mu\Omega+\partial_\mu\xi^\rho\partial_\rho\Omega-\partial_\mu\Lambda=\partial_\mu(\xi^\rho\partial_\rho\Omega-\Lambda).$$ There is no way that this vanishes identically unless we choose $\Lambda=\xi^\rho\partial_\rho\Omega=\xi^\rho A_\rho$ (up to a constant). We conclude that the $\Lambda$ must depend on $A$ as we found above. Then, isn't the statement in the exercise wrong? Or are we allowed to use a different gauge parameter for every field?

Answer 1

I don't have Henneaux's lectures on hand, but I assume the context is classical (not quantum) gauge theory.

The paradox comes from thinking that the gauge-transform parameter $\Lambda$ is not allowed to depend on the gauge field $A$ that is being transformed, but that is incorrect. Within any contractible patch, we can take the quantities $A_a$ (and also $\xi^a$) to be ordinary smooth functions, and then $\Lambda=\xi^a A_a$ is also a smooth function. The transform $A_a\to A_a+\partial_a\Lambda$ is a legal gauge transform for any smooth function $\Lambda$, including one of the form $\Lambda=\xi^a A_a$.

On a general manifold, the gauge field $A_a$ is defined only patchwise, and that's fine. A gauge transform $A_a\to A_a+\partial_a\Lambda$ is also defined patchwise, and setting $\Lambda=\xi^a A_a$ (defined patchwise) still gives a legal gauge transform because $A_a$ and $\partial_a\Lambda$ are both affected the same way by the transition functions that relate the different patches to each other.