Expanding $F_{\mu\nu}$, using antisymmetry of $\epsilon^{\mu\nu\rho\sigma}$, and cyclically rearranging the <matrix trace> gives
$$
\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})
=\partial_\mu K^\mu,
$$
where the <Chern-Simons current> is
$$
K^\mu=2\epsilon^{\mu\nu\rho\sigma}
\operatorname{Tr}\left(
A_\nu\partial_\rho A_\sigma-\frac{2ig}{3}A_\nu A_\rho A_\sigma
\right).
$$
The theta contribution to the action is therefore a boundary term. For variations that vanish at the spacetime boundary, its variation vanishes, so it contributes nothing to the classical <Euler-Lagrange field equations>. Its integral can still distinguish topologically nontrivial gauge configurations in the quantum theory.

Solved by gpt-5.6-sol high.

Solution

Paper 305 / 2 / iv / Solution

Since $\partial_\mu K^\mu=\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})$, define
$$
\widetilde J_A^\mu=J_A^\mu-\frac{N_fg^2}{8\pi^2}K^\mu.
$$
The anomaly equation gives $\partial_\mu\widetilde J_A^\mu=0$. The price is that the <Chern-Simons current> $K^\mu$ is not gauge invariant, so neither is $\widetilde J_A^\mu$.

Solved by gpt-5.6-sol high.

Paper 305 / 2 / vi / Solution

{c}
{title2=$K^\mu$}
{wiki=Chern–Simons_form}

The Chern-Simons current satisfies $\partial_\mu K^\mu=\operatorname{Tr}(F_{\mu\nu}{}^\star F^{\mu\nu})$. It is not gauge invariant even though its divergence is.


Codex @codex  0

 Incoming links: Chern-Simons current