Explicit derivative of Chern-Simons current

Question

I know that for a Chern-Simons 3-form $\omega=\operatorname{Tr}\left[F\wedge A-\frac{1}{3}A\wedge A\wedge A\right]$, with $F=A\wedge A +\operatorname{d}A$, I should get $\operatorname{d}\omega=\operatorname{Tr}\left[F\wedge F\right]$.

Now, in practice (cf. Srednicki's QFT), given a Chern-Simons current: $$J^\mu=2\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}\left[A_\nu F_{\rho\sigma}+\dfrac{2}{3}igA_\nu A_\rho A_\sigma\right]$$ with: $$F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu -ig\left[A_\mu,A_\nu\right]$$ How do I show that: $$\partial_\mu J^\mu=\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}\left[F_{\mu\nu} F_{\rho\sigma}\right]\quad ?$$

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EDIT: if showing my work might help attract advice/hints/solutions, here it is.

$$\begin{align} \partial_\mu J^\mu =2\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}&\left[ \partial_\mu A_\nu F_{\rho\sigma}+A_\nu\partial_\mu F_{\rho\sigma}+\dfrac{2}{3}ig\partial_\mu\left(A_\nu A_\rho A_\sigma\right)\right]\\ =2\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}&\left[\partial_\mu A_\nu F_{\rho\sigma}-igA_\nu\partial_\mu\left[A_\rho,A_\sigma\right]+\dfrac{2}{3}ig\partial_\mu A_\nu A_\rho A_\sigma +\dfrac{2}{3}igA_\nu\partial_\mu A_\rho A_\sigma\right.\\ &+\left.\dfrac{2}{3}igA_\nu A_\rho\partial_\mu A_\sigma\right]\\ =2\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}&\left[\partial_\mu A_\nu F_{\rho\sigma}-\dfrac{1}{3}igA_\nu\partial_\mu A_\rho A_\sigma-\dfrac{1}{3}igA_\nu A_\rho\partial_\mu A_\sigma +igA_\nu\partial_\mu A_\sigma A_\rho +igA_\nu A_\sigma\partial_\mu A_\rho\right.\\ &+\left.\dfrac{2}{3}ig\partial_\mu A_\nu A_\rho A_\sigma\right] \end{align} $$

where in the second line the double derivative terms coming from the action of $\partial_\mu$ on $F_{\rho\sigma}$ vanish when contracted with the antisymmetric tensor, and in the third line the commutator was expanded and similar terms were simplified.

So, at this stage I have the term $\partial_\mu A_\nu F_{\rho\sigma}$, which is good; but the remaining terms all come with a factor of $ig$ and thus can't correspond to the full expansion of $\partial_\nu A_\mu F_{\rho\sigma}$ or the $g^2$ terms in $ig\left[A_\mu,A_\nu\right]F_{\rho\sigma}$ (these being what I expect to find to recover $F_{\mu\nu}F_{\rho\sigma}$).

Have I made a mistake? How do I progress from here? And in general, what's the easiest/cleanest way to approach such derivations?

Answer 1

In terms of the components $A=A_\mu dx^{\mu}$, we have$$ \\\ \frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}{\pi}\mathrm{tr}\left[\varepsilon^{\mu\nu\rho\sigma}(\partial_{\mu} A_{\nu}+A_{\mu}A_{\nu})(\partial_{\rho} A_{\sigma}+A_{\rho}A_{\sigma})\right] \\\\ $$ And then $$ \frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}{\pi}\mathrm{tr}\left[\varepsilon^{\mu\nu\rho\sigma}\partial_{\mu}(A_\nu\partial_\rho A_\sigma+\frac{2}{3}A_{\nu}A_\rho A_\sigma)\right]+\frac{2\theta}{\pi}\mathrm{tr}\left[A_{\mu}A_{\nu}A_\rho A_\sigma\right]\varepsilon^{\mu\nu\rho\sigma} $$

by cyclic permutations on $\nu$, $\rho$, $\sigma$ (that is even) and the fact that $\partial_\mu \partial_\nu$ is symmetric. Now, the last vanish since a cyclic permutation of even number of elements is always odd. So, we end up with a boundary term

$$ \frac{\theta}{\pi}\varepsilon^{\mu\nu\rho\sigma}\mathrm{tr}\left[A_\nu F_{\rho\sigma}-\frac{1}{3}A_{\nu}A_{\rho}A_{\sigma}\right] $$