5
$\begingroup$

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: $$\partial_\mu J^\mu=\epsilon^{\mu\nu\rho\sigma}\operatorname{Tr}\left[F_{\mu\nu} F_{\rho\sigma}\right]\quad ?$$


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?

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.