I am trying to prove the identity: $$\text{Tr}\left\{\star F_{\mu\nu}F^{\mu\nu}\right\}=\partial^{\mu}K_{\mu} \tag{1}$$ where $K_{\mu}$ is given by: $$K_\mu=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{A^{\nu}F^{\rho\sigma}-\frac{2}{3}igA^{\nu}A^{\rho}A^{\sigma}\right\}\tag{2}$$ and $F$ and $\star F$ are the (non-abelian) field strength tensors defined by: $$\begin{align} F^{\mu\nu}&\equiv\partial^{\mu}A^{\nu}-\partial^{\mu}A^{\nu}+ig[A^{mu},A^{\nu}]\\ &=D^{\mu}A^\nu-D^{\nu}A^{\mu}, \end{align}\tag{3}$$ $$\star F^{\mu\nu}\equiv \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.\tag{4}$$

I have already seen this directly related question but after following the advice given there (namely expand the RHS of eq. 1 and use both the Bianchi identity for $F$ and dummy index relabeling) I still cannot prove the identity. Below is my work.

$$\begin{align*} &\partial^{\mu}\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{A^{\nu}F^{\rho\sigma}-\frac{2}{3}igA^{\nu}A^{\rho}A^{\sigma}\right\}\\ &=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})F^{\rho\sigma}+\underset{\text{Bianchi}}{\underbrace{A^{\nu}\require{cancel}\cancel{(\partial^{\mu}F^{\rho\sigma})}}}-\underset{\text{index relabeling + cyclic perm.}}{\underbrace{2ig(\partial^{\mu}A^{\nu})A^{\rho}A^{\sigma}}}\right\} \tag{5}\\ &\\ &=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})F^{\rho\sigma}-ig(\partial^{\mu}A^{\nu})[A^{\rho},A^{\sigma}]\right\}\tag{6}\\ &=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})(F^{\rho\sigma}-ig[A^{\rho},A^{\sigma}])\right\}\tag{7}\\ &=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{F^{\mu\nu}F^{\rho\sigma}\color{red}{-2igF^{\mu\nu}[A^{\rho},A^{\sigma}]-g^2[A^{\mu},A^{\nu}][A^{\rho},A^{\sigma}]}\right\}\tag{8}\\ \end{align*}$$

In order for the identity to be true the terms in red must vanish, which I cannot see to be true in general. Where have I gone wrong?

  • 1
    $\begingroup$ These calculations are typically much easier and cleaner when using differential forms, i.e., wedges and exterior derivatives, in an index-free notation. $\endgroup$ Jun 1, 2018 at 22:20
  • $\begingroup$ @AccidentalFourierTransform So I've heard and seen, which is why I plan on studying them starting today. $\endgroup$ Jun 1, 2018 at 22:25
  • $\begingroup$ That's definitely a good plan :-) $\endgroup$ Jun 1, 2018 at 22:25
  • 1
    $\begingroup$ You've got the wrong Bianchi identity. For nonabelian fields it reads $D_{[\mu} F_{\nu\rho]} = 0$ where $D_\mu$ is the gauge covariant derivative. $\endgroup$
    – Prahar
    Jun 2, 2018 at 5:03
  • $\begingroup$ @Prahar Oh my gosh I can't believe I mistook that, thank you! $\endgroup$ Jun 2, 2018 at 15:55

1 Answer 1


As alluded to in the comments, using forms is much easier here. We note the formulae, $$(\ast F)_{\mu\nu} F^{\mu\nu} = \ast \text{tr}[F \wedge F],\qquad F = d A + A \wedge A,\qquad dF + A \wedge F - F \wedge A = 0 .$$ The quantity $\ast K = \text{tr} [ A \wedge F - \frac{1}{3} A \wedge A \wedge A ]$ and $\nabla_\mu K^\mu = \ast d \ast K$. We are now ready to prove our result, which now reads $$ \text{tr}[ F \wedge F ] = d \ast K . $$ Start with the RHS \begin{align} d \ast K &= \text{tr}[d A \wedge F - A \wedge d F - d A \wedge A \wedge A] \\ &= \text{tr}[(F - A \wedge A ) \wedge F - A \wedge ( - A \wedge F + F \wedge A ) - (F - A \wedge A ) \wedge A \wedge A] \\ &= \text{tr}[F \wedge F - A \wedge A \wedge F + A \wedge A \wedge F- A \wedge F \wedge A - F \wedge A \wedge A + A \wedge A \wedge A \wedge A ]\\ &= \text{tr}[F \wedge F + A \wedge A \wedge A \wedge A ] . \\ \end{align} where we used the fact $\text{tr}[A \wedge F \wedge A + F \wedge A \wedge A] = 0 $ (verify!)

The final thing we have to show is that $\text{tr}[A \wedge A \wedge A \wedge A ] =0$. To see this, let us put in the explicit Lie algebra indices \begin{align} \text{tr}[A \wedge A \wedge A \wedge A ] &= \text{tr}[T_a T_b T_cT_d] A^a \wedge A^b \wedge A^c \wedge A^d \\ &= \text{tr}[ [ T_a , T_b ] [ T_c , T_d ] ] A^a \wedge A^b \wedge A^c \wedge A^d \\ &=f_{abe} f_{ecd} A^a \wedge A^b \wedge A^c \wedge A^d \\ &= - ( f_{cae} f_{ebd} + f_{bce} f_{ead} ) A^a \wedge A^b \wedge A^c \wedge A^d \\ &=- 2 f_{abe} f_{ecd} A^a \wedge A^b \wedge A^c \wedge A^d. \end{align} Here, we used the Jacobi identity and then did some index manipulation in the last step (do this yourself!). Thus, $\text{tr}[A \wedge A \wedge A \wedge A ] = 0$.

Hence $$ \text{tr} [ F \wedge F ] = d \ast K. $$

  • $\begingroup$ I'm not familiar with differential forms yet, but once I am I'll come back to this and work out everything you've suggested. Thanks! $\endgroup$ Jun 2, 2018 at 15:59
  • $\begingroup$ There's an easier way to see that $\text{tr}A^4=0$. Just use that the trace is cyclic, while the wedge product is antisymmetric. Omitting wedges, $$\begin{align}\text{tr}A^4&=\text{tr}[T_aT_bT_cT_d]A^aA^bA^cA^d\\ &=-\text{tr}[T_aT_bT_cT_d]\color{red}{A^bA^cA^dA^a}\\ &=-\text{tr}[\color{red}{T_bT_cT_dT_a}]A^bA^cA^dA^a\\ &=-\text{tr}A^4 \end{align}$$. $\endgroup$ Apr 29, 2019 at 20:41
  • $\begingroup$ (cont.) in fact by this same logic, the trace of an even product of odd-forms vanishes, i.e. $\text{tr}\,A^{2k}=0$ where $A$ is an odd-form and $k$ is an integer. $\endgroup$ May 6, 2019 at 20:38
  • $\begingroup$ And by the way, as you can tell, I came back to this and worked it out :). This is the statement that the Chern-Simons form $\omega_4$ is (locally, and also in fact globally) exact. $\endgroup$ May 6, 2019 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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