# Showing dual field strength tensor term in Lagrangian is a total derivative

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?

• These calculations are typically much easier and cleaner when using differential forms, i.e., wedges and exterior derivatives, in an index-free notation. – AccidentalFourierTransform Jun 1 '18 at 22:20
• @AccidentalFourierTransform So I've heard and seen, which is why I plan on studying them starting today. – Arturo don Juan Jun 1 '18 at 22:25
• That's definitely a good plan :-) – AccidentalFourierTransform Jun 1 '18 at 22:25
• 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. – Prahar Jun 2 '18 at 5:03
• @Prahar Oh my gosh I can't believe I mistook that, thank you! – Arturo don Juan Jun 2 '18 at 15:55

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 $$\text{tr} [ F \wedge F ] = d \ast K.$$
• 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}. – Arturo don Juan Apr 29 at 20:41
• (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. – Arturo don Juan May 6 at 20:38
• 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. – Arturo don Juan May 6 at 20:51