I want to show:$$ Tr\left (F\tilde{F} \right )=\partial_{\mu}K^{\mu }=\partial_{\mu}\left (\varepsilon _{\mu \nu \rho \sigma }Tr\left ( F_{\nu \varrho }A_{\sigma }-\frac{2}{3}A_{\nu }A_{\rho }A_{\sigma} \right )\right ).$$

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    $\begingroup$ Just expand the RHS! (And you'll have to use the Bianchi identity.) $\endgroup$ – José Figueroa-O'Farrill Jan 9 '14 at 23:14
  • $\begingroup$ @JoséFigueroa-O'Farrill Actually I would rather expand the LHS, think it is easier. But it is the same anyways. $\endgroup$ – Federico Carta Jan 10 '14 at 10:14
  • $\begingroup$ @FedericoCarta If you expand the RHS, it’s not obviously a total derivative. It requires use of Leibniz and then showing that what remains outside of the derivative is zero. If you expand the RHS and use Bianchi you get pretty much the LHS on the nose. $\endgroup$ – José Figueroa-O'Farrill Jan 10 '14 at 16:43
  • $\begingroup$ @JoséFigueroa-O'Farrill. I had never tried that way before, but today I did and you are right! Thanks $\endgroup$ – Federico Carta Jan 10 '14 at 16:46

I give you a hint, but I would not work the exercise all the way.

Start expanding the LHS.

Use the definition for the hodge dual field strenght. $$\tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$$

$$Tr\left(F\tilde{F}\right)=Tr\left(\frac{1}{2}F_{\mu\nu}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\right)=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma} Tr\left(F_{\mu\nu}F_{\rho\sigma}\right)$$

Now use the definition of the field strength in a non abelian Yang-Mills context $$F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu}A_{\mu}-ig[A_{\mu},A_{\nu}]$$

Upon working the products remember that anything symmetric contracted with $\epsilon^{\mu\nu\rho\sigma}$ vanishes. Therefore $\partial_{\mu}\partial_{nu}$ vanishes.

However terms of the form $A_{\nu}A_{\rho}$ are not symmetric under $\nu \mapsto \rho$ since they do not commute. To work out these terms, and those of the form $A_{\mu}A_{\nu}A_{\rho}$ remember that this is not equal $A_{\rho}A^{\nu}A^{\mu}$ because of course you are dealing with non abelian gauge fields. The trick here is to rename dummy indices. Remember they are contracted (with the Levi-Civita pseudotensor), so you can do it.

Doing this, you should find the RHS of the equation you wanted to prove.



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