How can I show non-Abelian CS term is a total derivative?

Question

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

Answer 1

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.

Try!