Show How Theta Term in QCD Lagrangian is a Total Derivative

[Kara386]

I'm trying to show that the theta term in the QCD Lagrangian, $\alpha G^a_{\mu\nu} \widetilde{G^a_{\mu\nu}}$, can be written as a total derivative, where
$\begin{equation} G^a_{\mu\nu} = \partial_{\mu} G^a_{\nu} - \partial_{\nu}G^a_{\mu}-gf_{bca}G^b_{\mu}G^c_{\nu} \end{equation}$
$\widetilde{G^a_{\mu\nu}} = \frac{1}{2} \epsilon^{\mu\nu\lambda\rho}G^a_{\lambda\rho}$
And $\epsilon$ is the Levi-Civita symbol.

The thing is, plenty of papers and books say that this is a total derivative, and give absolutely no indication of how to show that. Well, it's been suggested I should use the wedge product, but I've never encountered it and I thought brute force should work fine. So I expand $G^a_{\lambda\rho}$ using the first equation, multiply the two and expand the brackets:
$\frac{1}{2} \epsilon^{\mu\nu\lambda\rho} (\partial_{\lambda}G^a_{\rho}\partial_{\mu}G^a_{\nu}-\partial_{\lambda}G^a_{\rho}\partial_{\nu}G^a_{\mu}-\partial_{\lambda}G^a_{\rho}gf_{bca}G^b_{\mu}G^c_{\nu}-\partial_{rho}G^a_{\lambda}\partial_{\mu}G^a_{\nu}+\partial_{\rho}G^a_{\lambda}\partial_{\nu}G^a_{\mu}+\partial_{\rho}G^a_{\lambda}gf_{bca}G^b_{\mu}G^c_{\nu}-gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\mu}G^a_{\nu}+gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\nu}G^a_{\mu}+g^2f^2_{bca}G^b_{\lambda}G^c_{\rho}G^b_{\mu}G^c_{\nu})$
So that looks a mess, but because the Levi-Civita symbol is antisymmetric that equals
$\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}(4\partial_{\lambda}G^a_{\rho}\partial_{\mu}G^a_{\nu}+2\partial_{\rho}G^a_{\lambda}gf_{bca}G^b_{\mu}G^c_{\nu}+2gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\nu}G^a_{\mu}+g^2f^2_{bca}G^b_{\lambda}G^c_{\rho}G^b_{\mu}G^c_{\nu})$
What it's meant to be, somehow, is
$\partial_{\mu}\left(\alpha \epsilon^{\mu\nu\lambda\rho}G^a_{\nu}(G^a_{\lambda\rho} +\frac{1}{3}g f_{bca}G^b_{\lambda}G^c_{\rho}) \right)$
But I really, really can't see how you'd get there! I'm also a little confused about why the indices $\mu$ and $\nu$ are different in the definition of the dual, but the $a$ is kept the same - don't I need to relabel b,c,a?

I'd massively appreciate any pointers, I've been stuck on this for quite a while!

 

[ChrisVer]

Are you talking for eq5 here: https://arxiv.org/pdf/hep-ph/0607268.pdf ?
I think Weinberg's QFT book had some explanations... but it's been some time and I don't have the book now.

Finally I saw this: https://www.sciencedirect.com/science/article/pii/055032137490546X?via=ihub being cited when they claim that term to be a total divergence. Eq5 here https://arxiv.org/pdf/hep-ph/9606475.pdf

 

[MathematicalPhysicist]

Maybe this derivation appears in Srednicki's textbook or its solution manual.

 

[i0j1k2l3]

for those terms with no g, you can use ∂A∂B = ∂(A∂B) - A∂∂B，then you can find the second term ε^μναβ∂∂ should be zero.
for terms with g, you need ∂(ABC) = ∂ABC+A∂BC+AB∂C, and combining them with ε^μναβ f^abc, you will see the right hand three terms are equivalent with each other, that's why you get a 1/3 before AAA.
as for the last term with g², you will find it is zero after you use the property of ε^μναβ and f^abcf^aef.