# Chern-simons term to total derivative

I'm trying to prove $$Tr[G_{\mu\nu} \tilde{G}^{\mu\nu}]=2\epsilon^{\mu\nu\rho\sigma}\partial_{\mu}Tr[A_{\nu}G_{\rho\sigma}-\frac{2}{3}iA_{\nu}A_{\rho}A_{\sigma}]$$ expanding the L.H.S. I don't know the origin of the factor $$\frac{1}{3}$$ and how to derive the last term. There are many answers using differential forms, but I didn't study differential forms yet. Isn't it possible to prove the equation not knowing differential forms? If it is possible, please help me...

my notation of $$G_{\mu}$$ is $$G_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-ig[A_{\mu},A_{\nu}]$$ and the dual field strength is $$\tilde{G}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}$$

My calculation progress is below $$Tr[G_{\mu\nu} \tilde{G}^{\mu\nu}]=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-ig[A_{\mu},A_{\nu}])G_{\rho\sigma}$$ considering contraction to the indices of levi-civita symbol and relabeling the indices, \begin{align} Tr[G_{\mu\nu} \tilde{G}^{\mu\nu}]& = 2\epsilon^{\mu\nu\rho\sigma}Tr[(\partial_{\mu}A_{\nu}-igA_{\mu}A_{\nu})(\partial_{\rho}A_{\sigma}-igA_{\rho}A_{\sigma})]\\ & = 2\epsilon^{\mu\nu\rho\sigma}Tr[(\partial_{\mu}A_{\nu})(\partial_{\rho}A_{\sigma})-ig(\partial_{\mu}A_{\nu})A_{\rho}A_{\sigma}-igA_{\mu}A_{\nu}(\partial_{\rho}A_{\sigma})-g^2A_{\mu}A_{\nu}A_{\rho}A_{\sigma}] \end{align} The first term can be calculated like below $$(\partial_{\mu}A_{\nu})(\partial_{\rho}A_{\sigma})=\partial_{\mu}(A_{\nu}A_{\rho}A_{\sigma})-A_{\nu}\partial_{\mu}\partial_{\rho}A_{\sigma}$$ The last term $$A_{\nu}\partial_{\mu}\partial_{\rho}A_{\sigma}$$ is symmetric and producted by Levi-Civita which is antisymmetric, so it can vanish. $$Tr[A_{\mu}A_{\nu}A_{\rho}A_{\sigma}]$$ also vanishes because of the cycling property of trace and contraction to the Levi-Civita symbol.

As a result, $$Tr[G_{\mu\nu} \tilde{G}^{\mu\nu}]=2\epsilon^{\mu\nu\rho\sigma}Tr[\partial_{\mu}(A_{\nu}A_{\rho}A_{\sigma})-2ig(\partial_{\mu}A_{\nu})A_{\rho}A_{\sigma})]$$

and... I am stuck. Even I'm not sure whether my calculation is correct.

• For the sake of asking questions here, it's best to keep the question as self-contained as possible. Here that would mean defining you notations. What is $G_{\mu\nu}$ in terms of $A_\mu$, how are you defining $\tilde G_{\mu\nu}$, etc. Differential forms make life much easier, but without I recommend simply computing the derivative on the RHS and re-organizing things to produce the LHS. It's a lot of algebra and being careful with orderings (the $A$'s don't commute, but the trace is cyclic), but all manipulations are straightforward. Sep 24 at 6:40
• First of all, you made a typo in the line where you rewrote the derivatives. It should be $(\partial_\mu A_\nu) (\partial_\rho A_\sigma) = \partial(A_\nu \partial_\rho A_\sigma) - A_\nu \partial_\mu \partial_\rho A_\sigma$ Sep 24 at 10:06
• Sep 24 at 11:41

You are pretty far already. Just note that $$\epsilon^{\mu \nu \rho \sigma} \text{Tr} ((\partial_\mu A_\nu) A_\rho A_\sigma )= \epsilon^{\mu \nu \rho \sigma} \frac{1}{3} \partial_\mu \text{Tr} ( A_\nu A_\rho A_\sigma)$$. This follows immediately from the anti-symmetry of the Levi-Cevita-tensor and the cyclicatry of the trace. Next make use of the anti-symmetry again and only use the anti-symmetric part of $$A_\nu A_\rho$$. This way you get rid of the factor 2. Finally note that $$$$\text{Tr} ( [A_\nu, A_\rho] A_\sigma) = A_\nu^a A_\rho^b A_\sigma^c \text{Tr}(f^{abd} T^d T^c) = T_F f^{abc} A_\nu^a A_\rho^b A_\sigma^c$$$$ where $$T_F$$ is the normalization usually set to $$1/2$$ for SU(N) but the choice does not really matter as it cancels anyway. And then you are pretty much finished.
• I think I'm done! It's not much different from the reason why $Tr[A_{\mu}A_{\nu}A_{\rho}A_{\sigma}]$ can vanish. I don't know why I was stuck at it. Thank you very much. 😘