Chern Simons form: calculation

[SergejVictorov]

Hey folks, I'm trying to understand a couple of calculations in Chern-Simons theory and I'm stuck.

I want to prove the following in four spacetime dimensions:

Let G be a Lie Group with generators T_{a}. Further let
A=T_{a} A^{a}_{\mu} dx^{\mu}
be a connection and
F=dA+A \wedge A
its field strength. Now I want to prove
Tr F \wedge F= d Tr(A \wedge dA+\frac{2}{3}A\wedge A\wedge A)

First I tried the following
F\wedge F=dA\wedge dA+dA \wedge A\wedge A+A\wedge A\wedge dA+A\wedge A\wedge A\wedge A

d (A \wedge dA+\frac{2}{3}A\wedge A\wedge A)= dA\wedge dA+\frac{2}{3}dA\wedge A\wedge A +\frac{2}{3} A\wedge dA\wedge A+\frac{2}{3} A\wedge A \wedge dA

When I take the trace, I can make a A\wedge A\wedge dA into dA \wedge A\wedge A, but that leaves the questions of why the trace of A^4 has to vanish and where the factors of 2/3 come from.

Alternatively, I tried the following:

Let A_t=tA and
F_t=dA_t+A_t\wedge A_t=dt\wedgeA+t dA+t^2 A\wedge A

Now one can evaluate Tr F_t \wedge F_t and afterwards set t=1. This works out fine, but I still don't understand why the trace of A^4 has to vanish.

So for the moment, it all boils down to: why is Tr A^4=0?

Thank you for your help!

 

[weejee]

d (A \wedge dA+\frac{2}{3}A\wedge A\wedge A)= dA\wedge dA+\frac{2}{3}dA\wedge A\wedge A +\frac{2}{3} A\wedge dA\wedge A+\frac{2}{3} A\wedge A \wedge dA

The sign of the third term on the right hand side should be minus, due to antisymmetry of the exterior derivative. We also have
Tr\left[dA\wedge A\wedge A \right]= \epsilon^{ijkl}Tr\left[(\partial_{i}A_{j})A_{k}A_{l} \right]= \epsilon^{ijkl}Tr\left[A_{l}(\partial_{i}A_{j})A_{k} \right] = -\epsilon^{ijkl}Tr\left[A_{i}(\partial_{j}A_{k})A_{l} \right]=-Tr\left[A\wedge dA\wedge A \right],
in which the cyclic property of trace and the antisymmetry of the \displaystyle{\epsilon} tensor is used.

Tr\left[A^{4}\right]=0 can be proved similarly.