Differential of a form

Question

I'm learning about Chern-Simons theory and my differential geometry is a bit rusty.

The Chern-Simons 3-form is given by $\omega_3=tr(A\bigwedge\nolimits dA+\frac{2}{3}A\bigwedge\nolimits A \bigwedge\nolimits A)$ where $A$ is a connection 1-form.

I want to calculate $d\omega_3$.

We have $d(A\bigwedge\nolimits dA)=dA\bigwedge\nolimits dA$ since $d(dA)=0$.

Also, $d(A\bigwedge\nolimits A \bigwedge\nolimits A)=dA\bigwedge\nolimits A \bigwedge\nolimits A-A\bigwedge\nolimits dA \bigwedge\nolimits A+A\bigwedge\nolimits A \bigwedge\nolimits dA=3dA\bigwedge\nolimits A \bigwedge\nolimits A$

So I should have that $d\omega_3=tr(dA\bigwedge\nolimits dA+2dA\bigwedge\nolimits A \bigwedge\nolimits A)$

But I know the answer is supposed to be $d\omega_3=tr(dA\bigwedge\nolimits dA+2dA\bigwedge\nolimits A \bigwedge\nolimits A+A\bigwedge\nolimits A\bigwedge\nolimits A \bigwedge\nolimits A$)

So where have I gone wrong?

Answer 1

The problem is that your original formula for the Chern-Simons form is wrong. You should have the curvature $2$-form $F$ where you have $dA$. Those differ by a bracket term $A\wedge A$ (perhaps with some constant).