The Chern-Simons 3-form is given by

$\omega_3={\rm Tr} \left[ A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right]$

where $A$ is a connection one-form in the adjoint representation of a non-Abelian gauge group.

My differential geometry is rather rusty (and this is new to me too) hence my questions;

$A$ is a 1-form. By definition of the wedge product between a $p$ form $\alpha$ and $q$ form $\beta$ we have $\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$. So we should have $A\wedge A=-A\wedge A=0$.

Why is this note the case?

Next question; I want to calculate $d\omega_3$ Does the fact that everything is inside the trace effect my calculation? In other words does the differential operator pass through the trace and only act on the forms?

The Chern-Simons 3-form is given by

$\omega_3={\rm Tr} \left[ A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right]$

where $A$ is a connection one-form in the adjoint representation of a non-Abelian gauge group.

My differential geometry is rather rusty (and this is new to me too) hence my questions;

$A$ is a 1-form. By definition of the wedge product between a $p$ form $\alpha$ and $q$ form $\beta$ we have $\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$. So we should have $A\wedge A=-A\wedge A=0$.

Why is this not the case?

Next question; I want to calculate $d\omega_3$ Does the fact that everything is inside the trace effect my calculation? In other words does the differential operator pass through the trace and only act on the forms?