Follow up to this question.
$\omega$ is a one parameter family of connections on a principal bundle of frames $E$, associated to a complex vector bundle $V\to M$.
In the proof of the Chern Weil method in John Roe's book, one start with the curvature of a complex vector bundle $V\to M$ and then works in the associated frame bundle $E$. There (i.e. as forms on $E$) he gets the equality: $$\frac{d}{dt}\log \det(1+q\Omega)= d\sum_{l=0}^{\infty}(-1)^lq^{l+1}\text{tr}\{\Omega^l \omega'\}$$
The point is that we want to obtain such relation on $M$, in a certain sense we want to push down these forms to forms on $M$. The way to do that, according to the author, is to observe that the form on the LHS is the exterior derivative of an horizontal and invariant form on E. Hence since there is a $1-1$ correspondence between horizontal and invariant forms on $E$ and forms on $M$, we are done since we are allowed to push everything down.
Why is the form $$\sum_{l=0}^{\infty}(-1)^lq^{l+1}\text{tr}\{\Omega^l \omega'\}$$ horizontal and invariant?
According to a proposition in the book, the form $\Omega$ (i.e. the curvature) is horizontal (and $Ad$-equivariant), but $\omega$ is not horizontal at all (by definition), but it's still Ad-equivariant.
Since the trace is an invariant polynomial, it's enough to prove that $\Omega^l\omega'$ is horizontal. Problem is that I don't have much control on $\omega'$. Any suggestion? Does it suffice to have only $\Omega$ invariant?