Exterior Derivative Problem Suppose $\theta$ is a differential $1$-form defined on a manifold and with values in the Lie algebra of a Lie group $G$. 
On $M\times G$ define the $1$-form $ad(g)\theta$ where $\theta$ is extended by letting it be zero on the tangent space to $G$.
How do you compute the exterior derivative $d(ad(g)\theta)$?
BTW: For matrix Lie groups this is straightforward. What is the abstract calculation?
 A: $\newcommand{\Ad}{\mathrm{Ad}}
\newcommand{\ad}{\mathrm{ad}}
\newcommand{\lg}{\mathfrak g}$
I'll use the notations $\Ad : G \to L(\lg,\lg)$ and $\ad = d_e\Ad: \lg \to L(\lg,\lg)$ where $\lg$ is the Lie algebra of $G$.
Let's compute $d(\Ad_g \circ \theta)(v+\xi,w+\eta)$ for vectors $$v+\xi, w+\eta \in T_m M\oplus T_g G = T_{(m,g)}(M\times G).$$ Extend the vectors locally to fields. Then we have (noting that $\theta[v+\xi,w + \eta] = \theta[v,w]$ since $\theta$ is zero on $TG$)
$$ 
\begin{align}
d(\Ad_g \theta)(v+\xi, w+\eta) &= (v+\xi)(\Ad_g \theta (w+\eta)) - (w+\eta)(\Ad_g \theta (v + \xi)) - \Ad_g\theta[v+\xi,w + \eta]\\
&= (v (\Ad_g \theta(w)) - w(\Ad_g \theta(v)) - \Ad_g \theta[v,w]) + \xi(\Ad_g \theta(w)) - \eta(\Ad_g \theta(v)).
\end{align}
$$
Now, note that derivatives in the $TM$ directions commute with $\Ad_g$, so the first term is just $$\Ad_g (v \theta w - w \theta v - \theta[v,w]) = \Ad_g \circ d\theta(v,w).$$ 
For the remaining two terms note that the only $g$ dependence appears in $\Ad$, so by definition of $\ad$ they are simply $\ad_\xi \theta w - \ad_\eta \theta v.$ I'm unsure if there's a neater way to express this last portion: for now we've just got the expression
$$ d(\Ad_g \circ \theta)(v+\xi, w+\eta) = \Ad_g \circ d\theta (v+\xi, w+\eta) +  \ad_\xi \theta w - \ad_\eta \theta v.$$
