Exterior derivatives involving representations I have two questions regarding the exterior derivative of vector valued forms when representations are involved:
Suppose $V$ is a vector space, $M$ a smooth manifold and $\omega$ is a $V$ valued $k$-form on $M$. Ie, $\omega \in \Omega^k(M;V)$. Suppose furthermore that $\rho_1:G\rightarrow GL(V)$ is a representation for some Lie Group $G$ and $\rho_2:\mathfrak{g}\rightarrow GL(V)$ is the induced Lie algebra representation.
The function $\rho_1(g)\circ \omega$ could be considered as a $V$ valued $k$-form on $M$. If $d:\Omega^k(M;V)\rightarrow \Omega^{k+1}(M;V)$ is the exterior derivative for $V$ valued $k$-forms, then
$d(\rho(g)\circ \omega)=\rho(g)\circ d\omega$.
I am just wondering why this is true?
Furthermore, in some lecture notes I'm reading the author also writes for $\eta\in \Omega^1(M;\mathfrak{g})$ that
$d(\rho_2(\eta)\circ \omega)=\rho_2(d\eta)\circ \omega-\rho_2(\eta)\wedge d\omega$
I am completely in the dark as to how this equation makes sense. I don't understand how the RHS is a 2-form or where the RHS even comes from. How would you make sense of this equation?
 A: The first equation is true because $\rho(g)$ is a constant, so it commutes with the exterior derivative.  More explicitly, let $\{e_i\}$ be a basis for $V$.  Then you can write $\omega = \sum_i e_i \otimes \omega_i$ with $\omega_i \in \Omega^k(M)$.  Also, write $\rho(g) e_i = \sum_j g_i^j e_j$.  Then
$$
d(\rho(g)\omega) = d(\sum_{ij} g_i^j e_j \otimes \omega_i) = \sum_{ij} g_i^j e_j \otimes d\omega_i = \rho(g) d\omega.
$$
For the second equation you can do a similar thing (note that now $\eta$ is not constant and so you have to use the Leibniz rule).  Note that $\rho_2(\eta)\circ \omega$ combines the wedge product on forms with the action of $\mathfrak g$ on $V$ and so is in $\Omega^{k+1}(M;V)$.  So the right hand side should be a k+1 form, not a 2-form.
EDIT

Some more details for the second: write $\eta = \sum_i X_i \otimes \eta_i$ where $X_i \in \mathfrak g, \eta_i \in \Omega^1(M)$.  Write $\omega$ as before.  By definition, $\rho_2(\eta) \circ \omega = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes (\eta_i \wedge \omega_j)$.  So
$$
d(\rho_2(\eta) \circ \omega) = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes d(\eta_i \wedge \omega_j) \\
 = \sum_{ij} (\rho_2(X_i)(e_j)) \otimes (d\eta_i \wedge \omega_j - \eta_i \wedge d\omega_j) \\
= \sum_{ij} (\rho_2(X_i)(e_j)) \otimes d\eta_i \wedge \omega_j - \sum_{ij} (\rho_2(X_i)(e_j)) \otimes \eta_i \wedge d\omega_j \\
= \rho_2(d\eta) \circ \omega - \rho_2(\eta)\circ d\omega.
$$
Hope this clears it up.
