Wedge products of V-valued forms.

Question

Let $(P, \pi, M, G)$ be a principal $G$-bundle. Let $\omega$ be a Lie-Algebra valued one-form (connection one-form) on $P$. Then, the two-form $\omega \wedge \omega$ that comes in the connection two-form $\Omega := d \omega + \omega \wedge \omega$ is to be interpreted as $(\omega \wedge \omega)_p (X_p,Y_p) := [\omega(X_p), \omega(Y_p)]$. This makes sense to me.

Now, suppose further that $\theta$ is a $V$-valued one-form (for example the solder form), where V is a dim(M) dimensional representation space of the lie group $G$. How am I to interpret $\omega \wedge \theta$? I am referring to the '$\wedge$' that appears in the torsion "$ \Theta := d \theta + \omega \wedge \theta$".

How is the $\wedge$ to be interpreted in the expression $D \Theta = \Omega \wedge \theta$ (The second Bianchi identity for principal bundles)?

A side question: Suppose you have a representation $(\rho,V)$ of the lie group $G$, then the lie group acts on $V$ by simply $v \to \rho(g)v$. Can the Lie-Algebra $\mathfrak{g}$ aslo act on V? If so, how?

Answer 1

First of all, you should read this answer by Ivo Terek about vector-valued forms and their wedge products with respect to a given bilinear pairing $\mu$.

Now, you're given a smooth group homomorphism $\rho:G\to \text{GL}(V)$. By considering the tangent mapping at the identity, you obtain the mapping $T_e\rho:\mathfrak{g}\to \text{End}(V)$ (because $\text{GL}(V)$ is an open subset of the vector space $\text{End}(V)$, so its tangent space at any point is canonically identified with $\text{End}(V)$ itself). We therefore have a bilinear mapping $\mu:\mathfrak{g}\times V\to V$ in the only natural way possible: \begin{align} \mu(a,b)&:= [(T_e\rho)(a)](b) \end{align} Here, $(T_e\rho)(a)$ is an endomorphism of $V$, and it is being evaluated on a vector $b\in V$ to yield another vector in $V$. Another way of saying it is that $\mu$ is the composition $\mathfrak{g}\times V\to \text{End}(V)\times V\to V$, given by $(\text{evaluation on $V$})\circ (T_e\rho \times \text{id}_V)$.

So, $\omega$ is a $\mathfrak{g}$-valued 1-form and $\theta$ is a $V$-valued 1-form and $\mu$ is a bilinear pairing taking values in $V$, so we can define the wedge product $\omega\wedge_{\mu}\theta$ to obtain a $V$-valued 2-form. Same thing with $\Omega\wedge_{\mu}\theta$.