I'm probably missing something obvious, but I've been wondering what the motivation is for requiring the components $A\_\mu$ in a local trivialization of a gauge connection on a smooth principal $G$-bundle to lie in $\mathfrak{g}$, the Lie algebra of $G$. I can see that this gives a couple of nice properties; for example, in a local trivialization it ensures that under a gauge transformation $A'\_\mu=gA\_\mu g^{-1}+g\partial\_\mu g$ lies in $\mathfrak{g}$, and that the curvature form $F=dA+A\wedge A$ lies in $\mathfrak{g}$ (since $\mathfrak{g}$ is closed under the Lie bracket). But is there a more intrinsic or geometric reason that $A\_\mu$ must be in $\mathfrak{g}$? Thanks.



Gauge connections and Lie algebras?
First, I never liked working with principal bundles; vector bundles seem easier and more natural to me. Second, I never like thinking about abstract principal $G$-bundles. I prefer fixing a representation of $G$ and viewing the principal $G$ bundle as a reduced frame bundle associated with a vector bundle.


So let $E$ be a rank $k$ vector bundle and $F$ the bundle of arbitrary frames in $E$ (this is a principal $GL(k)$-bundle). Then $GL(k)$ acts on the right on $F$. Given a subgroup $G$ in $GL(k)$, let $F\_G$ be a subbundle of $F$ such that if $f \in F\_G$, then so is $f\cdot g$ for each $g \in G$.


The primary example is $E = T\_\*M$ and $F\_G$ is the bundle of orthonormal bases of the tangent space with respect to a Riemannian metric.


What is the critical property we want a $G$-connection to satisfy? Well, any connection allows you to parallel translate an arbitrary frame $f \in F$ along a curve. We'd like the $G$-connection to be such that if $f \in F\_G$, then the parallel translation remains in $F\_G$. This leads to the right definition of a $G$-connection.


