A gauge field *is* a connection on a principal bundle. I'll try to show here roughly how this formulation is related to the common construction in physics of the gauge fields in the process of making locally invariant a theory that only has global invariance.

Let $G$ be a Lie group, $G\times V\rightarrow V:(g,v)\mapsto gv$ a representation of $G$, $M$ the space-time manifold, $E\rightarrow M$ a principal $G$-bundle and $F\overset{\pi}{\rightarrow} M$ an associated bundle with fiber isomorphic to $V$. The space of gauge transformations $\Gamma(E)$ is equipped with a canonical action over the space of fields $\Gamma(F)$ as $(g,\phi)\mapsto g\phi$ where $(g\phi)(x)=g(x)\phi(x)$.

Observe that for a constant $g$, derivatives have the nice property that $\partial_\mu(g\phi)=g\partial_\mu\phi$. This isn't true for a general gauge transformation, so we want to construct an object similar to $\partial_\mu$ that satisfies the previous equation for any $g$. By adding a connection $A$ of the principal bundle $G$ (and using the Lie algebra representation $(T,v)\mapsto Tv$ corresponding to the representation of $G$ over $V$) we can define the **covariant derivative** $D_\mu$ to be $D_\mu\phi(x)=\partial_\mu\phi(x)+A_\mu\phi(x)$. Now we have our object $D_\mu$ for which the equation $D_\mu(g\phi)=gD_\mu\phi$ holds for any $g$.

The action functional is a map $S:\Gamma(F)\rightarrow\mathbb{R}$. It is said to have global invariance under $G$ if $S[\phi]=S[g\phi]$ for every constant gauge transformation $g$. If the condition $S[\phi]=S[g\phi]$ holds for any $g\in\Gamma(E)$ the action is said to be locally (gauge) invariant.

Suppose that we want to **construct locally invariant actions from globally invariant ones**. Usually, global invariance will depend on the fact that $\partial_\mu(g\phi)=g\partial_\mu\phi$, so an action with global invariance will not have in general local invariance. However, by replacing every $\partial_\mu$ by a $D_\mu$ in the expression for the action one obtains a new one which extends invariance to every gauge transformation because now $D_\mu(g\phi)=gD_\mu\phi$ is satisfied for all $g$.

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user coconut