Gauge group

Let $\pi\colon P\to M$ be a principal G-bundle with right action of $G$ on $P$.

The gauge group of $P$ is

the group of principal bundle automorphisms $\Phi\colon P\to P$ that cover the identity on the base:

Composition of automorphisms is the group operation.

Equivalent descriptions

1. (Equivariant maps into G) A gauge transformation can be encoded by a smooth map $u\colon P\to G$ satisfying the equivariance condition

   $$u(pg)=g^{-1}u(p)g,$$

   using the conjugation action of G on itself . From such a map, define

   $$\Phi_u(p):=p\cdot u(p).$$

   Then $\Phi_u$ is a principal bundle automorphism covering $\mathrm{id}_M$, and every element of $\mathcal{G}(P)$ arises uniquely this way.

2. (Sections of the adjoint bundle) Using the same conjugation action, form the adjoint bundle

   $$\mathrm{Ad}(P):=P\times_G G.$$

   The gauge group is naturally identified with the group of smooth sections of Ad(P) , with pointwise multiplication in the fiber.

Gauge transformations act on connections: if $\nabla$ is a principal connection on $P$, then any $\Phi\in\mathcal{G}(P)$ produces a new connection by pullback as in the gauge action on the space of connections . This is the global version of a gauge transformation in local trivializations.

Examples

1. Trivial principal bundle. If $P\cong M\times G$ is trivial, then $\mathcal{G}(P)\cong C^\infty(M,G),$ with multiplication pointwise in $G$. Under this identification, $g\in C^\infty(M,G)$ acts by $(x,h)\mapsto (x, g(x)h)$.
2. Abelian structure group. If $G$ is abelian, conjugation is trivial, so $\mathrm{Ad}(P)\cong M\times G$ for any $P$. Hence $\mathcal{G}(P)\cong C^\infty(M,G)$ even when $P$ is not trivial (for example, for circle bundles with $G=U(1)$).
3. Frame bundle viewpoint. If $E\to M$ is a rank-$n$ vector bundle and $P=\mathrm{Fr}(E)$ is its frame bundle , then $\mathcal{G}(P)$ identifies with the group of vector bundle automorphisms of $E$ covering $\mathrm{id}_M$ (equivalently, fiberwise $GL(n)$-valued “change of frame” fields).