Lemma: local curvature transforms by conjugation

Let $U\subseteq M$ be open, let $G$ be a Lie group with Lie algebra $\mathfrak g$ , and let

A\in\Omega^1(U;\mathfrak g)

be a local connection 1-form. Its local curvature is

F:=dA + A\wedge A\in\Omega^2(U;\mathfrak g),

Lemma (curvature transformation law)

Given a smooth map $g:U\to G$ , define the gauge-transformed local connection 1-form by

A^g := g^{-1}Ag + g^{-1}dg,

F^g:=dA^g + A^g\wedge A^g

be its curvature. Then

F^g = g^{-1}Fg.

Equivalently, $F$ transforms by the adjoint action of $G$ on $\mathfrak g$ .

Expand $F^g$ using $A^g=g^{-1}Ag+g^{-1}dg$ and the Leibniz rule:

the mixed terms involving $d(g^{-1}dg)$ cancel using the Maurer--Cartan identity for $g^{-1}dg$ ,
the remaining terms regroup into $g^{-1}(dA + A\wedge A)g$ .

This is the usual statement that curvature is tensorial under gauge transformations.

A direct consequence is that on overlaps $U_i\cap U_j$ with transition function $g_{ij}$ , the local curvature forms satisfy

F_j = g_{ij}^{-1}F_i g_{ij},

so invariant polynomials applied to $F_i$ glue to globally defined Chern--Weil forms.

Examples

Abelian groups (electromagnetism). If $G$ is abelian (for example $U(1)$ ), then $g^{-1}Fg=F$ , so the curvature 2-form is gauge invariant. In particular, the transformation reduces to $A^g=A+g^{-1}dg$ while $F^g=dA^g=dA=F$ .

Pure gauge connections have zero curvature. On a trivial bundle, if $A=g^{-1}dg$ is pure gauge, then $F=dA+A\wedge A=0$ . The lemma then gives $F^h=h^{-1}0\,h=0$ for any further gauge transformation $h$ .

Associated vector bundles (matrix conjugation). If $G$ acts on a vector space via a representation (see representation), the induced curvature on the associated vector bundle is a matrix-valued 2-form, and this lemma becomes the familiar rule “curvature matrices conjugate under change of frame.”

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