Local curvature formula

Let $\pi:P\to M$ be a principal G-bundle with structure group a Lie group $G$ and Lie algebra $\mathfrak g$ . Let $\omega$ be a principal connection on $P$ with curvature 2-form $\Omega\in\Omega^2(P;\mathfrak g)$ . For an open set $U\subset M$ and a local section $s:U\to P$ , define the local connection form and local curvature form by

A := s^*\omega \in \Omega^1(U;\mathfrak g), \qquad F := s^*\Omega \in \Omega^2(U;\mathfrak g).

Theorem (Local curvature formula). With the notation above,

F \;=\; dA \;+\; \tfrac12[A\wedge A],

where $d$ is the exterior derivative on $U$ . The bracket-wedge uses the Lie bracket on $\mathfrak g$ : for $\alpha\in\Omega^p(U;\mathfrak g)$ and $\beta\in\Omega^q(U;\mathfrak g)$ ,

[\alpha\wedge\beta](v_1,\dots,v_{p+q}) =\frac{1}{p!\,q!}\sum_{\sigma\in S_{p+q}}\mathrm{sgn}(\sigma)\, \big[\alpha(v_{\sigma_1},\dots,v_{\sigma_p}),\beta(v_{\sigma_{p+1}},\dots,v_{\sigma_{p+q}})\big].

Equivalent characterizations

Equivalently, this is the pullback along $s$ of the global structure equation $\Omega=d\omega+\tfrac12[\omega\wedge\omega]$ on $P$ .

Examples

Abelian structure group. If $G=U(1)$ (or any abelian Lie group), then the Lie bracket on $\mathfrak g$ is zero, hence $[A\wedge A]=0$ and the formula reduces to $F=dA$ .
Pure gauge has zero curvature. On a trivial bundle over $U$ , take $A=g^{-1}dg$ for a smooth map $g:U\to G$ . Then $F=dA+\tfrac12[A\wedge A]=0$ by the Maurer–Cartan equation.
Non-abelian commutator term. On $U\subset\mathbb R^2$ with coordinates $(x,y)$ , choose constant elements $X,Y\in\mathfrak g$ and set $A=X\,dx+Y\,dy$ . Then $dA=0$ but $[A\wedge A]=2[X,Y]\,dx\wedge dy$ , so $F=[X,Y]\,dx\wedge dy$ .