Maurer–Cartan equation for the left Maurer–Cartan form

The left Maurer–Cartan form on a Lie group satisfies the structure equation dθ + 1/2[θ∧θ] = 0.

Maurer–Cartan equation for the left Maurer–Cartan form

Let $G$ be a Lie group with Lie algebra $\mathfrak g$ . The left Maurer–Cartan form is the $\mathfrak g$ -valued $1$ -form $\theta_L\in\Omega^1(G;\mathfrak g)$ defined by

(\theta_L)_g := (dL_{g^{-1}})_g : T_gG \to T_eG\cong \mathfrak g.

Lemma (Maurer–Cartan equation). The form $\theta_L$ satisfies

d\theta_L + \tfrac12[\theta_L\wedge\theta_L]=0,

where $d$ is the exterior derivative and $[\theta_L\wedge\theta_L]$ uses the Lie bracket on $\mathfrak g$ .

This is the curvature-zero structure equation for the canonical flat connection on $G$ , and it is the algebraic reason that “pure gauge” potentials have vanishing curvature.

Examples

Matrix groups. If $G\subset \mathrm{GL}(n,\mathbb C)$ is a matrix Lie group, then $\theta_L=g^{-1}dg$ , and the identity becomes $d(g^{-1}dg) + (g^{-1}dg)\wedge (g^{-1}dg)=0.$
Abelian groups. For $G=\mathbb R^n$ (additive), the bracket on $\mathfrak g$ is zero and $\theta_L$ is constant, so the equation reduces to $d\theta_L=0$ .
Pure gauge curvature. If $A=g^{-1}dg$ on an open set $U$ , then substituting into the local curvature formula gives $F=0$ exactly because of the Maurer–Cartan equation.