Maurer–Cartan equation - Imported Knowlpedia Content

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ . Consider the left Maurer–Cartan form $\theta^L\in\Omega^1(G;\mathfrak{g})$ .

Define the bracket of $\mathfrak{g}$ -valued 1-forms by using the Lie bracket on $\mathfrak{g}$ : for tangent vectors $u,v\in T_gG$ set

[\theta^L\wedge \theta^L]_g(u,v) = [\theta^L_g(u),\,\theta^L_g(v)]\in\mathfrak{g},

and extend by bilinearity and antisymmetry. Then the Maurer–Cartan equation is the identity

\mathrm{d}\theta^L+\tfrac12[\theta^L\wedge\theta^L]=0,

where $\mathrm{d}$ is the exterior derivative applied componentwise.

Examples

Abelian Lie groups. If $\mathfrak{g}$ is abelian, the bracket term vanishes and the equation reduces to $\mathrm{d}\theta^L=0$ (and likewise for $\theta^R$ ). For $G=\mathbb{R}^n$ , this is just $\mathrm{d}(\mathrm{d}x^i)=0$ .
Matrix computation. For a matrix Lie group with $\theta^L=g^{-1}\mathrm{d}g$ , the equation becomes $\mathrm{d}(g^{-1}\mathrm{d}g)+ (g^{-1}\mathrm{d}g)\wedge(g^{-1}\mathrm{d}g)=0$ , which is the differential identity obtained by differentiating $g^{-1}g=I$ .
Structure constants viewpoint. If $\{e_i\}$ is a basis of $\mathfrak{g}$ with $[e_i,e_j]=c_{ij}^k e_k$ and $\theta^L=\theta^i e_i$ , the equation becomes $\mathrm{d}\theta^k + \tfrac12 c_{ij}^k\,\theta^i\wedge\theta^j=0$ , recovering the standard structure equations.

Remarks

With the same convention, the right Maurer–Cartan form $\theta^R$ satisfies

\mathrm{d}\theta^R-\tfrac12[\theta^R\wedge\theta^R]=0.

(These sign conventions match the standard matrix identities for $g^{-1}\mathrm{d}g$ and $\mathrm{d}g\,g^{-1}$ .)