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I've encountered the problem which I believe intends to address Cartan structure equations, while reading the book Modern Geometry by Dubrovin, Fomenko and Novikov. It goes like:

(Here the summation is understood to take place over repeated indices) Let $X_1, \dots, X_n$ be orthogonal vector fields in an n-dimensional Riemannian space, and denote by $\omega_1, \dots, \omega_n$ the dual basis of 1-forms: $\omega_i(X_j) = \delta_{ij}$ Define 1-forms $\omega_{ij} := \Gamma^{i}_{jk} \omega_k$ and 2-forms $\Omega_{ij} = \frac{1}{2}R_{ijkl} \omega_k \wedge \omega_l$.

When checking these relations \begin{eqnarray} d\omega_i &&= -\omega_j \wedge \omega_{ij} \\ d\omega_{ij} &&= \omega_{il} \wedge \omega_{lj} - \Omega_{ij} \\ d\Omega_{ij} &&= -\Omega_{il} \wedge \omega_{lj} + \omega_{il} \wedge \Omega_{lj}, \end{eqnarray} I made use of the intrinsic formula of Cartan's exterior derivative. In the first identity I am forced to assume that the connection $\Gamma^{k}_{ij}$ is torsion-free so that I can write $[ X_{k}, X_{l} ] = \nabla_{X_k}X_l - \nabla_{X_l}X_k$, although the problem didn't offer such condition.

I also have some problem dealing with the other two expressions in the calculations. Every insight will be appreciated.

Kelvin Hsu
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  • Presumably the connection here is the Levi-Civita connection of the Riemannian metric, which in particular is torsion-free. – Travis Willse Dec 01 '16 at 16:10
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    To derive the first equation, you need to assume that the connection is torsion-free. The second equation can be thought of as the equation of the Curvature in the Cartan formalism. The third equation is probably equivalent to the differential Bianchi identity. But E. Cartan had a different way of looking at these equations, which led to exterior differential systems, prolongation and Cartan connections. I know I did not help you with the calculations, but I hope my comment helps. – Malkoun Aug 08 '17 at 12:18
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    take a look at https://planetmath.org/cartanstructuralequations – janmarqz Nov 07 '19 at 18:40

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