Proof of Cartan's Structure Equations

Cartan's structure equations express the torsion and curvature of a connection in terms of connection and coframe forms, providing the most efficient computational framework for curvature calculations.

Statement

Theorem7.6Cartan's structure equations

Let $(M, g)$ be a Riemannian manifold with an orthonormal coframe $\{\theta^1, \ldots, \theta^n\}$ (i.e., $g = \delta_{ij}\theta^i \otimes \theta^j$ ) and connection 1-forms $\omega^i_j$ determined by the Levi-Civita connection. Then:

First structure equation (torsion):

d\theta^i + \omega^i_j \wedge \theta^j = 0.

Second structure equation (curvature):

\Omega^i_j = d\omega^i_j + \omega^i_k \wedge \omega^k_j,

where $\Omega^i_j$ are the curvature 2-forms. Additionally, metric compatibility gives $\omega_{ij} = -\omega_{ji}$ (skew-symmetry, where $\omega_{ij} = \delta_{ik}\omega^k_j$ ).

Proof

First structure equation. Let $\{e_1, \ldots, e_n\}$ be the orthonormal frame dual to $\{\theta^i\}$ , so $\theta^i(e_j) = \delta^i_j$ . The Levi-Civita connection gives $\nabla_X e_i = \omega^j_i(X) e_j$ . We need to show $d\theta^i = -\omega^i_j \wedge \theta^j$ .

By the formula for exterior derivative of a 1-form:

d\theta^i(X, Y) = X(\theta^i(Y)) - Y(\theta^i(X)) - \theta^i([X, Y]).

For the right side, $-(\omega^i_j \wedge \theta^j)(X,Y) = -\omega^i_j(X)\theta^j(Y) + \omega^i_j(Y)\theta^j(X)$ . Taking $X = e_k$ , $Y = e_l$ :

d\theta^i(e_k, e_l) = e_k(\delta^i_l) - e_l(\delta^i_k) - \theta^i([e_k, e_l]) = -\theta^i([e_k, e_l]).

And $-\omega^i_j(e_k)\delta^j_l + \omega^i_j(e_l)\delta^j_k = -\omega^i_l(e_k) + \omega^i_k(e_l)$ .

Torsion-freeness gives $\nabla_{e_k} e_l - \nabla_{e_l} e_k = [e_k, e_l]$ , so:

\theta^i([e_k, e_l]) = \theta^i(\nabla_{e_k} e_l) - \theta^i(\nabla_{e_l} e_k) = \omega^i_l(e_k) - \omega^i_k(e_l).

Therefore $d\theta^i(e_k, e_l) = -\omega^i_l(e_k) + \omega^i_k(e_l) = -(\omega^i_j \wedge \theta^j)(e_k, e_l)$ .

Skew-symmetry. Metric compatibility $0 = d(\delta_{ij}) = dg(e_i, e_j) = g(\nabla e_i, e_j) + g(e_i, \nabla e_j)$ gives $\omega_{ji} + \omega_{ij} = 0$ .

Second structure equation. Define the curvature 2-form by $\Omega^i_j(X,Y)e_i = R(X,Y)e_j$ . Then:

R(X,Y)e_j = \nabla_X\nabla_Y e_j - \nabla_Y\nabla_X e_j - \nabla_{[X,Y]}e_j.

Computing: $\nabla_X(\omega^i_j(Y)e_i) = X(\omega^i_j(Y))e_i + \omega^i_j(Y)\omega^k_i(X)e_k$ . Similarly for $\nabla_Y\nabla_X$ . The result is:

\Omega^i_j(X,Y) = d\omega^i_j(X,Y) + (\omega^i_k \wedge \omega^k_j)(X,Y). \quad \blacksquare

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Computational Example

ExampleCurvature of $S^2$ via structure equations

On $S^2$ with $\theta^1 = d\theta$ , $\theta^2 = \sin\theta \, d\phi$ . From the first structure equation $d\theta^2 = \cos\theta \, d\theta \wedge d\phi = -\omega^2_1 \wedge \theta^1$ , we read off $\omega^2_1 = -\cos\theta \, d\phi$ (and $\omega^1_2 = \cos\theta \, d\phi$ ).

The curvature: $\Omega^1_2 = d\omega^1_2 = -\sin\theta \, d\theta \wedge d\phi = -\sin\theta \, \theta^1 \wedge \frac{\theta^2}{\sin\theta} = -\theta^1 \wedge \theta^2$ .

So $K = 1$ (the Gaussian curvature of the unit sphere), recovering the expected result in just a few lines.

RemarkEfficiency of the method

Cartan's structure equations avoid computing Christoffel symbols (which involve $n^2(n+1)/2$ functions) and working directly with the curvature 2-forms (which leverage the skew-symmetry to reduce computation). This method is especially efficient for spaces with symmetry or for computing curvature of Lie groups.