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If $\omega_i$ are dual basis one forms corresponding to an orthonormal tetrad basis $e_i$, and given that the commutation coefficients $C_{ij}^k$ are defined by

\begin{equation} [e_i,e_j]=C_{ij}^k e_k \end{equation}

how do you prove the following equation

\begin{equation} \mathrm{d}\omega^a=-\frac{1}{2}C_{bc}^a\omega_b\wedge\omega_c \end{equation}

as given in equation 5.6, page 99 of "Relativity Demystified" by David McMahon, published by McGraw Hill, 2006, and where $\mathrm{d}$ is the exterior derivative operator?

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I will use the notation $\theta^i$ for the dual basis since $\omega$ is reserved for another important form.

Since $\mathrm{d}\theta^i$ is a 2-form, we may expand it in the basis $\theta^i$ itself: $$\mathrm{d}\theta^i=-\frac{1}{2}C^i{}_{jk}\theta^j\wedge\theta^k$$ But the first structure equation gives $$\mathrm{d}\theta^i=-\omega^i{}_j\wedge\theta^j$$ where $\omega^i{}_j=\Gamma^i{}_{kj}\theta^k$. Some algebra gives $C^i{}_{jk}=\Gamma^i{}_{jk}-\Gamma^i{}_{kj}$. Using $\nabla_XY-\nabla_YX=[X,Y]$ and the defintion of the Christoffel symbols as $\nabla_{e_i}e_j=\Gamma^k{}_{ij}e_k$ establishes the desired relation.

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