 Solution

Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.