= 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> gives
$$
0=d\eta_{ab}=g(\nabla e_a,e_b)+g(e_a,\nabla e_b)=\theta_{ba}+\theta_{ab}.
$$
Thus \b[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 $Y=e_b\omega^b(Y)$. Then
$$
\omega^a(\nabla_XY)=X\big(\omega^a(Y)\big)+\theta^a{}_b(X)\omega^b(Y).
$$
Subtract 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>:
$$
\omega^a\big(\nabla_XY-\nabla_YX-[X,Y]\big)=d\omega^a(X,Y)+(\theta^a{}_b\wedge\omega^b)(X,Y).
$$
The <Levi-Civita connection> is <torsion-free>, so <Cartan's first structure equation> is
$$
\boxed{d\omega^a+\theta^a{}_b\wedge\omega^b=0.}
$$
Use $R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$ and define its matrix of <curvature two-forms> by $R(X,Y)e_b=e_aR^a{}_b(X,Y)$. Applying two <covariant derivatives> to $e_b$ yields
$$
R^a{}_b(X,Y)=X\big(\theta^a{}_b(Y)\big)-Y\big(\theta^a{}_b(X)\big)-\theta^a{}_b([X,Y])
+\theta^a{}_c(X)\theta^c{}_b(Y)-\theta^a{}_c(Y)\theta^c{}_b(X).
$$
The 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> is
$$
\boxed{R^a{}_b=d\theta^a{}_b+\theta^a{}_c\wedge\theta^c{}_b.}
$$
The associated <Riemann curvature tensor> components satisfy $R^a{}_b=\tfrac12R^a{}_{bcd}\omega^c\wedge\omega^d$ with this convention.
For a position-dependent change of <pseudo-orthonormal frame>, write the frame row as $E'=E\Lambda$, with $\Lambda^T\eta\Lambda=\eta$. The dual <coframe> obeys $\omega'=\Lambda^{-1}\omega$. Applying the <Leibniz rule> to $\nabla(E\Lambda)$ gives
$$
\theta'=\Lambda^{-1}\theta\Lambda+\Lambda^{-1}d\Lambda.
$$
The <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 $R(X,Y)(E\Lambda)=(R(X,Y)E)\Lambda=E'\Lambda^{-1}R(X,Y)\Lambda$. Consequently
$$
\boxed{R'=\Lambda^{-1}R\Lambda,\qquad R'^a{}_b=(\Lambda^{-1})^a{}_cR^c{}_d\Lambda^d{}_b.}
$$
Equivalently, substituting the transformed connection into $d\theta'+\theta'\wedge\theta'$ 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.
Back to article page
