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\begin{document}

\( ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}(d\theta ^{2}+sin^{2}(\theta )d\phi ^{2}) \)

the cotangent space vielbeins are:(r\(>\)2M)

\( \theta ^{0}=(1-\frac{2M}{r})^{.5}dt \)

\( \theta ^{1}=(1-\frac{2M}{r})^{-.5}dr \)

\( \theta ^{2}=rd\theta  \)

\( \theta ^{3}=rsin(\theta )d\phi  \)

\( d\theta ^{0}=-\frac{M}{r^{2}}(1-\frac{2M}{r})^{-.5}dr \)\( \wedge dt \)\( =-\frac{M}{r^{2}}(1-\frac{2M}{r})^{-.5}\theta ^{1} \)\( \wedge \theta ^{0}=w^{0}_{b}\theta ^{b} \)

So \( w^{2}_{1}=w_{21}=-w_{12}=-w^{1}_{2}=(1-\frac{2M}{r})^{.5}d\theta  \)

\( w^{3}_{1}=-w^{1}_{3}=(1-\frac{2M}{r})^{.5}sin(\theta )d\phi  \)

\( w^{3}_{2}=-w^{2}_{3}=cos(\theta )d\phi  \)

\( w^{0}_{1}=-w_{01}=w_{10}=w^{1}_{0}=\frac{M}{r^{2}}dt \)

with the rest 0.

From here, you can get the curvature tensor via the second cartan
structure equation.

\end{document}