Choose
$$
\theta^0=\sqrt f\,dt,\qquad
\theta^1=\frac{dr}{\sqrt f},\qquad
\theta^2=r\,d\vartheta,\qquad
\theta^3=r\sin\vartheta\,d\varphi.
$$
Cartan's first structure equation gives the independent nonzero <connection 1-forms>
$$
\omega^0{}_1=\frac{f'}{2\sqrt f}\theta^0,\qquad
\omega^2{}_1=\frac{\sqrt f}{r}\theta^2,\qquad
\omega^3{}_1=\frac{\sqrt f}{r}\theta^3,\qquad
\omega^3{}_2=\frac{\cot\vartheta}{r}\theta^3,
$$
with the remaining forms obtained by metric antisymmetry. Cartan's second structure equation then gives
$$
\begin{aligned}
\mathcal R^0{}_1&=-\frac{f''}{2}\theta^0\wedge\theta^1,\\
\mathcal R^0{}_2&=-\frac{f'}{2r}\theta^0\wedge\theta^2,
&
\mathcal R^0{}_3&=-\frac{f'}{2r}\theta^0\wedge\theta^3,\\
\mathcal R^1{}_2&=-\frac{f'}{2r}\theta^1\wedge\theta^2,
&
\mathcal R^1{}_3&=-\frac{f'}{2r}\theta^1\wedge\theta^3,\\
\mathcal R^2{}_3&=\frac{1-f}{r^2}\theta^2\wedge\theta^3.
\end{aligned}
$$
These signs follow the curvature convention stated in Question 1.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-309/3/c/solution