Since $de^0=0$ and $de^i=\tfrac12\epsilon^i{}_{jk}e^j\wedge e^k$, <Cartan's first structure equation> gives
$$
\boxed{\omega^0{}_\mu=0,
\qquad \omega^i{}_j=\frac12\epsilon^i{}_{jk}e^k}.
$$
<Cartan's second structure equation> then gives
$$
\boxed{\Theta^0{}_\mu=0,
\qquad \Theta^i{}_j=\frac14e^i\wedge e^j}.
$$
Using $\Theta^i{}_j=\tfrac12R^i{}_{jkl}e^k\wedge e^l$ yields
$$
\boxed{R_{ijkl}=\frac14(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})},
$$
with no time-index curvature components. Contraction gives
$$
R_{00}=0,
\qquad R_{ij}=\frac12\delta_{ij},
\qquad R=\frac32,
$$
and therefore
$$
\boxed{G_{00}=\frac34,
\qquad G_{ij}=-\frac14\delta_{ij},
\qquad G_{0i}=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-309/4/c/solution