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

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- Solution c

Solution

Since

d e^{0} = 0

and

d e^{i} = \frac{1}{2} ϵ^{i}_{jk} e^{j} \land e^{k}

, Cartan's first structure equation gives

ω^{0}_{μ} = 0, ω^{i}_{j} = \frac{1}{2} ϵ^{i}_{jk} e^{k} .

Cartan's second structure equation then gives

Θ^{0}_{μ} = 0, Θ^{i}_{j} = \frac{1}{4} e^{i} \land e^{j} .

Using

Θ^{i}_{j} = \frac{1}{2} R^{i}_{jk l} e^{k} \land e^{l}

yields

R_{ijk l} = \frac{1}{4} (δ_{ik} δ_{j l} - δ_{i l} δ_{jk}),

with no time-index curvature components. Contraction gives

R_{00} = 0, R_{ij} = \frac{1}{2} δ_{ij}, R = \frac{3}{2},

and therefore

G_{00} = \frac{3}{4}, G_{ij} = - \frac{1}{4} δ_{ij}, G_{0 i} = 0 .

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