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

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- Solution c

Solution

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θ^{0} = f d t, θ^{1} = \frac{d r}{f}, θ^{2} = r d ϑ, θ^{3} = r sin ϑ d φ .

Cartan's first structure equation gives the independent nonzero connection 1-forms

ω^{0}_{1} = \frac{f ^{'}}{2 f} θ^{0}, ω^{2}_{1} = \frac{f}{r} θ^{2}, ω^{3}_{1} = \frac{f}{r} θ^{3}, ω^{3}_{2} = \frac{c o t ϑ}{r} θ^{3},

with the remaining forms obtained by metric antisymmetry. Cartan's second structure equation then gives

R^{0}_{1} R^{0}_{2} R^{1}_{2} R^{2}_{3} = - \frac{f ^{''}}{2} θ^{0} \land θ^{1}, = - \frac{f ^{'}}{2 r} θ^{0} \land θ^{2}, = - \frac{f ^{'}}{2 r} θ^{1} \land θ^{2}, = \frac{1 - f}{r ^{2}} θ^{2} \land θ^{3} . R^{0}_{3} R^{1}_{3} = - \frac{f ^{'}}{2 r} θ^{0} \land θ^{3}, = - \frac{f ^{'}}{2 r} θ^{1} \land θ^{3},

These signs follow the curvature convention stated in Question 1.

Solved by gpt-5.6-sol high.

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