 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/38c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

H_{α β} = \partial_{α} \partial_{β} ω_{p}, H = η^{α β} H_{α β} .

g_{α β} = e^{2 ω} η_{α β}

Γ_{α β}^{ρ} = δ_{α}^{ρ} \partial_{β} ω + δ_{β}^{ρ} \partial_{α} ω - η_{α β} \partial^{ρ} ω .

The assumed first-derivative condition makes

Γ_{α β}^{ρ} (p) = 0

. Differentiating once more gives, at

p

R_{α β γ δ} = e^{2 ω} (η_{α δ} H_{β γ} - η_{α γ} H_{β δ} - η_{β δ} H_{α γ} + η_{β γ} H_{α δ}) .

Its contractions are

R_{β δ} = - 2 H_{β δ} - η_{β δ} H, R = - 6 e^{- 2 ω} H .

Substitution into the defining expression for

C_{α β γ δ}

cancels every Hessian term, leaving

C_{α β γ δ} (p) = 0 .

Since

p

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