 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-309/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Put

r^{2} = y_{1}^{2} + y_{2}^{2}

g = e^{2 ω} δ, ω = - \frac{1}{2} lo g (1 + r^{2}) .

The base metric is flat, so its Ricci scalar is zero and its Laplacian is

Δ = \partial_{1}^{2} + \partial_{2}^{2}

R_{g} = - 2 e^{- 2 ω} Δ ω .

Differentiate explicitly:

\partial_{i} ω = - \frac{y _{i}}{1 + r ^{2}}, Δ ω = - \frac{2}{1 + r ^{2}} + \frac{2 r ^{2}}{( 1 + r ^{2} ) ^{2}} = - \frac{2}{( 1 + r ^{2} ) ^{2}} .

Since

e^{- 2 ω} = 1 + r^{2}

, the required Ricci scalar is

R_{g} (y_{1}, y_{2}) = \frac{4}{1 + y _{1}^{2} + y _{2}^{2}} .

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