 Solution

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

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

Use the numerator in the original PDF: it contains all

n + 1

squared differentials. The local TeX ends that numerator at

n - 1

, which would not give the stated Riemannian metric.

π (y) = (y_{1}, \dots, y_{n}, 1)

. Its differential sends a tangent vector

v

(v_{1}, \dots, v_{n}, 0)

, so

d X_{i} = d y_{i}

for

i \leq n

and

d X_{n + 1} = 0

. Consequently

g = π^{*} G = \frac{\sum _{i = 1}^{n} d y _{i}^{2}}{1 + \sum _{i = 1}^{n} y _{i}^{2}}, g_{ij} (y) = \frac{δ _{ij}}{1 + ∣ y ∣ ^{2}} .

The denominator is everywhere positive. For any nonzero tangent vector,

g_{y} (v, v) = \frac{∣ v ∣ ^{2}}{1 + ∣ y ∣ ^{2}} > 0,

so this is indeed a smooth positive-definite Riemannian metric on

U

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