 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/25i/a/solution

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

For a local parametrization

X (u^{1}, u^{2})

I = g_{ij} d u^{i} d u^{j}, g_{ij} = X_{i} \cdot X_{j} .

u^{1} = u, u^{2} = v

, this is

I = E d u^{2} + 2 F d u d v + G d v^{2}, E = X_{u}^{2}, F = X_{u} \cdot X_{v}, G = X_{v}^{2} .

After choosing an orientation, the Gauss map is

N = \frac{X _{u} \times X _{v}}{∣ X _{u} \times X _{v} ∣} .

II (v, w) = ⟨ S (v), w ⟩

, where the shape operator is

S = - d N

. In coordinates,

II = e d u^{2} + 2 f d u d v + g d v^{2}, e = N \cdot X_{uu}, f = N \cdot X_{uv}, g = N \cdot X_{vv} .

The Gaussian curvature and mean curvature are respectively the determinant and half the trace of

S

K = \frac{e g - f ^{2}}{EG - F ^{2}}, H = \frac{e G - 2 f F + g E}{2 ( EG - F ^{2} )} .

Changing the chosen normal reverses

II

and

H

but leaves

K

unchanged.

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