 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/37b/c/solution

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

\nabla_{T} (T \cdot S) = (\nabla_{T} T) \cdot S + T \cdot \nabla_{T} S = T \cdot \nabla_{S} T = \frac{1}{2} \nabla_{S} (T \cdot T) = 0.

Hence orthogonality at one point persists. Substituting the constant-curvature tensor into geodesic deviation gives

R^{α}_{β μν} T^{β} T^{μ} S^{ν} = K {T^{α} (T \cdot S) - S^{α} (T \cdot T)} = K S^{α},

because

T \cdot S = 0

and

T \cdot T = - 1

. Therefore

(\nabla_{T} \nabla_{T} S)^{α} - K S^{α} = 0

along the whole geodesic.

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