 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/37d/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Write

J^{α}_{μ} = \frac{\partial x ^{α}}{\partial x ^{μ}} .

V^{α} = J^{α}_{μ} V^{μ}

W^{α} = J^{α}_{μ} W^{μ}

, and

V^{β} \partial_{β} = V^{ν} \partial_{ν}

. Therefore

[V, W]^{α} = V^{ν} \partial_{ν} (J^{α}_{μ} W^{μ}) - W^{ν} \partial_{ν} (J^{α}_{μ} V^{μ}) = J^{α}_{μ} [V, W]^{μ} + (V^{ν} W^{μ} - W^{ν} V^{μ}) \frac{\partial ^{2} x ^{α}}{\partial x ^{ν} \partial x ^{μ}} .

The last term vanishes because its first factor is antisymmetric in

μ, ν

[V, W]^{α} = J^{α}_{μ} [V, W]^{μ} .

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