 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-49/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

L_{X} f = \frac{d}{d s}_{0} f (ϕ_{s} (p)) = X (f) .

Y

ϕ_{s}^{μ} (x) = x^{μ} + s X^{μ} (x) + O (s^{2}), (d ϕ_{s})^{- 1 μ}_{ν} = δ^{μ}_{ν} - s \partial_{ν} X^{μ} + O (s^{2}) .

Multiplying this inverse differential by

Y (ϕ_{s} (x))

gives

(ϕ_{s}^{*} Y)^{μ} = Y^{μ} + s (X^{ν} \partial_{ν} Y^{μ} - Y^{ν} \partial_{ν} X^{μ}) + O (s^{2}) .

Thus

L_{X} Y = [X, Y], [X, Y]^{μ} = X^{ν} \partial_{ν} Y^{μ} - Y^{ν} \partial_{ν} X^{μ} .

X (Y f) - Y (X f)

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