 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-313/3/a/solution

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

ρ = g^{- 1} d g .

Differentiating

g^{- 1} g = I

gives

d (g^{- 1}) = - g^{- 1} (d g) g^{- 1}

. Hence

d ρ = d (g^{- 1}) \land d g = - g^{- 1} d g \land g^{- 1} d g = - ρ \land ρ,

and therefore

d ρ + ρ \land ρ = 0 .

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