 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-3/12c/a/solution

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

Taking the symmetric part gives

S_{ij} = \frac{1}{2} (T_{ij} + T_{ji}) .

Contracting the antisymmetric part with the Levi-Civita symbol gives

V_{k} = \frac{1}{2} ε_{kij} T_{ij} .

Indeed,

\frac{1}{2} ε_{ijk} ε_{k ℓ m} T_{ℓ m} = \frac{1}{2} (T_{ij} - T_{ji})

, so

S_{ij} + ε_{ijk} V_{k} = T_{ij}

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