 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/38c/a/solution

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

Contract the defining formula in its first and third indices. In four dimensions,

C^{α}_{β α δ} = = R_{β δ} - \frac{1}{2} (4 R_{β δ} + g_{β δ} R - R_{β δ} - R_{β δ}) + \frac{1}{6} (4 g_{β δ} - g_{β δ}) R 0.

Thus

C^{α}_{β α δ} = 0 .

The Weyl tensor has the same algebraic symmetries as the Riemann curvature tensor: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the Trace-free property of the Weyl tensor.

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