 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-49/3/a/iii/solution

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

g^{a c} g_{c b} = δ^{a}_{b}

gives

δ g^{ab} = - h^{ab}

δ R = - R^{ab} h_{ab} + g^{ab} δ R_{ab} .

Put

h = g^{ab} h_{ab}

. The connection variation gives the contractions

g^{ab} C^{c}_{ab} = \nabla_{a} h^{c a} - \frac{1}{2} \nabla^{c} h, C^{c}_{a c} = \frac{1}{2} \nabla_{a} h .

g^{ab} δ R_{ab} = \nabla_{c} \nabla_{a} h^{c a} - \nabla_{c} \nabla^{c} h .

Renaming dummy indices gives

δ R = - R^{ab} h_{ab} + \nabla^{a} \nabla^{b} h_{ab} - \nabla^{c} \nabla_{c} (g^{ab} h_{ab}) .

No interchange of covariant derivatives on a tensor is needed in this derivation, so no hidden curvature-commutator term is discarded.

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