Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 49 / 3 / a

Write $h_{ab}=\delta g_{ab}$ and $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$. The difference between two affine connections is a tensor, so its infinitesimal change $C$ is a type $(1,2)$ tensor field. Both affine connections are torsion-free, giving $C^a{}_{bc}=C^a{}_{cb}$. Varying metric compatibility givesLower the first index of $C$. Add the versions with derivatives $b,c$ and subtract the one with derivative $d$; the lower-slot symmetry cancels the unwanted terms, leavingThereforeAll covariant derivatives use the original Levi-Civita connection. The PDF has ${\nabla }_b{h}_{dc}$ as its second term; the converted TeX's ${\nabla }_d{h}_{dc}$ is a transcription error.

Adopt the printed Riemann curvature tensor convention and contract $R_{ab}=R^c{}_{acb}$. At an arbitrary point choose normal coordinates for the original metric tensor. There the original connection coefficients vanish. Varying the coordinate curvature formula leavesAt that point these partial derivatives equal the covariant derivatives of the tensor $C$. Both sides are tensors, so the result is valid in every coordinate system:Contracting its first and third indices proves the Palatini identityThe normal-coordinate argument is applied independently at every point; it does not assume a flat background or set derivatives of the original connection to zero.

Variation of the inverse metric relation $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives $\delta g^{ab}=-h^{ab}$. Hence the metric variation of scalar curvature isPut $h=g^{ab}{h}_{ab}$. The connection variation gives the contractionsUsing metric compatibility and the Palatini identity,Renaming dummy indices givesNo interchange of covariant derivatives on a tensor is needed in this derivation, so no hidden curvature-commutator term is discarded.