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

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.

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

The pure gravitational action is invariant under diffeomorphisms. An infinitesimal diffeomorphism generated by a compactly supported vector field $\xi$ changes the metric by its tensor Lie derivative,Using symmetry of $E^{ab}$ and the variational expression already derived,As ${\xi }_b$ is arbitrary, the off-shell metric divergence identity isThis Noether identity follows for every metric tensor without imposing either the gravitational field equation or the matter equations. It is a consequence of the metric action's diffeomorphism invariance, not a conclusion requiring a long component calculation.

In four dimensions, Lovelock's theorem restricts a local natural symmetric divergence-free metric tensor with at most second derivatives of the metric to a constant linear combination of the Einstein tensor and the metric. A generic nonlinear f(R) gravity equation contains ${\nabla }_a{\nabla }_b{f}^{\prime }(R)$: the Ricci scalar already has second metric derivatives, so this term generally introduces fourth metric derivatives. It therefore violates the second-order hypothesis, although it remains covariant, symmetric and divergence-free.

The exception must be stated. For $f(R)=aR+b$, the same tensor reduces towhich is precisely of Lovelock form. In particular, $f(R)=R$ is a counterexample to a blanket claim that the theorem never applies. The intended exclusion concerns generic nonlinear $f$, with $f^{\prime \prime }$ not identically zero. Restricting attention to a special constant-curvature solution of a nonlinear theory does not turn its off-shell equations into a universally second-order metric tensor.

A four-dimensional vacuum Einstein solution with the fixed cosmological constant $\Lambda$ has $R_{ab}=\Lambda g_{ab}$ and $R=4\Lambda$, constant. Thus every derivative of $f^{\prime }(R)$ vanishes andThe metric tensor is nondegenerate, so this is zero precisely whenThis is the necessary and sufficient Einstein metric condition in f(R) gravity for the specified $\Lambda$, and works for every such Einstein metric, even when its Weyl tensor is nonzero. There is no need to divide by $f^{\prime }(4\Lambda )$; the degenerate case where both $f$ and $f^{\prime }$ vanish at that curvature is included. At $\Lambda =0$, the condition is simply $f(0)=0$.

If the intention is to demand the inclusion for every real $\Lambda$ simultaneously, the stronger functional condition is $R{f}^{\prime }(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=C{R}^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives $f(R)=C{R}^2$, including $C=0$. This stronger reading is distinct from fixing one cosmological constant.