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For $D>2$, rescale the string-frame metric by $g^{(E)}=e^{-4\phi /(D-2)}{g}^{(s)}$. The Weyl transformation of the Ricci scalar makes the coefficient of the Einstein-frame Ricci term independent of the dilaton. Indeed $g^{(s)}=e^{4\phi /(D-2)}{g}^{(E)}$ contributes factors $e^{2D\phi /(D-2)}$ from the volume, $e^{-2\phi }$ from the original coupling and $e^{-4\phi /(D-2)}$ from the Ricci term; their product is one. Derivative terms supply the transformed dilaton kinetic term. The two-dimensional case is excluded from this particular rescaling.

The Einstein-Hilbert action is the integral of the Ricci scalar against the metric volume form. Varying it with respect to the metric gives the vacuum Einstein field equations, up to a boundary term.

Every four-dimensional Einstein metric with $R_{ab}=\Lambda g_{ab}$ has constant Ricci scalar $R=4\Lambda$. The vacuum metric f(R) gravity tensor reduces to $(\Lambda f^{\prime }(4\Lambda )-f(4\Lambda )/2)g_{ab}$, giving the displayed necessary and sufficient condition for a fixed cosmological constant. Both function and derivative may vanish, so division by $f^{\prime }$ is unnecessary. Requiring this for all real $\Lambda$ and a smooth $f$ instead yields $f(R)=C{R}^2$.

For a spatially flat Friedmann-Lemaitre-Robertson-Walker metric, negative extrinsic curvature, and $K=-3H+\kappa$, the Hamiltonian constraint gives the displayed scalar perturbation equation. Indeed $K^2=9H^2-6H\kappa$ and $K^i{}_j{K}^j{}_i=3H^2-2H\kappa$ to first order. The spatial Ricci scalar is $4\Delta \Phi$; subtract the background equation $6H^2=16\pi G\bar{\rho }$ and divide by four.

For $h_{ab}=\delta g_{ab}$, the Ricci scalar varies asThe last two terms form a covariant divergence and hence contribute only a boundary term to a compactly supported variation of the Einstein-Hilbert action.

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.

With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.

Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is

Choose the asymptotic time translation: $N=1+O(r^{-1})$, $N^i=O(r^{-1})$, with their derivatives $O(r^{-2})$, in asymptotically Cartesian coordinates. Use the usual differentiable falloff $h_{ij}={\delta }_{ij}+O(r^{-1})$, ${\partial }_k{h}_{ij}=O(r^{-2})$, ${\pi }^{ij}=O(r^{-2})$, and analogous falloff for allowed variations; impose the standard additional conditions needed for finite bulk integrals. These variations need not fix the Arnowitt-Deser-Misner energy.

Only the spatial curvature term produces the leading nonvanishing surface variation. Applying the variation of the Ricci scalar and applying integration by parts to its two derivative terms yieldsThe terms involving derivatives of the asymptotic lapse function, and the shift boundary terms, vanish under these falloff conditions. The displayed surface integral need not vanish: its integrand is $O(r^{-2})$ and the area is $O(r^2)$.

Hamilton's equations require a Hamiltonian with a variation expressed by the bulk functional derivatives alone for the allowed variations. The unwanted term is exactly $-\delta E_{\mathrm{ADM}}$, so differentiability of the gravitational Hamiltonian at spatial infinity requiresIts variation cancels the surface term. On the constraint surface, $H_{\mathrm{total}}=E_{\mathrm{ADM}}$. Restoring units inserts $1/(16\pi G)$ in the surface integral; the paper has set this factor to one. The original PDF has the standard equation ${\partial }_t{h}_{ij}=\delta H/\delta {\pi }^{ij}$; the extra $t$ in the extracted TeX is a transcription error.

Contract the Gauss equation with $h^{ac}{h}^{bd}$. Since $h^{ab}=g^{ab}+n^a{n}^b$ and the antisymmetries of the Riemann curvature tensor eliminate the term containing four normals,The extrinsic terms contract to $K^2$ and $K_{ab}{K}^{ab}$. Hence the intrinsic Ricci scalar is