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For the minimally coupled scalar field, keep the metric tensor fixed and vary $\varphi$. A scalar's first covariant derivative equals its ordinary derivative, soApply integration by parts for tensor fields. The boundary contribution is proportional to $\delta \varphi$ and vanishes under the given boundary condition, leavingThe interior variation is arbitrary. Thus the scalar equation of motion isHere the d'Alembert operator contains the Levi-Civita connection in the second covariant derivative, even though the first derivative of the scalar field is ordinary.

Vary the inverse metric tensor while holding the matter fields fixed. Differentiating $g_{a\mu }{g}^{\mu b}={\delta }_a{}^b$ gives $\delta g_{ab}=-g_{a\mu }{g}_{b\nu }\delta g^{\mu \nu }$. The Jacobi determinant derivative formula consequently implies the inverse metric volume variationSince the matter Lagrangian in this question contains no derivatives of the metric tensor, its metric variation is algebraic. The product rule givesMultiplying by $-2/\sqrt{-g}$ proves the equivalence of the stress-energy definitions:The variations of the symmetric metric tensor are understood symmetrically. If the matter action contained metric derivatives, the definition would instead require a functional derivative and the corresponding integration by parts.

Contract the printed Riemann curvature tensor using $R_{ab}=R^{\mu }{}_{a\mu b}$. Only the derivative terms can contain second derivatives of the metric tensor:Substituting the Christoffel symbol formula, its second derivative part isSet $H_b=g_{b\lambda }{H}^{\lambda }=g^{\mu \nu }{\partial }_{\mu }{g}_{\nu b}-\frac{1}{2}{g}^{\mu \nu }{\partial }_b{g}_{\mu \nu }$. The first, second and fourth terms are the second derivative terms of $\frac{1}{2}({\partial }_a{H}_b+{\partial }_b{H}_a)$. Derivatives of the inverse metric tensor produce only products of first derivatives. Thuswhere $Q$ contains at most first metric derivatives. In harmonic coordinates, $H^{\lambda }$ vanishes throughout the chart, so $H_b$ and its derivatives vanish. The vacuum equations have the wave principal partThese are quasilinear partial differential equations with the principal part of a wave equation, because $g^{\mu \nu }$ has Lorentzian signature. This is the harmonic reduction of the vacuum Einstein equations.

First, the inverse metric tensor printed in the PDF has a missing factor: its lower-left entry must be $g^{i0}={\alpha }^{-2}{\beta }^i$, symmetric with $g^{0i}$. This correction follows either from symmetry or direct block multiplication and does not change the evolution equation used here.

Let $D_t={\partial }_t-{\beta }^k{\partial }_k$. Contract the spatial metric tensor evolution equation with ${\gamma }^{ij}$. Both shift-gradient terms contribute ${\partial }_k{\beta }^k$, givingThe Jacobi determinant derivative formula applies to every derivative of $\gamma =\det ({\gamma }_{ij})$ and gives $D_t\gamma =\gamma {\gamma }^{ij}{D}_t{\gamma }_{ij}$. Therefore the spatial determinant evolves asEquivalently, the spatial volume evolution identity is $D_t\sqrt{\gamma }=\sqrt{\gamma }({\partial }_k{\beta }^k-\alpha K)$, with $K$ the trace of the extrinsic curvature of a spatial hypersurface.

Use the equivalent density form of harmonic slicing from part (ii). With $\sqrt{-g}=\alpha \sqrt{\gamma }$, positive lapse function $\alpha$, and the corrected inverse metric tensor, its time component becomesThus $D_t(\sqrt{\gamma }/\alpha )=(\sqrt{\gamma }/\alpha ){\partial }_i{\beta }^i$, orThe spatial volume evolution identity gives $D_t\log \sqrt{\gamma }={\partial }_i{\beta }^i-\alpha K$. Subtraction yields $D_t\log \alpha =-\alpha K$, hence the harmonic lapse evolution isThe shift vector need not vanish; its transport term is essential in a general 3+1 decomposition of spacetime.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

The Levi-Civita connection is the unique affine connection that has both vanishing torsion tensor and metric compatibility. The affine connection decomposition shows exactly how an arbitrary affine connection departs from it: the contorsion tensor contributes the metric-compatible torsion correction, and the disformation tensor contributes the nonmetricity correction. The torsion and nonmetricity tensor, together with the metric tensor, determine the difference uniquely.

In the paper's convention the actual contorsion correction is $-K$, with the corrected half factor from part (i), and the disformation correction is $W$. Both corrections are tensors, although the separate connection coefficients $\Gamma$ and $S$ are not tensors. If both torsion and nonmetricity vanish, then $\boxed{\Gamma =S}$. Vanishing nonmetricity alone permits torsion, and vanishing torsion alone permits nonmetricity.

Choose a basis of $T_{p}M$ and use the affine exponential map to define affine normal coordinates. Its differential at zero is the identity, so the inverse function theorem gives a chart near $p$. By part (iii), radial geodesics have coordinates $x^{\mu }(s)=s{X}^{\mu }$. Their geodesic equations at $p$ imply ${\Gamma }^{\lambda }{}_{\mu \nu }(p)X^{\mu }{X}^{\nu }=0$ for every $X$. The polarization identity therefore gives ${\Gamma }^{\lambda }{}_{(\mu \nu )}(p)=0$.

The disformation tensor $W_{\mu \nu }{}^{\lambda }$ is symmetric in $\mu ,\nu$, whereas $T_{\mu \nu }{}^{\lambda }$ is antisymmetric. Also $T_{\nu }{}^{\lambda }{}_{\mu }=-T^{\lambda }{}_{\nu \mu }$. Taking the symmetric part of the corrected affine connection decomposition consequently gives the normal-coordinate identityThus the final identity in the PDF is correct after the half-factor repair in part (i). In general ${\Gamma }^{\lambda }{}_{\mu \nu }(p)=\frac{1}{2}{T}_{\mu \nu }{}^{\lambda }(p)$ in the paper's convention, rather than zero. Neither $S$ nor the first derivatives of the metric tensor need vanish in these affine normal coordinates. With the Levi-Civita connection, both torsion and nonmetricity vanish and the usual vanishing of the Christoffel symbols is recovered.

Let $\delta h_{\mu \nu }={\partial }_{\mu }{V}_{\nu }+{\partial }_{\nu }{V}_{\mu }$. Substitution into the linearized Levi-Civita connection cancels four terms and leavesTo first order in the metric perturbation, the products of Christoffel symbols are quadratic and may be discarded. Therefore the change in the Riemann curvature tensor isThis is the gauge invariance of the linearized Riemann tensor. Lowering the first index with the Minkowski metric does not change the conclusion; the change from using the full metric tensor instead is of second order because the background curvature is zero. HenceThe result is invariance at linear order about flat spacetime, not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of $V$ cancel in pairs by commuting partial derivatives.

Use geometrized units $G=c=1$ and metric signature $(-,+,+,+)$. Put $f=1-2M/r$ in the Schwarzschild metric. The Schwarzschild tortoise coordinate satisfies $d{r}_{\ast }/dr=f^{-1}$, soSubstituting $dt=dv-f^{-1}dr$ gives the Ingoing Eddington-Finkelstein coordinates:The radial metric tensor has determinant $-1$ and inverse components $g^{vv}=0$, $g^{vr}=1$, $g^{rr}=f$. Thus it is nondegenerate and analytic at $r=2M$. The same expression defines a Lorentzian metric for every $r>0$, extending the exterior across the future Schwarzschild event horizon into the black hole. It does not include the other exterior or the white hole of the full Kruskal spacetime. At $r=0$, the Kretschmann scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges, so this is a curvature singularity, not a removable coordinate singularity.

For the P(X, phi) scalar field theory specialized to $F(X)$, vary the inverse metric tensor while holding $\varphi$ fixed. With $q_a={\partial }_a\varphi$,Using $T_{ab}=-2(-g{)}^{-1/2}\delta S/\delta g^{ab}$ yieldsIn any local Lorentz frame, set $A=F^{\prime }(X)\ge 0$, $Y=X{F}^{\prime }(X)-F(X)\ge 0$, $s={\sum }_i{q}_i^2$, and $X=(q_0^2-s)/2$. ThenA direct calculation givesHence $\rho \ge \sqrt{T_{0i}{T}_{0i}}$ in every local Lorentz frame, proving the dominant energy condition. This proof covers timelike, null, spacelike, and zero $\nabla \varphi$ alike; it does not assume $X>0$ or use a perfect fluid in general relativity rest frame. The hypotheses need only hold at every attained value of $X$; the stated all-$X$ assumptions are sufficient. For example, $F=X-V_0$, $V_0\ge 0$, satisfies them.

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.

For the spherical coordinates in the question, the coordinate tangent vectors are mutually orthogonal, with lengths $1,r,r\sin \theta$. Their metric tensor is therefore $\mathrm{diag}(1,r^2,r^2{\sin }^2\theta )$ and the Jacobian determinant is $r^2\sin \theta$. Applying the divergence formula to the gradient givesFor example, the formula follows directly from $\Delta u=|\det h{|}^{-1/2}{\partial }_i(|\det h{|}^{1/2}{h}^{ij}{\partial }_{j}u)$ for this diagonal metric. These coordinates are valid away from the polar axis and angular seam; the Laplacian is a smooth geometric operator there as well, described by other charts. The coordinate volume form integrates over $1/2<r<2$, $0<\theta <\pi$, $0\le \varphi <2\pi$, with the usual periodic identification.

Put $q=r_0^2$ and $\eta =d\psi +\frac{1}{2}\cos \theta d\varphi$. Expanding the metric tensor in the stationary direction gives the useful exact identitiesTaking the exterior derivative givesThe fibered form of the metric tensor has determinant $\det g=-r^6{\sin }^2\theta /16$. Its inverse components needed for the flux areThereforesince $h(1-\alpha \Omega )=1$. This calculation includes the rotational mixed term; dropping it prematurely would miss the exact cancellation.

With the stipulated negative coordinate orientation, ${ϵ}_{tr\psi \theta \varphi }=-r^3\sin \theta /4$. The pullback of the Hodge star operator to the constant-$t,r$ three-dimensional spacelike submanifold is consequentlyThe coordinate ranges and standard sphere identifications giveThus the five-dimensional Komar mass, in the normalization given and $G_5=1$, isThe flux is already independent of $r$ in the vacuum region, so its limit at infinity has the same value. The negative orientation in the question is crucial for the positive sign; reversing that orientation reverses the flux.

For the zero-angular-momentum curves, the fiber term in the induced metric tensor vanishes along the tangent, leavingfor a null geodesic. In the exterior, a future geodesic has positive conserved energy $E=(f/h)\dot{t}$. Its equations areDefine the tortoise coordinate, separately on intervals bounded by horizon radii, byFor an outgoing curve $\dot{r}=+E\sqrt{h}$,for an ingoing curve $\dot{r}=-E\sqrt{h}$, $d(t+r_{\ast })/d\lambda =0$. Thus $u=t-r_{\ast }$ and $v=t+r_{\ast }$ are the respective constant labels. The derivative $\sqrt{h}/f$, with its sign, is required for continuation; taking $\sqrt{h}/|f|$ would give the wrong labels between the horizons.

For a nonextremal horizon $r=r_H$, the tortoise coordinate has leading behaviorAt the double root of the extremal black hole it instead has a pole. These singularities are coordinate effects at a regular Killing horizon; they are not the curvature singularity at $r=0$. To cross the future horizon, the ingoing coordinatesgive the regular induced metric tensorThis also fixes the future branch used in the causal argument below.