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Use the source's $+---$ metric signature and $x^0=\eta$, with conformal time derivatives denoted by primes. For this vector cosmological perturbation, the metric tensor and inverse metric through first order areSpatial indices on perturbations are raised with the background Kronecker delta. The Christoffel symbol formula isBecause $a=a(\eta )$ only, it givesFor the other requested Christoffel symbols, only the time derivative of the spatial metric tensor contributes:ThusFinally contract the first upper and first lower spatial indices, summing over $i$:Since $J_{ii}=2{\partial }_i{F}_i=0$ by the transverse vector field condition on $\mathbf{F}$,The useful Christoffel symbol traces are consequently ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}$ and ${\Gamma }_{\mu j}^{\mu }=0$ to first order: the first uses $J_{ii}^{\prime }=0$ and the second uses the contraction just proved.

Put $\mu =GM/c^2$, $A=1-2\mu /r$ and $B=1+2\mu /r$. For an affinely parametrized geodesic, use the unhalved quadratic Lagrangian $\mathcal{L}=-c^{2}A{\dot{t}}^2+B{\dot{\mathbf{r}}}^{,2}$. Its Euler-Lagrange equations are unchanged by this overall normalization. Since $t$ is a cyclic coordinate, $d(-2c^{2}A\dot{t})/d\tau =0$, equivalentlyThe spatial derivatives give $\partial \mathcal{L}/\partial {\dot{x}}^i=2B{\dot{x}}^i$ and $\partial \mathcal{L}/\partial x^i=-c^2{\dot{t}}^2{A}_i+{\dot{\mathbf{r}}}^{,2}{B}_i$, where $A_i={\partial }_{i}A$. Therefore

For the connection coefficients, take $x^0=ct$, so $g_{00}=-A$ and $g_{ij}=B{\delta }_{ij}$. Substituting the diagonal inverse metric into the Christoffel symbol formula gives all nonzero component types, including lower-index symmetry:All other component types vanish. If the time coordinate is $t$ rather than $ct$, replace ${\Gamma }^i{}_{00}$ by ${\Gamma }^i{}_{tt}=c^2{\Gamma }^i{}_{00}$; the displayed time–space mixed coefficients are unchanged. Spatial indices in these formulas are raised with the background Euclidean metric.

For the circular orbit let $u=\mu /R$ and $\Omega =d\varphi /dt=\omega /\dot{t}$. Both $A$ and $B$ are constant along it. The radial spatial equation gives $(1+u){\omega }^2=(GM/R^3){\dot{t}}^2$, so, treating the displayed metric coefficients algebraically, ${\Omega }^2=GM/[R^3(1+u)]$. The weak-field approximation securely determines the leading Keplerian orbit resultThe time normalization is $-c^2=-c^{2}A{\dot{t}}^2+B{R}^2{\omega }^2$, yielding $\dot{t}=1+3u/2+O(u^2)$ and $P_{\tau }/P_t=1-3u/2+O(u^2)$. One must distinguish the proper-time angular frequency in the supplied path from the coordinate-time angular frequency. The formal factor $1+u$ in the frequency equation comes from treating the truncated metric as exact; a complete relative post-Newtonian correction to the orbital period would also require the omitted higher-order temporal metric terms.

Use metric signature $(-,+,+,+)$, $x^0=ct$, and retain terms linear in $\Phi /c^2$ or $\Psi /c^2$. The inverse metric is $g^{00}=-(1-2\Phi /c^2)$ and $g^{ij}=(1+2\Psi /c^2){\delta }^{ij}$. In the Christoffel symbol formulathe differentiated metric perturbation is already first order, so the prefactor can be replaced by the flat inverse metric. For example ${\Gamma }^0{}_{ij}=\frac{1}{2}{\partial }_0{g}_{ij}=-{\delta }_{ij}{\partial }_0\Psi /c^2$ and ${\Gamma }^k{}_{00}=-\frac{1}{2}{\partial }_k{g}_{00}={\partial }_k\Phi /c^2$. Evaluating the remaining components givesThe lower-index symmetry follows from the torsion-free connection. Products of these first-order Christoffel symbols are beyond the retained weak-field approximation.

A pressureless relativistic fluid has stress-energy tensor $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$, with four-velocity normalized by $u^{\mu }{u}_{\mu }=-c^2$. In its rest frame, $u^i=0$, so $\boxed{T^0{}_0=-\rho c^2,T^i{}_j=0}$. Here $\rho$ is rest-frame mass density, or energy density divided by $c^2$.

For the linearized Ricci tensor and scalar, discard the quadratic connection terms. The contracted connection is ${\Gamma }^{\mu }{}_{\mu \nu }={\partial }_{\nu }(\Phi -3\Psi )/c^2$. Differentiation then givesThe Ricci scalar uses the zeroth-order inverse metric, $R=-R_{00}+{\delta }^{ij}{R}_{ij}$, because multiplying first-order curvature by another metric perturbation would again be second order. These are the linearized curvature of a two-potential weak-field metric in the paper's Riemann curvature tensor sign convention.

Write $\mathcal{H}=a^{\prime }/a$, with the prime denoting ${\partial }_0=c^{-1}{\partial }_{\eta }$. The metric and inverse metric are $g_{\mu \nu }=a^2{\eta }_{\mu \nu }$ and $g^{\mu \nu }=a^{-2}{\eta }^{\mu \nu }$. Since $a$ depends only on $x^0$, the Christoffel symbols reduce toConsequentlyFor the stated de Sitter spacetime scale factor, $\mathcal{H}=-1/(c\eta )$. The factor $c^{-1}$ is required because the prime is not differentiation with respect to $\eta$ itself.

A semicolon denotes the covariant derivative for the Levi-Civita connection. For example, the four lower indices of the Riemann curvature tensor contribute four connection terms:Here ${\partial }_{\eta }=\partial /\partial x^{\eta }$ is a partial derivative, ${\Gamma }_{b\eta }^a$ are Christoffel symbols, and repeated indices are summed from $0$ to $3$. Write $g_{\mu \nu }$ for the metric tensor, $g^{\mu \nu }$ for its inverse metric, $R_{\mu \nu }=g^{\lambda {\nu }^{\prime }}{R}_{\lambda \mu {\nu }^{\prime }\nu }$ for the Ricci tensor, and $R=g^{\mu \nu }{R}_{\mu \nu }$ for the Ricci scalar. This tensor contraction agrees with the paper's $R_{\mu \nu }=-R^a{}_{\mu \nu a}$.

By metric compatibility, the inverse metric may pass through each covariant derivative. Contracting the second Bianchi identity with $g^{\lambda \nu }$ giveswhere ${\nabla }_{\eta }$ is the same covariant derivative as the semicolon. Next contract with $g^{\mu \kappa }$. Antisymmetry of the first pair gives $g^{\mu \kappa }{R}_{\lambda \mu \kappa \eta }=-R_{\lambda \eta }$, soThus the contracted Bianchi identity is ${\nabla }^{\mu }{R}_{\mu \eta }=\frac{1}{2}{\partial }_{\eta }R$. Raising the remaining index and using metric compatibility provesThe Einstein tensor $G^{\mu \nu }$ therefore has identically zero covariant divergence, without imposing any field equation. This makes the Einstein field equations compatible with stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, where $T^{\mu \nu }$ is the stress-energy tensor. The connection terms describe how energy and momentum are compared at neighboring points; ordinary componentwise partial derivatives would not give a coordinate-independent conservation law in curved spacetime.

Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

Each lower index contributes a minus connection term to the covariant derivative of the metric tensor:Substitute the Christoffel symbols of the Levi-Civita connection and use the inverse metric identity $g_{\rho \nu }{g}^{\rho \sigma }={\delta }_{\nu }^{\sigma }$:Adding cancels the first and third terms and leaves exactly ${\partial }_{\lambda }{g}_{\mu \nu }$. HenceThis is metric compatibility: the Levi-Civita connection preserves the inner product under parallel transport.

For a Conformal Killing vector field, the same metric compatibility calculation givesThus its Conformal Killing equation is ${\nabla }_{(\mu }{X}_{\nu )}=\lambda g_{\mu \nu }$. Taking the tensor contraction with the inverse metric, and using $g^{\mu \nu }{g}_{\mu \nu }=4$, determines the conformal Killing factor:Parentheses include the factor $1/2$ in tensor index symmetrization.

In the order $(u,v,x,y)$ the metric tensor and inverse metric areOnly $g_{uu}$ has nonzero partial derivatives. Substitution into the formula for the Levi-Civita connection,therefore givesEvery other Christoffel symbol vanishes. For example $g^{ud}$ selects $d=v$, and the metric tensor is independent of $v$ with constant $g_{av}$; hence ${\Gamma }^u{}_{bc}=0$.

Use the paper's Riemann curvature tensor conventionFor the independent components,The quadratic Christoffel symbol terms vanish here: any potentially contributing intermediate $u$ index gives a factor ${\Gamma }^u{}_{ab}=0$, and the remaining possible products have a lower $v$ index and vanish. Since the symmetric matrix $M$ gives $F_{,ij}=2M_{ij}$, lowering the first index givesAll remaining nonzero components follow from the Riemann curvature tensor symmetriesThere are no other independent components: terms involving a $v$ index vanish, while nonzero terms require one $u$ in each antisymmetric pair.

The Ricci tensor is consequentlyThe Ricci scalar vanishes even when the trace is nonzero, since $g^{uu}=0$. With zero cosmological constant, the vacuum Einstein field equations reduce to $R_{ab}=0$. Therefore the necessary and sufficient condition isThis is the Ricci tensor of a Brinkmann plane wave: a nonzero, trace-free profile can have nonzero Riemann curvature tensor despite vanishing Ricci tensor.

A nondegenerate Lorentzian metric requires $M\ne 0$; take $M>0$ as the radius convention. The original coordinates describe $\lambda >0$, with $\lambda =0$ a candidate boundary rather than an ordinary point of that chart. The change of coordinate givesSubstituting and cancelling the two terms proportional to $d{\lambda }^2/{\lambda }^2$ producesThe $(v,\lambda )$ block of this metric tensor has determinant $-M^2$ and inverse metricAll coefficients are smooth through $\lambda =0$, and the block has one negative and one positive eigenvalue. Together with the positive round two-sphere factor this is a regular Lorentzian metric. It extends to a neighborhood with either sign of $\lambda$. The usual polar-coordinate problem at the poles of the two-sphere is removed by separate angular charts. Therefore $\lambda =0$ is only a coordinate singularity of the original chart. This is the degenerate-horizon patch of Bertotti-Robinson spacetime.

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

Use metric signature $(-,+,+,+)$ and units $c=1$. Write ${\dot{x}}^{\mu }=d{x}^{\mu }/d\lambda$ in this part. Varying the worldline einbein, without first fixing it, givesThe canonical four-momentum is $p_{\mu }=m{g}_{\mu \nu }{\dot{x}}^{\nu }/e$, so the same mass shell constraint is the mass shell $g^{\mu \nu }{p}_{\mu }{p}_{\nu }=-m^2$.

The Euler-Lagrange equation for the trajectory isMultiplying by $e$ and then by the inverse metric, and collecting the metric tensor derivatives into the Christoffel symbols, givesThis is a geodesic equation in a possibly nonaffine parameter.

Under a change of parameter, $e^{\prime }({\lambda }^{\prime })=e(\lambda )d\lambda /d{\lambda }^{\prime }$, so $ed\lambda$ is invariant. On the positive branch, the mass shell constraint identifies this invariant with the proper time increment $ds=\sqrt{-g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }}=ed\lambda$. Choosing ${\lambda }^{\prime }=s$ therefore sets $e^{\prime }=1$. The trajectory equation is then the affinely parametrized geodesic equation and its tangent has norm minus one. This is proper-time gauge for a massive worldline einbein. The mass shell constraint must still be imposed: fixing $e$ in the action before varying would lose it. A fixed parameter interval can retain a proper-time modulus; here changing to the actual proper-time interval is permitted.

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.

Use spatial indices $i,j=1,2,3$, set ${\varphi }_i={\partial }_i\varphi$, and assume $\varphi$ is smooth. The inverse metric has $g^{tt}=-c^{-2}{e}^{-2\varphi /c^2}$ and $g^{ij}={\delta }^{ij}$. The Christoffel symbols of the Levi-Civita connection follow from ${\Gamma }^{\alpha }{}_{\beta \gamma }=g^{\alpha \delta }({\partial }_{\beta }{g}_{\gamma \delta }+{\partial }_{\gamma }{g}_{\beta \delta }-{\partial }_{\delta }{g}_{\beta \gamma })/2$:All other components vanish. Although $g_{tt}=-c^2-2\varphi +O(c^{-2})$ diverges, its affine connection has a smooth limit on compact subsets:This is the Newtonian connection from an exponential lapse. It is torsion-free, and its geodesic equation, using $t$ as an affine parameter when $\dot{t}\ne 0$, is $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$. Thus $\varphi$ has the role of a Newtonian gravitational potential. There is no finite limiting nondegenerate Lorentzian metric in these fixed coordinates; $-g(c)/c^2$ tends to $d{t}^2$, a rank-one tensor. The limiting connection preserves $dt$ and the contravariant spatial tensor with components $h^{ij}={\delta }^{ij}$, $h^{t\mu }=0$, which describe the degenerate temporal/spatial structures of this limit.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincaré half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

For coordinates $x^{\alpha }$, the Christoffel symbols of the Levi-Civita connection arewhere $g^{\alpha \delta }$ is the inverse metric.

For the two-dimensional hyperbolic metric in polar coordinates,Only $g_{\varphi \varphi }$ has a nonzero derivative, namelyIt follows that the nonzero connection coefficients are

Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.