Subsidiary system of the dust reduction: Cartan identities, closed homogeneous residual system and vanishing of the residuals

The gauge gives ${\mathrm{\nabla }}_{U}U=0$; therefore the full conservation vector $q_j:={\mathrm{\nabla }}^i(r{U}_i{U}_j)=q{U}_j+r({\mathrm{\nabla }}_{U}U{)}_j$ vanishes, including at $r=0$. Since $S=\kappa r(UU-\gamma \mathrm{/}3)$ plus a parallel constant metric multiple, ${\mathrm{\nabla }}^j{S}_{lj}=-\kappa {\mathrm{\nabla }}_{l}r\mathrm{/}3={\mathrm{\nabla }}_l{\mathrm{tr}}_{\gamma }S$. The Weyl cyclic and trace identities now decompose $F$, with the reduced evolution projections zero, into the linear map $A$ defined by $A(P,Q{)}_{jkl}:=2U_j{P}_{[k}{U}_{l]}+h_{j[k}{P}_{l]}+Q_i(U_j{ϵ}^i{}_{kl}-{ϵ}^i{}_{j[k}{U}_{l]}),F=A(P,Q)\mathrm{.}$ Indeed the unreduced decomposition has the additional terms $-2P_{j[k}{U}_{l]}-Q_{ji}{ϵ}^i{}_{kl}-\kappa h_{j[k}{q}_{l]}\mathrm{/}2-\kappa q{h}_{j[k}{U}_{l]}\mathrm{/}3$; the symmetric tracefree parts are $P_{ab},Q_{ab}$, its traces are the conservation quantities, and the remaining antisymmetric parts are fixed by the cyclic identity. All these additional terms vanish by the reduced equations and gauge. Thus every Bianchi residual is represented by the two constraint vectors.

Cartan differentiation gives $D_{\omega }\mathrm{\Theta }=\mathrm{\Omega }\wedge \theta$ and $D_{\omega }\mathrm{\Omega }=0$ by $d^2=0$. Algebraic cyclic symmetry gives ${\bar{R}}^{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}\wedge \theta =0$, hence $D_{\omega }\mathrm{\Theta }=\mathrm{\Delta }\wedge \theta$. The reduced equations mean exactly ${\iota }_U\mathrm{\Theta }={\iota }_U\mathrm{\Delta }=0$, while the gauge gives ${\iota }_U\omega =0$. Contracting yields the exact torsion transport equation ${\mathcal{L}}_U{\mathrm{\Theta }}^i={\mathrm{\Delta }}^i{}_0\mathrm{.}$ The right side has endomorphism input index $0$, not a time form index. Its principal derivative is $e_0(T^i{}_{ab})$; converting the Lie derivative to frame components adds only coefficient times $T$, since $[U,e_a]=-{\mathrm{\Gamma }}_a{}^k{}_0{e}_k$. In particular $T^0{}_{ab}$ need not vanish.

For a covariant tensor $B$ define the curvature action $(K_{R,ab}B{)}_{i_1\cdots i_n}:=-{\sum }_{v=1}^n{R}^p{}_{i_{v}ab}{B}_{i_1\cdots i_{v-1}p{i}_{v+1}\cdots i_n}\mathrm{.}$ The antisymmetrized second covariant derivative, which also differentiates its lower derivative slot, is $[{\mathrm{\nabla }}_a,{\mathrm{\nabla }}_b]B=K_{\mathrm{\Omega },ab}B-T^p{}_{ab}{\mathrm{\nabla }}_{p}B$. Expanding the derivative-slot connection terms gives $[e_a,e_b]-({\mathrm{\Gamma }}_a{}^p{}_b-{\mathrm{\Gamma }}_b{}^p{}_a)e_p=-T^p{}_{ab}{e}_p$, proving this identity. Antisymmetry of $C$ in its first pair gives ${\mathrm{\nabla }}^j{\mathrm{\nabla }}^i{C}_{ijkl}=[{\mathrm{\nabla }}^j,{\mathrm{\nabla }}^i]C_{ijkl}\mathrm{/}2$. Move ${\mathrm{\nabla }}^j$ past ${\mathrm{\nabla }}_{[k}$ in the Schouten term and use step 1.1; the remaining scalar commutator is $\kappa {\mathrm{\nabla }}_{[k}{\mathrm{\nabla }}_{l]}r\mathrm{/}3=-\kappa T^p{}_{kl}{\mathrm{\nabla }}_{p}r\mathrm{/}6$. Therefore ${\mathrm{\nabla }}^j{F}_{jkl}=H_{kl}(T,\mathrm{\Delta })$, where the explicitly coefficient-linear map is $H_{kl}:=\frac{1}{2}{\gamma }^{ja}{\gamma }^{ib}(K_{\mathrm{\Delta },ab}C{)}_{ijkl}-{\gamma }^{ja}(K_{\mathrm{\Delta },a[k}S{)}_{l]j}-\frac{1}{2}{\gamma }^{ja}{\gamma }^{ib}{T}^p{}_{ab}{\mathrm{\nabla }}_p{C}_{ijkl}+{\gamma }^{ja}{T}^p{}_{a[k}{\mathrm{\nabla }}_p{S}_{l]j}-\frac{\kappa }{6}{T}^p{}_{kl}{\mathrm{\nabla }}_{p}r\mathrm{.}$ To see why no residual-free curvature term remains, substitute $\mathrm{\Omega }=\bar{R}+\mathrm{\Delta }$ into the commutators: pair interchange, trace-freeness and cyclic symmetry give $\frac{1}{2}{\gamma }^{ja}{\gamma }^{ib}(K_{\bar{R},ab}C{)}_{ijkl}-{\gamma }^{ja}(K_{\bar{R},a[k}S{)}_{l]j}=0.$ Expansion of $K$ gives four Weyl-slot terms and two Schouten-slot terms. The contracted first-pair Weyl-slot terms are Ricci contractions; the last-pair terms cancel the $CS$ contributions from the Schouten commutator using pair interchange and the cyclic identity. The remaining $SS$ terms are antisymmetric parts of the symmetric tensor $S^2$ and metric multiples of $\mathrm{tr}S$, hence zero. Every displayed term in $H$ consequently contains an undifferentiated $T$ or $\mathrm{\Delta }$.

Put ${\bar{B}}^i{}_{mjkl}:={\sum }_{\mathrm{c}\mathrm{y}\mathrm{c}(jkl)}{\mathrm{\nabla }}_j{\bar{R}}^i{}_{mkl}$. The algebraic dual identity, using conservation from step 1.1, is ${\bar{B}}^i{}_{mjkl}=\frac{1}{2}{ϵ}_{jkl}{}^p{F}_{prs}{ϵ}_m{}^{irs}\mathrm{.}$ For completeness split $\bar{R}=C+E+G$, where $E$ is its tracefree-Schouten part and $G$ its scalar-curvature part. Left and right Hodge duals agree on $C,G$ and have opposite sign on $E$, as follows by inserting ${\gamma }_{i[k}{S}_{l]j}-{\gamma }_{j[k}{S}_{l]i}$ and contracting the two volume tensors. Apply these identities to $\mathrm{\nabla }\bar{R}$ and contract the derivative with the right dual. The Weyl and Schouten derivative terms give $F$; the remaining trace terms are ${\mathrm{\nabla }}^j{S}_{lj}-{\mathrm{\nabla }}_l\mathrm{tr}S=\kappa q_l=0$. This proves the displayed identity without assuming vanishing torsion. Exterior differentiation has the separate torsion correction $(D_{\omega }{\bar{R}}^{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{)}^i{}_m(e_j,e_k,e_l)={\bar{B}}^i{}_{mjkl}+{\sum }_{\mathrm{c}\mathrm{y}\mathrm{c}(jkl)}{T}^n{}_{jk}{\bar{R}}^i{}_{mnl}\mathrm{.}$ Thus $D_{\omega }\mathrm{\Delta }=-D_{\omega }{\bar{R}}^{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}$, contracted with $U$, gives the exact curvature transport equation ${\mathcal{L}}_U{\mathrm{\Delta }}^i{}_m=-{\iota }_U(D_{\omega }{\bar{R}}^{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{)}^i{}_m\mathrm{.}$ Its right side is explicitly specified by the preceding formulas and $F=A(P,Q)$, and is linear in $T,P,Q$. The principal derivative is $e_0({\mathrm{\Delta }}^i{}_{mab})$; all frame conversion terms are algebraic in $\mathrm{\Delta }$.

Define the following linear lower-order maps of $v=(P,Q)$ by exact subtraction: $L(v{)}_a:={\mathrm{\nabla }}^{j}A(v{)}_{ja0}-e_0(P_a)-\frac{1}{2}{ϵ}_a{}^{bc}{e}_b(Q_c),$ $M(v{)}_a:=-\frac{1}{2}{ϵ}_a{}^{bc}{\mathrm{\nabla }}^{j}A(v{)}_{jbc}-e_0(Q_a)+\frac{1}{2}{ϵ}_a{}^{bc}{e}_b(P_c)\mathrm{.}$ Inserting $A$ shows that all derivatives of $P,Q$ cancel in these definitions; $L,M$ depend only linearly on $v$, with coefficients from $\mathrm{\Gamma }$. Indeed its rest-frame families are $A_{0a0}=P_a$, $A_{0ab}=Q_i{ϵ}^i{}_{ab}$, $A_{ba0}=-Q_i{ϵ}^i{}_{ba}\mathrm{/}2$, and $A_{bac}=h_{b[a}{P}_{c]}$. Contracting step 2.1 now gives the exact curl equations $e_0(P_a)+\frac{1}{2}{ϵ}_a{}^{bc}{e}_b(Q_c)=H_{a0}-L(v{)}_a,$ $e_0(Q_a)-\frac{1}{2}{ϵ}_a{}^{bc}{e}_b(P_c)=-\frac{1}{2}{ϵ}_a{}^{bc}{H}_{bc}-M(v{)}_a\mathrm{.}$ Together with steps 1.2--2.2 these constitute a closed linear homogeneous system in spatial $T$, spatial $\mathrm{\Delta }$, $P,Q$. Its coefficient functions are smooth functions of the fixed reduced solution and its first derivatives; no derivative of a residual occurs on a right side.

(Local cone energy.) Let $Z$ collect the independent spatial components of $(T,\mathrm{\Delta },P,Q)$ in the homogeneous residual system of step 3.1. Fix a point $p$ on a spacelike slice where $Z$ vanishes on an open disk in that slice. Parametrize that slice locally by $i(x)$ and define $\mathrm{\Phi }(t,x)={\exp }_{i(x)}(t{\mathfrak{n}}_{i(x)})$. The derivative of $\mathrm{\Phi }$ at $(0,x(p))$ maps the time direction to the unit normal and the spatial directions to $T_p\mathrm{\Sigma }$, hence is invertible. The inverse function theorem gives a chart after shrinking. Its $t$-curves are unit timelike geodesics, and the coordinate fields commute. Metric compatibility and torsion-freeness of the independently defined Levi--Civita connection of $g$, rather than of the tentative reduction connection, therefore give ${\mathrm{\partial }}_{t}g({\mathrm{\partial }}_t,{\mathrm{\partial }}_i)=g({\mathrm{\partial }}_t,{\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_t)=\frac{1}{2}{\mathrm{\partial }}_{i}g({\mathrm{\partial }}_t,{\mathrm{\partial }}_t)=0$; the cross terms vanish initially, so $g=-d{t}^2+{\sigma }_{ij}d{x}^{i}d{x}^j$ with $\sigma$ positive definite. (The geodesic initial-value problem and the exponential map of the metric-selected connection, The smooth inverse function theorem on manifolds, Metric compatibility, torsion freedom and the metric-selected connection class, Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields). In this chart the residual equations from steps 1.2--3.1 are a linear homogeneous first-order system $A^0{\mathrm{\partial }}_{t}Z+A^i{\mathrm{\partial }}_{i}Z+CZ=0$ with smooth coefficients. In the frame co-basis, for any covector $\zeta$ write $z=\zeta (U)$ and $w_a=\zeta (e_a)$. Its principal symbol on the transport variables is $zI$, and on $(P,Q)$ it is

The matrix $X(w)$ is skew-adjoint, so the displayed block and the full symbol are symmetric in the stated positive fibre product. This covector formula is tensorial in the principal part: evaluating it at $dt$ and each $d{x}^i$ gives the coordinate matrices $A^0,A^i$, so all those matrices are symmetric even though the Gaussian time vector need not equal the comoving vector $U$. The $(P,Q)$ block is bounded below by $z-\mathrm{\mid }w\mathrm{\mid }\mathrm{/}2$ because $\mathrm{\parallel }w\times \cdot \mathrm{\parallel }\le \mathrm{\mid }w\mathrm{\mid }$, while each transport block is $z$. Since $dt$ is future timelike in the Gaussian chart, $z=dt(U)>\mathrm{\mid }w\mathrm{\mid }$; thus $A^0$ is positive definite. On a compact coordinate cylinder choose $a>0$ and $b<\mathrm{\infty }$ with $A^0\ge aI$, $\mathrm{\parallel }{A}^i\mathrm{\parallel }\le b$, and bound the coefficient derivatives and $C$. Choose $\lambda >\sqrt{3}b\mathrm{/}a$, giving $\lambda A^0+{\nu }_i{A}^i\ge 0$ for every unit spatial conormal $\nu$. Shrink $\tau >0$ so the cone based on a disk within the zero slice, with radius $\lambda \tau$, lies in the cylinder. Put $j^0=Z^{\mathsf{T}}{A}^{0}Z$, $j^i=Z^{\mathsf{T}}{A}^{i}Z$. Direct differentiation gives ${\mathrm{\partial }}_t{j}^0+{\mathrm{\partial }}_i{j}^i=Z^{\mathsf{T}}({\mathrm{\partial }}_t{A}^0+{\mathrm{\partial }}_i{A}^i-C-C^{\mathsf{T}})Z\le C_0{j}^0$. On the shrinking balls $B_s=\{x:\mathrm{\mid }x-x(p)\mathrm{\mid }\le \lambda (\tau -s)\}$, the lateral flux is proportional to $Z^{\mathsf{T}}(\lambda A^0+{\nu }_i{A}^i)Z\ge 0$. The divergence theorem [F3] yields $E(s):={\int }_{B_s}{j}^{0}dx\le E(0)+C_0{\int }_0^{s}E(r)dr$. The base data give $E(0)=0$, so Gronwall [F3] and positivity of $A^0$ give $Z=0$ on the forward cone. Applying the same argument after reversing time gives zero on the backward cone. Their union contains an open double-cone neighborhood of $p$.

(Metric-timelike short cones for the whole residual system.) The principal symbol of the full system established in step 3.1 is the direct sum of $zI$ on every torsion and curvature transport component and $zI+J(w)\mathrm{/}2$ on $(P,Q)$, with $\mathrm{\parallel }J(w)\mathrm{\parallel }\le \mathrm{\mid }w\mathrm{\mid }$. All couplings between these blocks are zeroth order. Thus the symbol is nonnegative for every future-positive metric-causal conormal, since $z\ge \mathrm{\mid }w\mathrm{\mid }$, and its only propagation speeds in the comoving rest frame have magnitude at most $1\mathrm{/}2$. This is the full residual system, not the reduced electric/magnetic evolution block. At any point choose local coordinates whose coordinate frame there is $(U,e_1,e_2,e_3)$. Put $\lambda =3\mathrm{/}4$. At that point $\lambda A^0+{\nu }_i{A}^i$ is positive definite for every $\mathrm{\mid }\nu \mathrm{\mid }=1$, with lower bound at least $1\mathrm{/}4$, and each vector ${\mathrm{\partial }}_t+v^i{\mathrm{\partial }}_i$ with $\mathrm{\mid }v\mathrm{\mid }\le \lambda$ is future metric-timelike. Compactness of the unit sphere and continuity preserve both strict inequalities on a smaller relatively compact chart. The energy-current calculation of step 4.1 therefore applies on every sufficiently small straight truncated cone of speed $\lambda$ in this chart. Its lateral flux is nonnegative, and each straight segment from its base to its vertex is metric-timelike.

(A support predecessor on each close past slice.) Let $S$ be the closure in $\mathrm{\Omega }$ of the set where the full residual vector is nonzero; this zero/nonzero condition is independent of the chosen frame components. Step 4.1 gives a zero neighborhood of each point of $\mathrm{\Sigma }$, hence $S\cap \mathrm{\Sigma }=\mathrm{\varnothing }$. If $p\in S$, choose a chart of step 5.1 centered at $p$. For every sufficiently small $d>0$, the past base disk $t=t(p)-d$, $\mathrm{\mid }x-x(p)\mathrm{\mid }\le \lambda d$ meets $S$. Otherwise its compactness and closedness of $S$ give a slightly enlarged zero disk still inside the chart. The cone-energy argument based on that enlarged disk gives a zero neighborhood of the vertex $p$, contradicting $p\in S$. Thus there is a predecessor $p_{-}$ in that disk and in $S$, joined to $p$ by the straight future timelike coordinate segment. On a smaller chart the allowed $d$ and upper/lower comparisons between that segment's auxiliary Riemannian length and $d$ can be chosen uniformly for vertices in a compact subchart.

(A past-inextendible causal curve in the closed support.) Apply The auxiliary Riemannian metric h=g+2n-flat tensor n-flat is complete after a proper-exhaustion conformal factor, and its closed bounded sets are compact to the open spacetime $\mathrm{\Omega }$ and fix its complete proper auxiliary metric $h^{\mathrm{\prime }}$ on the component of $p$. For each integer $k\ge 1$ construct a backward polygonal timelike curve from $p$ with vertices in $S$ and every segment of $h^{\mathrm{\prime }}$-length at most $1\mathrm{/}k$, using step 6.1 and Dependent Choice. Here is the non-arrest argument. In each closed $h^{\mathrm{\prime }}$-ball of integer radius about $p$, the intersection with $S$ is compact. Cover it by finitely many compact subcharts of step 6.1; on these, choose positive step durations small enough to have segment length at most $1\mathrm{/}k$, and take the minimum of their positive lower length bounds. Until the accumulated length reaches that integer radius, the current vertices remain in that compact ball, and every step can therefore have length at least this minimum. Finitely many steps reach the next integer length level; then use the finite cover of the next ball. The resulting curve has infinite past $h^{\mathrm{\prime }}$-length and can be parametrized on $(-\mathrm{\infty },0]$ by that length, with endpoint $p$. Every point of the polygonal curve is within $1\mathrm{/}k$ of $S$ because its segment endpoint belongs to $S$. The time-reversed half-line statement of Basic limit curve lemma: arclength-parametrized causal curves have causal limit curves on compact parameter intervals and future-inextendible limits on half-lines applies: a subsequence converges uniformly on compact parameter intervals to a past-inextendible metric-causal curve in $\mathrm{\Omega }$ through $p$. Closedness of $S$ and the distance bound put its entire image in $S$. The same construction reversed in time gives a future-inextendible causal support curve when needed. This uses the already assumed Countable and Dependent Choice; it assumes neither strong causality of $\mathrm{\Omega }$ nor compactness of the complete support.

(Propagation on the relative metric domain.) If a point $q\in D_{\mathrm{\Omega }}^{+}(\mathrm{\Sigma })$ had a nonzero residual, then $q\in S$, and step 7.1 would give a past-inextendible causal curve in $\mathrm{\Omega }$ through $q$ whose image stays in $S$ and so misses $\mathrm{\Sigma }$. This contradicts the definition of $D_{\mathrm{\Omega }}^{+}(\mathrm{\Sigma })$. The time-reversed argument applies on $D_{\mathrm{\Omega }}^{-}(\mathrm{\Sigma })$. Thus all residuals vanish on their union. Vanishing torsion identifies the metric-compatible tentative connection with Levi--Civita, and vanishing $\mathrm{\Delta }$ identifies actual and prescribed curvature; the prescribed Ricci tensor is the Einstein-dust Ricci. Step 1.1 gives $F=0$, and the full Einstein-dust equation follows. No division by $r$ is used.