Dust initial data: curvature and connection from (h,K), and the tilted constrained initialization of the Bianchi residuals

(Admissible data and local normalization.) Locally let $\mathrm{\Sigma }$ be a smooth oriented three-manifold with smooth positive definite metric $h$, smooth symmetric tensor $K$, smooth ${\rho }_{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}\ge 0$, and constants $c>0,\kappa ,\mathrm{\Lambda }$. On $\mathcal{V}=\mathbb{R}N\oplus T\mathrm{\Sigma }$ define $\gamma (N,N)=1$, $\gamma (N,X)=0$, $\gamma (X,Y)=-h(X,Y)$. Future means positive $N$ coefficient. The original metric is $g=-\gamma$; $K(X,Y)=g({\mathrm{\nabla }}_{X}N,Y)$. A smooth future $\gamma$-unit section $U_0$ is prescribed. Define $\epsilon ={\rho }_{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{c}^2,T_0^{\gamma }=\epsilon U_0^{{\mathrm{♭}}_{\gamma }}\otimes U_0^{{\mathrm{♭}}_{\gamma }},{\mathcal{T}}_0=\kappa T_0^{\gamma }\mathrm{.}$ Here ${\mathcal{T}}_0$ is the scaled Einstein right-hand side and $T_0^{\gamma }$ uses the energy coefficient of the dust supplier. For $G(\gamma )-\mathrm{\Lambda }\gamma ={\mathcal{T}}_0$ the target Ricci tensor is $J=\kappa \epsilon U_0^{{\mathrm{♭}}_{\gamma }}\otimes U_0^{{\mathrm{♭}}_{\gamma }}-(\mathrm{\Lambda }+\frac{1}{2}\kappa \epsilon )\gamma \mathrm{.}$ Its trace is $-4\mathrm{\Lambda }-\kappa \epsilon$, and $J-\frac{1}{2}{\mathrm{tr}}_{\gamma }(J)\gamma ={\mathcal{T}}_0+\mathrm{\Lambda }\gamma$. No density division occurs, including at $\epsilon =0$; there $U_0$ remains the auxiliary gauge direction. Require the four corrected constraints for these data. Arbitrary $(h,K,U_0,\epsilon )$ need not satisfy them.

(Proper boost and full tensor transformation.) Choose an oriented local $h$-orthonormal frame $f_a$, $f_0=N$. Write uniquely $U_0=W(N+v^a{f}_a),W=\gamma (U_0,N)=(1-\mathrm{\mid }v{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2},\mathrm{\mid }v\mathrm{\mid }<1.$ The explicit boosted tetrad is $e_0=U_0,e_a=f_a+\frac{W^2}{W+1}{v}_a{v}^b{f}_b+W{v}_{a}N\mathrm{.}$ The coefficient $W^2\mathrm{/}(W+1)=(W-1)\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$ is smooth at $v=0$. Substitution gives $\gamma (e_A,e_B)={\eta }_{AB}$. The determinant is $+1$ by continuity from $v=0$ and invertibility; $e_0$ remains future. Thus the volume orientation is preserved. Additional spatial frame rotations belong to $SO(3)$ when using the same magnetic orientation. For $e_A=L_A{}^I{f}_I$ transform the entire covariant tensors: $R_{ABCD}^{\mathrm{\prime }}=L_A{}^I{L}_B{}^J{L}_C{}^K{L}_D{}^L{R}_{IJKL},C_{ABCD}^{\mathrm{\prime }}=L_A{}^I{L}_B{}^J{L}_C{}^K{L}_D{}^L{C}_{IJKL}\mathrm{.}$ The right dual is $C_{IJKL}^{\ast }=\frac{1}{2}{C}_{IJ}{}^{PQ}{ϵ}_{KLPQ}$. Lorentz invariance of the metric and the oriented volume tensor gives $(C^{\mathrm{\prime }}{)}_{ABCD}^{\ast }=L_A{}^I{L}_B{}^J{L}_C{}^K{L}_D{}^L{C}_{IJKL}^{\ast }$. Then define $E_{ab}^{\mathrm{\prime }}=C(U_0,e_a,U_0,e_b),B_{ab}^{\mathrm{\prime }}=C^{\ast }(U_0,e_a,U_0,e_b)\mathrm{.}$ A boost generally mixes electric and magnetic parts. Separate orthogonal transformations of $E$ and $B$ are false. Invertibility concerns the full Weyl tensor; its reconstruction lemma identifies these contractions with the reduction's prescribed curvature.

(Curvature construction in the normal frame, followed by the boost.) Use the normal tetrad $f_0=N$, $f_a\in T\mathrm{\Sigma }$ of step 1.2 throughout this construction: every unprimed component $R_{IJKL}$, $J_{IJ}$ and $K_{ab}=K(f_a,f_b)$ in this step is in that tetrad, and $R_{abcd}^h=h(R^h(f_c,f_d)f_b,f_a)$. The derivative $D$ is the Levi--Civita connection of $h$ on $T\mathrm{\Sigma }$. The boosted $e_a$ are not substituted into $D$ or $K$. The Gauss equation of [L1] gives $R_{abcd}^g=R_{abcd}^h+K_{ac}{K}_{bd}-K_{ad}{K}_{bc}$, and its Codazzi equation gives $R_{0abc}^g=(D_{f_c}K)(f_b,f_a)-(D_{f_b}K)(f_c,f_a)$. Therefore the lowered components for the reduction metric are explicitly $R_{abcd}^{\gamma }=-R_{abcd}^h-K_{ac}{K}_{bd}+K_{ad}{K}_{bc},R_{0abc}^{\gamma }=-(D_{f_c}K)(f_b,f_a)+(D_{f_b}K)(f_c,f_a)\mathrm{.}$ These supply the six independent spatial curvature components and eight mixed components, including their cyclic relation. The four Hamiltonian/momentum constraints fix $G_{00}^{\gamma }-\mathrm{\Lambda }=\kappa (T_0^{\gamma }{)}_{00}$ and ${\mathrm{Ric}}_{0a}^{\gamma }=\kappa (T_0^{\gamma }{)}_{0a}=J_{0a}$. They do not yet fix ${\mathrm{Ric}}_{00}^{\gamma }$: that component involves the as yet undetermined $R_{0a0b}^{\gamma }$. Indeed, with $Q_{\gamma }:={\sum }_{a,b}{R}_{abab}^{\gamma }$, the same contraction as in [L1], now with signature $(+---)$, gives $G_{00}^{\gamma }=-Q_{\gamma }\mathrm{/}2$, depending only on the already fixed spatial curvature. Define the six remaining symmetric entries by $R_{0a0b}^{\gamma }=J_{ab}+{\sum }_{c=1}^3{R}_{cacb}^{\gamma };$ then ${\mathrm{Ric}}_{ab}^{\gamma }=R_{0a0b}^{\gamma }-{\sum }_c{R}_{cacb}^{\gamma }=J_{ab}$. Now $S_{\gamma }={\mathrm{Ric}}_{00}^{\gamma }-{\sum }_a{J}_{aa}$, so $G_{00}^{\gamma }=({\mathrm{Ric}}_{00}^{\gamma }+{\sum }_a{J}_{aa})\mathrm{/}2$. The target identity of step 1.1 and the Hamiltonian constraint give $G_{00}^{\gamma }=\kappa (T_0^{\gamma }{)}_{00}+\mathrm{\Lambda }=(J_{00}+{\sum }_a{J}_{aa})\mathrm{/}2$. Subtraction yields ${\mathrm{Ric}}_{00}^{\gamma }=J_{00}$ only after the six spatial Ricci equations are imposed. Thus every Ricci component equals its target. Symmetry of $J_{ab}$ and of ${\sum }_c{R}_{cacb}^{\gamma }$ makes the six new entries symmetric, so the remaining algebraic curvature symmetries and cyclic identities hold; these fix all other entries and prove uniqueness. Set $S=J-\frac{1}{6}{\mathrm{tr}}_{\gamma }(J)\gamma$, subtract its fixed Kulkarni--Nomizu term to obtain $C$, and only then perform the full tensor basis change $R^{\mathrm{\prime }}=L^{\otimes 4}R$, $C^{\mathrm{\prime }}=L^{\otimes 4}C$ of step 1.2 and contract in the $e_0=U_0$ frame. Thus Gauss/Codazzi is used only with tangent arguments, while the proper boost supplies all rest-frame curvature components. No normal derivative of a constraint is needed.

(Explicit initial connection.) Let $D^h$ be the Levi--Civita connection and $S_K$ satisfy $h(S_{K}X,Y)=K(X,Y)$. In tangent directions define ${\mathrm{\nabla }}_X(aN+Y)=(X(a)+K(X,Y))N+a{S}_{K}X+D_X^{h}Y(X,Y\in T\mathrm{\Sigma })\mathrm{.}$ Differentiating $\gamma (aN+Y,bN+Z)=ab-h(Y,Z)$ verifies compatibility. Tangential torsion is zero because $D^h$ is torsion-free and $K$ symmetric. Differentiate the explicit boost to find every ${\mathrm{\nabla }}_X{e}_J$, hence $\omega (X)$, from $(h,K,U_0)$ and tangential derivatives. Specify $\omega (U_0)=0$ in this tetrad. Since $\mathcal{V}=T\mathrm{\Sigma }\oplus \mathbb{R}{U}_0$, these values determine the entire initial connection one-form. For the comoving chart $e_0={\mathrm{\partial }}_s=U_0$, $\mathrm{\Sigma }=\{s=0\}$, its conormal $\zeta =ds$ satisfies $\zeta (U_0)=1$ and initially $\zeta =N^{{\mathrm{♭}}_{\gamma }}\mathrm{/}W$. Therefore $\zeta (e_a)=v_a$. In particular $e_a=v_a{U}_0+X_a$ with $X_a\in T\mathrm{\Sigma }$, and ${\mathrm{\Gamma }}_a=\omega (e_a)=\omega (X_a)$. Rest-frame spatial directions must not be conflated with directions tangent to the tilted surface.

(Every initial zero quantity.) The torsion two-form is $\mathrm{\Theta }(X,Y)={\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y]$. Its tangential pullback is zero by the connection formula. The reduced frame transport equation imposes ${\iota }_U\mathrm{\Theta }=0$, meaning $T^i{}_{0a}=0$ for every output index $i$. A vector-valued two-form vanishing on tangent pairs and on every $(U,V)$ pair vanishes everywhere because $\mathcal{V}=T\mathrm{\Sigma }\oplus \mathbb{R}U$. Hence all $T$ vanish initially. Tangential connection curvature uses only tangential derivatives of its initial values. Corrected Gauss/Codazzi matches $R^{\mathrm{p}\mathrm{r}\mathrm{e}}$ on all tangent direction pairs and every endomorphism slot, so the tangential pullback of $\mathrm{\Delta }$ is zero. Reduced connection transport gives ${\iota }_U\mathrm{\Delta }=0$, meaning ${\mathrm{\Delta }}^i{}_{j0a}=0$ for every $i,j,a$. The same two-form argument gives all $\mathrm{\Delta }=0$. This uses reduced first jets, not a presumed Einstein development. The positive reduced time matrix below determines those jets uniquely. Density transport fixes $q=U(\epsilon )+\epsilon \chi =0$. Parallel gauge gives ${\mathrm{\nabla }}_{U}U=0$, hence $q_a=\epsilon ({\mathrm{\nabla }}_{U}U{)}_a=0$ in the corrected residual normalization. No division by $\epsilon$ occurs. The reduced Weyl equations give symmetric trace-free $P_{ab}=Q_{ab}=0$ at the initial first jet.

(Tilted differential Bianchi initialization.) By [L3], the explicit residual definition and $q_j=0$ give the identity ${\mathrm{cyclic}}_{jkl}{\mathrm{\nabla }}_j{R}^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{i}}{}_{mkl}=\frac{1}{2}{ϵ}_{jkl}{}^p{F}_{prs}{ϵ}_m{}^{irs}\mathrm{.}$ The identity uses left/right Hodge decomposition and contracted Schouten conservation. Pull back $D_{\omega }\mathrm{\Delta }=-D_{\omega }{R}^{\mathrm{p}\mathrm{r}\mathrm{e}}$ to three tangent slots. The pullback of $\mathrm{\Delta }$ is identically zero along the slice, so its intrinsic covariant exterior derivative vanishes. Torsion pullback zero removes the exterior/covariant correction. Thus the displayed cyclic tensor vanishes on tangent triples. Contracting three tangent directions into the volume tensor gives a nonzero multiple of the normal conormal ${\zeta }^p$; invertibility of the endomorphism-pair Hodge map yields ${\zeta }^p{F}_{prs}=0$. This uses only tangential derivatives, and assumes no transverse derivative of constraints or residuals zero. With $P_{ab}=Q_{ab}=q=q_a=0$, the corrected split is $F_{jkl}=2U_j{P}_{[k}{U}_{l]}+h_{j[k}{P}_{l]}+Q_i(U_j{ϵ}^i{}_{kl}-{ϵ}^i{}_{j[k}{U}_{l]})\mathrm{.}$ Here $P_a=F_{0a0}$, $Q_a=-\frac{1}{2}{ϵ}_a{}^{bc}{F}_{0bc}$, brackets have weight $1\mathrm{/}2$, $h_{ab}=-{\delta }_{ab}$ and ${ϵ}^i{}_{ab}=-{ϵ}_{iab}$. All four component families are $F_{0a0}=P_a,F_{0ab}=-Q_i{ϵ}_{iab},F_{ba0}=\frac{1}{2}{Q}_i{ϵ}_{iba},F_{bac}=\frac{1}{2}(-{\delta }_{ba}{P}_c+{\delta }_{bc}{P}_a)\mathrm{.}$ For $z=\zeta (U)>0$ and $w_a=\zeta (e_a)$, contract the $(a,0)$ components and take the negative half spatial dual of the $(a,c)$ components: $zP+\frac{1}{2}w\times Q=0,zQ-\frac{1}{2}w\times P=0.$ The exact matrix is $M=\left(\begin{array}{cc}zI& X(w)\mathrm{/}2\\ -X(w)\mathrm{/}2& zI\end{array}\right)$ with $X(w)b=w\times b$. Since $X(w{)}^T=-X(w)$ it is symmetric, and $⟨(P,Q),M(P,Q)⟩=z(\mathrm{\mid }P{\mathrm{\mid }}^2+\mathrm{\mid }Q{\mathrm{\mid }}^2)+P\cdot (w\times Q)\ge (z-\mathrm{\mid }w\mathrm{\mid }\mathrm{/}2)(\mathrm{\mid }P{\mathrm{\mid }}^2+\mathrm{\mid }Q{\mathrm{\mid }}^2)\mathrm{.}$ Timelike $\zeta$ gives $z>\mathrm{\mid }w\mathrm{\mid }$, hence $P=Q=0$; the four component families give every $F=0$. All initial quantities $(T,\mathrm{\Delta },F,q)$ required by the subsidiary theorem are therefore zero, by its explicit corrected dual identity. For reduced Weyl evolution the separately established curl bound is $\mathrm{\parallel }C(w)\mathrm{\parallel }\le \mathrm{\mid }w\mathrm{\mid }$. Its symmetric time matrix has lower bound $(z-\mathrm{\mid }w\mathrm{\mid })(\mathrm{\mid }E{\mathrm{\mid }}^2+\mathrm{\mid }B{\mathrm{\mid }}^2)$, while transport blocks have factor $z$. Initially $z=1,w=v$, so the evolution bound is $1-\mathrm{\mid }v\mathrm{\mid }>0$, and the residual bound is $1-\mathrm{\mid }v\mathrm{\mid }\mathrm{/}2>0$. These are distinct matrices. Arbitrary $\mathrm{\mid }v\mathrm{\mid }<1$, vacuum points and every $c>0$ remain allowed.