Propagation of the harmonic gauge and of the Einstein constraints: a constrained reduced solution is an Einstein solution

(Zero Cauchy data on $\mathrm{\Sigma }$.) Fix $p\in \mathrm{\Sigma }$ and a chart with $x^0$ the normal coordinate, $g_{00}=-1$, $g_{0i}=0$, $e_i={\mathrm{\partial }}_i$ at $p$. By [L1], ${\mathrm{\nabla }}_i{H}_0={\mathrm{\nabla }}_i{H}_j=0$ on $\mathrm{\Sigma }$ in tangential directions. Computing [L2] at $p$: ${\mathrm{\nabla }}^c{H}_c=g^{00}{\mathrm{\nabla }}_0{H}_0+g^{ij}{\mathrm{\nabla }}_i{H}_j=-{\mathrm{\nabla }}_0{H}_0$, so $E_{00}=\frac{1}{2}(2{\mathrm{\nabla }}_0{H}_0-g_{00}{\mathrm{\nabla }}^c{H}_c)=\frac{1}{2}(2{\mathrm{\nabla }}_0{H}_0-{\mathrm{\nabla }}_0{H}_0)=\frac{1}{2}{\mathrm{\nabla }}_0{H}_0$ and $E_{0i}=\frac{1}{2}({\mathrm{\nabla }}_0{H}_i+{\mathrm{\nabla }}_i{H}_0)=\frac{1}{2}{\mathrm{\nabla }}_0{H}_i$. Hypothesis 3 therefore gives ${\mathrm{\nabla }}_0{H}_0={\mathrm{\nabla }}_0{H}_i=0$ at every point of $\mathrm{\Sigma }$; together with hypothesis 2 and [L1] this says that $H$ and all its first derivatives vanish on $\mathrm{\Sigma }$.

(Local cone energy for the subsidiary wave equation.) Fix $p\in \mathrm{\Sigma }$ and a relatively compact normal-geodesic chart from [L5], with coordinates $(t,x)$ centered at $p$. Put $w_b=H_b$, $z_b={\mathrm{\partial }}_t{w}_b$ and $y_{ib}={\mathrm{\partial }}_i{w}_b$. The wave equation [L4] is equivalent on this chart to the first-order system ${\mathrm{\partial }}_t{w}_b-z_b=0,{\mathrm{\partial }}_t{y}_{ib}-{\mathrm{\partial }}_i{z}_b=0,{\mathrm{\partial }}_t{z}_b-h^{ij}{\mathrm{\partial }}_i{y}_{jb}=F_b(t,x,w,z,y),$ where $F$ is smooth and linear in $(w,z,y)$. Multiplying the middle equations by $h^{ai}$ gives symmetric spatial coefficient matrices and positive time matrix $A^0=\mathrm{diag}(I,I,h^{ij}I)$ on the independent variables $(w,z,y)$. On a compact coordinate cylinder choose $a>0$ and $b<\mathrm{\infty }$ such that $A^0\ge aI$ and $\mathrm{\parallel }{A}^i\mathrm{\parallel }\le b$; all coefficient derivatives and the lower-order matrix $C$ are bounded there. Choose $\lambda >\sqrt{3}b\mathrm{/}a$. Then $\lambda A^0+{\nu }_i{A}^i\ge 0$ for every Euclidean unit spatial conormal $\nu$, since $\mathrm{\parallel }{\nu }_i{A}^i\mathrm{\parallel }\le \sqrt{3}b$. Shrink $\tau >0$ so the double cone $\mathrm{\mid }t\mathrm{\mid }\le \tau$, $\mathrm{\mid }x-x(p)\mathrm{\mid }\le \lambda \tau$ lies in the chart. Step 1.1 and $H=0$ on $\mathrm{\Sigma }$ imply $w=z=y=0$ on its base disk at $t=0$. For $0\le s<\tau$, set $B_s=\{x:\mathrm{\mid }x-x(p)\mathrm{\mid }\le \lambda (\tau -s)\}$ and $E(s)={\int }_{B_s}{V}^{\mathsf{T}}{A}^{0}Vdx$, where $V=(w,z,y)$. For a symmetric system $A^0{\mathrm{\partial }}_{t}V+A^i{\mathrm{\partial }}_{i}V+CV=0$, its energy current $j^0=V^{\mathsf{T}}{A}^{0}V$, $j^i=V^{\mathsf{T}}{A}^{i}V$ satisfies ${\mathrm{\partial }}_t{j}^0+{\mathrm{\partial }}_i{j}^i=V^{\mathsf{T}}({\mathrm{\partial }}_t{A}^0+{\mathrm{\partial }}_i{A}^i-C-C^{\mathsf{T}})V\mathrm{.}$ The right side is bounded above by a constant times $j^0$ on the compact cone. The lateral outward flux on the shrinking cone is proportional to $V^{\mathsf{T}}(\lambda A^0+{\nu }_i{A}^i)V\ge 0$. Applying the divergence theorem [L6] gives $E(s)\le E(0)+C{\int }_0^{s}E(r)dr$. Since $E(0)=0$, Gronwall [L6] and positivity of $A^0$ imply $H=0$ on the forward cone. Reversing time changes each $A^i$ to $-A^i$ and preserves the same bounds, so the same argument gives vanishing on the backward cone. The two cone estimates give vanishing on the open set $W_p=\{(t,x):\mathrm{\mid }t\mathrm{\mid }<\tau ,\mathrm{\mid }x-x(p)\mathrm{\mid }<\lambda (\tau -\mathrm{\mid }t\mathrm{\mid })\}$, which contains $p$. Hence $W={\bigcup }_{p\in \mathrm{\Sigma }}{W}_p$ is an open neighborhood in $U$ on which $H=0$.

(The Einstein equation.) On $W$, $H=0$, so $E=0$ by [L2] (identity (a) with $H=0$); by the definition of the Einstein residual $E=G+\mathrm{\Lambda }g-\kappa T$, this is $G_{ab}+\mathrm{\Lambda }{g}_{ab}=\kappa T_{ab}$ on $W$.