A supplied complete maximal slice has zero extrinsic curvature in the zero-shift case

At $S$, $T=0$ gives $N=-g(T,n)=0$ and then $X=T-Nn=0$ directly. The smooth normal and orthogonal decomposition extend to the boundary by the supplied smooth spacelike boundary data. No null sheets are needed for these pointwise conclusions.

Orient $\bar{\mathrm{\Sigma }}$ by the smooth future normal $n$: a tangent basis $(v_1,v_2,v_3)$ is positive precisely when $(n,v_1,v_2,v_3)$ is positive in the oriented ambient spacetime. Since $n$ is everywhere transverse, this is a smooth orientation, including at $S$. For sufficiently large $R$, remove the open tail $\{r>R\}$ to obtain the compact smooth truncation ${\mathrm{\Omega }}_R$ with boundary $S\bigsqcup S_R$, where $S_R=\{r=R\}$. Compactness follows from the compact-core exterior convention; the end chart makes $S_R$ a smooth boundary disjoint from $S$. Restrict the induced orientation to ${\mathrm{\Omega }}_R$ and give its boundary the outward-normal-first orientation.

By [F3] and the geometric vacuum hypothesis, ${\mathrm{\Delta }}_{h}N=\mathrm{\mid }k{\mathrm{\mid }}_h^{2}N$. The potential $\mathrm{\mid }k{\mathrm{\mid }}_h^2$ is smooth and nonnegative. The exterior is connected with smooth compact boundary, $N$ is smooth up to that boundary, step 1.1 supplies its zero boundary values, and its uniform end limit is one. Thus the exterior maximum and boundary-control result gives $N>0$ inside and a strictly positive inward unit normal derivative at every point of $S$.

The field $Y$ in [F4] is smooth on compact ${\mathrm{\Omega }}_R$, so the divergence theorem applies there with countable choice supplied by [A1]. If ${\nu }_R$ is its outward unit normal on $S_R$, the identity becomes ${\int }_{{\mathrm{\Omega }}_R}N\mathrm{\mid }k{\mathrm{\mid }}_h^{2}d{\mu }_h={\int }_{S_R}h(Y,{\nu }_R)d{A}_h$, since $X=0$ makes the flux on $S$ zero. In an $h$-orthonormal frame, apply [F6] to each row of $(k_{ij})$ and the components of $X$, then sum: $\mathrm{\mid }Y{\mathrm{\mid }}_h^2={\sum }_i({\sum }_j{k}_{ij}{X}_j{)}^2\le ({\sum }_{i,j}{k}_{ij}^2)({\sum }_j{X}_j^2)=\mathrm{\mid }k{\mathrm{\mid }}_h^2\mathrm{\mid }X{\mathrm{\mid }}_h^2$. Taking nonnegative square roots gives $\mathrm{\mid }Y{\mathrm{\mid }}_h\le \mathrm{\mid }k{\mathrm{\mid }}_h\mathrm{\mid }X{\mathrm{\mid }}_h$. The bound $h-\delta =O(r^{-1})$ makes $h$ uniformly equivalent to the Cartesian metric for large $r$, and its restriction to $S_R$ therefore gives ${\mathrm{Area}}_h(S_R)\le C{R}^2$. The end estimates yield $\mid {\int }_{S_R}h(Y,{\nu }_R)d{A}_h\mid \le C{R}^2{R}^{-2}\log R{R}^{-1}\log R=C{R}^{-1}(\log R{)}^2\to 0$.

The integrand $N\mathrm{\mid }k{\mathrm{\mid }}_h^2$ is continuous and nonnegative by step 2.1. The compact truncations are nested and exhaust the hypersurface. For any fixed sufficiently large $R_0$ and all $R\ge R_0$, $0\le {\int }_{{\mathrm{\Omega }}_{R_0}}N\mathrm{\mid }k{\mathrm{\mid }}_h^{2}d{\mu }_h\le {\int }_{{\mathrm{\Omega }}_R}N\mathrm{\mid }k{\mathrm{\mid }}_h^{2}d{\mu }_h$, and the right side tends to zero by step 2.2. Every such fixed integral is therefore zero. A positive value of the integrand at an interior point would, by continuity and positivity of the Riemannian volume density, give a positive integral on a small coordinate neighborhood, a contradiction. Hence $N\mathrm{\mid }k{\mathrm{\mid }}_h^2=0$ throughout the interior; its strictly positive lapse factor forces $k=0$ there. Smoothness up to $S$ then gives $k=0$ on the boundary as well.