The same maximal slice has the differentiated article end gauge and a limiting old time

Reconstruct the spacetime tensor from its ADM fields: $g_{00}=-N^2+h(X,X)$, $g_{0i}=h_{ij}{X}^j$ and $g_{ij}=h_{ij}$. The preceding lapse, shift and metric estimates give all displayed differentiated bounds. Only spatial coordinates were changed after tau was chosen in [F3], so $\tau =0$ is still the original supplied maximal hypersurface.

On a large tau-constant sphere, stationarity gives ${\mathrm{\nabla }}_0{T}_r=-\frac{1}{2}{\mathrm{\partial }}_r{g}_{00}=m\mathrm{/}{r}^2+O(r^{-3})$. Its future section normal is ${\mathrm{\partial }}_{\tau }+O(r^{-1})$, its outward unit normal is radial up to order $r^{-1}$, and its area is $4\pi r^2+O(r)$. The normalized Komar limit is therefore m. The same calculation in the original regular chart gives M0. The vacuum identity in [F4] and Stokes in [F5] equate these integrals on large homologous spheres in the same stationary end, even when the spheres lie in different time sections. Thus $m=m_K=M_0$.

Construct the quotient metric on the spatial orbit labels, without invoking a global quotient theorem. On the common stationary end $g_{00}<0$. The tensor $Q=g-T^{\mathrm{♭}}\otimes T^{\mathrm{♭}}\mathrm{/}g(T,T)$ annihilates $T$ and is stationary. For a spatial vector $V$, its value is $Q(V,V)=g(V^{\perp },V^{\perp })$, where $V^{\perp }=V-g(V,T)T\mathrm{/}g(T,T)$. Its coefficients on the old labels are $q_{ij}^{\mathrm{o}\mathrm{l}\mathrm{d}}=g_{ij}^{\mathrm{o}\mathrm{l}\mathrm{d}}-g_{0i}^{\mathrm{o}\mathrm{l}\mathrm{d}}{g}_{0j}^{\mathrm{o}\mathrm{l}\mathrm{d}}\mathrm{/}{g}_{00}^{\mathrm{o}\mathrm{l}\mathrm{d}}$; define $q^{\mathrm{n}\mathrm{e}\mathrm{w}}$ by the same formula in the new chart. The old estimates in [F3] and step 1.1 give $q^{\mathrm{o}\mathrm{l}\mathrm{d}}=\delta +O({\rho }^{-1})$ and $q^{\mathrm{n}\mathrm{e}\mathrm{w}}=\delta +O(r^{-1})$. Thus both are smooth positive-definite metrics sufficiently far out, to which [F6] applies. Substituting $dt=d\tau +du$ cancels the $du$ terms in this formula, and the spatial tensor transformation in [F11] then gives $q^{\mathrm{o}\mathrm{l}\mathrm{d}}=f^{\ast }{q}^{\mathrm{n}\mathrm{e}\mathrm{w}}$. This defines the Riemannian metric on the stationary orbit labels and proves the pullback identity used here.

Uniform comparison of each quotient metric with its Euclidean metric and the pullback identity give constants $0<a\le b<\mathrm{\infty }$ with $a\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }D{f}_{x}v\mathrm{\mid }\le b\mathrm{\mid }v\mathrm{\mid }$ on a sufficiently far-out common end, and hence a uniform bound for $D(f^{-1})$. Since both charts describe the same end outside compact inner parts, points tending to its infinity have both $\rho \to \mathrm{\infty }$ and $r\to \mathrm{\infty }$. Choose a fixed old coordinate sphere beyond the inner parts. Integrating the upper bound for $Df$ along its outward radial rays gives $r\le b\rho +C$, since the image of the initial sphere is compact. Applying the same argument to $f^{-1}$ from a fixed new coordinate sphere gives $\rho \le a^{-1}r+C^{\mathrm{\prime }}$. After increasing the radii, $r$ and $\rho$ are therefore comparable. This uses the existing overlapping end charts, rather than postulating a Euclidean asymptotic formula for $f$.

The mixed tensor identity for this same coordinate change is $g_{0a}^{\mathrm{n}\mathrm{e}\mathrm{w}}(f(x)){\mathrm{\partial }}_i{f}^a=g_{0i}^{\mathrm{o}\mathrm{l}\mathrm{d}}+g_{00}^{\mathrm{o}\mathrm{l}\mathrm{d}}{\mathrm{\partial }}_{i}u$. The old mixed component is $O({\rho }^{-2})$ by [F3], while the new one is $O(r^{-2})$ by step 1.1. Step 3.1 and $g_{00}^{\mathrm{o}\mathrm{l}\mathrm{d}}\to -1$ therefore give $Du=O({\rho }^{-2})$. For each unit vector $\omega$, integration of $\mathrm{\mid }{\mathrm{\partial }}_{\rho }u(\rho \omega )\mathrm{\mid }\le C{\rho }^{-2}$ from $\rho$ to infinity gives a limit $c(\omega )$ and a uniform error at most $C\mathrm{/}\rho$. Any two points on the radius-$\rho$ sphere can be joined there by an arc of length at most $\pi \rho$; integrating the same gradient bound along it gives oscillation at most $C\pi \mathrm{/}\rho$. Passing to infinity shows that all $c(\omega )$ coincide. Hence $u=c+O({\rho }^{-1})$, and a constant old-time translation sets $c$ to zero. The adapted time already has the identical slice at zero. Only first derivatives in the old chart were used in this comparison.