Mass-one end coordinates on the unchanged maximal section

Keep the original time $t$ and put $x=p(y)=f^{-1}(y)$, $J^i{}_a={\mathrm{\partial }}_a{p}^i$. The inverse estimates in [F2] imply $r\asymp R$, $J=I+O_6(R^{-1})$ and $D^{j}p=O(R^{-j})$ for $2\le j\le 7$. For any $u=O_6(r^{-b})$, repeated chain differentiation of $u\circ p$ has factors $D^{\mathrm{\ell }}u$ and $D^{j_1}p,\dots ,D^{j_{\mathrm{\ell }}}p$ with $\sum j_i=k\le 6$. Bounded first derivatives and the stronger bounds for $j_i\ge 2$ give $D^k(u\circ p)=O(R^{-b-k})$. The tensor transformation law at fixed $t$ is $G_{00}=g_{00}\circ p$, $G_{0a}=(g_{0i}\circ p)J^i{}_a$, $G_{ab}=(g_{ij}\circ p)J^i{}_a{J}^j{}_b$. Therefore $G_{0a}=O_6(R^{-2})$ and $G_{ab}={\delta }_{ab}+O_6(R^{-1})$. The scalar $G_{00}=g(T,T)$ is independent of stationary time gauge, so the normalized chart in [F2] gives $G_{00}=-1+2m\mathrm{/}R+O_6(R^{-2})$ with its full differentiated remainder. This last remainder is not inferred merely from $f-x=O(\log r)$.

On a far tail put $V=G_{00}$ and $a_a=G_{0a}\mathrm{/}V$. Then $V$ stays negative and away from zero, and [F3] gives $a=O_6(R^{-2})$. The decomposition in [F5] gives the metric of the original $t=0$ section as $h_{\mathrm{\Sigma }}=h_{\mathrm{q}\mathrm{u}\mathrm{o}\mathrm{t}}+Va\otimes a$, so $h_{\mathrm{\Sigma }}-h_{\mathrm{q}\mathrm{u}\mathrm{o}\mathrm{t}}=O_6(R^{-4})$. Thus neither this connection estimate nor its differentiated version requires taking the section $t^{\mathrm{\prime }}=0$.

To compute the ADM limit in this same $y$ gauge, use the analytic compact quotient data in [F5]. Write $s=\mathrm{\mid }\xi \mathrm{\mid }=R^{-1}$, $\omega =y\mathrm{/}R$ and $Q=I-2\omega \otimes \omega$. By [F4], ${\gamma }_{ij}(\xi )={\delta }_{ij}+O(s^2)$; analyticity gives $f_{\mathrm{\Omega }}(\xi )=1+c\cdot \xi +O(s^2)$ for a fixed vector $c$. The inversion differential is $D_y\xi =s^{2}Q$, with $Q^{T}Q=I$. Since $\gamma ={\mathrm{\Omega }}^2\bar{h}$, transforming back gives $\bar{h}=f_{\mathrm{\Omega }}^{-2}{Q}^T\gamma Q=\delta -2(c\cdot \omega )\delta \mathrm{/}R+O_1(R^{-2})$; the differentiated remainder follows by differentiating the convergent analytic series after inversion. An orthogonal adjustment of $y$ merely rotates $c$. Dividing by $-V=1-2m\mathrm{/}R+O_1(R^{-2})$ and using step 2.1 yields $h_{\mathrm{\Sigma },ab}=(1+u){\delta }_{ab}+O_1(R^{-2})$, where $u=2(m-c\cdot \omega )\mathrm{/}R$. This allows the possible dipole term of this conformal gauge instead of identifying it with the distinct multipole gauge of source 0.

For $(1+u)\delta$, the ADM integrand contracted with the Euclidean outward normal is $-2{\mathrm{\partial }}_{R}u=4(m-c\cdot \omega )\mathrm{/}{R}^2$. The $O_1(R^{-2})$ tensor remainder contributes $O(R^{-3})$ pointwise and hence $O(R^{-1})$ after integration. On the unit sphere $\int \omega d\omega =0$ by antipodal symmetry and $\int d\omega =4\pi$. Consequently the ADM integral divided by $16\pi$ tends to $m$. This is a direct calculation on the unchanged section.

Under $\widehat{t}=t\mathrm{/}m$, $\widehat{x}=y\mathrm{/}m$ and $\widehat{g}=m^{-2}g$, the new components equal $G_{\mu \nu }(m\widehat{x})$, since both coordinate differentials contribute a factor $m$. Step 1.1 and the chain rule give exactly the three stated $O_6$ mass-one component bounds, and $mT={\mathrm{\partial }}_{\widehat{t}}$. On a coordinate sphere the derivative in the ADM integrand contributes $m$, while Euclidean area contributes $m^{-2}$ relative to the old sphere of radius $m\widehat{R}$; its unit radial vector is unchanged. Thus the ADM limit is $m^{-1}$ times the limit in step 4.1, hence one. By [F1] the same section remains maximal and Ricci-flatness is preserved.

Apply only the spatial cutoff diffeomorphism in [F2] to the original section chart. It fixes a neighborhood of $S$ and agrees with $y$ on the far end; no time graph is introduced. If its radial coordinate is $q$, choose $B>A>1$ sufficiently large that $B\mathrm{/}m>A$. A smooth positive derivative can equal $1$ near $A$ and $1\mathrm{/}m$ near $B$ and have integral $B\mathrm{/}m-A$: choose short endpoint collars with total integral smaller than this positive number, then adjust a positive interior profile to give the remaining integral. Integrating and using $\rho (q)=q$ below $A$, $\rho (q)=q\mathrm{/}m$ above $B$, gives a smooth increasing radial diffeomorphism fixing radius one. The chart $z\mapsto \rho (\mathrm{\mid }z\mathrm{\mid })z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ therefore retains the sphere at radius one and agrees with $y\mathrm{/}m$ sufficiently far out. All claims concern the original section; no leading angular-momentum term in its time gauge was used.