Extending normalized end coordinates while preserving the section

On a sufficiently far tail $T$ is timelike. In either chart the quotient metric is $h_{ij}=g_{ij}-g_{0i}{g}_{0j}\mathrm{/}{g}_{00}$, so $h=\delta +O_6(r^{-1})$ in the original chart and $h^{\mathrm{\prime }}=\delta +O_{\mathrm{\infty }}(\mathrm{\mid }y{\mathrm{\mid }}^{-1})$ in the normalized chart. These represent the same Riemannian metric by [F2]. That supplier bounds $Df,D{f}^{-1}$ and compares $r,\mathrm{\mid }y\mathrm{\mid }$. Apply [F4] to its unique quotient Levi–Civita connection. Its symbols satisfy $\mathrm{\Gamma }=O_5(r^{-2})$ and ${\mathrm{\Gamma }}^{\mathrm{\prime }}=O_{\mathrm{\infty }}(\mathrm{\mid }y{\mathrm{\mid }}^{-2})$; the inverse-matrix derivative estimates follow by differentiating $h^{-1}h=I$. In particular the transformation law rearranges to ${\mathrm{\partial }}_i{\mathrm{\partial }}_j{f}^a={\mathrm{\Gamma }}_{ij}^k{\mathrm{\partial }}_k{f}^a-{\mathrm{\Gamma }}^{\mathrm{\prime }}{}_{bc}^a(f){\mathrm{\partial }}_i{f}^b{\mathrm{\partial }}_j{f}^c$, giving $D^{2}f=O(r^{-2})$.

Integrate the derivative of $Df(r\omega )$ along each fixed ray: its norm is at most $C{r}^{-2}$, so [F6] gives a limit $Q(\omega )$ and $\mathrm{\mid }Df(r\omega )-Q(\omega )\mathrm{\mid }\le C\mathrm{/}r$. Here is a local construction for any two unit directions $u,v$. If they are not parallel, the coordinate formula in [F10] gives a nonzero $u\times v$ perpendicular to both (vanishing of all its coordinate minors would make $u,v$ proportional); normalize it to a unit vector $w$. If they are parallel, choose a coordinate vector not parallel to $u$ and normalize its cross product with $u$, again obtaining $w$ perpendicular to both. For $a=u$ and $a=v$ the rational arc ${\gamma }_a(s)=r((1-s^2)a+2sw)\mathrm{/}(1+s^2)$, $0\le s\le 1$, joins $ra$ to $rw$. Since $a,w$ are orthonormal, direct expansion gives $\mathrm{\mid }{\gamma }_a(s)\mathrm{\mid }=r$, while the quotient rule in [F6] gives $\mathrm{\mid }{\gamma }_a^{\mathrm{\prime }}(s)\mathrm{\mid }=2r\mathrm{/}(1+s^2)\le 2r$. Each arc therefore has speed integral at most $2r$. Concatenate ${\gamma }_u$ with the reversed ${\gamma }_v$; the total is at most $4r$. The chain rule and fundamental theorem in [F6], applied componentwise to $Df\circ {\gamma }_a$ and the bound from step 1.1, give $\mathrm{\mid }Df(ru)-Df(rv)\mathrm{\mid }\le 4C\mathrm{/}r$ (enlarging the fixed finite-dimensional norm constant if needed), so all limits are a single matrix $Q_0$. The pullback identity $h=(Df{)}^T(h^{\mathrm{\prime }}\circ f)Df$ now gives $Q_0^T{Q}_0=I$. Integrating $Df-Q_0=O(r^{-1})$ radially from one fixed large sphere gives $f-Q_{0}x=O(\log r)$ uniformly in angle; bounded values on that fixed sphere also follow by the same angular path estimate from one anchor. No orientation of $Q_0$ is presumed.

Set $A=Q_0^T$ and $z=Ay$. The identity $A(v\times w)=(\det A)(Av\times Aw)$ follows by expanding the determinant definition of the cross product, for any orthogonal $A$. The normalized spatial metric is conjugated by $A$, the mixed metric covector transforms by $A$, and $\mathrm{\mid }z\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }$. Hence its leading mixed coefficient is $-2(S_A\times z)\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^3$, where $S_A=(\det A)AS$, and all differentiated remainder orders and the mass are unchanged. For the ADM check, a test rotation $B\times z$ pulls back to $(\det A)(A^{T}B)\times y$. The outward normal and positive area measure transform isometrically; the flux convention in [F1] therefore gives $J_z(B)=(\det A)(A^{T}B)\cdot S=B\cdot S_A$. Thus the spin convention is preserved as an axial vector, even if $\det A=-1$, and $\mathrm{\mid }{S}_A\mathrm{\mid }=\mathrm{\mid }S\mathrm{\mid }$. Rename $z$ as $y$ and $A\circ f$ as $f$; then $Df=I+O(r^{-1})$, $f-x=O(\log r)$.

Differentiate the Christoffel identity of step 1.1 at most five times. Inductively, after $k$ differentiations its right-hand side contains only derivatives of $f$ through order $k+1$, together with derivatives of $\mathrm{\Gamma }$ through order $k$ and of ${\mathrm{\Gamma }}^{\mathrm{\prime }}$ through order $k$. The bounded first derivatives and the induction estimates $D^{j}f=O(r^{-j})$ for $j\ge 2$ bound every term by $O(r^{-k-2})$. Thus $D^{k+2}f=O(r^{-k-2})$ for $0\le k\le 5$. The same argument with charts reversed gives the inverse bounds through order seven. Now put ${\phi }_i=g_{0i}\mathrm{/}{g}_{00}$ in each chart. The original metric gives $\phi =O_6(r^{-2})$ and the normalized metric gives ${\phi }^{\mathrm{\prime }}=O_{\mathrm{\infty }}(\mathrm{\mid }y{\mathrm{\mid }}^{-2})$. Equating the two representations of $g(T,\cdot )\mathrm{/}g(T,T)$ gives $(dF{)}_i={\phi }_i-{\phi }_a^{\mathrm{\prime }}(f){\mathrm{\partial }}_i{f}^a$. In each of its derivatives of order $k\le 6$, the chain and product rules use derivatives of $f$ of order at most $k+1\le 7$. The bounds just proved and radius comparison give ${\mathrm{\partial }}^k(dF)=O(r^{-2-k})$. This derives $dF=O_6(r^{-2})$ from the stationary one-form, rather than from spatial connection estimates alone. Normalize $F=O(r^{-1})$ by the constant adjustment in [F2].

On the common far end choose ${\chi }_R$ from [F5] and set $f_R=x+{\chi }_R(f-x)$, extending by the identity to all of ${\mathbb{R}}^3$. This is smooth because the cutoff has an inner plateau and its nonzero part lies inside the domain of $f$. On $R\le r\le 2R$, $D{f}_R-I={\chi }_R(Df-I)+(f-x)\otimes d{\chi }_R=O(\log R\mathrm{/}R)$; outside it is either zero or $O(r^{-1})$. Choose $R$ so $\sup \mathrm{\parallel }D{f}_R-I\mathrm{\parallel }<1\mathrm{/}2$. For $q_R=f_R-I$, integration along the straight segment from $a$ to $b$ gives $\mathrm{\mid }{q}_R(a)-q_R(b)\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }a-b\mathrm{\mid }$, hence $\mathrm{\mid }{f}_R(a)-f_R(b)\mathrm{\mid }\ge \frac{1}{2}\mathrm{\mid }a-b\mathrm{\mid }$. Thus $f_R$ is injective and its derivative is invertible everywhere.

Since $q_R=O(\log r)$, $\mathrm{\mid }{f}_R(x)\mathrm{\mid }\to \mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. The inverse image of a compact Euclidean set is therefore bounded and is closed by continuity and closedness of compact sets, hence compact by [F7]. So $f_R$ is proper. Its target ${\mathbb{R}}^3$ is connected, since straight segments join every pair of points and [F7] applies. The proper local-diffeomorphism theorem gives surjectivity; the smooth inverse function theorem gives local smooth inverses, which agree by injectivity and yield a global smooth inverse. The map fixes the ball $r\le R$ pointwise. Bijectivity then prevents any exterior point from mapping into that ball, so its restriction is a diffeomorphism of $\{r>1\mathrm{/}2\}$, fixing a neighborhood of $S_0$.

The supplied product satisfies [F3], so $t$ is Cauchy temporal and $dt$ is timelike throughout $E$. The original metric bounds imply its inverse tends uniformly to the Minkowski inverse on the end. Thus the conditional spacelike-graph hypotheses of [F2] hold, with $g^{00}\le -1\mathrm{/}2$ and uniformly bounded inverse components sufficiently far out. Choose $R$ as in step 4.2 and large enough for that supplier. It gives $u_R={\chi }_{R}F$ with $\sup \mathrm{\mid }{u}_R\mathrm{\mid }\le C\mathrm{/}R$ and $\sup \mathrm{\mid }d{u}_R\mathrm{\mid }\le C\mathrm{/}{R}^2$, and proves that $t+u_R=0$ is smooth and spacelike. In particular only first derivatives are needed for this conclusion; step 4.1 separately establishes the stronger differentiated transition estimates. Applying [F3] with displacement $-u_R$ makes this graph $C^{\mathrm{\prime }}$ Cauchy and gives its diffeomorphism with $C$ by product projection. The choice assumptions required by these suppliers are included in ZFC as recorded in [F3].

Where $r\le R$ the graph is exactly the old section. Attach the unchanged part of ${\mathrm{\Sigma }}_0$ there; equality on the overlap makes the ambient section extension smooth, with literal equality on a neighborhood of the explicitly supplied sphere $S_0$. This proves the bifurcation-sphere conclusion whenever $S_0$ is that sphere, without inferring existence of a bifurcation surface from a stationary product. Transport $f_R\circ x$ to $C^{\mathrm{\prime }}$ using product projection and keep it on the unchanged inner extension. Step 5.1 makes it a global spatial chart on that extension, fixing $S_0$; in $r\ge 2R$ it equals $y$, while $t+u_R=t+F=t^{\mathrm{\prime }}$. Thus the new section is precisely $t^{\mathrm{\prime }}=0$ in the normalized end. These constructions assert no maximality of the new section.