Orthogonal flow coverage gives a global static product and the needed end bounds

By completeness and [F3], $\mathrm{\Phi }:\mathbb{R}\times \mathcal{E}\to \mathcal{E}$ is smooth, ${\mathrm{\Phi }}_{s+t}={\mathrm{\Phi }}_s\circ {\mathrm{\Phi }}_t$ and ${\mathrm{\Phi }}_t^{-1}={\mathrm{\Phi }}_{-t}$. Differentiating the group law using [F7] gives $d{\mathrm{\Phi }}_t(T)=T$. Since ${\mathcal{L}}_{T}g=0$, [F5] gives ${\mathrm{\Phi }}_t^{\ast }g=g$. Every point lies on an orbit through $p\in \mathrm{\Sigma }$, where $g(T,T)=-N(p{)}^2<0$. This equality persists along the orbit, so $T$ is nowhere zero and timelike everywhere. It remains future-directed along that orbit by continuity from $T_p=N(p)n_p$. A nonzero period would therefore give a nonconstant closed future timelike curve $\gamma :[0,P]\to \mathcal{E}$, $P>0$, based at $p$. Choose a point $q\mathrm{\ne }p$ on this curve and the open neighborhood $\mathcal{E}\setminus \{q\}$ of $p$. Strong causality supplies $V$ containing $p$ inside this neighborhood. Since both endpoints of $\gamma$ are in $V$, [F2] forces its whole image into $V$, contradicting its passage through $q$. Thus no orbit has a nonzero period.

By [F6], the inclusion $i:\mathrm{\Sigma }\hookrightarrow \mathcal{E}$ is a smooth embedding. In the product charts of [F3], the map $(t,p)\mapsto (t,i(p))$ is smooth, so its composition with the smooth global flow gives the smooth map $F(t,p)={\mathrm{\Phi }}_t(i(p))$. Identifying $T_{(0,p)}(\mathbb{R}\times \mathrm{\Sigma })$ with $\mathbb{R}\times T_p\mathrm{\Sigma }$, and $T_p\mathrm{\Sigma }$ with its image under $d{i}_p$, [F7] gives its differential as $d{F}_{(0,p)}(a,v)=a{T}_p+v$, an isomorphism by transversality. The group law and [F7] give $d{F}_{(t,p)}(a,v)=d{\mathrm{\Phi }}_t{\mathrm{\mid }}_p(a{T}_p+v)$, again an isomorphism because ${\mathrm{\Phi }}_t$ is a diffeomorphism. Hence [F4] makes $F$ a local diffeomorphism. Orbit coverage makes it surjective. If $F(t,p)=F(s,q)$, then $p,q\in \mathrm{\Sigma }$ lie on the same orbit, so the unique intersection hypothesis gives $p=q$; step 1.1 then gives $t=s$. Thus $F$ is bijective. Its set-theoretic inverse agrees on every local image with the smooth local inverse from [F4], and is therefore smooth everywhere. This proves that $F$ is a diffeomorphism.

For $v,w\in T_p\mathrm{\Sigma }$, the differential formula and the isometry property give $g(dF({\mathrm{\partial }}_t),dF({\mathrm{\partial }}_t))=g(T_p,T_p)=-N(p{)}^2$, $g(dF({\mathrm{\partial }}_t),dF(0,v))=g(T_p,v)=0$ and $g(dF(0,v),dF(0,w))=h_p(v,w)$. By the pullback definition these are exactly $F^{\ast }g=-N^{2}d{t}^2+h$, with the coefficients independent of $t$.

In the end chart use $\tau$ for the original stationary time coordinate, so $T={\mathrm{\partial }}_{\tau }$, whereas $t$ denotes the product parameter of $F$. The tangent vectors to the supplied graph are ${\mathrm{\partial }}_i+u_i{\mathrm{\partial }}_{\tau }$. Their orthogonality to $T$ gives $g_{0i}+g_{00}{u}_i=0$. For sufficiently large $r$, $-g_{00}\ge 1\mathrm{/}2$, so $u_i=-g_{0i}\mathrm{/}{g}_{00}$. The given differentiated bounds, the product rule and differentiation of this reciprocal give ${\mathrm{\partial }}^{\beta }{u}_i=O(r^{-2-\mathrm{\mid }\beta \mathrm{\mid }})$ for $\mathrm{\mid }\beta \mathrm{\mid }\le 2$. In particular the claimed derivatives of $u$ of orders one through three have the asserted decay.

The radial derivative bound $\mathrm{\mid }{\mathrm{\partial }}_{r}u(r\omega )\mathrm{\mid }\le C{r}^{-2}$ is uniform in $\omega \in S^2$. Integration along a ray shows that $L(\omega )={\lim }_{r\to \mathrm{\infty }}u(r\omega )$ exists and $\mathrm{\mid }u(r\omega )-L(\omega )\mathrm{\mid }\le C\mathrm{/}r$. Any two points on a radius-$r$ sphere can be joined on it by an arc of length at most $\pi r$, so the gradient bound also gives $\mathrm{\mid }u(r\omega )-u(r{\omega }^{\mathrm{\prime }})\mathrm{\mid }\le C\pi \mathrm{/}r$. Taking the limit shows that all $L(\omega )$ equal one constant $L$. The coordinate shift $\tau \mapsto \tau -L$ therefore gives $u=O(r^{-1})$ without changing the stationary metric components or $T={\mathrm{\partial }}_{\tau }$. Finally the induced components are $h_{ij}=g_{ij}+g_{0i}{u}_j+g_{0j}{u}_i+g_{00}{u}_i{u}_j={\delta }_{ij}+O_2(r^{-1})$, because the correction terms are $O_2(r^{-4})$. On the graph, $N^2=-g(T,T)=-g_{00}=1-2M_0\mathrm{/}r+O_2(r^{-2})$. The positive square root, differentiated up to order two, yields $N=1-M_0\mathrm{/}r+O_2(r^{-2})$. No maximal-graph elliptic regularity theorem is needed for this improvement once orthogonality is already known.