A decaying stationary time gauge gives bounded section displacement

First check the needed pointwise linear algebra in an orthonormal Lorentz basis. If $n=(n_0,a)$ is timelike, then $n_0^2>\mathrm{\mid }a{\mathrm{\mid }}^2$; a vector $v=(v_0,z)$ perpendicular to $n$ has $v_0=(a\cdot z)\mathrm{/}{n}_0$. The elementary identity $\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{\mid }z{\mathrm{\mid }}^2-(a\cdot z{)}^2={\sum }_{i<j}(a_i{z}_j-a_j{z}_i{)}^2\ge 0$ gives $g(v,v)\ge (1-\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}{n}_0^2)\mathrm{\mid }z{\mathrm{\mid }}^2>0$ for nonzero $v$. For a spatial vector $v$ in either stationary chart, its horizontal lift is $\widehat{v}=v-g(T,v)T\mathrm{/}g(T,T)$, perpendicular to $T$, and $h(v,v)=g(\widehat{v},\widehat{v})$. The lift is nonzero for nonzero spatial $v$ because $T={\mathrm{\partial }}_t$. Thus the quotient coordinate metrics are Riemannian. The tensor annihilates $T$, so a change of time section does not change its spatial pullback.

By the pullback formula, $h_x(v,v)=h_{f(x)}^{\mathrm{\prime }}(D{f}_{x}v,D{f}_{x}v)$. Choose common constants $0<\lambda \le \mathrm{\Lambda }$ for the two uniform Euclidean comparisons. Then $\lambda \mathrm{\mid }D{f}_{x}v{\mathrm{\mid }}^2\le \mathrm{\Lambda }\mathrm{\mid }v{\mathrm{\mid }}^2$. Applying the same identity to $f^{-1}$ gives the inverse bound. Consequently both derivatives have a uniform bound $L$.

Here is an explicit angular path estimate. For unit vectors $a,b$ with $a\cdot b\ge 0$, the path $q(s)=((1-s)a+sb)\mathrm{/}\mathrm{\mid }(1-s)a+sb\mathrm{\mid }$, $0\le s\le 1$, lies on the unit sphere. Its denominator is at least $1\mathrm{/}\sqrt{2}$, and differentiating gives $\mathrm{\mid }{q}^{\mathrm{\prime }}\mathrm{\mid }\le 2$. For arbitrary unit $a,b$, take $w=(a+b)\mathrm{/}\mathrm{\mid }a+b\mathrm{\mid }$ when $a+b\mathrm{\ne }0$; then $a\cdot w,b\cdot w\ge 0$. If $b=-a$, take a unit $w$ perpendicular to $a$, obtained by projecting and normalizing a coordinate vector not parallel to $a$. The two normalized segments through $w$ therefore join $a$ to $b$ with total speed integral at most $4$. After scaling they give a path on the radius-$s$ sphere with speed integral at most $4s$. Fix an anchor on a sufficiently large sphere in the $x$ chart, join it angularly to the direction of $x$, and continue radially to $x$. The chain rule, componentwise fundamental theorem and integral inequalities give $\mathrm{\mid }f(x)\mathrm{\mid }\le A+Br$. Apply the identical construction to $f^{-1}$ in the other end chart to obtain $\mathrm{\mid }x\mathrm{\mid }\le A^{\mathrm{\prime }}+B^{\mathrm{\prime }}{r}^{\mathrm{\prime }}$. On the common far end both radii tend to infinity, since these are charts of the same end; increasing the compact cut removes the additive constants and proves comparability.

Equating the two representations of the stationary one-form and using $d{t}^{\mathrm{\prime }}=dt+dF$ gives $dF=\phi -f^{\ast }{\phi }^{\mathrm{\prime }}$. In components, $(f^{\ast }{\phi }^{\mathrm{\prime }}{)}_i={\phi }_j^{\mathrm{\prime }}(f(x)){\mathrm{\partial }}_i{f}^j$. The derivative bound and radius comparability imply $\mathrm{\mid }dF\mathrm{\mid }\le C{r}^{-2}$, after increasing a uniform constant.

For a fixed unit $\omega$, set $q_{\omega }(s)=F(s\omega )$. The chain rule gives $\mathrm{\mid }{q}_{\omega }^{\mathrm{\prime }}(s)\mathrm{\mid }\le C{s}^{-2}$. The fundamental theorem and order bounds give $\mathrm{\mid }{q}_{\omega }(v)-q_{\omega }(u)\mathrm{\mid }\le C(1\mathrm{/}u-1\mathrm{/}v)$ for $v>u$ sufficiently large; the integral of $s^{-2}$ is computed from the derivative of $-s^{-1}$. Thus the proper tail integrals of $q_{\omega }^{\mathrm{\prime }}$ satisfy the improper-integral Cauchy criterion. Their convergence and the fundamental theorem give a limit $L(\omega )$ with $\mathrm{\mid }F(r\omega )-L(\omega )\mathrm{\mid }\le C\mathrm{/}r$. On the radius-$r$ angular path constructed in step 3.1, integrating $dF$ gives $\mathrm{\mid }F(r{\omega }_1)-F(r{\omega }_2)\mathrm{\mid }\le 4C\mathrm{/}r$. Taking $r$ to infinity shows $L({\omega }_1)=L({\omega }_2)$ for every pair. Subtract this common constant; the bound $\mathrm{\mid }F(r\omega )\mathrm{\mid }\le C\mathrm{/}r$ is uniform in angle.

Use the one-dimensional function $b$ of [F4] to set ${\chi }_R(x)=1-b(r\mathrm{/}R)$ and $u_R={\chi }_{R}F$ on the end, extending $u_R$ by zero inside. Choose $R$ so the end formula is defined on a neighborhood of $r\ge R$; the plateau of $b$ makes this extension smooth. The power and chain rules give $\mathrm{\mid }dr\mathrm{\mid }=1$, so $\mathrm{\mid }d{\chi }_R\mathrm{\mid }\le B_0\mathrm{/}R$, with $B_0$ a bound for $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }$ on $[1,2]$. On $R\le r\le 2R$, the product rule gives $\mathrm{\mid }{u}_R\mathrm{\mid }\le C\mathrm{/}R$ and $\mathrm{\mid }d{u}_R\mathrm{\mid }\le \mathrm{\mid }dF\mathrm{\mid }+\mathrm{\mid }F\mathrm{\mid }\mathrm{\mid }d{\chi }_R\mathrm{\mid }\le C(1+B_0)\mathrm{/}{R}^2$. Outside this annulus $u_R$ is either zero or $F$, and the bounds of steps 4.1 and 5.1 give the same global estimates. Every fixed compact interior is unchanged once $R$ is large.

Now assume the additional stationary-product and inverse-metric hypotheses of the conditional conclusion. Put ${\alpha }_s=dt+sd{u}_R$ for $0\le s\le 1$. On $r\ge R$ write ${\epsilon }_R=C_2\mathrm{/}{R}^2$, so $\mathrm{\mid }{\mathrm{\partial }}_i{u}_R\mathrm{\mid }\le {\epsilon }_R$. Expanding in the original coordinates yields $g^{-1}({\alpha }_s,{\alpha }_s)=g^{00}+2s{g}^{0i}{\mathrm{\partial }}_i{u}_R+s^2{g}^{ij}{\mathrm{\partial }}_i{u}_R{\mathrm{\partial }}_j{u}_R\le -\kappa +6K_0{\epsilon }_R+9K_0{\epsilon }_R^2$. For sufficiently large $R$ this is at most $-\kappa \mathrm{/}2$, uniformly in $s$ and on the end. On $r\le R$ it equals $g^{-1}(dt,dt)<0$. Hence the entire deformation stays timelike, and its time-cone choice is the one of $dt$. Uniform convergence to the Minkowski inverse supplies the displayed end hypotheses, for example with $g^{00}\le -1\mathrm{/}2$ and bounded inverse components beyond a fixed radius.

For a level value $c$, set $H(p)=c-u_R(p)$, smooth by step 6.1. By [F6], ${\mathrm{\Gamma }}_H\subset C\times \mathbb{R}$ is an embedded three-dimensional submanifold. The factor swap $S:C\times \mathbb{R}\to \mathbb{R}\times C$, $S(p,t)=(t,p)$, is a diffeomorphism: in the product charts of [F6] it and its inverse are coordinate permutations. If $\psi$ is a slice chart for ${\mathrm{\Gamma }}_H$ on an open set $V$, then $\psi \circ S^{-1}$ on $S(V)$ is a smooth slice chart for $S({\mathrm{\Gamma }}_H)$, since it has the same coordinate image and sends $S({\mathrm{\Gamma }}_H)\cap S(V)$ to the same coordinate plane. Thus the level set $S({\mathrm{\Gamma }}_H)=\{(c-u_R(p),p):p\in C\}$ is embedded in the required order of product factors. Its parametrization is smooth and injective, with smooth inverse the spatial projection; its differential is $v\mapsto (-d{u}_R(v),v)$ and has rank three, so its tangent space is exactly $\ker (dt+d{u}_R)$. The vector $n=(dt+d{u}_R{)}^{\mathrm{♯}}$ is timelike by step 7.1, and this kernel is $n^{\perp }$ by the musical definition; step 1.1 proves positivity of its induced metric. This is exactly the spacelike-hypersurface definition. The graph agrees with the original level where $u_R=0$ and with the normalized $t^{\mathrm{\prime }}$ level where $u_R=F$. The unconditional decay and cutoff conclusions were established before imposing these additional graph hypotheses.