Bounded boost times give a uniform Gaussian collar cone

The coordinate geodesic equations have jointly smooth coefficients in position, velocity and $q$; at the endpoints of $[0,1]$, smooth family means smooth extension to a parameter neighborhood. Thus their solutions and ${\mathrm{\Psi }}_q$ are jointly smooth near the zero section. At $(0,y)$ the differential in $(\tau ,x)$ is $(u,V)\mapsto u{\bar{n}}_q+V$, an isomorphism because the reference tangent hyperplane is spacelike. Apply the inverse theorem to $(q,\tau ,x)\mapsto (q,{\mathrm{\Psi }}_q(\tau ,x))$. Compactness of $[0,1]\times S$ supplies a common domain of definition and local invertibility. For global injectivity on a sufficiently small tube, otherwise select collisions at radii tending to zero. Passing to subsequences, their parameters and base points converge; equality of images forces both limiting base points to be the same point of $S$. Both preimages then lie in one augmented inverse neighborhood, a contradiction. This is the compact normal-tube argument with the parameter retained. Hence the local inverses agree to give a common smooth Gaussian family. The identical argument gives a common jointly smooth normal tube ${\mathrm{\Xi }}_q(y,a,b)$ about $S$. Shrink both tubes to have compactly contained smaller domains; when the event data are supplied, keep their images in $O$.

Write $V={\mathrm{\partial }}_{\tau }{\mathrm{\Psi }}_q$ and let $J$ be a variation in a reference base coordinate. Each normal geodesic has constant squared speed $-1$. Torsion freedom gives ${\mathrm{\nabla }}_{V}J={\mathrm{\nabla }}_{J}V$, whence ${\mathrm{\partial }}_{\tau }{g}_q(V,J)=g_q(V,{\mathrm{\nabla }}_{J}V)=\frac{1}{2}J{g}_q(V,V)=0$. The pairing is initially zero, so remains zero. This proves the Gaussian metric form and the assertions about its timelike fibres and gradient. Since $K=0$ on $S$, the constant curves there are its integral curves for all time. Finite iteration of the local flow neighborhoods from F4 along these constant curves, followed by a finite cover of $S$, gives a smooth flow on a neighborhood of $[-C,C]\times S$ for any fixed finite $C$. Uniqueness identifies it with the maximal local flow. Continuity and compactness then put all bounded-time images of a small common reference collar in the Gaussian tube.

Now impose the event-boundary hypotheses. For each fixed $q$, F6 identifies the two branch germs at $S$ with the null exponential sheets ${\mathrm{\Gamma }}_{q,\pm }(y,t)={\mathrm{\Xi }}_q(y,\pm t,t)$, with the indicated future/past labels. These sheets are jointly smooth in $(q,y,t)$ on one common small interval by step 1.1. They remain in the corresponding event boundaries throughout that interval, not merely on a possibly $q$-dependent germ: for each fixed $(q,y)$ continue from $t=0$ along the affine null geodesic. At a finite endpoint of its maximal interval of membership inside the common tube, closedness of the ambient event boundary puts the limiting point on the same branch. Smoothness of that branch makes the limiting geodesic tangent its null radical, since it is a limit of radical tangents. The generator theorem in F6 and geodesic uniqueness in F3 continue it in the branch past that endpoint, a contradiction. The same argument holds in either direction of $t$. This uses no uniform derivative bound for the initially supplied horizons.

At $p\in S$, set $A(s)=d{\varphi }_s^K{\mathrm{\mid }}_p$. Differentiation of the flow law gives $A^{\mathrm{\prime }}(s)=B_{p}A(s)$, $A(0)=I$, where $B_p(v)={\mathrm{\nabla }}_v^{(q)}K$ since $K_p=0$. The flow fixes $S$ pointwise, so its tangent differential is the identity. On the specified ordered normal basis the given matrix is $\kappa \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $\kappa >0$, F2 applies directly; for $\kappa <0$ apply it to $-K$ at time $-s$; for $\kappa =0$, the equation on the normal plane is $A^{\mathrm{\prime }}=0$. In every case the result is $d{\varphi }_s^K{\bar{\nu }}_q=\cosh (\kappa s){\bar{\nu }}_q+\sinh (\kappa s){\bar{n}}_q$.

Project ${\mathrm{\Gamma }}_{q,\pm }$ onto the reference hypersurface. At $t=0$ its differential in $(t,y)$ is $(u,Y)\mapsto u{\bar{\nu }}_q+Y$, an isomorphism, while the time derivative is $\pm 1$. The augmented inverse theorem, compactness and the collision argument from step 1.1 therefore give jointly smooth inverse projections on a common two-sided reference collar. They define graphs $f_{q,\pm }$ with $f_{q,\pm }(0,y)=0$ and ${\mathrm{\partial }}_R{f}_{q,\pm }(0,y)=\pm 1$. These graphs exhaust the event boundaries on a common smaller Gaussian cylinder: a second boundary point on a Gaussian fibre would be timelike related to its existing graph point, contradicting achronality from F6. Their jointly smooth second radial derivatives are bounded on a compact smaller cylinder, and F5 gives $f_{q,\pm }=\pm R+O(R^2)$ with one error constant. Thus uniform branch control has been derived from the smooth exponential maps and the event-boundary geometry.

Compose the Gaussian coordinate functions with ${\varphi }_s^K{c}_q(r,y)$. At $r=0$ both ${\tau }_q$ and $R_q$ vanish, and step 3.1 gives their radial derivatives $\sinh (\kappa s)$ and $\cosh (\kappa s)$. The compositions are jointly smooth on a neighborhood of a compact smaller cylinder $[0,1]\times [-C,C]\times S\times [0,r_0]$. Thus their second radial derivatives have one finite bound $M$ by F5. Apply its $n=1$ Taylor estimate in $r$, holding $(q,s,y)$ fixed, to obtain the asserted uniform quadratic errors. In particular, writing the errors as $E_{\tau },E_R$, one has $\mathrm{\mid }{E}_{\tau }\mathrm{\mid }+\mathrm{\mid }{E}_R\mathrm{\mid }\le M{r}^2$ after enlarging $M$.

Put $T=\tanh (\mathrm{\mid }\kappa \mathrm{\mid }C)<1$ and $B=\cosh (\mathrm{\mid }\kappa \mathrm{\mid }C)$. For $\mathrm{\mid }s\mathrm{\mid }\le C$, $\mathrm{\mid }\sinh (\kappa s)\mathrm{\mid }\le T\cosh (\kappa s)$ and $1\le \cosh (\kappa s)\le B$. Choose $r_0$ smaller so $M{r}_0\le 1\mathrm{/}2$. Then $r\mathrm{/}2\le R_q\le (B+1\mathrm{/}2)r$, giving radial comparison. Moreover $\mathrm{\mid }{\tau }_q\mathrm{\mid }\le T{R}_q+(1+T)M{r}^2$. Choose $T<\theta <1$ and shrink once more so $2(1+T)M{r}_0\le \theta -T$. Dividing by $R_q\ge r\mathrm{/}2$ for $r>0$ proves the cone bound; at $r=0$ both sides vanish. This includes $C=0$ and $\kappa =0$.

For each $q$, F6 and the positive reference collar in ${\mathcal{E}}_q$ select $I_q^{-}(E_q)$ as the past side of the upper graph and $I_q^{+}(E_q)$ as the future side of the lower graph near $S$. These side identities extend throughout the common smaller cylinder: off its graph, each connected side of a boundary has constant membership in its chronological set, which is open by F7, and the local identity supplies a point on each side. There are no further boundary points by step 3.2. The selected component on $R>0$ is consequently exactly $f_{q,-}<{\tau }_q<f_{q,+}$. If $M_H$ bounds the graph remainder constants, shrink further so $M_{H}R<1-\theta$. Then $f_{q,-}<-\theta R\le {\tau }_q\le \theta R<f_{q,+}$ for every bounded-flow image with $R>0$, by step 5.1; hence these images lie on the selected exterior side as well.