Boundary angle bound and normal-graph cone

Write ${\mathrm{\Phi }}_q(x)={\varphi }_{u_q^K(x)}^K(x)$ on the reference collar. The complete reference $K$ product makes this an embedding in the interior, and [F3] gives its individual smooth extension to $S$. Apply the fixed-boundary normal lemma in [F5]: the variable-height derivative term is multiplied by $K=0$ on $S$. Thus the future candidate normal is exactly the boost of the reference normal, and its pairing is $\cosh (\kappa u_q^K)$. The bound in [F3] proves the stated uniform boundary angle estimate.

Apply the uniform Gaussian-flow lemma with the fixed time bound $C_{\delta }$. The smooth background family and compact $S$ give one chart and one Taylor remainder bound for all bounded times and all $q$. Its pointwise ratio estimate yields a strict cone constant ${\theta }_{\delta }<1$, while its radial comparison gives constants $c_1,c_2>0$ such that $c_1{\rho }_q(x)\le {\rho }_q({\pi }_q^{\perp }{\mathrm{\Phi }}_q(x))\le c_2{\rho }_q(x)$ on a common reference collar. This substitution requires no derivative estimate for $u_q^K$, since it uses only its bounded value at each point.

Verify the action hypotheses for the fixed original-metric tube in [F6]. Since $T{\mathrm{\mid }}_S\equiv ̸0$ and $K{\mathrm{\mid }}_S=0$, the identity $K=T+{\mathrm{\Omega }}_{H}Z$ forces ${\mathrm{\Omega }}_H\mathrm{\ne }0$. Regularity makes $T$ tangent to both transverse horizon branches, hence to their intersection $S$; therefore $Z{\mathrm{\mid }}_S=-{\mathrm{\Omega }}_H^{-1}T{\mathrm{\mid }}_S$ is tangent to $S$. Local existence for $Z{\mathrm{\mid }}_S$ and ambient uniqueness in [F7] make the periodic $Z$ action preserve $S$. Its Killing flow preserves the metric: differentiating its pullback gives the pullback of ${\mathcal{L}}_{Z}g=0$. Along the flow, the differential carries a future timelike vector to a continuous timelike vector starting in the future component, so it cannot leave that component. Thus this circle action preserves future orientation, and the invariant-normal-coordinate lemma applies with the commuting fields, fixed sphere and positive boost rate in [F2] and [F6].

The invariant frame need not be tangent-adapted to the reference section. Let $(n,\nu )=({\bar{n}}_1,{\bar{\nu }}_1)$ be its future/outward orthonormal normal frame on $S$. Since $(e_0,e_1)$ is also future/outward, a smooth real function $\beta (y)$ satisfies $\nu =\sinh \beta e_0+\cosh \beta e_1$: explicitly $\beta =\mathrm{arsinh}(-g(\nu ,e_0))$, with the positive $e_1$ coefficient fixed by the outward orientation. Projection of the reference section in the tube to $(v,y)$ has timelike kernel $\mathbb{R}{e}_0$ at $S$, so its differential on the spacelike section is an isomorphism. By [F8] it gives local smooth graphs $u=w(v,y)$ with $w(0,y)=0$ and ${\mathrm{\partial }}_{v}w(0,y)=\tanh \beta (y)$. The exterior side has positive $v$ derivative, since its boundary tangent is a positive multiple of $\nu$ modulo $TS$ and $\cosh \beta >0$. These are graphs of the germ of the embedded reference section, not arbitrary other sheets: local uniqueness, compactness of $S$ and [F10] allow a common shrinking and make the local graph germs agree. Thus one obtains $w$ for $0\le v<ϵ$ on all of $S$. This verifies the available orientation condition without asserting $\nu =e_1$.

Normal projection is locally invertible on each spacelike candidate because its kernel is a timelike Gaussian fibre. On $S$ its composite with the reference flow parametrization is the identity and has radial derivative $\cosh (\kappa u_q^K)>0$ and tangential identity. Hence each fixed $q$ has an individual one-sheet boundary germ, but no uniform inverse-function radius has been presumed. Restrict the target to $0<R<r_{\delta }<c_1{\rho }_0\mathrm{/}2$, where ${\rho }_0$ is a common reference radius from step 1.2. Radial comparison bounds preimages of every compact target set inside a compact reference subannulus, so projection is proper onto this common target. The finite-sheet theorem then gives a finite covering.

Carry out [F6]'s projection-and-division construction with this actual slope. Set $h(v,y)={\int }_0^1{\mathrm{\partial }}_{v}w(tv,y)dt$. The continuous integrand is integrable by [F9], and Newton–Leibniz gives $w(v,y)=vh(v,y)$ and $h(0,y)={\mathrm{\partial }}_{v}w(0,y)$. In surface charts, iterate differentiation under the integral sign on compact coordinate rectangles to obtain all parameter derivatives. Their joint continuity follows from [F10] applied to the compact $t$ interval and the uniform integral-error bound in [F9]. Therefore $h$ is smooth up to zero, also seen using the local smooth extensions in [F8]. By spacelikeness, $\mathrm{\mid }h(0,y)\mathrm{\mid }=\mathrm{\mid }\tanh \beta (y)\mathrm{\mid }<1$; compact extrema in [F10] give $m={\max }_S\mathrm{\mid }h(0,y)\mathrm{\mid }<1$, and the tube lemma gives $\mathrm{\mid }h(v,y)\mathrm{\mid }<\theta <1$ on a common smaller collar, for any fixed $m<\theta <1$.

Put $R(v,y)=v\sqrt{1-h(v,y{)}^2}$. It is smooth and ${\mathrm{\partial }}_{v}R(0,y)=\sqrt{1-h(0,y{)}^2}\ge \sqrt{1-m^2}>0$. By [F10] shrink once to have ${\mathrm{\partial }}_{v}R>c=\frac{1}{2}\sqrt{1-m^2}$ throughout. Fix ${ϵ}_1>0$ inside this collar. The mean value theorem in [F11] gives $R({ϵ}_1,y)>c{ϵ}_1$ and strict increase of $R(\cdot ,y)$; the intermediate value theorem gives a unique inverse $v=V(r,y)$ for every $0\le r<r_0=c{ϵ}_1$. The differential of $(v,y)\mapsto (R(v,y),y)$ is invertible, so [F8] makes its inverse smooth, including at zero by the smooth extensions; uniqueness patches these inverses. Set $a(r,y)=\mathrm{arctanh}h(V(r,y),y)$. Then $\mathrm{\mid }a\mathrm{\mid }\le A=\mathrm{arctanh}\theta$, and the original reference collar is $x(r,y)=\mathrm{\Xi }(y,r\sinh a(r,y),r\cosh a(r,y))$. Here $a$ is rapidity and $b=a\mathrm{/}\kappa$ is the boost-time height in the supplier's notation. Neither the positive-lapse monotonic-crossing conclusion of that supplier nor its stronger tangent alignment is invoked.

For fixed $q$, sufficiently small target radius forces all preimages into its individual injective boundary germ by the radial comparison in step 1.2. The germ supplies a fibre with one point. The finite sheet number is constant on the connected target annulus, hence is one on the entire common collar. The last helper in [F5] gives the smooth normal graph and its smooth individual boundary extension. Its values obey the cone inequality already proved. This establishes the graph for the candidate portion parametrized by ${\mathrm{\Phi }}_q$.

Now prove local saturation in that tube. Set $r=(v^2-u^2{)}^{1\mathrm{/}2}$ and $\eta =\mathrm{arctanh}(u\mathrm{/}v)$ on $v>\mathrm{\mid }u\mathrm{\mid }$, so $u=r\sinh \eta$, $v=r\cosh \eta$. Choose $r_{\ast }>0$ so that all $x(r,y)$ with $0\le r\le r_{\ast }$ lie in the closed source collar $0\le {\rho }_q\le {\rho }_0$ used in steps 1.2 and 2.1, for every $q\in [\delta ,1]$. This follows from joint smoothness of the reference distance collars, compactness of $[\delta ,1]\times S$, and ${\rho }_q(x(0,y))=0$. Choose $b>0$ so that the closed box $\mathrm{\mid }u\mathrm{\mid }\le b$, $\mathrm{\mid }v\mathrm{\mid }\le b$ over $S$ lies in the original tube and its exterior-quadrant neighborhood. Finally choose $0<\epsilon <\min \{r_{\ast },b\}$ with $\epsilon \cosh A<b$. All these choices depend on the background and $\delta$, not on a candidate.

Let $p=\mathrm{\Xi }(y,u,v)$ be any original exterior point with $0<v<\epsilon$. Then $\mathrm{\mid }u\mathrm{\mid }<v$ and $0<r\le v<r_{\ast }$. Its rapidity ${\eta }_p$ is finite. Join $x(r,y)$ to $p$ by keeping $(r,y)$ fixed and varying rapidity monotonically from $a(r,y)$ to ${\eta }_p$. On this entire segment, $v=r\cosh \eta \le \max \{r\cosh a(r,y),v(p)\}<b$ and $\mathrm{\mid }u\mathrm{\mid }<v$, since $\cosh$ is even and increasing with absolute value. Thus the segment remains in the actual local tube, where the exact field formula in [F6] makes it a $K$ integral curve with elapsed time $s=({\eta }_p-a(r,y))\mathrm{/}\kappa$. The complete field in [F1] extends this same local field, so uniqueness of integral curves in [F7], continued along the finite segment, identifies it with its complete orbit from $x(r,y)$. Hence $p={\varphi }_s^{K}x(r,y)$. The open normal box $\mathrm{\mid }u\mathrm{\mid }<\epsilon$, $\mathrm{\mid }v\mathrm{\mid }<\epsilon$ is an ambient neighborhood of $S$, and its entire exterior part is therefore in the saturation of the chosen source collar. This argument permits arbitrarily large finite $\mathrm{\mid }s\mathrm{\mid }$ as $p$ approaches either null branch; no uniform time bound is used.

If a candidate point lies in the exterior neighborhood from step 5.1, it lies on the complete $K$ orbit of some $x$ in the chosen source collar. The unique candidate intersection on that orbit in [F3] forces it to be ${\mathrm{\Phi }}_q(x)$. Consequently no additional near-boundary candidate portion is omitted. Put $a={\rho }_0$ for the closed source radius used in steps 1.2 and 2.1, and write its jointly smooth boundary-distance parametrization as $c_q(\rho ,\eta )$. Choose $0<b<c_{1}a\mathrm{/}2$ below the common target radius in step 3.2. That step's one-sheet normal graph is exactly the portion of ${\mathrm{\Phi }}_q(c_q([0,a]\times S))$ projecting to $0\le R_q<b$, including its zero boundary on $S$; the radial bounds are those of step 1.2. Joint smoothness of the Gaussian family and compactness of $[\delta ,1]\times S$ allow one ambient open neighborhood $U$ of $S$ inside the normal box from step 5.1 and all the Gaussian domains, with target radial coordinates less than $b$ for every $q$. Every candidate point in $U$ is therefore in this graph, and every graph point is a candidate point. Reducing $b$ further preserves the image equality by restriction; shrink $U$ again using the same joint smoothness to preserve capture. The cone bound and individual smooth boundary extension from step 3.2 are unchanged. Finally the Gaussian proper-time normal is $-{\mathrm{grad}}_{g_q}{\tau }_q={\bar{n}}_q$ on $S$, so step 1.1 gives the last boundary angle estimate.

For the separate background conclusion, retain only the standing original data in [A1]. Steps 1.3 and 1.4 construct the invariant frame and the original reference graph in it from those background data, allowing its actual smooth spacelike tilt. Steps 2.2 and 3.1 give $u=r\sinh a(r,y)$, $v=r\cosh a(r,y)$ with smooth bounded rapidity $a$, using no candidate height or candidate section. Put $b=a\mathrm{/}\kappa$ and $\eta =\mathrm{arctanh}(u\mathrm{/}v)\mathrm{/}\kappa$. Since $\kappa >0$ is fixed, $b$ is smooth and bounded and the exact boost-field formula gives $K={\mathrm{\partial }}_{\eta }$. These steps did not use the candidate premises in [F3] or the candidate conclusions in steps 1.1, 1.2, 2.1, 3.2, 4.1, 5.1 and 6.1. Thus they prove the stated background result independently of existence at any candidate parameter.