Closed Ernst form and existence of normalized potential

Put $U=M^{\mathrm{e}\mathrm{n}\mathrm{d}}$, $P=I_M^{-}(U)$ and $Q=I_M^{+}(U)$. The end is a nonempty connected open product, so [F3] gives connected nonempty $E=P\cap Q$ with $U\subset E$. Its supplied smooth spacelike Cauchy section $C$ is connected, and the generic Cauchy-flow diffeomorphism gives $E\cong \mathbb{R}\times C$. Since $E$ is open, $C$ is open in ${\mathrm{\Sigma }}_0$. Every relative boundary point of $C$ in ${\mathrm{\Sigma }}_0$ is an ambient closure point outside $E$, so the one-sided-boundary result puts it in $S$. The set $S$ is disjoint from $E$, since it is contained in the boundary of each of the open sets $P,Q$. Hence $C$ is nonempty and clopen in ${\mathrm{\Sigma }}_0\setminus S$.

In the chart $\chi$, the complement of $S$ is the disjoint union of the clopen radial shells $A_{-}:=\{1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<1\}$ and $A_{+}:=\{\mathrm{\mid }x\mathrm{\mid }>1\}$. Each is path connected: join two points radially to a common radius within its interval, join their directions on the unit sphere using [F5], and reverse the second radial segment. Thus both shells are connected by [F5]. Connectedness of $C$ prevents it from meeting both shells. Its intersection with the one it meets is nonempty and clopen there, so [F5] makes that intersection the whole shell. Consequently ${\chi }^{-1}(C)$ is exactly one of $A_{-},A_{+}$. Selecting which shell contains the physical end is unnecessary for the following exactness argument.

By [F6] all the choice requirements below hold. By [F7], $P$ and $Q$ are open unions of chronological point pasts and futures; transitivity makes $P$ a past set and $Q$ a future set. At $p\in {\mathrm{\partial }}_{M}P$, let a short smooth future timelike arc $\gamma$ pass through $p=\gamma (0)$. For $t<0$ the closure rule gives $\gamma (t)\in I_M^{-}(p)\subset P$. For $t>0$, membership $\gamma (t)\in \bar{P}$ would give $p\in I_M^{-}(\gamma (t))\subset P$, impossible since $P$ is open and $p\in {\mathrm{\partial }}_{M}P$. Thus the positive-time side is outside $\bar{P}$. Reversing time, at $p\in {\mathrm{\partial }}_{M}Q$ the positive-time side is in $Q$ and the negative-time side is outside $\bar{Q}$.

The polar maps in [F3] identify either shell smoothly with its radial interval times $S^2$: the same stereographic-coordinate formulas in that proof are smooth for every $r>0$. For $A_{+}$ retain $R=r$. For $A_{-}$ set $R=(2r-1{)}^{-1}>1$, with smooth inverse $r=(1+R^{-1})\mathrm{/}2$. These explicit maps identify either possible $C$ with $(1,\mathrm{\infty })\times S^2$. Compose with the generic product from step 1.1 to obtain a specified diffeomorphism $E\cong \mathbb{R}\times (1,\mathrm{\infty })\times S^2$. This product need not be generated by the stationary field, and no spatial identification with the asymptotic chart has been made.

Work near $S$ where the two boundaries are smooth and null by [A1]. A null tangent hyperplane has nonnegative induced metric, so it contains no timelike vector; hence $N$ is transverse to each branch. At each boundary point choose coordinates $(y,t)$ with ${\mathrm{\partial }}_t$ equal to $N$ at that point, and shrink so ${\mathrm{\partial }}_t$ remains future timelike. For a local defining function $h$ from [F8], ${\mathrm{\partial }}_{t}h\mathrm{\ne }0$ at the point. Applying the inverse function theorem to $(y,t)\mapsto (y,h(y,t))$ writes the boundary as a graph $t=a(y)$ in a smaller box. Each vertical segment through this graph is an arc in step 2.2. Thus $r_i=a(y)-t$ is positive precisely on $P$, zero on its boundary and negative outside $\bar{P}$ in a box for ${\mathrm{\partial }}_{M}P$; $r_i=t-a(y)$ has the corresponding signs for $Q$. On the boundary, $d{r}_i$ is nonzero on every timelike vector. Convexity of the future timelike cone in [F6] implies that $d{r}_i(N)$ has the sign of $d{r}_i({\mathrm{\partial }}_t)$, negative for $P$ and positive for $Q$: otherwise it vanishes on a vector on their timelike line segment.

The ambient complex one-form $\eta$ is smooth and closed by [F1]. The product in step 3.1 verifies the exact hypothesis of [F2], so there is a smooth primitive on $E$. The original end has $T={\mathrm{\partial }}_t$ and exactly the differentiated vacuum metric bounds in [A1]. Thus the normalization result in [F4] applies in that end's own chart and shifts the primitive by one complex constant to $\sigma =f+i\psi \to 1$, with $f=-g(T,T)$. Its stationary end conclusion makes the limit uniform in the time coordinate as well as in angle. This application is independent of the auxiliary coordinates used in step 3.1.

For each branch separately, take finitely many smaller coordinate balls covering the compact sphere $S$, with closures inside boxes of step 3.2, and choose the bumps $b_i$ of [F8] equal to one on the smaller balls and supported inside their boxes. Extend $b_i$ and $b_i{r}_i$ by zero outside each box; the extensions are smooth since their supports stay inside the boxes. On the neighborhood $W=\{{\sum }_i{b}_i>0\}$ define $r=({\sum }_i{b}_i{r}_i)\mathrm{/}({\sum }_i{b}_i)$. All active $r_i$ have the same sign, determined by membership in the chronological set, its boundary or the complement of its closure; hence $r$ has those same signs and zeros. On the boundary, differentiating the weights contributes zero because each $r_i$ there is zero. Thus $dr(N)$ is a weighted average of strictly negative numbers for $P$, or of strictly positive numbers for $Q$, and is nonzero. Intersect the two resulting neighborhoods to obtain functions $u,v$ near $S$ with $P=\{u>0\}$, $Q=\{v>0\}$ and the respective boundaries as their zero sets. Transversality in [A1] makes $du,dv$ independent along $S$, with common kernel $TS$.

The compact embedded sphere $S$ is closed in the Hausdorff manifold $M$. Take a tubular map $\mathrm{\Phi }$ and projection $\pi$ from [F9], restricting its image to the neighborhood of step 4.2, and consider $G=(\pi ,u,v)$. Along $S$, $d\pi$ is the identity on $TS$, while $du,dv$ have common kernel $TS$, so $dG$ is invertible. By [F8], $G$ is a local diffeomorphism near $S$. Choose the smooth normal-bundle metric of [F9]. Compactness of the zero section gives a uniform ${\epsilon }_0>0$ such that the tube $\{(p,w):p\in S,\mathrm{\parallel }w\mathrm{\parallel }<{\epsilon }_0\}$ is in the domain of $\mathrm{\Phi }$ and $G$ is a local diffeomorphism throughout its image: finite bundle charts and equivalence of their continuous fibre norms give this uniform shrinking. There is $0<\epsilon <{\epsilon }_0$ on whose tube image $G$ is injective. Otherwise choose unequal pairs $\mathrm{\Phi }(p_n,w_n)$, $\mathrm{\Phi }(q_n,z_n)$ with equal $G$ images and fibre norms less than $\min \{{\epsilon }_0\mathrm{/}2,1\mathrm{/}n\}$. Equality of the $\pi$ coordinates gives $p_n=q_n$; a subsequence tends to some $p\in S$. Both fibre vectors tend to zero in a bundle chart, so both image points tend to $p$ and eventually lie in one inverse chart for $G$, a contradiction. Thus $G$ on that tube image is a diffeomorphism onto an open neighborhood of $S\times \{(0,0)\}$ in $S\times {\mathbb{R}}^2$. A finite product-neighborhood cover of $S\times \{(0,0)\}$ gives $\delta >0$ for which its image contains $S\times (-\delta ,\delta {)}^2$.

Set $O=G^{-1}(S\times (-\delta ,\delta {)}^2)$ inside the injective tube of step 5.1. In these coordinates $O$ retracts smoothly to $S$ by $(p,u,v)\mapsto (p,(1-\tau )u,(1-\tau )v)$, $0\le \tau \le 1$. By step 4.2 and $E=P\cap Q$, its overlap with $E$ is exactly $S\times (0,\delta {)}^2$. Both $O$ and this nonempty overlap are path connected: move the two interval coordinates along straight segments and join the sphere coordinates by a path from [F5]. Hence they are connected and $E\cup O$ is connected, since $E$ is connected and the overlap is nonempty. The retraction and [F10] give $H_{\mathrm{d}\mathrm{R}}^1(O)\cong H_{\mathrm{d}\mathrm{R}}^1(S^2)=0$.

By [F1] the form $\eta$ is smooth and closed on the ambient manifold. By [F10] and step 6.1, its real and imaginary parts on $O$ have smooth real primitives, giving a smooth complex primitive $s_O$. The normalized exterior primitive ${\sigma }_E$ of step 4.1 satisfies $d(s_O-{\sigma }_E)=0$ on the connected overlap. By [F11] their difference is one complex constant. Subtract it from $s_O$ and glue the two agreeing functions on their open overlap. This gives a smooth primitive $\sigma$ on $E\cup O$ extending the normalized ${\sigma }_E$. Every ingredient used here follows from the local geometric hypotheses in [A1]; no compatibility of the radial section chart with the end chart has entered.

Antisymmetry gives $d\sigma (T)=2\mathcal{F}(T,T)=0$, so the extended primitive is stationary wherever defined. Also $d(\mathrm{Re}\sigma -f)=0$ by [F1]; connectedness and [F11], together with its vanishing on $E$, imply $\mathrm{Re}\sigma =f$ throughout $E\cup O$. On this connected domain any two primitives differ by a complex constant by [F11], and the common end limit forces that constant to vanish. The exterior normalization and the horizon extension therefore agree on their overlap and are unique as stated.