Selected ingoing RN extension and time orientation

In product coordinates the $v,r$ block is $\left(\begin{array}{cc}-f& 1\\ 1& 0\end{array}\right),$ with determinant $-1$ and inverse $\left(\begin{array}{cc}0& 1\\ 1& f\end{array}\right)\mathrm{.}$ Its eigenvalues have opposite signs because their product is negative. The sphere block $r^2\gamma$ is positive definite for $r>0$, so $g_{\mathrm{i}\mathrm{n}}$ is a smooth nondegenerate Lorentzian metric on the stated product. The two-form is smooth since $r^{-2}$ is smooth on $(0,\mathrm{\infty })$ and $\omega$ is global.

The product fields ${\mathrm{\partial }}_v,{\mathrm{\partial }}_r$ are global and smooth, hence so are $\mathrm{\ell },n,N$. Direct contraction gives $g(\mathrm{\ell },\mathrm{\ell })=g(n,n)=0$, $g(\mathrm{\ell },n)=-1$, $g(N,N)=-2$, and $g(\mathrm{\ell },N)=g(n,N)=-1$. Thus $N$ is an everywhere timelike field and fixes the asserted time orientation; [F3] then makes both nonzero null fields $\mathrm{\ell },n$ future-directed.

On any component where $f\mathrm{\ne }0$, substitute $dv=dt+f^{-1}dr$ into $-fd{v}^2+2dvdr$. The mixed terms cancel and the radial coefficient is $f^{-1}$, giving $-fd{t}^2+f^{-1}d{r}^2$. Also $dv\wedge dr=dt\wedge dr$, so ${\mathcal{F}}_{\mathrm{i}\mathrm{n}}$ becomes the static two-form in [F1]. The substitution is defined separately on each component where $f\mathrm{\ne }0$; it does not extend $r_{\ast }$ through a root.

On $f>0$, ${\mathrm{\partial }}_t={\mathrm{\partial }}_v$ and $g({\mathrm{\partial }}_v,N)=-1-f\mathrm{/}2<0$, so the exterior orientations agree. On $f<0$, the static-coordinate vector is ${\mathrm{\partial }}_r{\mathrm{\mid }}_t={\mathrm{\partial }}_r{\mathrm{\mid }}_v+f^{-1}{\mathrm{\partial }}_v$, and $g({\mathrm{\partial }}_r{\mathrm{\mid }}_t,N)=1\mathrm{/}2-1\mathrm{/}f>0$; hence the static definition's ${\mathrm{\partial }}_r{\mathrm{\mid }}_t$ is past-directed under the selected ingoing orientation. This proves the stated convention boundary and makes no identification of the two interior time orientations.

Fix $t_0\in \mathbb{R}$. For each $r\in J$, let $C_r$ be the chain obtained by radially projecting the eight outward-oriented triangular faces of the octahedron with vertices $\pm e_1,\pm e_2,\pm e_3$ onto the unit sphere and then embedding that sphere as $t=t_0$, radius $r$. Each face lies in an affine plane not containing the origin, so its radial projection extends smoothly to a neighbourhood of its simplex. Order the six vertices once, parameterize each face in that order, and include it with the sign that makes its radial image outward-oriented. Each common edge is then the same ordered singular 1-simplex in its two incident faces, with opposite boundary coefficients. Thus $C_r$ is a finite smooth singular 2-cycle covering $S_r$ once with its outward orientation. The face interiors partition the sphere, and the polar area form is $\omega =\sin \theta d\theta \wedge d\varphi$; the finitely many face edges have zero 2-dimensional integral. Therefore ${\int }_{C_r}\omega ={\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi \mathrm{.}$ Pulling [F1]'s $\mathcal{F}$ back to $C_r$ removes the electric term and leaves $q_m\omega$, so the first flux is $4\pi q_m$. The inverse static metric has $g^{tt}=-1\mathrm{/}f$, $g^{rr}=f$, and the oriented volume form is $r^2\sin \theta dt\wedge dr\wedge d\theta \wedge d\varphi$. In the component formula of [F2], $(dt\wedge dr{)}^{tr}=g^{tt}{g}^{rr}=-1$ and ${\omega }^{\theta \varphi }=1\mathrm{/}(r^4\sin \theta )$. The oriented volume component is ${ϵ}_{\theta \varphi tr}=r^2\sin \theta$, so that formula gives ${\star }_g(dt\wedge dr)=-r^2\omega ,{\star }_g\omega =r^{-2}dt\wedge dr,$ for either sign of $f$. Therefore ${\star }_g\mathcal{F}=q_e\omega +q_m{r}^{-2}dt\wedge dr$; its sphere pullback integrates to $4\pi q_e$.

On $U_N=\{r+z>0\}$ and $U_S=\{r-z>0\}$, respectively, $(1-\cos \theta )d\varphi =\frac{xdy-ydx}{r(r+z)},(1+\cos \theta )d\varphi =\frac{xdy-ydx}{r(r-z)}\mathrm{.}$ The denominators are nonzero on the indicated patches, so both one-forms are smooth there. Differentiation gives $d[(1-\cos \theta )d\varphi ]=d[-(1+\cos \theta )d\varphi ]=\omega$ and $d(-q_e{r}^{-1}dt)=-q_e{r}^{-2}dt\wedge dr$. Thus both potentials differentiate to [F1]. Subtracting them gives ${\mathcal{A}}_S-{\mathcal{A}}_N=-2q_{m}d\varphi$ on the overlap.

If a global smooth one-form $\mathcal{A}$ with $d\mathcal{A}=\mathcal{F}$ existed and $q_m\mathrm{\ne }0$, apply [F3] to the oriented sphere cycle $S_r$: $4\pi q_m={\int }_{S_r}\mathcal{F}={\int }_{S_r}d\mathcal{A}={\int }_{\mathrm{\partial }{S}_r}\mathcal{A}=0,$ since $\mathrm{\partial }{S}_r=0$. This contradiction rules out a global potential when $q_m\mathrm{\ne }0$.

If $q_m=0$, the one-form $\mathcal{A}=-q_e{r}^{-1}dt$ is smooth and global on ${\mathbb{R}}_t\times J\times S^2$, and its exterior derivative is $\mathcal{F}$. This proves the exact criterion.

Multiplying $f$ by $r^2$ gives $r^{2}f=r^2-2Mr+Q^2\mathrm{.}$ If $M^2\ge Q^2$, the quadratic formula gives the stated roots. Because $0<Q^2\le M^2$ and $M>0$, the nonnegative root $s=\sqrt{M^2-Q^2}$ satisfies $s<M$: otherwise $s^2\ge M^2$, contradicting $s^2=M^2-Q^2<M^2$. Thus both $M+s$ and $M-s$ are positive. If $M^2<Q^2$, then $r^{2}f=(r-M{)}^2+(Q^2-M^2)>0$ for every $r>0$, so there is no positive root.

The inverse $v,r$ block in [F1] gives $g^{-1}(dr,dr)=g^{rr}=f$. Thus a positive root gives a null level hypersurface. The metric coefficients are independent of $v$, hence ${\mathcal{L}}_{\xi }g=0$ and $\xi$ is Killing by [F3]. At a root, ${\xi }^{\mathrm{♭}}=-fdv+dr=dr$ and $dr(\xi )=0$, so $\xi$ is null, tangent and normal to that level. The product field ${\mathrm{\partial }}_v$ is nonzero everywhere on the level, so its nonzero-generator part is the whole level, nonempty and dense. This verifies the Killing-horizon conditions in [L1].

On the root set, ${\xi }^{\mathrm{♭}}=dr$ and $d(g(\xi ,\xi ))=d(-f)=-f^{\mathrm{\prime }}(r_H)dr\mathrm{.}$ Comparing with the normalization in [L1] gives ${\kappa }_H=f^{\mathrm{\prime }}(r_H)\mathrm{/}2$. Factoring $f=(r-r_{+})(r-r_{-})\mathrm{/}{r}^2$ and differentiating at either root gives $f^{\mathrm{\prime }}(r_{\pm })=\frac{r_{\pm }-r_{\mp }}{r_{\pm }^2},$ which proves the stated formula, including its negative sign at the inner simple root. If $M^2=Q^2$, then $r_{+}=r_{-}=M$ and the formula gives zero.

If $M^2<Q^2$, step 1.1 gives $f>0$ for every $r>0$. Therefore $g^{-1}(dr,dr)>0$ and no positive constant-$r$ level is null. This rules out only the radial-root horizons under discussion, not other global horizon notions.

In the ingoing chart the $v,r$ block is $\left(\begin{array}{cc}-f& 1\\ 1& 0\end{array}\right)$, with determinant $-1$ for every $r>0$. Its inverse is $\left(\begin{array}{cc}0& 1\\ 1& f\end{array}\right)$, so the block has one positive and one negative eigenvalue. The sphere block $r^2\gamma$ is positive definite, and the two-form in [F1] is smooth. Therefore every positive root is a regular crossing without using a static coordinate at that root.

Assume $M^2>Q^2$. Put $s=\sqrt{M^2-Q^2}$. Since $Q^2>0$, $s^2=M^2-Q^2<M^2$ and $s,M\ge 0$, so $s<M$ and $r_{-}=M-s>0$. Set $r_{+}=M+s$, $d=r_{+}-r_{-}>0$, and $x=r-r_{+}$. Then $f(r)=\frac{(r-r_{+})(r-r_{-})}{r^2}=xb(r),b(r):=\frac{r-r_{-}}{r^2},2\kappa =b(r_{+})=\frac{d}{r_{+}^2}>0.$ The explicit algebraic factorization $2\kappa -b(r)=\frac{(r-r_{+})(dr-r_{+}{r}_{-})}{r_{+}^2{r}^2}$ shows that $p(r):=\frac{1}{f(r)}-\frac{1}{2\kappa (r-r_{+})}=\frac{dr-r_{+}{r}_{-}}{2\kappa r_{+}^2(r-r_{-})}$ is smooth near $r_{+}$ after shrinking the interval so $r-r_{-}\mathrm{\ne }0$. Define $h(r):={\int }_{r_{+}}^{r}p(s)ds$. By [F4], $h^{\mathrm{\prime }}=p$; repeated differentiation of this identity shows $h$ is smooth, and $h(r_{+})=0$.

Define $K(r):=\frac{r-r_{+}}{{\mathrm{\ell }}_{\ast }}{e}^{2\kappa h(r)}\mathrm{.}$ It is smooth and $K^{\mathrm{\prime }}(r_{+})={\mathrm{\ell }}_{\ast }^{-1}>0$. By [F3], it has a smooth local inverse $r=\rho (z)$ near $z=0$. After shrinking again, $K^{\mathrm{\prime }}(r)>0$ and $\frac{f(r)K^{\mathrm{\prime }}(r)}{K(r)}=2\kappa (r\mathrm{\ne }{r}_{+}),$ which follows by differentiating $K$ and using the definition of $h$. The identity extends in the equivalent form $f{K}^{\mathrm{\prime }}=2\kappa K$ to $r=r_{+}$.

For the original ingoing region define $V=e^{\kappa v}$ and $U=-K(r)e^{-\kappa v}$. The Jacobian of $(v,r)\mapsto (U,V)$ is $\kappa K^{\mathrm{\prime }}(r)>0$, so this is a smooth local coordinate change across the future crossing. Since $UV=-K(r)$, the inverse radial coordinate is $r=\rho (-UV)$. Differentiation gives $dv=\frac{dV}{\kappa V},dr=-\frac{V}{K^{\mathrm{\prime }}(r)}dU-\frac{U}{K^{\mathrm{\prime }}(r)}dV\mathrm{.}$ Substitute these in $-fd{v}^2+2dvdr$; the $d{V}^2$ coefficient cancels by $f{K}^{\mathrm{\prime }}=2\kappa K$ and $K=-UV$, leaving $-\frac{1}{\kappa K^{\mathrm{\prime }}(\rho (-UV))}(dU\otimes dV+dV\otimes dU)\mathrm{.}$ This coefficient is smooth and nonzero at $U=V=0$. The sphere term $\rho (-UV{)}^2\gamma$ is smooth and positive definite, so the radial block retains Lorentz signature throughout a small rectangle. To make this a local extension chart, choose $\epsilon >0$ so $\rho (-UV)>0$ for $(U,V)\in (-\epsilon ,\epsilon {)}^2$. The original ingoing model maps the open subset whose image satisfies $V>0$, $\mathrm{\mid }U\mathrm{\mid },\mathrm{\mid }V\mathrm{\mid }<\epsilon$ smoothly and bijectively onto that open half-rectangle, by the nonzero Jacobian above. Adjoin the full rectangle and identify that half with the old subset. A sequence in the overlap tending inside the new rectangle but outside the half-rectangle can only approach $V=0$; there $v={\kappa }^{-1}\log V$ tends to $-\mathrm{\infty }$, so it has no limit at any finite old point. The other edges lie outside the open rectangle. Thus the local gluing separates the new points from all old points, and the displayed coordinates and metric give a smooth Hausdorff local extension.

The electric two-form transforms as $-\frac{q_e}{r^2}dv\wedge dr=-\frac{q_e}{\kappa r^2{K}^{\mathrm{\prime }}(r)}dU\wedge dV,$ whose coefficient is smooth because $r_{+}>0$, $\kappa >0$, and $K^{\mathrm{\prime }}(r_{+})>0$; $q_m\omega$ is unchanged. Since $dv\wedge dr=(\kappa K^{\mathrm{\prime }}(r){)}^{-1}dU\wedge dV$ with positive coefficient, the product orientation extends as $dU\wedge dV\wedge \omega$. Thus the two-form extends smoothly over both axes. The vector field transforms to ${\mathrm{\partial }}_v=\kappa (V{\mathrm{\partial }}_V-U{\mathrm{\partial }}_U)\mathrm{.}$ The extended metric coefficients depend only on $UV$, which is invariant under this vector field; also ${\mathcal{L}}_{{\mathrm{\partial }}_v}dU=-\kappa dU$ and ${\mathcal{L}}_{{\mathrm{\partial }}_v}dV=\kappa dV$, so the radial metric tensor is invariant. By [F5] the extended field is Killing. It vanishes at $U=V=0$, where the induced sphere metric is $r_{+}^2\gamma$. Each null axis is a smooth null hypersurface: on $U=0$ its generator has covector $-V\mathrm{/}{K}^{\mathrm{\prime }}(r_{+})dU$, and on $V=0$ it has covector $U\mathrm{/}{K}^{\mathrm{\prime }}(r_{+})dV$. Thus the Killing field is tangent, null and normal, nonzero away from the common sphere; that portion is nonempty and dense on each axis. The axes meet precisely at the spacelike sphere where the extended field vanishes.

On each quadrant $UV\mathrm{\ne }0$, define $t:=\frac{1}{2\kappa }\log \mid \frac{V}{U}\mid ,r=\rho (-UV)\mathrm{.}$ These are smooth coordinates there: $r=\rho (-UV)$ and the logarithmic ratio independently determine the product and quotient of the nonzero coordinates, so their Jacobian is nonzero. Differentiating gives $dr=-\frac{VdU+UdV}{K^{\mathrm{\prime }}(r)},dt=\frac{1}{2\kappa }(\frac{dV}{V}-\frac{dU}{U}),dt\wedge dr=\frac{1}{\kappa K^{\mathrm{\prime }}(r)}dU\wedge dV\mathrm{.}$ Put $\alpha =dU\mathrm{/}U$ and $\eta =dV\mathrm{/}V$. Since $UV=-K(r)$, the displayed differentials give $dr=(K\mathrm{/}{K}^{\mathrm{\prime }})(\alpha +\eta )$ and $dt=(\eta -\alpha )\mathrm{/}(2\kappa )$. Substituting in the static radial metric $-fd{t}^2+f^{-1}d{r}^2$ of [F1], and using $f{K}^{\mathrm{\prime }}=2\kappa K$ from step 2.1, the $\alpha \otimes \alpha$ and $\eta \otimes \eta$ coefficients cancel and each cross coefficient is $f\mathrm{/}(2{\kappa }^2)=K\mathrm{/}(\kappa K^{\mathrm{\prime }})$. Therefore the radial tensor is $\frac{K}{\kappa K^{\mathrm{\prime }}}(\alpha \otimes \eta +\eta \otimes \alpha )=-\frac{1}{\kappa K^{\mathrm{\prime }}}(dU\otimes dV+dV\otimes dU),$ using $K=-UV$. The sphere term remains $r^2\gamma$. The two-form satisfies $-q_e{r}^{-2}dt\wedge dr+q_m\omega =-q_e(\kappa r^2{K}^{\mathrm{\prime }}{)}^{-1}dU\wedge dV+q_m\omega$, exactly its expression from step 4.1. Thus each quadrant is a static RN chart with $f\mathrm{\ne }0$. Since $\kappa K^{\mathrm{\prime }}(r)>0$, its orientation agrees with $dt\wedge dr\wedge \omega$. The exact static-domain interface [L1] makes the geometric Einstein and Maxwell residuals vanish on all four quadrants. The metric, field and inverse metric are smooth on the full extension chart, so the residual tensors are continuous; a continuous tensor vanishing on a dense subset vanishes on its closure. This proves the stated conditional equation extension to both axes and the bifurcation sphere.

If $M^2=Q^2$, then $r_H=M$ and $f(r)=(1-M\mathrm{/}r{)}^2$. The determinant and smooth field in [F1] remain regular there and $f^{\mathrm{\prime }}(M)=0$, so $\kappa =0$. The construction in steps 1.2-3.1 requires $\kappa >0$ and makes no extremal bifurcation-chart claim.

The sphere coordinates $X,Y,Z$ and one-form $\beta =XdY-YdX$ are smooth globally. Since $r>0$, $\mathrm{\Sigma }=r^2+a^2{Z}^2>0$; every coefficient in the displayed product tensors is therefore smooth, and ${\mathcal{F}}_{\mathrm{i}\mathrm{n}}$ is smooth by exterior differentiation. In polar charts, $\beta ={\sin }^2\theta d\psi$ and $\gamma =d{\theta }^2+{\sin }^2\theta d{\psi }^2$.

Away from the axes and roots, put $\beta ={\sin }^2\theta d\psi$ and set $dv=d{x}^0+\frac{R}{\mathrm{\Delta }}dr,d\psi =d\varphi +\frac{a}{\mathrm{\Delta }}dr\mathrm{.}$ This triangular coordinate change has Jacobian determinant $1$. Substitution in the product metric gives exactly the BL metric in [F1]. Also $dv-a{\sin }^2\theta d\psi =d{x}^0-a{\sin }^2\theta d\varphi +\frac{\mathrm{\Sigma }}{\mathrm{\Delta }}dr\mathrm{.}$ Thus ${\mathcal{A}}_{\mathrm{i}\mathrm{n}}$ differs from the BL potential by the radial one-form $-qr{\mathrm{\Delta }}^{-1}dr$, whose exterior derivative is zero because its coefficient depends only on $r$. The two field strengths agree. Finally $dv\wedge dr\wedge \omega =d{x}^0\wedge dr\wedge \omega$, so this change preserves the stated orientation.

On $0<\theta <\pi$, the determinant in coordinates $(v,r,\theta ,\psi )$ is $-{\mathrm{\Sigma }}^2{\sin }^2\theta$: it is the determinant $-{\mathrm{\Sigma }}^2$ from [F1], multiplied by $(dy\mathrm{/}d\theta {)}^2={\sin }^2\theta$; the coordinate change in step 1.2 has Jacobian $1$. This determinant is nonzero when $\mathrm{\Delta }\mathrm{\ne }0$ and extends by continuity to $\mathrm{\Delta }=0$, since the product metric coefficients are smooth by step 1.1. At a polar axis $\beta =0$ as a covector on the sphere tangent space; the radial block is $\left(\begin{array}{cc}-f_0& 1\\ 1& 0\end{array}\right)$ with determinant $-1$, and the angular block is $\mathrm{\Sigma }\gamma$, positive definite. So the metric is nondegenerate at the axes too. On a large-$r$ pole-free region it has signature $(-,+,+,+)$ by [F1] and step 1.2. A continuous nondegenerate symmetric form has locally constant signature; the product manifold is connected, so this signature holds everywhere.

The field $N$ is smooth and global by the product construction. Direct contraction gives $g_{\mathrm{i}\mathrm{n}}(N,N)=-(f_0^2+f_0+1)=-((f_0+\frac{1}{2}{)}^2+\frac{3}{4})<0.$ Thus $N$ is everywhere timelike, and by [F6] it selects the stated smooth time orientation on the whole product.

By [F3] the Einstein and Maxwell residual tensors vanish on the pole-free BL domains. The coordinate and field identities in step 1.2 carry that zero residual to the ingoing formulas on the dense set $\mathrm{\Delta }\mathrm{\ne }0$, $0<\theta <\pi$. By steps 1.1, 2.1 and 3.1, the metric and field are smooth and nondegenerate on the full product; [F4-F5] therefore make each residual tensor continuous, including at the axes and roots. A continuous tensor zero on a dense subset is zero on its closure. This proves the stated conditional equation continuation exactly when the interface [F3] is established, without extending beyond the selected ingoing model.