Regular RN crossings and local bifurcation extension

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.