Fully verified electric Kerr-Newman electrovacuum

On a pole-free patch, [F1] and commutation of smooth mixed partials give $d\mathcal{F}=d^2\mathcal{A}=0$. On the full-sphere exterior, the global potential in [F5] gives the same identity everywhere.

Put $w=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$. For any smooth vector field $J$, the vector-current density identity in [F3] and the coordinate formula ${\mathrm{\nabla }}_i{J}^i={\mathrm{\partial }}_i{J}^i+{\mathrm{\Gamma }}^i{}_{ik}{J}^k$ imply ${\mathrm{\Gamma }}^i{}_{ik}=w^{-1}{\mathrm{\partial }}_{k}w$ by comparing arbitrary $J$. For the antisymmetric field, the tensor derivative expands as ${\mathrm{\nabla }}_i{\mathcal{F}}^{ij}={\mathrm{\partial }}_i{\mathcal{F}}^{ij}+{\mathrm{\Gamma }}^i{}_{ik}{\mathcal{F}}^{kj}+{\mathrm{\Gamma }}^j{}_{ik}{\mathcal{F}}^{ik}$. The last term vanishes by symmetry of the connection in $i,k$ and antisymmetry of the field. Thus ${\mathrm{\nabla }}_i{\mathcal{F}}^{ij}=w^{-1}{\mathrm{\partial }}_i(w{\mathcal{F}}^{ij})$. In the rational chart $(x^0,r,y=\cos \theta ,\varphi )$, [F2] gives $w=\mathrm{\Sigma }$, so its four zero Maxwell-density residuals give $v^j:={\mathrm{\nabla }}_i{\mathcal{F}}^{ij}=0$ there. To relate this divergence to $d({\star }_g\mathcal{F})$ without assuming the divergence equation, fix any pole-free point and use the point-normal chart supplied by the direct prerequisite Point-normal and exponential coordinate charts of the metric-selected connection, choosing its orientation positive (if necessary, reverse one spatial coordinate, which preserves the normal form). At the point, $g_{ab}={\eta }_{ab}$, ${\mathrm{\Gamma }}^a{}_{bc}=0$, and $\mathrm{\partial }{g}_{ab}=0$. Differentiating $g^{ac}{g}_{cb}={\delta }_b^a$ gives $\mathrm{\partial }{g}^{ab}=0$ there, and $w=\sqrt{\mathrm{\mid }\det (g_{ab})\mathrm{\mid }}$ gives $\mathrm{\partial }w=0$. Put $E_i:=-{\mathcal{F}}_{0i}$ and $B_i:=\frac{1}{2}{ϵ}_{ijk}{\mathcal{F}}_{jk}$ as local component abbreviations only, with ${ϵ}_{123}=+1$, and define $(\mathrm{curl}B{)}_k:={ϵ}_{kij}{\mathrm{\partial }}_i{B}_j$. The Hodge component formula in [F3] gives ${\mathcal{F}}^{0i}=E_i$, ${\mathcal{F}}^{ij}={ϵ}_{ijk}{B}_k$, $({\star }_g\mathcal{F}{)}_{0i}=B_i$, and $({\star }_g\mathcal{F}{)}_{ij}={ϵ}_{ijk}{E}_k$ at the point. Since the inverse-metric and volume derivatives vanish there, differentiating the Hodge formula contributes no extra terms. The coordinate formulas for divergence and exterior derivative therefore give $v^0=-{\mathrm{\partial }}_i{E}_i$, $v^k={\mathrm{\partial }}_0{E}_k-(\mathrm{curl}B{)}_k$, $d({\star }_g\mathcal{F}{)}_{123}={\mathrm{\partial }}_i{E}_i$, and $d({\star }_g\mathcal{F}{)}_{0ij}={ϵ}_{ijk}({\mathrm{\partial }}_0{E}_k-(\mathrm{curl}B{)}_k)$. Also $({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{123}=v^0$ and $({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{0ij}=-{ϵ}_{ijk}{v}^k$. Hence $d({\star }_g\mathcal{F})=-{\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g$ at this point, and therefore at every pole-free point. As $v=0$, $d({\star }_g\mathcal{F})=0$ on that region. In the full-sphere extension, this three-form is smooth by [F5] and vanishes on the dense pole-free set, so it vanishes at the poles by continuity.

The sixteen metric-field residuals in [F2] are exactly zero on every pole-free patch, using the metric, orientation and curvature convention in [F3] and [A2]. In the full-sphere extension both sides of the geometric Einstein equation are smooth by [F5]; equality on the dense pole-free region therefore extends to the poles by continuity. This proves $\mathrm{Ric}=2Q(\mathcal{F})$ for the length-normalized form. The tensor conversion follows from [L1] as established below.

Steps 1.1, 1.2 and 1.3 give the three equations in [F4] on every pole-free chart under [A1]. For the full-sphere exterior, [F5] supplies the smooth time-oriented Lorentzian datum, and the same equations hold at the poles by the continuity arguments in steps 1.2 and 1.3. Thus both stated domains are geometric electrovac data. No horizon or maximal-extension claim is made.

With the positive constant $\alpha$ in the Statement, linearity gives $dF={\alpha }^{-1}d\mathcal{F}=0$ and $d({\star }_{g}F)={\alpha }^{-1}d({\star }_g\mathcal{F})=0$. Trace-freeness of the quadratic tensor follows by contracting its formula: $g^{ab}{Q}_g(\mathcal{F}{)}_{ab}={\mathcal{F}}_{ac}{\mathcal{F}}^{ac}-\frac{4}{4}{\mathcal{F}}_{cd}{\mathcal{F}}^{cd}=0$. Taking the trace of the Ricci equation in step 1.3 gives $S=0$, hence $G=\mathrm{Ric}$. By [L1], $2Q_g(\mathcal{F})={\kappa }_E{\mu }_0^{-1}{Q}_g(F)={\kappa }_{E}T$, proving every normalized equation in the Statement. This is algebra for the explicitly defined tensor $T$; identifying it as a Hilbert tensor requires the separate stated action hypotheses.