Nonharmonic conformastatic data fail the electrovac equation

The function in [A1] is smooth and strictly positive. Its derivatives are $H_i=2\lambda x^i$, $q_H=4{\lambda }^2\mathrm{\mid }x{\mathrm{\mid }}^2$, and ${\mathrm{\Delta }}_{\delta }H=6\lambda >0$. The potential ${\mathcal{A}}_{\sigma }=(H^{-1}-1)d{x}^0$ is smooth on all of $D$, and its field is $\mathcal{F}=d{\mathcal{A}}_{\sigma }$, with ${\mathcal{F}}_{i0}=-H_i\mathrm{/}{H}^2$. Since $d^2=0$, it satisfies $d\mathcal{F}=0$. The metric is Lorentzian by [F2].

In the orthonormal frame [F2], the field components satisfy $\mathcal{F}(n,e_i)=H\cdot H^{-1}{\mathcal{F}}_{0i}=H_i\mathrm{/}{H}^2$, so ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=-2q_H\mathrm{/}{H}^4$ and $Q(n,n)=q_H\mathrm{/}(2H^4)$. Also ${\mathcal{F}}^{i0}=H_i\mathrm{/}{H}^2$ and $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=H^2$ by [F1, F2], hence ${\mathrm{\partial }}_i(\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}{\mathcal{F}}^{i0})={\mathrm{\partial }}_i{H}_i={\mathrm{\Delta }}_{\delta }H=6\lambda$. Using ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha i}=2H_i\mathrm{/}H$ and ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha 0}=0$ from [F3], together with symmetry of ${\mathrm{\Gamma }}^0{}_{ac}$ and antisymmetry of ${\mathcal{F}}^{ac}$, gives ${\mathrm{\nabla }}_a{\mathcal{F}}^{a0}={\mathrm{\partial }}_i(H_i\mathrm{/}{H}^2)+(2H_i\mathrm{/}H)(H_i\mathrm{/}{H}^2)={\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^2=6\lambda \mathrm{/}{H}^2\mathrm{\ne }0$. Define the smooth vector field $v$ by $v^b:={\mathrm{\nabla }}_a{\mathcal{F}}^{ab}$. Fix auxiliary constants ${\mu }_0>0$ and $c>0$, using $t=x^0\mathrm{/}c$, and define the smooth vector field $j:=-{\mu }_0^{-1}v$. Then ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}=-{\mu }_0{j}^b$ by construction, and $d\mathcal{F}=0$ by step 1.1, so all hypotheses of [F5] hold on the smooth, boundaryless, oriented and time-oriented chart manifold from [F1, F2]. The auxiliary $j$ is used only to apply this conditional mathematical identity. By [F5], $d({\star }_g\mathcal{F})=-{\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. Contraction with the nowhere-zero volume form is injective, and the computed $v^0=6\lambda \mathrm{/}{H}^2$ is nonzero. Hence $d({\star }_g\mathcal{F})\mathrm{\ne }0$, violating the source-free Maxwell equation in [F4]. If this residual were interpreted as a matter current, that would require an additional sourced model; none is assumed here. The Ricci table [F3] gives $\mathrm{Ric}(n,n)=H^2{\mathrm{Ric}}_{00}=q_H\mathrm{/}{H}^4-{\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^3$. Therefore $\mathrm{Ric}(n,n)-2Q(n,n)=-{\mathrm{\Delta }}_{\delta }H\mathrm{/}{H}^3=-6\lambda \mathrm{/}{H}^3\mathrm{\ne }0$. This also violates the necessary Ricci equation [F4], so these smooth conformastatic data are not geometric electrovac. This counterexample uses the arbitrary-positive-$H$ supplier directly; the harmonic-only theorem is not applied outside its hypotheses.