Positive symmetric Maxwell field-strength reduction

On a sufficiently small neighborhood the positive spatial metric makes each Gram–Schmidt denominator strictly positive. Hence the adapted orthonormal frame and its inverse coframe are smooth functions of $g$; differentiating the frame introduces only $g$ and $\mathrm{\partial }g$. In that frame the six components of $dF$ and $d{\star }_{g}F$ with one normal slot are exactly the six evolution equations. On a slice with induced metric $h$ and orientation inherited from the oriented tetrad, let ${\mathrm{v}\mathrm{o}\mathrm{l}}_h$ be the corresponding Riemannian volume form. The all-spatial restrictions are $F{\mathrm{\mid }}_{\mathrm{\Sigma }}={\iota }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_h$ and $({\star }_{g}F){\mathrm{\mid }}_{\mathrm{\Sigma }}={\iota }_e{\mathrm{v}\mathrm{o}\mathrm{l}}_h$: in the oriented orthonormal tetrad their components are $F_{ij}={ϵ}_{ijk}{b}_k$ and $({\star }_{g}F{)}_{ij}={ϵ}_{ijk}{e}_k$. Here $({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_h)(X,Y):={\mathrm{v}\mathrm{o}\mathrm{l}}_h(v,X,Y)$. In oriented coordinates, ${\mathrm{v}\mathrm{o}\mathrm{l}}_h=\sqrt{\det (h_{ij})}d{x}^1\wedge d{x}^2\wedge d{x}^3$, so $d_{\mathrm{\Sigma }}({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_h)={\mathrm{\partial }}_i(\sqrt{\det h}{v}^i)d{x}^1\wedge d{x}^2\wedge d{x}^3=(D_i{v}^i){\mathrm{v}\mathrm{o}\mathrm{l}}_h$, using ${\mathrm{\Gamma }}^i{}_{ik}={\mathrm{\partial }}_k\log \sqrt{\det h}$ from the Levi-Civita formula and determinant differentiation. Thus their all-spatial components vanish exactly when the two Gauss divergences vanish.

Freeze the tetrad at a point and use its timelike vector as the time direction. To make the Hodge-to-divergence conversion explicit, set ${\epsilon }_{abcd}:=({\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{abcd}$ and raise its indices with $g^{-1}$. In a positively oriented chart ${\epsilon }_{0123}=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$; differentiating the determinant gives ${\mathrm{\Gamma }}^a{}_{ac}={\mathrm{\partial }}_c\log \sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$, and the covariant derivative formula for a top form then gives ${\mathrm{\nabla }}_a{\epsilon }_{bcde}=0$. The Hodge formula $({\star }_{g}F{)}_{cd}=\frac{1}{2}{\epsilon }_{cdef}{F}^{ef}$, torsion-freeness and metric-compatibility, and ${\epsilon }^{bacd}{\epsilon }_{cdef}=-2({\delta }_e^b{\delta }_f^a-{\delta }_f^b{\delta }_e^a)$ give the exact identity ${\mathrm{\nabla }}_a{F}^{ab}=\frac{1}{3!}{\epsilon }^{bacd}(d{\star }_{g}F{)}_{acd}=\frac{1}{\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}}{\mathrm{\partial }}_a(\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}{F}^{ab})\mathrm{.}$ For the first equality, expand $(d{\star }_{g}F{)}_{acd}=3{\mathrm{\nabla }}_{[a}({\star }_{g}F{)}_{cd]}$ and contract; the second follows from the same Christoffel trace identity and ${\mathrm{\Gamma }}^b{}_{ac}{F}^{ac}=0$. Thus $d{\star }_{g}F=0$ is equivalent to ${\mathrm{\nabla }}_a{F}^{ab}=0$, and the $(0,i,j)$ components of $dF=0$, with $F_{0i}=-e_i$ and $F_{ij}={ϵ}_{ijk}{b}_k$, give $n(b)+\mathrm{curl}e=0$, while the spatial components of this divergence equation give $n(e)-\mathrm{curl}b=0$. Hence the spatial symbol in the convention $n(U)+A^j{\mathsf{e}}_j(U)=0$ is $S(\xi )$. Since $C(\xi {)}^T=-C(\xi )$, its two off-diagonal blocks are transposes and $S(\xi {)}^T=S(\xi )$. Also $\mathrm{\mid }C(\xi )v{\mathrm{\mid }}^2=\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }v{\mathrm{\mid }}^2-(\xi \cdot v{)}^2$, so $\mathrm{\parallel }C(\xi )\mathrm{\parallel }=\mathrm{\mid }\xi \mathrm{\mid }$ and $\mathrm{\parallel }S(\xi )\mathrm{\parallel }=\mathrm{\mid }\xi \mathrm{\mid }$.

The identity $C(\xi {)}^{2}v=\xi (\xi \cdot v)-\mathrm{\mid }\xi {\mathrm{\mid }}^{2}v$ shows that $-C(\xi {)}^2$ has eigenvalues $0,\mathrm{\mid }\xi {\mathrm{\mid }}^2,\mathrm{\mid }\xi {\mathrm{\mid }}^2$. Squaring $S(\xi )$ therefore gives eigenvalues $0,0,\mathrm{\mid }\xi {\mathrm{\mid }}^2,\mathrm{\mid }\xi {\mathrm{\mid }}^2,\mathrm{\mid }\xi {\mathrm{\mid }}^2,\mathrm{\mid }\xi {\mathrm{\mid }}^2$; because $S$ is symmetric and has trace zero, its eigenvalues are $0,0,\mathrm{\mid }\xi \mathrm{\mid },\mathrm{\mid }\xi \mathrm{\mid },-\mathrm{\mid }\xi \mathrm{\mid },-\mathrm{\mid }\xi \mathrm{\mid }$. For any $z\in {\mathbb{R}}^6$, $z^{T}P(\zeta )z\ge (\zeta (n)-\mathrm{\mid }{\zeta }_{\perp }{\mathrm{\mid }}_h)\mathrm{\mid }z{\mathrm{\mid }}^2$, proving positivity on the stated future covector cone. Taking $\zeta =ds$ gives $ds(n)={\alpha }^{-1}$ and ${\zeta }_{\perp }=0$, hence the positive time matrix ${\alpha }^{-1}{I}_6$. The shift contributes only scalar multiples of $I_6$ to spatial matrices, so those matrices remain symmetric.

The frame-derivative, Hodge-coefficient, and commutator terms contain at most first derivatives of $g$ and no derivatives of $(e,b)$; they are smooth on the open state set where $g$ is Lorentzian, $\alpha >0$, and $h>0$. This proves the claimed lower-order form and completes the conditional reduction proof.