SI and geometrized Einstein–Maxwell parameters

Contracting the displayed definition of $\mathcal{Q}$ gives $g^{ab}{\mathcal{Q}}_{ab}=F_{ac}{F}^{ac}-\frac{1}{4}(g^{ab}{g}_{ab})F_{cd}{F}^{cd}=F_{cd}{F}^{cd}-F_{cd}{F}^{cd}=0$, because the metric dimension is four. Thus $\mathcal{Q}$ is trace-free.

Each term of $\mathcal{Q}(g,F)$ contains exactly two factors linear in $F$, so for every scalar $s$, $\mathcal{Q}(g,sF)=s^2\mathcal{Q}(g,F)$. In particular, $\mathcal{Q}(g,\mathcal{F})={\alpha }^2\mathcal{Q}(g,F)$.

Since $2{\alpha }^2=\kappa \mathrm{/}{\mu }_0$ by [F1], the right-hand side $(\kappa \mathrm{/}{\mu }_0)\mathcal{Q}(g,F)$ equals $2\mathcal{Q}(g,\mathcal{F})$; replacing it in the first equation proves equivalence of the first two equations in both directions.

By [F2], tracing the second equation gives $-S+4\mathrm{\Lambda }=0$, since $g^{ab}{G}_{ab}=S-\frac{1}{2}(4)S=-S$ and the right-hand side is trace-free by step 1.1. Hence $S=4\mathrm{\Lambda }$.

Substituting $S=4\mathrm{\Lambda }$ into $G_{ab}+\mathrm{\Lambda }{g}_{ab}$ gives ${\mathrm{Ric}}_{ab}-2\mathrm{\Lambda }{g}_{ab}+\mathrm{\Lambda }{g}_{ab}={\mathrm{Ric}}_{ab}-\mathrm{\Lambda }{g}_{ab}$, proving the second equation implies the third. Conversely, tracing the third equation and using step 1.1 gives $S-4\mathrm{\Lambda }=0$; substituting this into $G_{ab}$ recovers the second equation.

In dimension four, $\det ({\tilde{g}}_{ab})={\mathrm{\Omega }}^8\det (g_{ab})$. Since $\mathrm{\Omega }>0$, [F2] therefore gives ${\mathrm{vol}}_{\tilde{g}}={\mathrm{\Omega }}^4{\mathrm{vol}}_g$. By [F1], ${\star }_{\tilde{g}}F={\star }_{g}F$.

If $dF=0$, then the same two-form $F$ still obeys $dF=0$ for $\tilde{g}$, because the exterior derivative does not depend on the metric.

The lower tensor satisfies ${\tilde{F}}_b{}^c={\mathrm{\Omega }}^{-2}{F}_b{}^c$ and ${\tilde{F}}_{cd}{\tilde{F}}^{cd}={\mathrm{\Omega }}^{-4}{F}_{cd}{F}^{cd}$. Substitution in its definition gives ${\tilde{\mathcal{Q}}}_{ab}={\mathrm{\Omega }}^{-2}{\mathcal{Q}}_{ab}$, hence ${\tilde{T}}_{ab}={\mathrm{\Omega }}^{-2}{T}_{ab}$. Raising both indices with ${\tilde{g}}^{-1}={\mathrm{\Omega }}^{-2}{g}^{-1}$ gives ${\tilde{T}}^{ab}={\mathrm{\Omega }}^{-6}{T}^{ab}$.

Using step 1.1 and linearity of contraction in its vector argument, ${\iota }_{\tilde{j}}{\mathrm{vol}}_{\tilde{g}}={\iota }_{{\mathrm{\Omega }}^{-4}j}({\mathrm{\Omega }}^4{\mathrm{vol}}_g)={\iota }_j{\mathrm{vol}}_g$. Thus $d({\star }_{\tilde{g}}F)=d({\star }_{g}F)={\mu }_0{\iota }_j{\mathrm{vol}}_g={\mu }_0{\iota }_{\tilde{j}}{\mathrm{vol}}_{\tilde{g}}$ whenever the original sourced equation holds.

Since $\tilde{g}({\mathrm{\Omega }}^{-1}n,{\mathrm{\Omega }}^{-1}n)=g(n,n)=-1$, the rescaled vector is unit timelike. Contracting the lower-tensor identity twice with $\tilde{n}={\mathrm{\Omega }}^{-1}n$ yields $\tilde{T}(\tilde{n},\tilde{n})={\mathrm{\Omega }}^{-4}T(n,n)$.

Conversely, $\tilde{j}={\mathrm{\Omega }}^{-4}j$ implies $j={\mathrm{\Omega }}^4\tilde{j}$. Reversing the contraction equality in step 2.1 and using ${\star }_{g}F={\star }_{\tilde{g}}F$ gives the original sourced equation; the homogeneous equation is unchanged by step 1.2.

Raising the two indices of $F$ contributes ${\mathrm{\Omega }}^{-4}$ and the volume form contributes ${\mathrm{\Omega }}^4$, so $F_{ab}{F}^{ab}{\mathrm{vol}}_g$ is invariant pointwise. Integration over the same domain preserves equality whenever defined, as do compactly supported action differences.

In a chart, ${\mathrm{\partial }}_a{\tilde{g}}_{bc}={\mathrm{\Omega }}^2({\mathrm{\partial }}_a{g}_{bc}+2u_a{g}_{bc})$ and ${\tilde{g}}^{ad}={\mathrm{\Omega }}^{-2}{g}^{ad}$. Substituting these into [F3] cancels the common ${\mathrm{\Omega }}^2$ and gives ${\tilde{\mathrm{\Gamma }}}^a{}_{bc}-{\mathrm{\Gamma }}^a{}_{bc}=C^a{}_{bc}:={\delta }_b^a{u}_c+{\delta }_c^a{u}_b-g_{bc}{u}^a$, where $u^a=g^{ab}{u}_b$. Iterating the derivative formula in [F6] gives $D^m(r^{-1})=(-1{)}^{m}m!r^{-m-1}$ for every integer $m\ge 0$ and $r>0$; all these derivatives are continuous, so $r\mapsto r^{-1}$ is smooth on $(0,\mathrm{\infty })$. Its composition with $\mathrm{\Omega }$ is smooth by [F7], and $d\mathrm{\Omega }$ is a smooth one-form by [F5], so $u$ and the displayed components $C^a{}_{bc}$ are smooth. Finally, define $A(X,Y):={\tilde{\mathrm{\nabla }}}_{X}Y-{\mathrm{\nabla }}_{X}Y$ for smooth local vector fields. The affine-connection rules in [F3] give $A(fX,Y)=fA(X,Y)$ and $A(X,fY)=fA(X,Y)$ for every smooth function $f$, while its components are the smooth differences $C^a{}_{bc}$; hence $A$ is a smooth tensor of type $(1,2)$ with those components.

By [F5], $0=d(1)=d({\mathrm{\Omega }}^{-1}\mathrm{\Omega })={\mathrm{\Omega }}^{-1}d\mathrm{\Omega }+\mathrm{\Omega }d({\mathrm{\Omega }}^{-1})$, so $d({\mathrm{\Omega }}^{-1})=-{\mathrm{\Omega }}^{-2}d\mathrm{\Omega }$. Therefore $du=d({\mathrm{\Omega }}^{-1})\wedge d\mathrm{\Omega }+{\mathrm{\Omega }}^{-1}{d}^2\mathrm{\Omega }=-{\mathrm{\Omega }}^{-2}d\mathrm{\Omega }\wedge d\mathrm{\Omega }=0$. Torsion freedom in [F3] gives $(du{)}_{ab}={\mathrm{\nabla }}_a{u}_b-{\mathrm{\nabla }}_b{u}_a$, hence $H_{ab}=H_{ba}$.

Apply the coordinate curvature formula [F8] to $g$ and $\tilde{g}$ and subtract, using $\tilde{\mathrm{\Gamma }}=\mathrm{\Gamma }+C$ from step 1.1. Since $C$ is a tensor, [F4] gives $({\mathrm{\nabla }}_{c}C{)}^a{}_{db}-({\mathrm{\nabla }}_{d}C{)}^a{}_{cb}={\mathrm{\partial }}_c{C}^a{}_{db}-{\mathrm{\partial }}_d{C}^a{}_{cb}+{\mathrm{\Gamma }}^a{}_{ce}{C}^e{}_{db}-{\mathrm{\Gamma }}^a{}_{de}{C}^e{}_{cb}+{\mathrm{\Gamma }}^e{}_{db}{C}^a{}_{ce}-{\mathrm{\Gamma }}^e{}_{cb}{C}^a{}_{de}$; the two covariant-derivative terms acting on the lower curvature-direction slots $d,c$ cancel because ${\mathrm{\Gamma }}^e{}_{cd}={\mathrm{\Gamma }}^e{}_{dc}$, while the lower $b$-slot contributions are the last two displayed terms. These are exactly the partial-derivative and mixed $\mathrm{\Gamma }C$ terms in the subtraction, and the remaining terms are $C^a{}_{ce}{C}^e{}_{db}-C^a{}_{de}{C}^e{}_{cb}$. Therefore ${\tilde{R}}^a{}_{bcd}-R^a{}_{bcd}={\mathrm{\nabla }}_c{C}^a{}_{db}-{\mathrm{\nabla }}_d{C}^a{}_{cb}+C^a{}_{ce}{C}^e{}_{db}-C^a{}_{de}{C}^e{}_{cb}$.

Contracting $a$ with $c$ as in [F2], the derivative terms are ${\mathrm{\nabla }}_a{C}^a{}_{db}-{\mathrm{\nabla }}_d{C}^a{}_{ab}$. The definition of $C$ gives $C^a{}_{ab}=4u_b$ and, using $\mathrm{\nabla }g=0$ from [F4], ${\mathrm{\nabla }}_a{C}^a{}_{db}=H_{db}+H_{bd}-g_{db}\theta =2H_{bd}-g_{db}\theta$ by step 2.1. Thus their difference is $-2H_{bd}-g_{bd}\theta$.

The first quadratic contraction is $C^a{}_{ae}{C}^e{}_{db}=4u_e({\delta }_d^e{u}_b+{\delta }_b^e{u}_d-g_{db}{u}^e)=8u_b{u}_d-4g_{bd}s$. Expanding the second contraction into its nine terms gives $C^a{}_{de}{C}^e{}_{ab}=u_d{u}_b+u_d{u}_b-g_{db}s+4u_d{u}_b+u_d{u}_b-u_d{u}_b-u_d{u}_b-g_{db}s+u_d{u}_b=6u_b{u}_d-2g_{bd}s$. Their difference is $2u_b{u}_d-2g_{bd}s$. Combining this with step 3.1 proves the Ricci transformation in the Statement.

Contracting the Ricci transformation with ${\tilde{g}}^{ab}={\mathrm{\Omega }}^{-2}{g}^{ab}$ gives $\tilde{S}={\mathrm{\Omega }}^{-2}(S-2\theta -4\theta +2s-8s)={\mathrm{\Omega }}^{-2}(S-6\theta -6s)$.

By definition, ${\tilde{G}}_{ab}={\stackrel{}{\mathrm{Ric}}}_{ab}-\frac{1}{2}\tilde{S}{\tilde{g}}_{ab}$. Substituting the Ricci formula from step 4.1 and $\tilde{S}{\tilde{g}}_{ab}=(S-6\theta -6s)g_{ab}$ from step 5.1 cancels the scalar-curvature terms and leaves $G_{ab}-2H_{ab}+2u_a{u}_b+2g_{ab}\theta +g_{ab}s$. This is the claimed Einstein-tensor formula.

Since $\lambda$ is constant, $(g^{\mathrm{\prime }}{)}^{ab}={\lambda }^{-2}{g}^{ab}$ and ${\mathrm{\partial }}_c{g}_{ab}^{\mathrm{\prime }}={\lambda }^2{\mathrm{\partial }}_c{g}_{ab}$. Substitution in the Christoffel formula [F3] gives ${\mathrm{\Gamma }}^{\mathrm{\prime }}[g^{\mathrm{\prime }}]{}^a{}_{bc}=\mathrm{\Gamma }[g{]}^a{}_{bc}$; hence the two curvature operators and their lower-index Ricci tensors agree by [F3]. Tracing with $(g^{\mathrm{\prime }}{)}^{-1}$ gives $S[g^{\mathrm{\prime }}]={\lambda }^{-2}S[g]$, so $G_{ab}(g^{\mathrm{\prime }})=G_{ab}(g)$.

In the definition of $\mathcal{Q}$, $F_{ac}^{\mathrm{\prime }}=\lambda F_{ac}$ while $F_b^{\mathrm{\prime }}{}^c=(g^{\mathrm{\prime }}{)}^{cd}{F}_{bd}^{\mathrm{\prime }}={\lambda }^{-1}{F}_b{}^c$, so the first quadratic term is unchanged. Also $F_{cd}^{\mathrm{\prime }}{F}^{\mathrm{\prime }cd}={\lambda }^{-2}{F}_{cd}{F}^{cd}$ and $g_{ab}^{\mathrm{\prime }}={\lambda }^2{g}_{ab}$, so the trace term is unchanged. Therefore ${\mathcal{Q}}_{ab}(g^{\mathrm{\prime }},F^{\mathrm{\prime }})={\mathcal{Q}}_{ab}(g,F)$.

We have ${\mathrm{\Lambda }}^{\mathrm{\prime }}{g}_{ab}^{\mathrm{\prime }}={\lambda }^{-2}\mathrm{\Lambda }{\lambda }^2{g}_{ab}=\mathrm{\Lambda }{g}_{ab}$. Together with steps 1.1 and 1.2, both sides of the Einstein equation are unchanged. Equivalently, ${\mathcal{F}}^{\mathrm{\prime }}=\alpha F^{\mathrm{\prime }}=\lambda \mathcal{F}$ and the normalized equation in [F1] has the same quadratic tensor under $(g,\mathcal{F})\mapsto (g^{\mathrm{\prime }},{\mathcal{F}}^{\mathrm{\prime }})$.

Apply [F2] with the constant conformal factor $\mathrm{\Omega }=\lambda$: $(g,F,j)$ maps to $(g^{\mathrm{\prime }},F,{\lambda }^{-4}j)$. The differential-form Maxwell equations are linear in the pair $(F,j)$, so multiplying both entries of this solution by $\lambda$ gives $(g^{\mathrm{\prime }},F^{\mathrm{\prime }},j^{\mathrm{\prime }})=(g^{\mathrm{\prime }},\lambda F,{\lambda }^{-3}j)$, which also solves them. This proves the sourced Maxwell transformation; when $j=0$, the transformed source remains zero.

Exterior differentiation and the Hodge map are real-linear by [F3], so $d{F}_{\epsilon }=\epsilon d{F}_1=0$ and $d({\star }_g{F}_{\epsilon })=\epsilon d({\star }_g{F}_1)=0$. Thus the source-free Maxwell equations remain exactly satisfied on the unchanged $g$.

Each term of $\mathcal{Q}$ is quadratic in $F$, so $\mathcal{Q}(g,F_{\epsilon })={\epsilon }^2\mathcal{Q}(g,F_1)$ and the associated lower tensor is ${\epsilon }^2{T}^{(1)}$ by the definition in the Statement.

Substituting the assumed vacuum equation $G(g)+\mathrm{\Lambda }g=0$ and step 1.2 into the residual definition gives $\mathcal{R}(g,F_{\epsilon };\mathrm{\Lambda })=-\kappa {\epsilon }^2{T}^{(1)}$ exactly.

If $K=\mathrm{\varnothing }$, the stated convention makes both norms in the claimed equality zero. Suppose $K\mathrm{\ne }\mathrm{\varnothing }$. Each component $f=S_{ab}{\mathrm{\mid }}_K$ is continuous; the relative open sets $f^{-1}((-n,n))$, for positive integers $n$, cover $K$. Compactness gives a finite subcover, and the largest of its indices bounds $\mathrm{\mid }f\mathrm{\mid }$ on $K$. Thus every component supremum, and hence their finite maximum, is finite. The norm is homogeneous, so step 2.1 and $\kappa {\epsilon }^2\ge 0$ give $\mathrm{\parallel }\mathcal{R}{\mathrm{\parallel }}_K=\kappa {\epsilon }^2\mathrm{\parallel }{T}^{(1)}{\mathrm{\parallel }}_K$. If $\epsilon \mathrm{\ne }0$ and $T^{(1)}$ is not identically zero on $K$, then $K$ is nonempty and some component of $T^{(1)}$ is nonzero at a point of $K$; step 2.1 makes the residual nonzero there, so the fixed pair fails the coupled equation on $K$. No estimate comparing it with a different metric follows from this identity.