Compact charged-scalar action variations at classical finite regularity

Fix the compact support $K$ of the directions. Use [F2]'s finite smooth localization near $K$, independently of the variation parameter. In each chart the localized density difference and its parameter derivative are continuous on a fixed compact parameter-rectangle product: every density is a smooth algebraic expression in the nondegenerate metric, its derivatives through order two, and the matter fields through order one. The families in [F1] have the required joint derivatives. The one-variable fundamental theorem gives the difference quotient as the average of those parameter derivatives along the parameter segment. Uniform continuity on this compact product makes that quotient converge uniformly to the derivative at zero. Monotonicity and the absolute-value estimate [F6], compared with the constant ${\sup }_Q\mathrm{\mid }f\mathrm{\mid }$, give $\mathrm{\mid }{\int }_{Q}f\mathrm{\mid }\le \mathrm{vol}(Q){\sup }_Q\mathrm{\mid }f\mathrm{\mid }$; this passes the limit through each finite Riemann integral. Thus the relative actions can be differentiated by differentiating their densities, even when the background global actions diverge.

Let $W$ be any compactly supported $C^1$ vector field. Write $w=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$. The Christoffel formula and determinant derivative give ${\mathrm{\Gamma }}^a{}_{ab}={\mathrm{\partial }}_b\log w$, hence ${\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g={\mathrm{\partial }}_a(w{W}^a)d^{4}x$ in a positive chart. For each finite cutoff ${\chi }_i$ in [F2], $w{\chi }_i{W}^a$ extends by zero to a $C^1$ compactly supported coefficient on a bounding rectangle. Its partial derivatives and all sections are continuous, so [F3]–[F5] integrate its coordinate divergence to zero: the opposite-face values vanish. Summing over $i$ gives the same result for $W$, because ${\sum }_i{\chi }_i=1$ and ${\sum }_{i}d{\chi }_i=0$ near its support. Therefore ${\mathcal{I}}_M(({\mathrm{\nabla }}_a{W}^a){\mathrm{v}\mathrm{o}\mathrm{l}}_g)=0$ for this $C^1$ field. No smoothness approximation or unproved rough Stokes assertion is used.

Put $h_{ab}=\delta g_{ab}=-g_{ac}{g}_{bd}{q}^{cd}$. Differentiation of $g^{ac}{g}_{cb}={\delta }_b^a$ and of the determinant gives $\delta w=-\frac{1}{2}w{g}_{ab}{q}^{ab}$. Differentiating the coordinate formula for the Christoffel symbols and collecting the terms containing $h$ gives $\delta {\mathrm{\Gamma }}^a{}_{bc}=\frac{1}{2}{g}^{ad}({\mathrm{\nabla }}_b{h}_{cd}+{\mathrm{\nabla }}_c{h}_{bd}-{\mathrm{\nabla }}_d{h}_{bc})\mathrm{.}$ To differentiate Ricci, use $R_{ab}={\mathrm{\partial }}_c{\mathrm{\Gamma }}^c{}_{ab}-{\mathrm{\partial }}_b{\mathrm{\Gamma }}^c{}_{ac}+{\mathrm{\Gamma }}^c{}_{cd}{\mathrm{\Gamma }}^d{}_{ab}-{\mathrm{\Gamma }}^c{}_{bd}{\mathrm{\Gamma }}^d{}_{ac}$. The differentiated quadratic terms are exactly the connection terms that complete the two derivatives to $\delta R_{ab}={\mathrm{\nabla }}_c\delta {\mathrm{\Gamma }}^c{}_{ab}-{\mathrm{\nabla }}_b\delta {\mathrm{\Gamma }}^c{}_{ac}$. Metric compatibility in this coordinate connection then gives $\delta ((R-2\mathrm{\Lambda })w)=w(G_{ab}+\mathrm{\Lambda }{g}_{ab})q^{ab}+{\mathrm{\partial }}_a(w{V}^a),V^a=g^{bc}\delta {\mathrm{\Gamma }}^a{}_{bc}-g^{ab}\delta {\mathrm{\Gamma }}^c{}_{bc}\mathrm{.}$ All expressions exist classically: $g,h$ are $C^3$, $\delta \mathrm{\Gamma }$ is $C^2$, and $V$ is a compactly supported $C^2$ vector field. Steps 1.1–1.2 therefore yield the Einstein–Hilbert first variation $c^3(16\pi G_N{)}^{-1}{\mathcal{I}}_M((G+\mathrm{\Lambda }g{)}_{ab}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$.

At fixed metric, $\delta F_{ab}={\mathrm{\partial }}_a{\alpha }_b-{\mathrm{\partial }}_b{\alpha }_a=2{\mathrm{\nabla }}_{[a}{\alpha }_{b]}$; torsion-freeness follows directly from symmetry of the Christoffel formula. Thus $\delta (F_{ab}{F}^{ab})=4F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. The $C^1$ compact vector $W^a=F^{ab}{\alpha }_b$ satisfies ${\mathrm{\nabla }}_a{W}^a=({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b+F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. Steps 1.1–1.2 give ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=({\mu }_{0}c{)}^{-1}{\mathcal{I}}_M(({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. At fixed lower $F$, direct inverse-metric differentiation gives $\delta (F_{cd}{F}^{cd})=2F_{ac}{F}_b{}^c{q}^{ab}$. Combining it with $\delta w$ from step 2.1 gives ${\delta }_g{S}_{\mathrm{E}\mathrm{M}}=-(2c{)}^{-1}{\mathcal{I}}_M(T_{ab}^{\mathrm{E}\mathrm{M}}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. This uses only $F\in C^1$ and the declared metric class.

For $\eta =\delta \phi$, direct differentiation gives ${\delta }_{\phi }{L}_{\phi }=2\mathrm{Re}(\overline{{\mathcal{D}}_a\eta }{P}^a+V^{\mathrm{\prime }}\bar{\eta }\phi )$. The product rule is ${\mathrm{\nabla }}_a(\bar{\eta }{P}^a)=\overline{{\mathcal{D}}_a\eta }{P}^a+\bar{\eta }{\mathcal{D}}_a{P}^a$: the opposite gauge signs cancel. Its real and imaginary parts are $C^1$ compact vector fields. Integrating their divergence by step 1.2 yields ${\delta }_{\phi }{S}_{\phi }=(K_{\phi }\mathrm{/}c)\mathrm{Re}{\mathcal{I}}_M(\bar{\eta }{E}_{\phi }{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. At fixed $g,\phi$, ${\delta }_A{P}_a=-i\lambda \phi {\alpha }_a$, so ${\delta }_A{L}_{\phi }=-2\lambda \mathrm{Im}(\bar{\phi }{P}^a){\alpha }_a$, giving the claimed $j$. At fixed lower $A$ and fixed $\phi$, ${\delta }_g{L}_{\phi }=\mathrm{Re}(\overline{P_a}{P}_b)q^{ab}$; the volume term then gives the displayed $T^{\phi }$. These calculations require no derivative beyond $\phi \in C^2$ and $A\in C^2$.

Add steps 2.1, 3.1 and 3.2. The equality $(2c{)}^{-1}=[c^3\mathrm{/}(16\pi G_N)]{\kappa }_E$ gives exactly the total first variation in the Statement, with its three independent directions. Its coefficients are continuous; the metric coefficient is symmetric. To see that testing detects a nonzero continuous component, take a positive chart where that component has one strict sign. A nonnegative smooth bump supported inside that chart, positive on a smaller closed rectangle, has a strictly positive signed pairing with the component times $w>0$. Such a bump can be taken as $e^{-1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2)}$ inside a coordinate ball and zero outside. The resulting integral is nonzero: on the smaller rectangle the continuous sign-adjusted coefficient has a positive lower bound $a>0$, so its integral is at least $a$ times that rectangle volume by [F6]. Cutting a containing rectangle along the smaller rectangle faces and using coordinate-slice additivity in [F6] shows the full integral is at least that positive value, since the integrand is nonnegative on the remainder. A one-form test isolates any connection component; diagonal and symmetrized off-diagonal metric tests isolate every symmetric component. Scalar tests $\eta =f$ and $\eta =if$ isolate the real and imaginary parts of $E_{\phi }$. Therefore stationarity for all these tests forces all three coefficients to vanish pointwise. Conversely their pointwise vanishing makes every first variation zero. The identity $dF=0$ in [F1] supplies the remaining homogeneous equation.

The charged product rule gives ${\mathrm{\nabla }}_a{j}^a=K_{\phi }\lambda \mathrm{Im}(\overline{P_a}{P}^a+\bar{\phi }{\mathcal{D}}_a{P}^a)$. The first contraction is real and $V^{\mathrm{\prime }}$ is real, giving the stated current identity. Differentiating $T^{\phi }$ and using ${\mathrm{\partial }}_b{L}_{\phi }=2\mathrm{Re}(\overline{P^a}{\mathcal{D}}_b{P}_a)+2V^{\mathrm{\prime }}\mathrm{Re}(\bar{\phi }{P}_b)$ gives ${\mathrm{\nabla }}_a(T^{\phi }{)}^a{}_b=K_{\phi }\mathrm{Re}(\overline{E_{\phi }}{P}_b+\overline{P^a}({\mathcal{D}}_a{P}_b-{\mathcal{D}}_b{P}_a))\mathrm{.}$ The symmetric Christoffel terms cancel in the difference. Expanding $P=\mathrm{\partial }\phi -i\lambda A\phi$ and commuting the continuous second scalar partial derivatives gives ${\mathcal{D}}_a{P}_b-{\mathcal{D}}_b{P}_a=-i\lambda F_{ab}\phi$. Its contribution to the real part is $-K_{\phi }\lambda F_{ab}\mathrm{Im}(\bar{\phi }{P}^a)=F_{ba}{j}^a$. This proves the two scalar off-shell identities using only first derivatives of $P$ and $C^2$ scalar partials.

Since $F=dA$ with $A\in C^2$, its homogeneous identity is ${\mathrm{\nabla }}_a{F}_{bc}+{\mathrm{\nabla }}_b{F}_{ca}+{\mathrm{\nabla }}_c{F}_{ab}=0$. Contraction with $F^{ac}$ gives $2F^{ac}{\mathrm{\nabla }}_a{F}_{bc}=\frac{1}{2}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})$. The product rule in the explicit Maxwell tensor therefore yields ${\mathrm{\nabla }}_a(T^{\mathrm{E}\mathrm{M}}{)}^a{}_b={\mu }_0^{-1}[({\mathrm{\nabla }}_a{F}^{ac})F_{bc}+F^{ac}{\mathrm{\nabla }}_a{F}_{bc}-\frac{1}{4}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})]={\mu }_0^{-1}{F}_{ba}{\mathrm{\nabla }}_c{F}^{ca}\mathrm{.}$ These are continuous identities for $F\in C^1$. If the sourced Maxwell equation holds, the last expression is $-F_{ba}{j}^a$, which cancels the scalar force term in step 4.2 when $E_{\phi }=0$. This proves total-stress conservation and current conservation before any Einstein equation. No smooth-only stress or Noether theorem is invoked at the classical regularity.