Sourced Maxwell sign and current conservation

The quotient map is open: for any open $U\subseteq {\mathbb{R}}^3$, $q^{-1}(q(U))={\bigcup }_{\lambda \in \mathrm{\Lambda }}(U+\lambda )$ is open, so $q(U)$ is open by the quotient topology. For any $x\in {\mathbb{R}}^3$, choose $0<r<\frac{1}{2}{\min }_i{L}_i$ and let $B=B(x,r)$ be the Euclidean ball. If $y,z\in B$ and $q(y)=q(z)$, then $y-z\in \mathrm{\Lambda }$; each coordinate has absolute value less than $2r<L_i$, so each lattice coordinate is zero and $y=z$. Thus $q{\mathrm{\mid }}_B$ is a continuous open bijection from $B$ to the open set $q(B)$, hence a homeomorphism by [F1]. These neighborhoods give local Euclidean charts.

Every class in ${\mathbb{T}}_L^3$ has a representative in the compact box $C:={\prod }_i[0,L_i]$, since in each coordinate an integer multiple of $L_i$ can be subtracted to leave a value in $[0,L_i)$. Hence $q(C)={\mathbb{T}}_L^3$. The quotient map is continuous by [F1], so [F4] makes its image compact.

If $q(x)\mathrm{\ne }q(y)$, then $x-y\mathrm{\notin }\mathrm{\Lambda }$. Among the lattice vectors with $\mathrm{\mid }x-y-\lambda \mathrm{\mid }\le 1$, each integer coordinate is bounded by $(\mathrm{\mid }x-y\mathrm{\mid }+1)\mathrm{/}{L}_i$, so there are finitely many. Their distances from $x-y$ are all positive, and every other distance exceeds $1$; therefore $\delta :={\inf }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }x-y-\lambda \mathrm{\mid }>0$. Choose $0<ϵ<\delta \mathrm{/}3$. If $q(B(x,ϵ))$ and $q(B(y,ϵ))$ met, some $x^{\mathrm{\prime }}\in B(x,ϵ)$ and $y^{\mathrm{\prime }}\in B(y,ϵ)$ would satisfy $x^{\mathrm{\prime }}-y^{\mathrm{\prime }}\in \mathrm{\Lambda }$, implying $\mathrm{\mid }x-y-(x^{\mathrm{\prime }}-y^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }y-y^{\mathrm{\prime }}\mathrm{\mid }<2ϵ<{\inf }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }x-y-\lambda \mathrm{\mid }$, a contradiction. These quotient neighborhoods are disjoint, so the torus is Hausdorff. Let $\mathcal{B}$ be the countable rational-box base of [F3]. Since $q$ is open by step 1.1, $\{q(B):B\in \mathcal{B}\}$ is a countable base on ${\mathbb{T}}_L^3$: if $p\in V$ with $V$ open, choose $x\in q^{-1}(p)$ and a rational box $B$ with $x\in B\subseteq q^{-1}(V)$; then $p\in q(B)\subseteq V$. Together with step 1.1 this proves the topological three-manifold conditions in [F2].

The charts $q{\mathrm{\mid }}_B^{-1}$ from step 1.1 form a smooth atlas. In these charts the local representative of $q$ is the identity, so $q$ is a smooth local diffeomorphism. On an overlap, fix lifts $x,y$ of one point with $y-x=\lambda \in \mathrm{\Lambda }$. The distance from $\lambda$ to any other lattice vector is positive because each lattice spacing $L_i$ is positive. Restrict both chart neighborhoods so their lifts are within less than one quarter of that distance from $x$ and $y$. Any pair of lifts of one point in this restricted overlap must then differ by the same $\lambda$; otherwise the triangle inequality would put two distinct lattice vectors less than half their positive separation apart. Thus the transition is locally a fixed translation, smooth by [F5]. The atlas generates the smooth structure of [F2]. Since every chart has open image in ${\mathbb{R}}^3$, there is no boundary. The translation transitions have derivative the identity, so the local one-forms $d{x}^i$ and vector fields ${\mathrm{\partial }}_i$ agree on overlaps and define global smooth fields. A smooth $\mathrm{\Lambda }$-periodic function is constant on quotient fibres; in each chart its representative is a translate of the original smooth function, so it descends smoothly. The product is Hausdorff: $\mathbb{R}$ is Hausdorff because distinct real coordinates have disjoint intervals of radius less than half their distance, and ${\mathbb{T}}_L^3$ is Hausdorff by step 2.1. For distinct product points, separate a differing coordinate by disjoint open neighborhoods and take their products with the full other factor. It is second-countable because the products of a countable rational-interval base of $\mathbb{R}$ and the countable base of step 2.1 form a countable base. Product chart images are open in ${\mathbb{R}}^4$, so they give a boundaryless topological four-manifold. Their transitions are products of identity maps and the smooth translations just proved, hence form a smooth structure. In these charts the displayed $g,N,o$ have the same constant components, so they patch globally; $g$ has signature $(-,+,+,+)$, $N$ is timelike, and $o$ never vanishes. The metric formula [F6] then gives ${\mathrm{\Gamma }}^a{}_{bc}=0$ in every product chart. This proves all claims.

Fix $p\in M$. By the local tetrad theorem in [F3], choose a future-directed orthonormal tetrad at $p$; if necessary reverse one spatial leg to make the basis positively oriented relative to $o$. This leaves its orthonormality and future-directed timelike leg unchanged. Apply the point-normal exponential-chart theorem of [F3] to this basis. In this chart the inverse metric and volume coefficients have zero first derivatives at $p$: differentiating $g^{ac}{g}_{cb}={\delta }^a{}_b$ shows $\mathrm{\partial }{g}^{ab}(p)=0$, and differentiating $w=\sqrt{\mathrm{\mid }\det (g_{ab})\mathrm{\mid }}$ gives $\mathrm{\partial }w(p)=0$ when $\mathrm{\partial }{g}_{ab}(p)=0$. Thus the first derivatives of the coefficient map in the Hodge formula [F2] vanish at $p$. Define the local component functions $E_i=-c{F}_{0i}$ and $B_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$. Raising indices with $\eta$ and using [F2] gives $F^{0i}=E_i\mathrm{/}c$, $F^{i0}=-E_i\mathrm{/}c$, $F^{ij}={ϵ}_{ijk}{B}_k$, $({\star }_{g}F{)}_{0i}=B_i$, and $({\star }_{g}F{)}_{ij}={ϵ}_{ijk}{E}_k\mathrm{/}c$ at $p$.

Since $\mathrm{\Gamma }(p)=0$, the divergence components at $p$ are $v^0={\mathrm{\partial }}_i{F}^{i0}=-\mathrm{div}E\mathrm{/}c$ and $v^k={\mathrm{\partial }}_0{F}^{0k}+{\mathrm{\partial }}_i{F}^{ik}=c^{-2}{\mathrm{\partial }}_t{E}_k-(\mathrm{curl}B{)}_k$. The coordinate formula for $d$ and the vanishing first derivatives of the Hodge coefficients in step 1.1 give $(d{\star }_{g}F{)}_{123}={\mathrm{\partial }}_i{E}_i\mathrm{/}c$ and $(d{\star }_{g}F{)}_{0ij}={ϵ}_{ijk}(c^{-2}{\mathrm{\partial }}_t{E}_k-(\mathrm{curl}B{)}_k)$. In the same oriented chart, $({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{123}=v^0$ and $({\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{0ij}=-{ϵ}_{ijk}{v}^k$. These four independent components prove $d({\star }_{g}F)=-{\iota }_v{\mathrm{v}\mathrm{o}\mathrm{l}}_g$ at $p$. Since $p$ was arbitrary and both sides are smooth three-forms, the identity holds on $M$.

Because $\mathrm{\nabla }$ is torsion-free by [F1], its symmetric connection coefficients cancel under total antisymmetrization, so $(dF{)}_{abc}=3{\mathrm{\partial }}_{[a}{F}_{bc]}=3{\mathrm{\nabla }}_{[a}{F}_{bc]}$. In the normal chart of step 1.1, substitution of $F_{0i}=-E_i\mathrm{/}c$ and $F_{ij}={ϵ}_{ijk}{B}_k$ gives $(dF{)}_{123}=\mathrm{div}B$ and $(dF{)}_{0ij}={ϵ}_{ijk}({\mathrm{\partial }}_t{B}_k+(\mathrm{curl}E{)}_k)\mathrm{/}c$. These are pointwise component identities for the local chart functions; they do not make $E_i,B_i$ global fields on $M$.

Substituting $v^b={\mathrm{\nabla }}_a{F}^{ab}=-{\mu }_0{j}^b$ into step 2.1 yields $d({\star }_{g}F)=-{\iota }_{-{\mu }_{0}j}{\mathrm{v}\mathrm{o}\mathrm{l}}_g={\mu }_0{\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. Conversely, if this form equation holds, step 2.1 gives ${\iota }_{v+{\mu }_{0}j}{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. At each point contraction with a nonvanishing four-form is an isomorphism from the four-dimensional tangent space to the space of three-covectors, so $v+{\mu }_{0}j=0$. This proves the equivalence in the Statement.

Apply $d$ to the form equation in step 3.1. By [F2], $d^2({\star }_{g}F)=0$; since ${\mu }_0$ is a nonzero constant, $d({\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g)=0$. The exact identity in [F4] then gives $({\mathrm{\nabla }}_a{j}^a){\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. The volume form is nowhere zero, hence ${\mathrm{\nabla }}_a{j}^a=0$ throughout $M$. This uses the sourced equation only and does not invoke $dF=0$.

Under the additional flat-chart hypothesis [A2], the normal-coordinate component identities hold at every point in the same Cartesian chart. The time component of $v=-{\mu }_{0}j$ is $-\mathrm{div}E\mathrm{/}c=-{\mu }_{0}c\rho$, giving $\mathrm{div}E=\rho \mathrm{/}{ϵ}_0$. Its spatial components are $c^{-2}{\mathrm{\partial }}_{t}E-\mathrm{curl}B=-{\mu }_{0}J$, or $\mathrm{curl}B-c^{-2}{\mathrm{\partial }}_{t}E={\mu }_{0}J$. The two formulas from step 2.2 give $\mathrm{div}B=0$ and ${\mathrm{\partial }}_{t}B+\mathrm{curl}E=0$. This is the stated flat-coordinate reduction.

Interpret a compactly supported variation as a smooth family equal to its background off one fixed compact set $K$ for all sufficiently small parameters. Connection variations $A+s\alpha$ have $K=\mathrm{supp}\alpha$ and $F(s)=F+sd\alpha$ in the fixed sector. For a prescribed metric direction $q$, put $h_{ab}=-g_{ac}{g}_{bd}{q}^{cd}$ and $g(s)=g+sh$. If $K=\mathrm{\varnothing }$, then $h=0$ and the metric is unchanged. Otherwise, to choose one parameter interval valid on all of $K$, fix $p\in K$ and make a constant linear coordinate change so that $g(p)$ has matrix $\eta =\mathrm{diag}(-1,1,1,1)$. By continuity choose a relatively compact chart neighborhood on which $\mathrm{\parallel }g(x)-\eta {\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1\mathrm{/}2$ and $\mathrm{\parallel }h(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ is bounded. Choose ${\delta }_p>0$ so $\mathrm{\mid }s\mathrm{\mid }\mathrm{\parallel }h(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1\mathrm{/}2$ there for $\mathrm{\mid }s\mathrm{\mid }<{\delta }_p$. Then $A(x,s):=g(x)+sh(x)$ is positive on the fixed three-dimensional spatial coordinate subspace and negative on the time coordinate line. It is nondegenerate: if $A(x,s)v=0$, then $v=-{\eta }^{-1}(A(x,s)-\eta )v$, contradicting $\mathrm{\parallel }{\eta }^{-1}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$ and $\mathrm{\parallel }A(x,s)-\eta {\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1$. Thus its inertia has three positive and one negative directions, exactly $(-,+,+,+)$. Such neighborhoods cover $K$; compactness gives a finite subcover and the minimum of their parameter bounds works on all of $K$. Outside $K$, $h=0$ and the metric is unchanged. Thus the action-density difference $\omega (s)-\omega (0)$ is a smooth top form supported in a common $K$, even when neither background action is finite. Its integral in [F12] is the relative action; additivity of integrals makes successive relative differences consistent. No subtraction of infinite action values is involved.

Fix the chart cover and partition $({\rho }_i)$ of [F12] independently of $s$. Local finiteness gives a neighborhood of each point meeting only finitely many supports; a finite such neighborhood cover of $K$ shows that only finitely many ${\rho }_i$ meet $K$. Each ${\rho }_i(\omega (s)-\omega (0))$ has compact support inside its chart. Extend its coefficient by zero and enclose that support in a coordinate rectangle. On a closed parameter interval about zero the coefficient derivative is uniformly continuous on the product rectangle; the mean-value argument in [F3] makes its difference quotients converge uniformly. The estimate $\mathrm{\mid }{\int }_{Q}u\mathrm{\mid }\le \mathrm{vol}(Q){\sup }_Q\mathrm{\mid }u\mathrm{\mid }$ from [F7] permits differentiation inside this chart integral. Summing finitely many terms proves $\delta {\int }_M(\omega (s)-\omega (0))={\int }_M\delta \omega$. Choice independence is exactly [F12].

For any smooth compactly supported vector field $W$, the same partition has only finitely many terms meeting its support. Each ${\rho }_{i}W$ is compactly supported inside one chart. In that chart its density divergence is ${\mathrm{\partial }}_a(w{\rho }_i{W}^a)$ by [F4]. Zero extension is smooth, and FTC/Fubini on a containing rectangle gives zero integral, because its coefficients vanish near every rectangle face. The global integral agrees with this chart integral by [F12]. On a neighborhood of $\mathrm{supp}W$, ${\sum }_i{\rho }_i=1$ and ${\sum }_{i}d{\rho }_i=0$, so ${\sum }_i{\mathrm{\nabla }}_a({\rho }_i{W}^a)={\mathrm{\nabla }}_a{W}^a$; off the support both sides vanish. Finite additivity therefore gives ${\int }_M{\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. This proves the compact divergence rule without requiring the support to fit one chart.

For a connection variation in [A1], $\delta F=d\alpha$; in components $\delta F_{ab}=2{\mathrm{\nabla }}_{[a}{\alpha }_{b]}$ because the Levi-Civita Christoffel symbols are symmetric in their lower indices. Holding $g$ fixed gives $\delta (F_{ab}{F}^{ab})=2F^{ab}\delta F_{ab}=4F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. Differentiating the coefficient in the action therefore yields ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=-({\mu }_{0}c{)}^{-1}{\int }_M{F}^{ab}{\mathrm{\nabla }}_a{\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$.

Let $q^{ab}=\delta g^{ab}$ and hold $F_{ab}$ fixed. From $F_{cd}{F}^{cd}=g^{ca}{g}^{db}{F}_{cd}{F}_{ab}$, symmetry of $q$ and antisymmetry of $F$ give $\delta (F_{cd}{F}^{cd})=2F_{ac}{F}_b{}^c{q}^{ab}$. The volume variation in [F3] then gives ${\delta }_g(F_{cd}{F}^{cd}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)=(2F_{ac}{F}_b{}^c-\frac{1}{2}{g}_{ab}{F}_{cd}{F}^{cd})q^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g\mathrm{.}$ By [F2], this is the variation of the same action density as its wedge-form representative. Multiplying by $-1\mathrm{/}(4{\mu }_{0}c)$ and comparing with the definition in part (b) gives the displayed $T_{\mathrm{E}\mathrm{M},ab}$. The variation is local and compactly supported, so no boundary term is discarded.

Set $W^a:=F^{ab}{\alpha }_b$. The product rule gives ${\mathrm{\nabla }}_a{W}^a=({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b+F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. By step 3.1, ${\int }_M{\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$, for arbitrary compact support of $\alpha$. Substitution into step 3.2 gives ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=({\mu }_{0}c{)}^{-1}{\int }_M({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$, proving the connection-variation formula.

Adding [A2] to step 4.1 gives ${\delta }_A(S_{\mathrm{E}\mathrm{M}}+S_{\mathrm{m}})[\alpha ]=({\mu }_{0}c{)}^{-1}{\int }_M({\mathrm{\nabla }}_a{F}^{ab}+{\mu }_0{j}^b){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. If ${\mathrm{\nabla }}_a{F}^{ab}+{\mu }_0{j}^b=0$, this integral vanishes for every allowed $\alpha$. Conversely, if it vanishes for every smooth compactly supported one-form supported in one chart rectangle, fix a point and suppose a component $K^{b_0}:={\mathrm{\nabla }}_a{F}^{a{b}_0}+{\mu }_0{j}^{b_0}$ is nonzero there. By continuity, shrink to a coordinate ball with compact closure where $K^{b_0}$ has one strict sign and $w>0$. In coordinates centered at that point define $\varphi (x)=\exp (-1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2))$ for $\mathrm{\mid }x\mathrm{\mid }<r$ and $\varphi (x)=0$ for $\mathrm{\mid }x\mathrm{\mid }\ge r$. Every derivative approaching $\mathrm{\mid }x\mathrm{\mid }=r$ is a finite sum of a polynomial in $1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2)$ times the same exponential, and $s^{-m}{e}^{-1\mathrm{/}s}\to 0$ as $s\downarrow 0$ for every $m\ge 0$; hence $\varphi$ is a smooth nonnegative compactly supported function, positive inside the ball. Set $\alpha =\varphi d{x}^{b_0}$ in the chart and extend it by zero outside the compact support; this gives an admissible global one-form in [A1]. The bump and rectangle are explicitly constructed, so the ambient convention [F5] introduces no arbitrary choice. The product $K^{b_0}\varphi w$ is continuous and supported in a compact rectangle containing the bump support, so [F8] makes its full integral well-defined. At the center, $\mathrm{\mid }{K}^{b_0}\varphi w\mathrm{\mid }>0$; continuity gives a smaller closed nondegenerate rectangle $R$ on which this coefficient has one strict sign and absolute value at least some $\eta >0$. By [F8] it is Riemann integrable there. Applying [F11] to its sign-adjusted coefficient shows that the absolute value of the integral over $R$ is at least $\eta \mathrm{vol}(R)>0$. On the entire bump support the product has the same sign, since $K^{b_0}$ has one sign on the ball and $\varphi ,w$ are nonnegative; therefore its full integral over the support cannot vanish. This contradicts the assumed stationarity integral. Thus every component $K^b$ vanishes at every point. Conversely its pointwise vanishing makes the variation zero for every compactly supported global one-form by step 4.1. This proves the stated global stationarity equivalence under [A2]–[A3].

When the Euler–Lagrange equation in step 5.1 holds, apply [F6] to obtain $d({\star }_{g}F)={\mu }_0{\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. This uses the conditional current interface of [A1] and the exact current-sign identity proved in the earlier item; it adds no global potential or matter-action claim.

By [F1], $d({\star }_{g}F)={\mu }_0{\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. Integrate this three-form over the finite smooth singular three-chain $\mathrm{\Sigma }$. By [F2], ${\int }_{\mathrm{\Sigma }}d({\star }_{g}F)={\int }_{\mathrm{\partial }\mathrm{\Sigma }}{\star }_{g}F$, so ${\mu }_0{\int }_{\mathrm{\Sigma }}{\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g={\int }_{\mathrm{\partial }\mathrm{\Sigma }}{\star }_{g}F$. Divide by ${\mu }_{0}c>0$ and use the definition of $Q$ to obtain (a)'s formula. The chain $\mathrm{\Sigma }$ was fixed arbitrarily, and no family of chains is selected, as [F4] records.

Let $V$ be a finite smooth singular three-chain. Apply [F2] to the two-form $F$ on $V$ and use [A1]: ${\int }_{\mathrm{\partial }V}F={\int }_{V}dF=0$. This proves (b) for bounding surfaces; the argument supplies no primitive or flux statement for a nonbounding cycle.

If $\mathrm{\partial }\mathrm{\Sigma }=0$, the right side of step 1.1 is the integral over the zero two-chain, which is zero by finite additivity in [F2]. Hence $Q(\mathrm{\Sigma })=0$. This conclusion uses a cycle, not merely a closed surface whose homology class is unspecified.

For the four-chain $C$, [F1] gives ${\mathrm{\nabla }}_a{j}^a=0$ throughout its domain. By [F3], ${\mathrm{\Phi }}_j(\mathrm{\partial }C)={\int }_C({\mathrm{\nabla }}_a{j}^a){\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. Additivity of the finite-chain integral and $\mathrm{\partial }C={\mathrm{\Sigma }}_2-{\mathrm{\Sigma }}_1+B$ give ${\mathrm{\Phi }}_j({\mathrm{\Sigma }}_2)-{\mathrm{\Phi }}_j({\mathrm{\Sigma }}_1)+{\mathrm{\Phi }}_j(B)=0$. Divide by $c>0$ and apply the definition of $Q$ to obtain (c). The proof integrates only finite smooth chains, so no boundary-at-infinity term is present.