Adopted trivial-potential classical charged scalar model — Examples

Since $\phi \equiv 0$, every derivative of $\phi$ is zero and $P_a={\mathcal{D}}_a\phi =0$. Hence $E_{\phi }={\mathcal{D}}_a{P}^a-V^{\mathrm{\prime }}(0)\phi =0$ and $j^a=K_{\phi }\lambda \mathrm{Im}(\bar{\phi }{P}^a)=0$. Also $g^{ab}\overline{P_a}{P}_b+V(\mathrm{\mid }\phi {\mathrm{\mid }}^2)=V(0)=0$, so the defining formula gives $T_{ab}^{\phi }=0$.

Substitution of $j=0$ and $T^{\phi }=0$ into [F1] leaves exactly ${\mathrm{\nabla }}_a{F}^{ab}=0$, $dF=0$, and $G_{ab}+\mathrm{\Lambda }{g}_{ab}={\kappa }_E{T}_{ab}^{\mathrm{E}\mathrm{M}}$, the displayed assumptions. Thus the zero extension solves the full classical model. For compatible zero-scalar Cauchy data, [L1] supplies a local development. In its temporal gauge, ${\mathrm{\partial }}_0=n$ initially, so ${\pi }_0={\pi }_n=0$ there; the charged spatial compatibility gives ${\pi }_i={\mathcal{D}}_i{\phi }_0=0$ as well. Hence the real scalar state $z=(\mathrm{Re}\phi ,\mathrm{Im}\phi ,\mathrm{Re}{\pi }_0,\mathrm{Im}{\pi }_0,\mathrm{Re}{\pi }_i,\mathrm{Im}{\pi }_i)$ has zero initial data.

In the admissible coordinates, the scalar first-order equations have ${\mathrm{\partial }}_0\phi ={\pi }_0$ and the $\pi$-block $A_{\pi }^0{\mathrm{\partial }}_0\pi +{\sum }_k{A}_{\pi }^k{\mathrm{\partial }}_k\pi +\text{zero-order terms}=0,$ where $A_{\pi }^0=\left(\begin{array}{cc}-g^{00}& 0\\ 0& (g^{ij})\end{array}\right),A_{\pi }^k=\left(\begin{array}{cc}-2g^{0k}& -g^{kj}\\ -g^{ik}& 0\end{array}\right)\mathrm{.}$ These are real symmetric matrices, with $A_{\pi }^0$ positive definite because $g^{00}<0$ and $(g^{ij})$ is positive definite. Adding the scalar transport equation gives symmetric matrices $A^0=I\oplus A_{\pi }^0$ and $A^k=0\oplus A_{\pi }^k$ on the real state $z$, so the full scalar system is $A^0{\mathrm{\partial }}_{0}z+{\sum }_k{A}^k{\mathrm{\partial }}_{k}z+C_{0}z=0$. The remaining terms can be written $C_{0}z$ with $C_0$ bounded on each compact solution cylinder: they are linear in $\phi ,{\pi }_0,{\pi }_i$, with coefficients formed from $\mathrm{\Gamma }(g)$, $A_i$, $F_{0i}$ and $V^{\mathrm{\prime }}(\mathrm{\mid }\phi {\mathrm{\mid }}^2)$. The metric and matter regularity in [L1] gives $g\in C^3$ and $A,\phi \in C^2$ in the finite-Sobolev branch at $s\ge 5$; hence $A^0,A^k$ are continuously differentiable and $C_0$ is continuous there. Writing $e=z^{\mathsf{T}}{A}^{0}z$ and $f^k=z^{\mathsf{T}}{A}^{k}z$, symmetry and the system equation give ${\mathrm{\partial }}_{0}e+{\sum }_k{\mathrm{\partial }}_k{f}^k=z^{\mathsf{T}}({\mathrm{\partial }}_0{A}^0+{\sum }_k{\mathrm{\partial }}_k{A}^k-C_0-C_0^{\mathsf{T}})z\mathrm{.}$ The matrix on the right is bounded, while $A^0\ge aI$ for some $a>0$ on the compact cylinder. On the periodic torus, integrating this identity gives $E^{\mathrm{\prime }}\le CE$ for $E=\int e$, since the spatial fluxes cancel at opposite periodic faces. In a local chart, choose a coordinate cube $Q_0$ around the initial point and shrink each side at a speed $\sigma$ satisfying $\sigma a>{\max }_k\sup \mathrm{\parallel }{A}^k\mathrm{\parallel }$. Differentiating the integral over the moving cube and integrating each spatial derivative by the one-variable fundamental theorem of calculus gives $E^{\mathrm{\prime }}(t)\le CE(t)-{\sum }_{\text{faces }S}{\int }_S{z}^{\mathsf{T}}(\sigma A^0+{\nu }_k{A}^k)zdS\le CE(t),$ where $\nu =\pm {\mathrm{\partial }}_k$ is the outward coordinate normal on each face; the face matrices are positive definite by the choice of $\sigma$. In either domain $E(0)=0$ and $E\ge 0$, so integrating the inequality and applying [F2] gives $E=0$ forward and backward in time. Positivity of $A^0$ and continuity of $z$ then give $\phi =\pi =0$ near every initial point. This proves the zero branch for finite-Sobolev data without using smooth geometric uniqueness; for smooth data, [L1] additionally identifies the branch up to its stated isometry and gauge.