Conformal construction of constrained initial data

For each integer $n\ge 1$, induction by the product rule gives $(t^n{)}^{\mathrm{\prime }}=n{t}^{n-1}$, and the quotient rule on $t>0$ gives $(t^{-n}{)}^{\mathrm{\prime }}=-n{t}^{-n-1}$; iterating this identity shows that every derivative of $t\mapsto t^{-2}$ is a constant multiple of a negative integer power, hence is continuous on $(0,\mathrm{\infty })$. The map $t\mapsto t^4$ is a polynomial. Thus both maps are smooth there; [F5] gives their $C^k$ regularity at every finite order and [F6] makes their composites with $\varphi$ smooth. Hence $h={\varphi }^4\tilde{h}$ is smooth, symmetric and positive definite, because for every nonzero $v\in T_p\mathrm{\Sigma }$, $h_p(v,v)=\varphi (p{)}^4{\tilde{h}}_p(v,v)>0$ by [F2]. The same rules show that $K={\varphi }^{-2}\tilde{\sigma }+(\tau \mathrm{/}3)h$ is smooth and symmetric. The Riemannian connection theorem [F3] supplies both $\tilde{D}$ and $D$.

In each chart $\det ({\tilde{h}}_{ij})>0$ and ${\tilde{h}}^{ij}=\mathrm{adj}(\tilde{h}{)}_{ij}\mathrm{/}\det (\tilde{h})$; the adjugate entries are polynomials in the smooth metric components [F1], and the reciprocal is smooth by the same quotient-rule induction as in the earlier proof step 1.1 and finite-order algebra closure [F5]. Thus ${\tilde{h}}^{ij}$ is smooth; $h^{ij}={\varphi }^{-4}{\tilde{h}}^{ij}$ is smooth by step 1.1. Formula [F7] therefore makes the Christoffel symbols smooth, so the displayed Laplacian is defined for smooth $f$. By [F13] it is the trace of the covariant derivative of $df$, so its value is independent of the chart. By [F8] curvature is tensorial, and by [F9, F10] its Ricci and scalar contractions do not depend on the chart. The same basis-independent contractions define the traces and squared norms; all derivative terms use the connections supplied by [F3].

For data induced by an embedded spacelike hypersurface, the convention in [F11] fixes the sign of $K$, and [F12] gives precisely the stated Hamiltonian and momentum residuals when $\mathrm{\Lambda }=0$ and stress-energy vanishes. The abstract data definition itself makes no embedding or development assertion.

From [F1], $h^{ij}={\varphi }^{-4}{\tilde{h}}^{ij}$ and $h^{ij}{K}_{ij}={\varphi }^{-6}{\tilde{h}}^{ij}{\tilde{\sigma }}_{ij}+\tau =\tau$. Writing $A_{ij}:={\varphi }^{-2}{\tilde{\sigma }}_{ij}$, its trace is zero and $\mathrm{\mid }A{\mathrm{\mid }}_h^2={\varphi }^{-12}\mathrm{\mid }\tilde{\sigma }{\mathrm{\mid }}_{\tilde{h}}^2$: each of the two inverse metrics contributes ${\varphi }^{-4}$ and the two $A$ factors contribute ${\varphi }^{-2}$. The cross term between $A$ and $(\tau \mathrm{/}3)h$ vanishes by the trace-free condition, while $\mathrm{\mid }(\tau \mathrm{/}3)h{\mathrm{\mid }}_h^2=3(\tau \mathrm{/}3{)}^2={\tau }^2\mathrm{/}3$. Hence $\mathrm{\mid }K{\mathrm{\mid }}_h^2={\varphi }^{-12}\mathrm{\mid }\tilde{\sigma }{\mathrm{\mid }}_{\tilde{h}}^2+\frac{{\tau }^2}{3}\mathrm{.}$

Put $u:=2\log \varphi$, so $h=e^{2u}\tilde{h}$, and let $C^k{}_{ij}:=\mathrm{\Gamma }[h{]}^k{}_{ij}-{\tilde{\mathrm{\Gamma }}}^k{}_{ij}$. Substitute $h_{ij}=e^{2u}{\tilde{h}}_{ij}$ and $h^{ij}=e^{-2u}{\tilde{h}}^{ij}$ in [F4]; using ${\mathrm{\partial }}_i(e^{2u}{\tilde{h}}_{j\mathrm{\ell }})=e^{2u}(2u_i{\tilde{h}}_{j\mathrm{\ell }}+{\mathrm{\partial }}_i{\tilde{h}}_{j\mathrm{\ell }})$ then gives $C^k{}_{ij}={\delta }_i^k{u}_j+{\delta }_j^k{u}_i-{\tilde{h}}_{ij}{u}^k,u_i:={\tilde{D}}_{i}u,u^k:={\tilde{h}}^{k\mathrm{\ell }}{u}_{\mathrm{\ell }}\mathrm{.}$ By the affine-connection Leibniz rule, $D_{{\mathrm{\partial }}_i}(f{\mathrm{\partial }}_a)=({\mathrm{\partial }}_{i}f){\mathrm{\partial }}_a+f{\mathrm{\Gamma }}^c{}_{ia}{\mathrm{\partial }}_c$. Applying this in the curvature definition [F1] and using $[{\mathrm{\partial }}_i,{\mathrm{\partial }}_j]=0$ gives $R[\mathrm{\Gamma }{]}^a{}_{bij}={\mathrm{\partial }}_i{\mathrm{\Gamma }}^a{}_{jb}-{\mathrm{\partial }}_j{\mathrm{\Gamma }}^a{}_{ib}+{\mathrm{\Gamma }}^a{}_{i\mathrm{\ell }}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jb}-{\mathrm{\Gamma }}^a{}_{j\mathrm{\ell }}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ib}$. The difference of two affine connections is tensorial, since the derivative terms in their Leibniz rules cancel. For this type $(1,2)$ tensor, ${\tilde{D}}_i{C}^a{}_{jb}={\mathrm{\partial }}_i{C}^a{}_{jb}+{\tilde{\mathrm{\Gamma }}}^a{}_{i\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{jb}-{\tilde{\mathrm{\Gamma }}}^{\mathrm{\ell }}{}_{ij}{C}^a{}_{\mathrm{\ell }b}-{\tilde{\mathrm{\Gamma }}}^{\mathrm{\ell }}{}_{ib}{C}^a{}_{j\mathrm{\ell }}\mathrm{.}$ Substituting $\mathrm{\Gamma }[h]=\tilde{\mathrm{\Gamma }}+C$ into the curvature formula, the terms linear in $C$ are ${\tilde{D}}_i{C}^a{}_{jb}-{\tilde{D}}_j{C}^a{}_{ib}$: the two terms containing ${\tilde{\mathrm{\Gamma }}}^{\mathrm{\ell }}{}_{ij}$ cancel because the lower Christoffel indices are symmetric, and the remaining mixed terms pair with the displayed two covariant-derivative terms. The quadratic terms in $C$ remain, giving $R[h{]}^a{}_{bij}={\tilde{R}}^a{}_{bij}+{\tilde{D}}_i{C}^a{}_{jb}-{\tilde{D}}_j{C}^a{}_{ib}+C^a{}_{i\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{jb}-C^a{}_{j\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{ib}$. Contracting the output and first direction slots gives $\mathrm{Ric}(h{)}_{ij}={\stackrel{}{\mathrm{Ric}}}_{ij}+{\tilde{D}}_k{C}^k{}_{ij}-{\tilde{D}}_j{C}^k{}_{ik}+C^k{}_{k\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{ij}-C^k{}_{j\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{ik}\mathrm{.}$ Directly contracting the displayed $C$ gives ${\tilde{D}}_k{C}^k{}_{ij}=2{\tilde{D}}_i{u}_j-{\tilde{h}}_{ij}\tilde{\mathrm{\Delta }}u$, $C^k{}_{ik}=3u_i$, and ${\tilde{D}}_j{C}^k{}_{ik}=3{\tilde{D}}_j{u}_i$. The two quadratic terms expand to $C^k{}_{k\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{ij}=6u_i{u}_j-3{\tilde{h}}_{ij}\mathrm{\mid }du{\mathrm{\mid }}_{\tilde{h}}^2$ and $C^k{}_{j\mathrm{\ell }}{C}^{\mathrm{\ell }}{}_{ik}=5u_i{u}_j-2{\tilde{h}}_{ij}\mathrm{\mid }du{\mathrm{\mid }}_{\tilde{h}}^2$. By [F4] the symbols ${\tilde{\mathrm{\Gamma }}}^k{}_{ij}$ are symmetric in $i,j$; hence ${\tilde{D}}_i{u}_j-{\tilde{D}}_j{u}_i={\mathrm{\partial }}_i{\mathrm{\partial }}_{j}u-{\mathrm{\partial }}_j{\mathrm{\partial }}_{i}u=0$ by [F7]. Substitution gives $\mathrm{Ric}(h{)}_{ij}={\stackrel{}{\mathrm{Ric}}}_{ij}-{\tilde{D}}_i{u}_j+u_i{u}_j-(\tilde{\mathrm{\Delta }}u+\mathrm{\mid }du{\mathrm{\mid }}_{\tilde{h}}^2){\tilde{h}}_{ij}\mathrm{.}$ Tracing with $h^{ij}={\varphi }^{-4}{\tilde{h}}^{ij}$ gives $R(h)={\varphi }^{-4}(\tilde{R}-4\tilde{\mathrm{\Delta }}u-2\mathrm{\mid }du{\mathrm{\mid }}_{\tilde{h}}^2)$. Since $u=2\log \varphi$, [F8, F9] give $u_i=2{\varphi }^{-1}{\mathrm{\partial }}_i\varphi$. Differentiating once more by [F10] gives ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}u=2{\varphi }^{-1}{\mathrm{\partial }}_i{\mathrm{\partial }}_j\varphi -2{\varphi }^{-2}({\mathrm{\partial }}_i\varphi )({\mathrm{\partial }}_j\varphi )$. Substitution in the coordinate Laplacian yields $\tilde{\mathrm{\Delta }}u=2{\varphi }^{-1}\tilde{\mathrm{\Delta }}\varphi -2{\varphi }^{-2}\mathrm{\mid }d\varphi {\mathrm{\mid }}_{\tilde{h}}^2,$ with the minus sign applying to the second term. [F8, F9, F10] also give $\mathrm{\mid }du{\mathrm{\mid }}_{\tilde{h}}^2=4{\varphi }^{-2}\mathrm{\mid }d\varphi {\mathrm{\mid }}_{\tilde{h}}^2$. The gradient-square terms cancel, so $R(h)={\varphi }^{-5}(-8\tilde{\mathrm{\Delta }}\varphi +\tilde{R}\varphi )\mathrm{.}$

Raise both indices of $A_{ij}$ with $h$ to get $A^{ij}={\varphi }^{-10}{\tilde{\sigma }}^{ij}$. Apply the tensor-product derivative rule [F6] to $A=A^{kj}{\mathrm{\partial }}_k\otimes {\mathrm{\partial }}_j$: $(D_{i}A{)}^{kj}={\mathrm{\partial }}_i{A}^{kj}+\mathrm{\Gamma }[h{]}^k{}_{i\mathrm{\ell }}{A}^{\mathrm{\ell }j}+\mathrm{\Gamma }[h{]}^j{}_{i\mathrm{\ell }}{A}^{k\mathrm{\ell }}$, with the same formula for $\tilde{D}$. Subtracting cancels the ordinary derivative and gives $D_i{A}^{ij}={\tilde{D}}_i{A}^{ij}+C^i{}_{ik}{A}^{kj}+C^j{}_{ik}{A}^{ik}\mathrm{.}$ Metric compatibility [F5, F6] gives $D_i{h}^{k\mathrm{\ell }}=0$ and ${\tilde{D}}_i{\tilde{h}}^{k\mathrm{\ell }}=0$: apply the dual-pairing rule to $h^{k\mathrm{\ell }}{h}_{\mathrm{\ell }m}={\delta }_m^k$ and use $Dh=0$ (and likewise for $\tilde{h}$). The chain and quotient rules give ${\mathrm{\partial }}_i{\varphi }^{-1}=-{\varphi }^{-2}{\mathrm{\partial }}_i\varphi$; applying the product rule to ten factors and using [F11] gives ${\mathrm{\partial }}_i{\varphi }^{-10}=10{\varphi }^{-9}(-{\varphi }^{-2}{\mathrm{\partial }}_i\varphi )=-10{\varphi }^{-11}{\mathrm{\partial }}_i\varphi$. Now $C^i{}_{ik}=6{\varphi }^{-1}{\mathrm{\partial }}_k\varphi$, and, because $\tilde{\sigma }$ is symmetric and trace-free, $C^j{}_{ik}{\tilde{\sigma }}^{ik}=4{\varphi }^{-1}{\mathrm{\partial }}_i\varphi {\tilde{\sigma }}^{ij}$; ${\tilde{D}}_i({\varphi }^{-10}{\tilde{\sigma }}^{ij})$ contributes $-10{\varphi }^{-11}{\mathrm{\partial }}_i\varphi {\tilde{\sigma }}^{ij}+{\varphi }^{-10}{\tilde{D}}_i{\tilde{\sigma }}^{ij}$, so the three conformal gradient terms cancel. By metric compatibility and the TT condition, ${\tilde{D}}_i{\tilde{\sigma }}^{ij}={\tilde{h}}^{j\mathrm{\ell }}{\tilde{h}}^{ik}{\tilde{D}}_i{\tilde{\sigma }}_{k\mathrm{\ell }}={\tilde{h}}^{j\mathrm{\ell }}({\mathrm{div}}_{\tilde{h}}\tilde{\sigma }{)}_{\mathrm{\ell }}=0.$ Therefore $D_i{A}^{ij}=0$. Lowering the free index with the metric-compatible connection gives $h_{j\mathrm{\ell }}{D}_i{A}^{i\mathrm{\ell }}=D_i(h_{j\mathrm{\ell }}{A}^{i\mathrm{\ell }})=D_i{A}^i{}_j=h^{ik}{D}_i{A}_{kj}=0$, so ${\mathrm{div}}_{h}A=0$. The pure trace part contributes $h^{ik}{D}_i((\tau \mathrm{/}3)h_{kj})-D_j\tau =0$, because $Dh=0$ by [F5, F6] and $\tau$ is constant by [F1]. Therefore $\mathcal{M}(h,K)=0$.

By [F3] and steps 1.1 and 1.2, $\mathcal{H}(h,K)={\varphi }^{-5}(-8\tilde{\mathrm{\Delta }}\varphi +\tilde{R}\varphi )+\frac{2}{3}{\tau }^2-{\varphi }^{-12}\mathrm{\mid }\tilde{\sigma }{\mathrm{\mid }}_{\tilde{h}}^2\mathrm{.}$ Multiplying by ${\varphi }^5>0$ is reversible and gives exactly the scalar equation in the Statement. The momentum residual is already zero by step 2.1, so simultaneous satisfaction of both vacuum constraints is equivalent to that scalar equation. This is a deduction about the constraint equations only; it does not assert a solution of the nonlinear equation for arbitrary free data.

For a constant $\varphi >0$, $\tilde{\mathrm{\Delta }}\varphi =0$. Substituting $\tilde{R}=0$ and $\mathrm{\mid }\tilde{\sigma }{\mathrm{\mid }}_{\tilde{h}}^2=A$ in [F2], its equation becomes $-A{\varphi }^{-7}+(2\mathrm{/}3){\tau }^2{\varphi }^5=0$. Multiplication by ${\varphi }^7>0$ is reversible, so this is equivalent to ${\varphi }^{12}=3A\mathrm{/}(2{\tau }^2)$.

The right side $3A\mathrm{/}(2{\tau }^2)$ is positive because $A>0$ and $\tau \mathrm{\ne }0$. By [F3], it has exactly one positive twelfth root, and by [F4] that root is $(3A\mathrm{/}(2{\tau }^2){)}^{1\mathrm{/}12}$. Thus this is the unique positive constant solution.

For a fixed smooth real direction $v$, its periodic lift is continuous on the nonempty compact box $[0,2\pi {]}^3$ by [F4, F7], so [F8] gives a finite bound $V:=\sup \mathrm{\mid }v\mathrm{\mid }$. If $V>0$, choose $\mathrm{\mid }ϵ\mathrm{\mid }<{\varphi }_0\mathrm{/}(2V)$; if $V=0$, positivity is immediate. In either case ${\varphi }_0+ϵv>0$ for all such directions and sufficiently small $\mathrm{\mid }ϵ\mathrm{\mid }$. Every term in $c$ is positive because $A>0$, ${\varphi }_0>0$ and $\tau \mathrm{\ne }0$. Since ${\tilde{h}}_{ij}=L_{\ast }^2{\delta }_{ij}$, the Christoffel formula in the conformal-data definition gives ${\tilde{\mathrm{\Gamma }}}^k{}_{ij}=0$ in the local periodic charts. The curvature definition then gives $\tilde{R}=0$, and the Laplacian formula gives $\tilde{\mathrm{\Delta }}=L_{\ast }^{-2}{\sum }_i{\mathrm{\partial }}_i^2$. Differentiating the scalar expression [F2] at the constant ${\varphi }_0$ gives $-8\tilde{\mathrm{\Delta }}v$ from its linear Laplacian term, zero from $\tilde{R}=0$, $7A{\varphi }_0^{-8}v$ from $-A{\varphi }^{-7}$, and $(10\mathrm{/}3){\tau }^2{\varphi }_0^{4}v$ from $(2\mathrm{/}3){\tau }^2{\varphi }^5$. Thus its directional linearization is $\mathcal{A}=-8\tilde{\mathrm{\Delta }}+c$. For each $i$, periodicity makes the integral of ${\mathrm{\partial }}_i(v{\mathrm{\partial }}_{i}v)$ over one period zero; expanding that derivative gives $\int v{\mathrm{\partial }}_i^{2}vd\mu =-\int ({\mathrm{\partial }}_{i}v{)}^{2}d\mu$. Summing over $i$ therefore yields $\int v\mathcal{A}vd\mu =8L_{\ast }^{-2}\int \mathrm{\mid }{\mathrm{\nabla }}_{0}v{\mathrm{\mid }}^{2}d\mu +c\int v^{2}d\mu >0$ for every nonzero smooth real $v$: choose a point where $v\mathrm{\ne }0$; continuity gives a coordinate box of positive torus measure on which $\mathrm{\mid }v\mathrm{\mid }$ is bounded below by a positive constant, so $\int v^{2}d\mu >0$. Here ${\mathrm{\nabla }}_{0}v=({\mathrm{\partial }}_{1}v,{\mathrm{\partial }}_{2}v,{\mathrm{\partial }}_{3}v)$.

For any $v\in H^{s+2}$, the weak-derivative Fourier identity [F5] gives $\widehat{\mathcal{A}v}(k)=a_k\widehat{v}(k)$ with $a_k=8L_{\ast }^{-2}\mathrm{\mid }k{\mathrm{\mid }}^2+c$. Since $c>0$, every $a_k>0$, and $a_k\ge m(1+\mathrm{\mid }k{\mathrm{\mid }}^2)$ for $m:=\min \{8L_{\ast }^{-2},c\}>0$; also $a_k\le M(1+\mathrm{\mid }k{\mathrm{\mid }}^2)$ for $M:=\max \{8L_{\ast }^{-2},c\}$. The upper bound proves $N_s(\mathcal{A}v)\le M{N}_{s+2}(v)$. Norm equivalence in [F5] therefore makes the weak operator bounded for the derivative Sobolev norms. For real $f\in H^s$, [F6] gives $\widehat{f}(-k)=\overline{\widehat{f}(k)}$. Since $a_{-k}=a_k$ is real, $\widehat{v}(k):=\widehat{f}(k)\mathrm{/}{a}_k$ has the same conjugate symmetry and therefore [F6] gives it a unique complex $L^2$ representative $v$. The coefficient definition gives $\widehat{\bar{v}}(k)=\overline{\widehat{v}(-k)}=\widehat{v}(k)$, so injectivity of the coefficient isometry [F6] yields $\bar{v}=v$ in $L^2$. The lower bound gives $N_{s+2}(v{)}^2={\sum }_k\frac{(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{s+2}}{a_k^2}\mathrm{\mid }\widehat{f}(k){\mathrm{\mid }}^2\le m^{-2}{N}_s(f{)}^2\mathrm{.}$ The lower bound makes the coefficient sequence square-summable with weight $(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{s+2}$; [F5] then gives $v\in H^{s+2}$, with $\mathcal{A}v=f$. If $\mathcal{A}v=0$, then $a_k\widehat{v}(k)=0$ for every $k$, hence all coefficients vanish and $v=0$. This proves surjectivity, injectivity and the asserted inverse bound for every integer $s\ge 2$.