Conformal reduction of the vacuum constraint equations

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.