First-order curvature of the diagonal perturbation h_00 = -2 phi, h_ij = -2 phi delta_ij for the explicit potential phi = |x|^2/2

(Admissibility.) In the coordinate frame the components of $\lambda h$ are $\lambda h_{00}=-2\lambda \phi$, $\lambda h_{ij}=-2\lambda \phi {\delta }_{ij}$ and $0$ otherwise, so the matrix of $g(\lambda )=\eta +\lambda h$ in the coordinate basis is $\mathrm{diag}(-1,1,1,1)+E(\lambda )$ with $E(\lambda )=-2\lambda \phi I_4$, the entries of which satisfy $\mathrm{\mid }E(\lambda {)}_{ij}\mathrm{\mid }=2\mathrm{\mid }\lambda \mathrm{\mid }\phi <2\cdot \frac{1}{8}\cdot \frac{1}{2}=\frac{1}{8}$ on $\mathrm{\Omega }$, because $\phi <\frac{1}{2}$ there; all components are stationary and smooth, so $g(\lambda )$ is a smooth symmetric component field. By [L5] $g(\lambda )$ has inertia $(3,1,0)$ for every $0<\mathrm{\mid }\lambda \mathrm{\mid }<{\epsilon }_1$, and $q\equiv 0$ is smooth; hence $({\epsilon }_1,h,q)$ satisfies clause (a) of [L1].

(First-order connection.) By [L2] and [L3], ${\mathrm{\Gamma }}^{(1)i}{}_{00}=\frac{1}{2}{\eta }^{ia}(2{\mathrm{\partial }}_0{h}_{a0}-{\mathrm{\partial }}_a{h}_{00})$. The time derivatives vanish and ${\mathrm{\partial }}_a{h}_{00}=-2{\mathrm{\partial }}_a\phi$, so ${\mathrm{\Gamma }}^{(1)i}{}_{00}={\eta }^{ia}{\mathrm{\partial }}_a\phi ={\mathrm{\partial }}^i\phi =x^i$. Similarly ${\mathrm{\Gamma }}^{(1)i}{}_{0j}=\frac{1}{2}{\eta }^{ia}({\mathrm{\partial }}_0{h}_{aj}+{\mathrm{\partial }}_j{h}_{a0}-{\mathrm{\partial }}_a{h}_{0j})=0$ and ${\mathrm{\Gamma }}^{(1)i}{}_{j0}=0$ because the spatial components $h_{k0}=h_{0k}$ vanish and ${\eta }^{i0}=0$. For the spatial pairs ${\mathrm{\Gamma }}^{(1)i}{}_{jk}=\frac{1}{2}{\eta }^{ia}({\mathrm{\partial }}_j{h}_{ak}+{\mathrm{\partial }}_k{h}_{aj}-{\mathrm{\partial }}_a{h}_{jk})=\frac{1}{2}{\delta }^{ia}(-2{\mathrm{\partial }}_j\phi {\delta }_{ak}-2{\mathrm{\partial }}_k\phi {\delta }_{aj}+2{\mathrm{\partial }}_a\phi {\delta }_{jk})=-{\delta }_k^i{x}_j-{\delta }_j^i{x}_k+{\delta }_{jk}{x}^i$. For the components with upper index $0$, ${\mathrm{\Gamma }}^{(1)0}{}_{bc}=\frac{1}{2}{\eta }^{0a}({\mathrm{\partial }}_b{h}_{ca}+{\mathrm{\partial }}_c{h}_{ba}-{\mathrm{\partial }}_a{h}_{bc})=-\frac{1}{2}({\mathrm{\partial }}_b{h}_{c0}+{\mathrm{\partial }}_c{h}_{b0}-{\mathrm{\partial }}_0{h}_{bc})$, since ${\eta }^{0a}$ is nonzero only at $a=0$: for $(b,c)=(0,0)$ and for $b,c$ both spatial this vanishes ($h_{i0}=0$ for spatial $i$ and all component time derivatives vanish), while ${\mathrm{\Gamma }}^{(1)0}{}_{0i}={\mathrm{\Gamma }}^{(1)0}{}_{i0}=-\frac{1}{2}{\mathrm{\partial }}_i{h}_{00}={\mathrm{\partial }}_i\phi =x_i$.

(Trace data.) By [L2], $h={\eta }^{ab}{h}_{ab}={\eta }^{00}{h}_{00}+{\eta }^{ij}{h}_{ij}=2\phi -6\phi =-4\phi$, and the mixed components $h^c{}_b={\eta }^{ca}{h}_{ab}$ are $h^0{}_0={\eta }^{00}{h}_{00}=2\phi$, $h^i{}_j={\eta }^{ik}{h}_{kj}=-2\phi {\delta }_j^i$, $h^0{}_i={\eta }^{0a}{h}_{ai}=0$ and $h^i{}_0={\eta }^{ik}{h}_{k0}=0$.

(First-order Ricci tensor.) By [L3] and step 1.3: for the $(00)$ component, ${\mathrm{\partial }}_c{\mathrm{\partial }}_0{h}^c{}_0=0$ because all components of $h^c{}_0$ are stationary, and $-\mathrm{\square }{h}_{00}-{\mathrm{\partial }}_0{\mathrm{\partial }}_{0}h=2\mathrm{\square }\phi =2\mathrm{\Delta }\phi =6$ (the field $h$ is stationary), so ${\mathrm{Ric}}_{00}^{(1)}=\frac{1}{2}(0+0+6+0)=3$. For the $(0i)$ component, ${\mathrm{\partial }}_c{\mathrm{\partial }}_0{h}^c{}_i=0$ and ${\mathrm{\partial }}_c{\mathrm{\partial }}_i{h}^c{}_0=0$ because the fields $h^c{}_i$ are stationary and $h^c{}_0$ vanishes for $c\mathrm{\ne }0$ and is stationary for $c=0$; and $\mathrm{\square }{h}_{0i}={\mathrm{\partial }}_0{\mathrm{\partial }}_{i}h=0$ because $h_{0i}=0$ and $h$ is stationary; so ${\mathrm{Ric}}_{0i}^{(1)}=0$. For the $(ij)$ component, ${\mathrm{\partial }}_c{\mathrm{\partial }}_i{h}^c{}_j={\mathrm{\partial }}_k{\mathrm{\partial }}_i(-2\phi {\delta }_j^k)=-2{\delta }_{ij}$, likewise ${\mathrm{\partial }}_c{\mathrm{\partial }}_j{h}^c{}_i=-2{\delta }_{ij}$, $-\mathrm{\square }{h}_{ij}=2{\delta }_{ij}\mathrm{\Delta }\phi =6{\delta }_{ij}$ and $-{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}h=4{\mathrm{\partial }}_i{\mathrm{\partial }}_j\phi =4{\delta }_{ij}$, so ${\mathrm{Ric}}_{ij}^{(1)}=\frac{1}{2}(-2-2+6+4){\delta }_{ij}=3{\delta }_{ij}$.

(Scalar curvature.) By [L3] and [L2], ${\mathrm{\partial }}_a{\mathrm{\partial }}_b{h}^{ab}={\mathrm{\partial }}_i{\mathrm{\partial }}_j{h}^{ij}={\mathrm{\partial }}_i{\mathrm{\partial }}_j(-2\phi {\delta }^{ij})=-2\mathrm{\Delta }\phi =-6$, since $h^{00}=-2\phi$, $h^{0i}=0$, $h^{ij}=-2\phi {\delta }^{ij}$ and all time derivatives vanish; and $\mathrm{\square }h=\mathrm{\square }(-4\phi )=-4\mathrm{\Delta }\phi =-12$. Hence $S^{(1)}=-6+12=6$. As an arithmetic check on the defining relation $S=g^{ab}{\mathrm{Ric}}_{ab}$ of The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors, whose first-order expansion at $\lambda =0$ is $S^{(1)}={\eta }^{ab}{\mathrm{Ric}}_{ab}^{(1)}$ because $\mathrm{Ric}(0)=0$: the values of step 2.1 give $-{\mathrm{Ric}}_{00}^{(1)}+{\delta }^{ij}{\mathrm{Ric}}_{ij}^{(1)}=-3+9=6$, in agreement.

(Einstein tensor and cross-check.) By [L2] the trace reversal of step 1.3 is ${\bar{h}}_{00}=h_{00}-\frac{1}{2}{\eta }_{00}h=-2\phi -\frac{1}{2}(-1)(-4\phi )=-4\phi$, ${\bar{h}}_{ij}=-2\phi {\delta }_{ij}-\frac{1}{2}{\delta }_{ij}(-4\phi )=0$ and ${\bar{h}}_{0i}=0$; that is, ${\bar{h}}_{ab}=-4\phi {\delta }_{a0}{\delta }_{b0}$. The closed formula of [L3] then gives: for $(00)$, ${\mathrm{\partial }}^c{\mathrm{\partial }}^d{\bar{h}}_{cd}={\eta }^{00}{\eta }^{00}{\mathrm{\partial }}_0{\mathrm{\partial }}_0{\bar{h}}_{00}=0$ and ${\mathrm{\partial }}^c{\mathrm{\partial }}_{(0}{\bar{h}}_{0)c}=0$ by stationarity, so $G_{00}^{(1)}=-\frac{1}{2}\mathrm{\square }(-4\phi )=2\mathrm{\Delta }\phi =6$; for $(0i)$ and $(ij)$ every term of the closed formula involves a component ${\bar{h}}_{0\cdot }$ or a time derivative, hence vanishes, so $G_{0i}^{(1)}=G_{ij}^{(1)}=0$. Finally, the values satisfy $G_{ab}^{(1)}={\mathrm{Ric}}_{ab}^{(1)}-\frac{1}{2}{\eta }_{ab}{S}^{(1)}$: $3-\frac{1}{2}(-1)(6)=6$ for $(00)$, $3{\delta }_{ij}-\frac{1}{2}{\delta }_{ij}\cdot 6=0$ for $(ij)$ and $0-0=0$ for $(0i)$, which is the first-order expansion at $\lambda =0$ of the defining relation $G=\mathrm{Ric}-\frac{1}{2}Sg$ of The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors, using $S(0)=0$ and $g(0)=\eta$.