A first-order vacuum family whose exact metric has nonzero Ricci tensor at order epsilon^2: a harmonic diagonal perturbation with phi = x^1

(The family.) The fields $h$ and $q\equiv 0$ are smooth and stationary, and for $0<\mathrm{\mid }\lambda \mathrm{\mid }<{\epsilon }_1=1\mathrm{/}4$ the exact metric matrix is $\mathrm{diag}(-p,q,q,q)$. By [L6], both $p$ and $q$ are positive everywhere on $\mathrm{\Omega }$, so [L5] gives the Lorentzian signature $(-,+,+,+)$ directly. Thus every $g(\lambda )$ is a smooth Lorentzian metric on $\mathrm{\Omega }$ and $({\epsilon }_1,h,q)$ is a flat linearized family.

(First-order vacuum.) By [L1], $h={\eta }^{ab}{h}_{ab}=2\phi -6\phi =-4\phi$, so ${\bar{h}}_{00}=-2\phi -\frac{1}{2}(-1)(-4\phi )=-4\phi$ and ${\bar{h}}_{0i}=0$, ${\bar{h}}_{ij}=-2\phi {\delta }_{ij}-\frac{1}{2}{\delta }_{ij}(-4\phi )=0$; that is, ${\bar{h}}_{ab}=-4\phi {\delta }_{a0}{\delta }_{b0}$. With $\mathrm{\square }\phi =\mathrm{\Delta }\phi =0$ and stationarity, the closed formula of [L2] gives $G_{00}^{(1)}=-\frac{1}{2}\mathrm{\square }(-4\phi )=2\mathrm{\Delta }\phi =0$, while ${\mathrm{\partial }}^c{\mathrm{\partial }}^d{\bar{h}}_{cd}={\eta }^{00}{\eta }^{00}{\mathrm{\partial }}_0{\mathrm{\partial }}_0{\bar{h}}_{00}=0$ and every divergence term involves ${\bar{h}}_{0\cdot }$ components or time derivatives; the $(0i)$ and $(ij)$ components contain only such terms as well. Hence $G_{ab}^{(1)}(h)=0$; by the linearity of [L2], $G_{ab}^{(1)}(\lambda h)=\lambda G_{ab}^{(1)}(h)=0$ for every $\lambda$.

(Exact Christoffel symbols.) In the coordinate frame $g(\lambda )=\mathrm{diag}(-p,q,q,q)$, so $g^{00}=-1\mathrm{/}p$ and $g^{ii}=1\mathrm{/}q$; the general formula of [L3] and [L6] give, and all other components vanish: ${\mathrm{\Gamma }}^0{}_{01}={\mathrm{\Gamma }}^0{}_{10}={\mathrm{\partial }}_{1}p\mathrm{/}(2p)=\lambda \mathrm{/}p$; ${\mathrm{\Gamma }}^1{}_{00}={\mathrm{\partial }}_{1}p\mathrm{/}(2q)=\lambda \mathrm{/}q$; ${\mathrm{\Gamma }}^1{}_{11}={\mathrm{\partial }}_{1}q\mathrm{/}(2q)=-\lambda \mathrm{/}q$; ${\mathrm{\Gamma }}^1{}_{22}={\mathrm{\Gamma }}^1{}_{33}=-{\mathrm{\partial }}_{1}q\mathrm{/}(2q)=\lambda \mathrm{/}q$; ${\mathrm{\Gamma }}^2{}_{12}={\mathrm{\Gamma }}^2{}_{21}={\mathrm{\Gamma }}^3{}_{13}={\mathrm{\Gamma }}^3{}_{31}={\mathrm{\partial }}_{1}q\mathrm{/}(2q)=-\lambda \mathrm{/}q$. (For example ${\mathrm{\Gamma }}^1{}_{22}=\frac{1}{2}{g}^{11}({\mathrm{\partial }}_2{g}_{12}+{\mathrm{\partial }}_2{g}_{21}-{\mathrm{\partial }}_1{g}_{22})=\frac{1}{2}(1\mathrm{/}q)(-{\mathrm{\partial }}_{1}q)$, since only the ${\mathrm{\partial }}_1$ derivative of the diagonal entries is nonzero; all components with a time derivative vanish by stationarity.)

(Exact Ricci component.) By [L3] and step 1.3, $R^a{}_{0a0}={\mathrm{\partial }}_a{\mathrm{\Gamma }}^a{}_{00}-{\mathrm{\partial }}_0{\mathrm{\Gamma }}^a{}_{a0}+{\mathrm{\Gamma }}^a{}_{ae}{\mathrm{\Gamma }}^e{}_{00}-{\mathrm{\Gamma }}^a{}_{0e}{\mathrm{\Gamma }}^e{}_{a0}$ and ${\mathrm{Ric}}_{00}=R^a{}_{0a0}$. The terms are: ${\mathrm{\partial }}_a{\mathrm{\Gamma }}^a{}_{00}={\mathrm{\partial }}_1(\lambda \mathrm{/}q)=2{\lambda }^2\mathrm{/}{q}^2$ by [L6]; ${\mathrm{\partial }}_0{\mathrm{\Gamma }}^a{}_{a0}=0$; ${\mathrm{\Gamma }}^a{}_{a1}={\mathrm{\Gamma }}^0{}_{01}+{\mathrm{\Gamma }}^1{}_{11}+{\mathrm{\Gamma }}^2{}_{21}+{\mathrm{\Gamma }}^3{}_{31}=\lambda \mathrm{/}p-3\lambda \mathrm{/}q$, so ${\mathrm{\Gamma }}^a{}_{ae}{\mathrm{\Gamma }}^e{}_{00}={\mathrm{\Gamma }}^a{}_{a1}{\mathrm{\Gamma }}^1{}_{00}={\lambda }^2\mathrm{/}(pq)-3{\lambda }^2\mathrm{/}{q}^2$; and ${\mathrm{\Gamma }}^a{}_{0e}{\mathrm{\Gamma }}^e{}_{a0}={\mathrm{\Gamma }}^0{}_{01}{\mathrm{\Gamma }}^1{}_{00}+{\mathrm{\Gamma }}^1{}_{00}{\mathrm{\Gamma }}^0{}_{10}=2{\lambda }^2\mathrm{/}(pq)$. Summing, ${\mathrm{Ric}}_{00}=2{\lambda }^2\mathrm{/}{q}^2+{\lambda }^2\mathrm{/}(pq)-3{\lambda }^2\mathrm{/}{q}^2-2{\lambda }^2\mathrm{/}(pq)=-{\lambda }^2\mathrm{/}{q}^2-{\lambda }^2\mathrm{/}(pq)=-{\lambda }^2(p+q)\mathrm{/}(p{q}^2)=-2{\lambda }^2\mathrm{/}(p{q}^2)$ since $p+q=2$ by [L6].

(Refutation.) At the origin $x=0$ one has $p=q=1$, so step 2.1 gives ${\mathrm{Ric}}_{00}(g(\lambda ))=-2{\lambda }^2\mathrm{\ne }0$ for every $0<\mathrm{\mid }\lambda \mathrm{\mid }<{\epsilon }_1$; hence the exact Ricci tensor does not vanish on $\mathrm{\Omega }$. If the exact Einstein tensor vanished, then contracting $G_{ab}={\mathrm{Ric}}_{ab}-\frac{1}{2}S{g}_{ab}$ of [L3] with $g^{ab}$ would give $S=-g^{ab}{G}_{ab}=0$ and then ${\mathrm{Ric}}_{ab}=G_{ab}+\frac{1}{2}S{g}_{ab}=0$, contradicting ${\mathrm{Ric}}_{00}\mathrm{\ne }0$. So $G_{ab}(g(\lambda ))\mathrm{\ne }0$ on $\mathrm{\Omega }$ although $G_{ab}^{(1)}(\lambda h)=0$ by step 1.2, which refutes the statement. The failure is of order ${\lambda }^2$, consistent with the quadratic remainder bound of First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clause (f).