First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder

(Christoffel expansion and remainder bound.) By [L2], ${\mathrm{\partial }}_{\lambda }{g}_{ab}(\lambda ,p){\mathrm{\mid }}_{\lambda =0}=h_{ab}(p)$, and the metric components are smooth in $(\lambda ,p)$; by [L1] the Christoffel symbols are smooth functions of the components and their first derivatives, hence $(\lambda ,p)\mapsto {\mathrm{\Gamma }}^a{}_{bc}(\lambda ,p)$ is smooth. The chain rule gives ${\mathrm{\Gamma }}^{(1)a}{}_{bc}={\mathrm{\partial }}_{\lambda }{\mathrm{\Gamma }}^a{}_{bc}{\mathrm{\mid }}_0=\frac{1}{2}{\eta }^{ad}({\mathrm{\partial }}_b{h}_{cd}+{\mathrm{\partial }}_c{h}_{bd}-{\mathrm{\partial }}_d{h}_{bc})$: the term differentiating the factor $g^{ad}(\lambda )$ vanishes at $\lambda =0$ because ${\mathrm{\partial }}_e{g}_{cd}(0)=0$. For the bound, if $K=\mathrm{\varnothing }$ choose $C_K=0$, since all estimates in (a)-(c) are vacuous. Otherwise fix the nonempty compact $K$; its product with $[-{\epsilon }_1\mathrm{/}2,{\epsilon }_1\mathrm{/}2]$ is closed and bounded in ${\mathbb{R}}^5$, hence compact by [L4]. Note that ${\mathrm{\partial }}_{\lambda }^2{\mathrm{\Gamma }}^a{}_{bc}(\lambda ,p)$ is continuous on the compact set $\bar{K}\times [-{\epsilon }_1\mathrm{/}2,{\epsilon }_1\mathrm{/}2]$, hence bounded there by a constant $C_K$ by [L4]; applying [L3] to $\lambda \mapsto {\mathrm{\Gamma }}^a{}_{bc}(\lambda ,p)$ for each fixed $p\in K$ gives the estimate of (a).

(First-order coefficient of the inverse.) Differentiating the identity ${\sum }_c{g}^{ac}(\lambda ,p)g_{cb}(\lambda ,p)={\delta }^a{}_b$ at $\lambda =0$ and using $g_{cb}(0)={\eta }_{cb}$, ${\mathrm{\partial }}_{\lambda }{g}_{cb}(0)=h_{cb}$ gives ${\sum }_c({\mathrm{\partial }}_{\lambda }{g}^{ac}(0)){\eta }_{cb}+h^a{}_b=0$, hence ${\mathrm{\partial }}_{\lambda }{g}^{ab}(0)=-{\eta }^{ac}{h}_{cd}{\eta }^{db}$.

(Gauge transformation algebra.) Compute with the definitions of Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge clause (c)-(d): the trace of $h^{\mathrm{\prime }}$ is $h^{\mathrm{\prime }}={\eta }^{ab}(h_{ab}-{\mathrm{\partial }}_a{\xi }_b-{\mathrm{\partial }}_b{\xi }_a)=h-2{\mathrm{\partial }}^c{\xi }_c$, so ${\bar{h}}^{\mathrm{\prime }}{}_{ab}=h_{ab}^{\mathrm{\prime }}-\frac{1}{2}{\eta }_{ab}{h}^{\mathrm{\prime }}=h_{ab}-{\mathrm{\partial }}_a{\xi }_b-{\mathrm{\partial }}_b{\xi }_a-\frac{1}{2}{\eta }_{ab}h+{\eta }_{ab}{\mathrm{\partial }}^c{\xi }_c={\bar{h}}_{ab}-{\mathrm{\partial }}_a{\xi }_b-{\mathrm{\partial }}_b{\xi }_a+{\eta }_{ab}{\mathrm{\partial }}^c{\xi }_c\mathrm{.}$ Contracting with ${\mathrm{\partial }}^a$ gives ${\mathrm{\partial }}^a{\bar{h}}^{\mathrm{\prime }}{}_{ab}={\mathrm{\partial }}^a{\bar{h}}_{ab}-{\mathrm{\partial }}^a{\mathrm{\partial }}_a{\xi }_b-{\mathrm{\partial }}^a{\mathrm{\partial }}_b{\xi }_a+{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c={\mathrm{\partial }}^a{\bar{h}}_{ab}-\mathrm{\square }{\xi }_b,$ because the last two terms cancel by [L3]: ${\mathrm{\partial }}^a{\mathrm{\partial }}_b{\xi }_a={\mathrm{\partial }}_b{\mathrm{\partial }}^a{\xi }_a={\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c$. Hence ${\mathrm{\partial }}^a{\bar{h}}^{\mathrm{\prime }}{}_{ab}=0$ holds exactly when $\mathrm{\square }{\xi }_b={\mathrm{\partial }}^a{\bar{h}}_{ab}$.

(Ricci expansion.) By [L1], ${\mathrm{Ric}}_{bd}(\lambda )={\mathrm{\partial }}_a{\mathrm{\Gamma }}^a{}_{db}(\lambda )-{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{ab}(\lambda )+{\sum }_e({\mathrm{\Gamma }}^a{}_{ae}{\mathrm{\Gamma }}^e{}_{db}-{\mathrm{\Gamma }}^a{}_{de}{\mathrm{\Gamma }}^e{}_{ab})(\lambda )$ in the chart, so differentiating at $\lambda =0$ and using $\mathrm{\Gamma }(0)=0$ gives ${\mathrm{Ric}}_{bd}^{(1)}={\mathrm{\partial }}_a{\mathrm{\Gamma }}^{(1)a}{}_{db}-{\mathrm{\partial }}_d{\mathrm{\Gamma }}^{(1)a}{}_{ab}$. By step 1.1, ${\mathrm{\Gamma }}^{(1)a}{}_{ab}=\frac{1}{2}{\eta }^{ae}({\mathrm{\partial }}_a{h}_{eb}+{\mathrm{\partial }}_b{h}_{ea}-{\mathrm{\partial }}_e{h}_{ab})=\frac{1}{2}({\mathrm{\partial }}^e{h}_{eb}+{\mathrm{\partial }}_{b}h-{\mathrm{\partial }}^e{h}_{eb})=\frac{1}{2}{\mathrm{\partial }}_{b}h$, so ${\mathrm{\partial }}_d{\mathrm{\Gamma }}^{(1)a}{}_{ab}=\frac{1}{2}{\mathrm{\partial }}_d{\mathrm{\partial }}_{b}h$. Substituting the formula of step 1.1 into the first term gives ${\mathrm{Ric}}_{bd}^{(1)}=\frac{1}{2}({\mathrm{\partial }}_c{\mathrm{\partial }}_d{h}^c{}_b+{\mathrm{\partial }}_c{\mathrm{\partial }}_b{h}^c{}_d-\mathrm{\square }{h}_{bd}-{\mathrm{\partial }}_b{\mathrm{\partial }}_{d}h),$ which is the formula of (b) after renaming $b,d$ as $a,b$. The remainder bound of (b) follows from [L3] as in step 1.1, since $\lambda \mapsto {\mathrm{Ric}}_{bd}(\lambda ,p)$ is $C^2$ and its second derivative is bounded on $\bar{K}\times [-{\epsilon }_1\mathrm{/}2,{\epsilon }_1\mathrm{/}2]$ by [L4].

(Scalar and Einstein expansion.) Since $S(\lambda )={\sum }_{a,b}{g}^{ab}(\lambda ){\mathrm{Ric}}_{ab}(\lambda )$ and ${\mathrm{Ric}}_{ab}(0)=0$, differentiation at $0$ with steps 1.2 and 2.1 gives $S^{(1)}={\eta }^{ab}{\mathrm{Ric}}_{ab}^{(1)}=\frac{1}{2}({\mathrm{\partial }}_c{\mathrm{\partial }}^b{h}^c{}_b+{\mathrm{\partial }}_c{\mathrm{\partial }}^a{h}^c{}_a-\mathrm{\square }h-\mathrm{\square }h)={\mathrm{\partial }}_a{\mathrm{\partial }}_b{h}^{ab}-\mathrm{\square }h$. Since $G_{ab}(\lambda )={\mathrm{Ric}}_{ab}(\lambda )-\frac{1}{2}S(\lambda )g_{ab}(\lambda )$ and $S(0)=0$, $G_{ab}^{(1)}={\mathrm{Ric}}_{ab}^{(1)}-\frac{1}{2}{\eta }_{ab}{S}^{(1)}$. Substituting the formulas of step 2.1 and of $S^{(1)}$ and expanding with ${\bar{h}}_{ab}=h_{ab}-\frac{1}{2}{\eta }_{ab}h$: $\mathrm{\square }{\bar{h}}_{ab}=\mathrm{\square }{h}_{ab}-\frac{1}{2}{\eta }_{ab}\mathrm{\square }h$; ${\mathrm{\partial }}^c{\mathrm{\partial }}_{(a}{\bar{h}}_{b)c}=\frac{1}{2}({\mathrm{\partial }}_a{\mathrm{\partial }}_c{h}^c{}_b+{\mathrm{\partial }}_b{\mathrm{\partial }}_c{h}^c{}_a)-\frac{1}{2}{\mathrm{\partial }}_a{\mathrm{\partial }}_{b}h$; and ${\mathrm{\partial }}^c{\mathrm{\partial }}^d{\bar{h}}_{cd}={\mathrm{\partial }}^c{\mathrm{\partial }}^d{h}_{cd}-\frac{1}{2}\mathrm{\square }h$. Collecting terms gives exactly ${\mathrm{Ric}}_{ab}^{(1)}-\frac{1}{2}{\eta }_{ab}{S}^{(1)}=-\frac{1}{2}\mathrm{\square }{\bar{h}}_{ab}+{\mathrm{\partial }}^c{\mathrm{\partial }}_{(a}{\bar{h}}_{b)c}-\frac{1}{2}{\eta }_{ab}{\mathrm{\partial }}^c{\mathrm{\partial }}^d{\bar{h}}_{cd}$, the formula of (c). The remainder bounds follow as in step 2.1.

(Quadratic remainder in length-scaled jet coordinates.) If $K=\mathrm{\varnothing }$, the convention in (f) gives $B_K(A)=0$; take ${\delta }_0=1$ and $C_K=0$, and the signature and residual assertions at $p\in K$ are vacuous. Assume now $K\mathrm{\ne }\mathrm{\varnothing }$. Fix $L>0$ and let $U(p)=(A,L\mathrm{\partial }A,L^2{\mathrm{\partial }}^{2}A)$ in the finite-dimensional real vector space of symmetric-matrix second jets (symmetric in the matrix indices and in the second-derivative indices), with maximum-entry norm $m(U)$. Write $b_j$ for the real supremum on $K$ of the maximum magnitude of the order-$j$ derivatives of $A$, finite by [L4]. Then $B_K(A)=b_0+L{b}_1+L^2{b}_2$ and ${\sup }_{K}m(U)=\max \{b_0,L{b}_1,L^2{b}_2\}\le B_K(A)\le 3{\sup }_{K}m(U)$. Define $F_{ab}(U)=L^2{G}_{ab}(\eta +A)$ by the chart formulas of [L1]-[L2], using the jet encoded by $U$. These formulas make $F$ rational smooth on the open jet domain where $\eta +A$ is invertible. Choose $0<d<\rho$ with $\rho$ from [L5], and let $Q=\{U:m(U)\le d\}$. In independent jet coordinates this is a closed Euclidean box, hence compact by [L4]; all its zeroth-order matrices are Lorentzian by [L5], so $F$ is smooth on an open neighborhood of $Q$. Its finitely many second partial derivatives are continuous and bounded there by [L4]; summing their bounds gives $H<\mathrm{\infty }$ such that $\mathrm{\mid }{D}^2{F}_{ab}(U)[V,V]\mathrm{\mid }\le Hm(V{)}^2$ on $Q$, uniformly in the finitely many $a,b$. Also $F(0)=0$ and the pointwise differentiation of step 3.1 gives $DF(0)[U]=L^2{G}_{ab}^{(1)}(A)$. This is a local jet calculation: for each fixed $p$ continuity gives an invertible metric for sufficiently small parameter near $0$, so no globally uniform family $\eta +tA$ is assumed to obtain the derivative. Take ${\delta }_0=d\mathrm{/}2$. If $B_K(A)\le {\delta }_0$, then $m(U(p))\le d\mathrm{/}2$ and $tU(p)\in Q$ for every $p\in K$, $0\le t\le 1$; $\eta +A$ is Lorentzian at every such $p$ by [L5] and on an open neighborhood of $K$ by smoothness and openness. Applying [L3] to $t\mapsto F_{ab}(tU(p))$ yields $\mathrm{\mid }{G}_{ab}(\eta +A)(p)-G_{ab}^{(1)}(A)(p)\mathrm{\mid }\le \frac{Hm(U(p){)}^2}{2L^2}\le \frac{H{B}_K(A{)}^2}{2L^2}\mathrm{.}$ Thus (f) holds with $C_K=H\mathrm{/}2$. The compact Taylor domain is the finite-dimensional jet box, and the Einstein tensor is evaluated on the open Lorentzian neighborhood just established.

(Harmonic-gauge reduction.) The first two displays of (d) are step 1.3. If $\mathrm{\square }{\xi }_b={\mathrm{\partial }}^a{\bar{h}}_{ab}$ then ${\mathrm{\partial }}^a{\bar{h}}^{\mathrm{\prime }}{}_{ab}=0$, so in the formula of step 3.1 applied to ${\bar{h}}^{\mathrm{\prime }}$ both divergence terms vanish and $G_{ab}^{(1)}(h^{\mathrm{\prime }})=-\frac{1}{2}\mathrm{\square }{\bar{h}}^{\mathrm{\prime }}{}_{ab}$, proving the last display of (d).

(Gauge invariance.) Let $H_{ab}:={\mathrm{\partial }}_a{\xi }_b+{\mathrm{\partial }}_b{\xi }_a$, so that $h_{ab}^{\mathrm{\prime }}=h_{ab}-H_{ab}$. By step 2.1, ${\mathrm{Ric}}_{ab}^{(1)}(H)=\frac{1}{2}({\mathrm{\partial }}_c{\mathrm{\partial }}_a{H}^c{}_b+{\mathrm{\partial }}_c{\mathrm{\partial }}_b{H}^c{}_a-\mathrm{\square }{H}_{ab}-{\mathrm{\partial }}_a{\mathrm{\partial }}_{b}H)\mathrm{.}$ Substituting $H_{ab}={\mathrm{\partial }}_a{\xi }_b+{\mathrm{\partial }}_b{\xi }_a$, so $H=2{\mathrm{\partial }}^c{\xi }_c$, and using $H^c{}_b={\mathrm{\partial }}^c{\xi }_b+{\mathrm{\partial }}_b{\xi }^c$: ${\mathrm{\partial }}_c{\mathrm{\partial }}_a{H}^c{}_b={\mathrm{\partial }}_a\mathrm{\square }{\xi }_b+{\mathrm{\partial }}_a{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c$, ${\mathrm{\partial }}_c{\mathrm{\partial }}_b{H}^c{}_a={\mathrm{\partial }}_b\mathrm{\square }{\xi }_a+{\mathrm{\partial }}_a{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c$, $\mathrm{\square }{H}_{ab}={\mathrm{\partial }}_a\mathrm{\square }{\xi }_b+{\mathrm{\partial }}_b\mathrm{\square }{\xi }_a$ and ${\mathrm{\partial }}_a{\mathrm{\partial }}_{b}H=2{\mathrm{\partial }}_a{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c$; the sum is ${\mathrm{\partial }}_a\mathrm{\square }{\xi }_b+{\mathrm{\partial }}_b\mathrm{\square }{\xi }_a+2{\mathrm{\partial }}_a{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c-{\mathrm{\partial }}_a\mathrm{\square }{\xi }_b-{\mathrm{\partial }}_b\mathrm{\square }{\xi }_a-2{\mathrm{\partial }}_a{\mathrm{\partial }}_b{\mathrm{\partial }}^c{\xi }_c=0$, so ${\mathrm{Ric}}_{ab}^{(1)}(H)=0$. By linearity of the formula of step 2.1 in its argument, ${\mathrm{Ric}}^{(1)}(h-H)={\mathrm{Ric}}^{(1)}(h)$; by step 3.1, $S^{(1)}(H)={\eta }^{ab}{\mathrm{Ric}}_{ab}^{(1)}(H)=0$ and $G_{ab}^{(1)}(H)={\mathrm{Ric}}_{ab}^{(1)}(H)-\frac{1}{2}{\eta }_{ab}{S}^{(1)}(H)=0$, so $G^{(1)}(h-H)=G^{(1)}(h)$. This proves (e).