Lemma: Homogeneous reduction of the cubic source

Let ${\gamma }_{0\mu }=0$ and ${\gamma }_{ij}={\gamma }_{ij}(t)$ on an open set $W\subset M_T$. Then on $W$,

with spatial traces and contractions taken with ${\delta }_{ij}$, and ${{\mathcal{H}}_1}^{00}[\gamma ]=0$.

Proof. On $W$ the metric is $q=\mathrm{d}{t}^2-h_{ij}(t)\mathrm{d}{x}^i\mathrm{d}{x}^j$ with $h=\delta -\kappa \gamma (t)$. The nonzero Christoffel symbols of XI-A's convention are ${{\Gamma }^0}_{ij}=\frac{1}{2}{\dot{h}}_{ij}$ and ${{\Gamma }^i}_{0j}=K^i{}_j:=\frac{1}{2}{h}^{ik}{\dot{h}}_{kj}$. Direct substitution into $R_{\mu \nu }={\partial }_{\rho }{{\Gamma }^{\rho }}_{\mu \nu }-{\partial }_{\nu }{{\Gamma }^{\rho }}_{\mu \rho }+{{\Gamma }^{\rho }}_{\rho \lambda }{{\Gamma }^{\lambda }}_{\mu \nu }-{{\Gamma }^{\rho }}_{\nu \lambda }{{\Gamma }^{\lambda }}_{\mu \rho }$ gives $R_{0i}=0$, $R_{00}=-\dot{K}-K^i{}_j{K}^j{}_i$ with $K=K^i{}_i$, and $h^{ij}{R}_{ij}=\dot{K}+K^2$, so that $R=R_{00}-h^{ij}{R}_{ij}$ and

Since $q^{00}=1$ and $q^{0i}=0$, $G^{00}=G_{00}$ and $G^{0i}=0$ for every $\kappa$; hence ${J_3}^{0i}=0$. Expanding $h^{ik}={\delta }^{ik}+O(\kappa )$ and $\dot{h}=-\kappa \dot{\gamma }$ gives $K^i{}_j=-\frac{1}{2}\kappa {\dot{\gamma }}_{ij}+O({\kappa }^2)$, $K=-\frac{1}{2}\kappa \mathrm{tr}\dot{\gamma }+O({\kappa }^2)$, and $\sqrt{-q}=1+O(\kappa )$, so $\sqrt{-q}{G}^{00}=\frac{1}{8}{\kappa }^2[(\mathrm{tr}\dot{\gamma }{)}^2-{\dot{\gamma }}_{ij}{\dot{\gamma }}_{ij}]+O({\kappa }^3)$. Thus ${{\mathcal{H}}_1}^{00}=0$ and ${J_3}^{00}=2{{\mathcal{H}}_2}^{00}=q(\dot{\gamma },\dot{\gamma })$. The signature does not affect $G_{\mu \nu }$; the Friedmann check $G_{00}=3H^2$ for $h=a(t{)}^2\delta$ confirms the sign.