= Solution
For the expanding empty model, $\dot a=H_0$ in the preceding normalization. If $t_B$ denotes the origin of the expanding slicing,
$$
\boxed{a(t)=H_0(t-t_B),\qquad q_0=-\frac{a\ddot a}{\dot a^2}=0,\qquad t_0-t_B=H_0^{-1}.}
$$
The constant-time spatial slices are negatively curved <hyperbolic spaces>, with curvature radius $R_{\rm curv}=a/\sqrt{-k}=c(t-t_B)$ and three-dimensional scalar curvature $6k/a^2$. In the usual simply connected description they are copies of hyperbolic three-space. This is the <Milne model>: its four-dimensional spacetime is actually flat. Setting $T=(t-t_B)\cosh\chi$ and $R=c(t-t_B)\sinh\chi$ converts the metric to the interior of a future light cone in <Minkowski spacetime>. Its zero-scale-factor boundary is a slicing boundary, not a curvature singularity.
Add separately conserved <pressureless matter>, with present mass-equivalent density $\rho_{M0}$. Define the <cosmological density parameters>
$$
\Omega_M=\frac{8\pi G\rho_{M0}}{3H_0^2},\qquad \Omega_K=-\frac{kc^2}{H_0^2},\qquad\Omega_M+\Omega_K=1.
$$
Then
$$
\boxed{H^2=H_0^2(\Omega_Ma^{-3}+\Omega_Ka^{-2}),\qquad q(a)=\frac{\Omega_M}{2(\Omega_M+\Omega_Ka)},\qquad q_0=\frac{\Omega_M}{2}.}
$$
Curvature and dust have equal magnitudes at $a_K=\Omega_M/\Omega_K$, giving
$$
\boxed{1+z_K=\frac{\Omega_K}{\Omega_M},\qquad z_K=\frac{\Omega_K}{\Omega_M}-1.}
$$
For the almost-empty open case $0<\Omega_M<1/2$, curvature domination begins at positive redshift. Equality is today for $\Omega_M=1/2$ and in the future for $1/2<\Omega_M<1$. With $\Omega_M=0$, curvature dominates at all times and there is no finite matter–curvature equality; with $\Omega_K=0$, curvature never dominates.
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