\quad H \equiv \frac{1}{a} \frac{{{\rm d}}a}{{{\rm d}}T},$$ and where $\rho=\rho_0(a_0/a)^3$ is the energy density of a pressureless fluid. A subscript $0$ indicates that the quantity is evaluated today. It is convenient to introduce the cosmological parameters $$\Omega_{\rm m}=\frac{8\pi G \rho_0}{3 H_0^2},
\quad
\Omega_K=-\frac{K}{a_0^2H_0^2},
\quad
\Omega_{\Lambda}=\frac{\Lambda}{3 H_0^2},$$ in terms of which the Friedmann equation takes the form $$\left(\frac{H}{H_0}\right)^2 = \Omega_{\rm m} \left( \frac{a_0}{a} \right)^3 + \Omega_{K} \left( \frac{a_0}{a} \right)^2 +\Omega_{\Lambda} .$$

### The “hole”—Kottler geometry

Inside the hole, the geometry is described by the extension of the Schwarzschild metric to the case of a nonzero cosmological constant, known as the Kottler solution [@Kot18; @Wey19] (see e.g. Ref. [@Per04] for a review). In spherical coordinates $(r,\theta,{\varphi})$, it reads $$\begin{aligned}
\label{eq1c}
{{\rm d}}s^2 &= -A(r) \, {{\rm d}}t^2 + A^{-1}(r) \, {{\rm d}}r^2 + r^2{{\rm d}}\Omega^2,\\
&\text{with} \quad A(r) \equiv 1 - \frac{r{_{\text{S}}}}{r} - \frac{\Lambda\,r^2}{3},\end{aligned}$$ and where $r{_{\text{S}}}\equiv 2\,GM$ is the Schwarzschild radius associated with the mass $M$ at the center of the hole. It is easy to check that the above metric describes a static spacetime. The corresponding Killing vector $\xi^\mu = \delta^\mu_0$ has norm $g_{\mu\nu}\xi^\mu\xi^\nu=A(r)$ and is therefore timelike as long as $A>0$. Hence, there are two cases:

1.  If $9(GM)^2\Lambda>1$, 