A question about black hole interior I am reading Maldacena's paper "Eternal black holes in anti-de Sitter".
In the first paragraph, he wrote something about the black hole interior (in AdS):

The regions close to the spacetime singularities can be viewed as big-bang or big-crunch cosmologies (which are homogeneous but not isotropic).

Here I understand that why the regions can be viewed as big-bang or big-crunch cosmologies. But I do not understand why they are not isotropic. If we look at the metric, there are no $\theta$- and $\phi$-dependencies. And it seems that the spacetimes inside the black holes are also isotropic. What did I miss?
 A: Let us write the metric inside the Schwarzschild black hole in a form more suitable for cosmology:
$$
ds^2 = -d\tau^2+a(\tau)^2(d\theta^2 + \sin^2 \theta\, d\phi^2)+ b(\tau)^2 d\xi^2.
$$ 
Here the $\xi$ variable is related to Schwarzschild's time outside the horizon, while the $\tau$ is connected to Schwarzschild's outside radial coordinate via $\sqrt{|g_{rr}|}dr = d \tau$. This class of cosmological spacetimes is called Kantowski–Sachs cosmologies (and a typical feature of these cosmologies is geodesical incompleteness, we would need a different chart to describe the outside of a black hole). 
Let us consider a spatial slice $\tau=\mathrm{const}$ of constant cosmological time. This slice has a topology $S_2 \times \mathbb{R} $, a product of 2-sphere and a real line. The isometries of the solution act on this space transitively: the cosmology is homogeneous. But it is not isotropic: different directions within this slice are not equivalent. For instance the direction given by $\partial_\xi$ is clearly singled out: the space is infinite along the $\xi$ coordinate and it is clearly finite in any direction orthogonal to it, additionally the curvature tensor is anisotropic.
As for “there are no $θ$- and $ϕ$-dependencies”, this does not imply isotropic cosmology since we cannot interpret $θ$ and $ϕ$ coordinates as angles on celestial sphere seen at some spacetime point, these are simply spatial coordinates. Consequently $SO(3)$ isometries do not correspond to isotropy group of any spacetime 
point (all the isometries that keep a given point fixed). Instead isotropy group of any point is just $SO(2)$, corresponding to rotations of a sphere $\tau=\mathrm{const}$, $\xi=\mathrm{const}$ keeping this point fixed.
