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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?

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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.

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  • $\begingroup$ Thanks for your answer but I am not sure I fully agree with it. I now understand the difference between “there are no θ- and ϕ-dependencies” and isotropy. But I still cannot see why the spacetime you wrote down is not isotropic. You argue that within the slice $\tau=const$, directions are not equivalent by taking the example on $\partial_\xi$. $\endgroup$ – Wein Eld Aug 5 '19 at 6:09
  • $\begingroup$ At the slice $\tau$=const, we have three spatial coordinates $\xi$,$\theta$ and $\phi$. Certainly $\xi$ is special compared with the other two because it is the radial coordinate while $\theta$ and $\phi$ are the angular coordinates. And I do not understand what do you mean by "the space is infinite along the $\xi$ coordinate and it is clearly finite in any direction orthogonal to it". I assume the directions orthogonal to $\partial_\xi$ are $\partial_\theta$ and $\partial_\phi$ (because there are no terms like $d\xi d\theta$ in the metric). $\endgroup$ – Wein Eld Aug 5 '19 at 6:24
  • $\begingroup$ But these coordinates are angular coordinates, shouldn't it be natural that along with them the space is compact? If this logic validates, then the spatial hyperbolic slice in flat spacetime is also not isotropic. Is that true? $\endgroup$ – Wein Eld Aug 5 '19 at 6:28
  • $\begingroup$ these coordinates are angular coordinates … like I said, these are not angles. You could interpret $\theta$ and $\phi$ as angles only if there is a point near which the spatial metric behaves like $dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)+\text{h.o.t.}$ (h.o.t. -higher order terms). There is no such point on the slice $\tau=\text{const}$, $a(t)$ is constant on this slice. $\endgroup$ – A.V.S. Aug 5 '19 at 6:49

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