# 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?

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.
• 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 $\partial_\xi$. – Wein Eld Aug 5 '19 at 6:09
• 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). – Wein Eld Aug 5 '19 at 6:24
• 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. – A.V.S. Aug 5 '19 at 6:49