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What is the topology of a black hole spacetime? What about its horizon and its singularity?

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    $\begingroup$ This seems to have arisen from arguments in comments on an answer to this question and I would like to understand the correct resolution of the controversy. Thanks for making a separate question. $\endgroup$
    – G. Smith
    Jan 25 '20 at 6:20
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    $\begingroup$ @safesphere I don't think you can deduce the topology from Schwarzschild coordinates, because they are bad coordinates at $r=2M$. Of course, in the end the result is the same. And I'm pretty sure the manifold is simply connected. $\endgroup$
    – Javier
    Jan 25 '20 at 13:23
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    $\begingroup$ It is a general theorem of topology that a Cartesian product of two simply connected spaces is simply connected. And both of $R^2$ and $S^2$ are simply connected, so their product $R^2 \times S^2$ is indeed simply connected as well. $\endgroup$
    – Lee Mosher
    Jan 25 '20 at 15:13
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    $\begingroup$ @safesphere: The topology of an astrophysical black hole is not R4. This spacetime is a 2-to-1 cover (see: en.wikipedia.org/wiki/Covering_space) of both R4 and R2×S2 over the coordinate space. If you would like to write an answer explaining your reasoning, I'd be interested in seeing it. Re some of the other issues, we've been over some of this quite a few times, and I don't think this comment thread is the best place to try to clear that up. $\endgroup$
    – user4552
    Jan 27 '20 at 16:01
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    $\begingroup$ It’s a sad commentary on PSE that the OP, one of the dozen most valuable contributors in the entire history of this site, and an author of a textbook on General Relativity, decided to delete their account. $\endgroup$
    – G. Smith
    Jan 29 '20 at 23:06
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This is my self-answer based on looking around at papers and on the web. I've noted some issues that I'm uncertain about, and would be grateful if others could help with these. Thanks, safesphere and Javier, for helpful comments and for the clarification provided by this question on math.SE.

The topology of the maximal extension of the Schwarzschild spacetime, if you don't identify points in regions I and III, is $R^2\times S^2$. It seems to me that the most straightforward way to see this is to consider the embedding diagram of an Einstein-Rosen bridge, which is topologically a cylinder, and then consider the two suppressed coordinates, which are an angle and a time. See Misner, Thorne, and Wheeler, p. 837, fig 31.5a. For a more formal proof, see Monte 2011. This topology is simply connected. It's equivalent to $R^4$ with a line removed, which, as Javier pointed out in comments, makes sense if you "use ingoing Eddington-Finkelstein coordinates, in which the spacetime is just $R^4$ minus the line $r=0$."

For the plain old Schwarzschild metric, consisting only of regions I and II, Javier's comments have convinced me that the topology is also $R^2\times S^2$. This seems right because in the Penrose diagram, every point represents a two-sphere, and, unlike the Penrose diagram for Minkowski space, this one does not have a point at $r=0$ that is not a two-sphere.

For a black hole that forms by gravitational collapse, I think the topology is $R^4$, simply because there are theorems (Geroch, Borde) in GR that forbid topology change. (It's true that there are gravitational collapse solutions, such as the Oppenheimer-Snyder dust solution, that violate the causality assumptions of these theorems and do have topology change, but I don't think this kind of behavior is what most people think happens in real-world gravitational collapse.) This seems counterintuitive, because it would imply that I+II has a different topology than the actual astrophysical spacetime.

What specialists seem to find interesting is the topology of a time-slice of the horizon, and this is what they seem to mean when they talk about "the topology of a black hole." There is a well-known theorem by Hawking that says that a time-slice of the horizon has to have the topology $S^2$. This theorem uses global arguments and assumes the dominant energy condition. There are a bunch of other variations on this theorem, such as versions that are local rather than global, that assume a different energy condition, and that talk about the apparent horizon.

The topology of the singularity/-ies is not well defined. When we say, for example, that the singularity is a spacelike surface, we're actually talking about a set of specialized definitions that allow us to apply such words to a singularity, which is not a point-set on the manifold, and can't be measured by the metric. We can't even say how many dimensions the singularity has. For more on this, see Is a black hole singularity a single point? .

References

Borde, 1994, "Topology Change in Classical General Relativity," http://arxiv.org/abs/gr-qc/9406053

Geroch 1967, http://adsabs.harvard.edu/abs/1967JMP.....8..782G

Monte, 2011, "What is the topology of a Schwarzschild black hole?," https://arxiv.org/abs/1111.5790

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  • $\begingroup$ It seems to me that the geometry of regions I+III should also be $\mathbb{R}^2\times S^2$, since in the Penrose diagram you have a two-dimensional region (homeomorphic to the plane) in which each point is a sphere. Or use ingoing Eddington-Finkelstein coordinates, in which the spacetime is just $\mathbb{R}^4$ minus the line $r=0$. $\endgroup$
    – Javier
    Jan 25 '20 at 3:25
  • $\begingroup$ $\mathbb{R}^4$ with a line removed is simply connected: if you have a loop around the line, you can move it into the fourth dimension and shrink it to a point. And the argument does not apply to Minkowski spacetime, because its Penrose diagram has a boundary, the line $r=0$, in which each point is not a sphere but just a point. This fills the line-shaped hole in $\mathbb{R}^2\times S^2$. $\endgroup$
    – Javier
    Jan 25 '20 at 13:42
  • $\begingroup$ @safesphere No idea, really. We don't usually consider closed loops in spacetime, so I don't know where simple connectedness would come into play. The topology of space, on the other hand, is more interesting. $\endgroup$
    – Javier
    Jan 25 '20 at 17:09
  • $\begingroup$ How can $r=0$ be a line if we can’t even say how many dimensions the singularity has? $\endgroup$
    – G. Smith
    Jan 25 '20 at 19:28
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    $\begingroup$ It’s a sad commentary on PSE that the OP, one of the dozen most valuable contributors in the entire history of this site, and an author of a textbook on General Relativity, decided to delete their account. $\endgroup$
    – G. Smith
    Jan 29 '20 at 23:02
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Singularity:

Suppose we have the unit disc in $\mathbb R^2$ with a singularity at $r=0$. Then we would remove the point at $r=0$ and get the punctured unit disc $0<r<1$. Now we can apply a well-behaved (no singularities) coordinate transform $r\mapsto r+1$ to get a ring $(1,2)\times S_1$. This transforms the singularity from a point to the circle $r=1$, i.e. transforms the singularity from a single point to $S_1$.

You can even squeeze$^1$ the circle $r=1$ (figure-0 so to speak) to a line (figure-1), or you can squeeze it in such a way that it becomes a figure-8 or a circular line with an appendix$^3$ attached to it (figure-6): Even if we only allowed coordinate transforms that fix the dimension of the interesting part of the boundary, the topology is not well-defined.

Hence, the dimension and topology of the singularity depends on the choice of coordinates and has no physical meaning. To discuss the structure of a singularity, we have to fix the coordinate system, or at least make sure that coordinates in question belong to the same class.

These 2-dimensional examples generalize easily to 4 dimensions.

The intuitive idea behind "topology of a singularity" is this: Our coordinates are homeomorphic to an open subset$^2$ of $\mathbb R^4$, and the singularity is imagined as connected part of the boundary. Whilst this can be made mathematically rigorous, the result is not invariant even under smooth changes of coordinates.

$^1$ Again by smooth change of coordinates. For example, $z\mapsto z+1/z$ applied to $\{z\in \mathbb C \,/\, |z|\geqslant 1\}$ maps the circle $|z|=1$ onto the real line $\mathbb R \cap |z|\leqslant 1$. This map is smooth for all $|z|>1$.

$^2$ For each point in the set, there exists a subset engulfing / surrounding that point.

$^3$ With finite or infinite length at your disposal.

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