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General relativity is defined on a base manifold which, viewed as a topological space, is simply connected (which means there's no holes). However, we know that inside a black hole there's a singularity, a point of infinite density and curvature.

The way this was explained to me was thinking of it as a literal hole on the fabric of space-time, a point in which nothing is defined. This, however yields a serious problem. How is it possible that a theory defined on a simply connected space predicts holes on it? Wouldn't the prediction of holes in a space assumed to be simply connected prove the theory conceptually wrong?

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  • $\begingroup$ It has not been proven that black holes contain singularities because centrifugal forces resulting from angular momentum would allow particles to travel paths without end. $\endgroup$
    – Wookie
    Commented Apr 21 at 9:59
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    $\begingroup$ From physics.stackexchange.com/a/144458/123208 "A singularity in GR is like a piece that has been cut out of the manifold. It's not a point or point-set at all. Because of this, formal treatments of singularities have to do a lot of nontrivial things to define stuff that would be trivial to define for a point set." Also note that the singularity is never in the past light-cone of any observer, even an observer inside the event horizon. $\endgroup$
    – PM 2Ring
    Commented Apr 21 at 10:24
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    $\begingroup$ @PM2Ring That last statement is the essence of the strong cosmic censorship conjecture, which is not yet fully proven (especially for charged black holes). $\endgroup$
    – TimRias
    Commented Apr 21 at 12:29
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    $\begingroup$ You are questioning fundamental properties of a theory while making multiple errors. (1) GR is defined on a metric manifold. It doesn’t have to be simply connected. For example, a flat universe can be a 3-torus, which is not simply connected. (2) Simply connected does not mean no holes per se. Again, in a flat 3-torus, you can connect any 2 points with a straight line without encountering any holes. (3) A singularity is not a point. In the simplest Schwarzschild case it is an infinitely long coordinate line being a moment of time. (4) A singularity has no density, much less infinite density. $\endgroup$
    – safesphere
    Commented Apr 21 at 14:08
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    $\begingroup$ @JanG I’ve deleted my last couple comments as misdirected, sorry. Yes, the space and time coordinates flip only in the vacuum Schwarzschild solution, because it is eternal. My comments above referred to this solution. I agree that this solution does not exist in reality, because the universe is not eternal in the past. The point of my initial comment is not that the theory is perfect, but that the OP cannot hope to find a flaw in it by using severely flawed arguments. $\endgroup$
    – safesphere
    Commented Apr 25 at 2:54

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For physical reasons we usually assume that the spacetime manifold is connected, but it is not necessary to assume that it is simply connected. However, simply connected manifolds do have some nice properties that allow certain proofs or theorems which fail in non-simply connected manifolds.

One common class is regarding the existence of geodesics connecting two nearby points, meaning points close enough that the effects of curvature are negligible. In non-simply connected manifolds it is possible that there is no geodesic path connecting two such events because all paths must turn to avoid a small hole.

The Einstein field equations, the core of general relativity, can be well defined on a manifold with a hole. So despite the problems with geodesics, the curvature, the metric, and the stress energy tensor can still work.

However, having said all the above, the Schwarzschild spacetime is simply connected. The singularity in the Schwarzschild spacetime is spacelike, meaning that it is a moment in time, specifically the end of time.

The topology of the Schwarzschild spacetime is $R^2 \times S^2$. Both $R^2$ and $S^2$ are simply connected. So their product is simply connected also.

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  • $\begingroup$ Even for a simply connected manifold it is not guaranteed that any two points be connected by a geodesic. (E.g. take the submanifold of the euclidean plane described by a (fat) letter C). Additional assumptions/restrictions are required. $\endgroup$
    – TimRias
    Commented Apr 21 at 16:45
  • $\begingroup$ @TimRias good point. I will need to revise that part $\endgroup$
    – Dale
    Commented Apr 21 at 16:57
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    $\begingroup$ +1, great answer. For completeness, I would like to add that in GR we usually assume spacetime to be connected (without further adjectives), since we wouldn't have access to other connected components anyway and hence they would be unphysical. $\endgroup$
    – Níckolas Alves
    Commented Apr 21 at 17:21
  • $\begingroup$ @NíckolasAlves thanks for the suggestion. It is nice to get two good suggestions that actually improve the answer. $\endgroup$
    – Dale
    Commented Apr 21 at 18:09

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