In most general relativity texts, the singularity is treated as a point removed from the manifold, to avoid having to deal with the infinite curvature of the Ricci scalar.

But in the case of a more realistic scenario (let's say a spherically symmetric collapse, for instance), does this idea really hold up? We start with a, let's say, (partial) Cauchy surface, with topology $\Bbb R^3$. The matter collapses, leading to a black hole. I'm not quite sure of the topology of a non-maximally extended black hole, especially one from stellar collapse (in particular I'm not sure if the singularity matters all that much since I have a suspiscion it might not affect the topology, as it may be at timelike infinity). If the topology of the spacelike hypersurface is of the form $\Bbb R^3 \setminus \{0\}$, does that not violate theorems on topology change? As I'm pretty sure it will not violate conditions of that theorem on time orientability or closed timelike curves, does the metric have a degenerate point?

I know that there are also ways to treat singularities without removing the point (ie with generalized functions), and that the problem probably doesn't happen in quantum gravity, but is it a problem when it comes to classical general relativity?

  • 1
    $\begingroup$ You can't get a "realistic" singularity out of GR, which doesn't describe matter, it doesn't even describe spacetime, it only describes the distortion of spacetime by matter and the movement of that matter in its own distortion, but we don't even know up to what scale the theory is valid (and it's not clear there is a non-suicidal way of finding out). $\endgroup$
    – CuriousOne
    May 31, 2016 at 11:16
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    $\begingroup$ Once again @CuriousOne has posted the answer in a comment. The fact that a singularity appears in the theory is more an indicator of a failure of the theory than it is an indicator of what the center of a black hole is like. Many theorists think there probably isn't a physical singularity. $\endgroup$
    – Asher
    May 31, 2016 at 12:41

2 Answers 2


The singularity for a Schwarzschild metric is a spatial surface. It is a surface where the Weyl curvature diverges. So technically there is no topology change. For the Kerr blackhole the singularity is a ring and things are a bit different. However, the inner horizon is a Cauchy horizon where photons of arbitrary wavelength pile up. So this might in fact be the physical singularity and the ring singularity is sort of mathematical fiction.

  • $\begingroup$ I was gonna ask about the topology change for the Kerr metric, but then I recalled that it has CTCs in it. So I suppose that's why it could do this in a "perfect" collapse? But does the Kerr black hole from collapse actually have CTCs, I seem to recall that it doesn't. $\endgroup$
    – Slereah
    May 31, 2016 at 12:03

In this case I would rather forget about possible no go theorems on topology changes. They are more related to how we mathematically model the problem, than to the physics.

The physical fact is that matter is collapsing in a initially approximately flat space. General relativity is unable to explain in full generality what happen next, indeed the prediction is that a trapped surface will form, therefore a singularity and an event horizon concealing it. The singularity gives you infinities. In general, infinity in physics means very big numbers, but this time the theory pretend that this a real mathematical infinity, and this is nonsense. It's like taking seriously the infinite attraction of a point particles in newtonian gravity.

This is a clear signal that the theory is incomplete and a more fundamental theory of quantum gravity is needed.

  • $\begingroup$ This doesn't answer the question. $\endgroup$
    – user4552
    Aug 12, 2017 at 21:56

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