# Topologically, is a curvature singularity just a hole?

Topologically speaking, a hole can be introduced into a manifold and it will still be a manifold, e.g. remove points within a 2-sphere of some radius from the cartesian plane and you'll still have a manifold.

Penrose's singularity theorems prove the existence (mathematically) of incomplete null geodesics, but not of curvature singularities, per say. So Im wondering if, in my naive view, maybe a topological perspective is better suited to describe curvature singularities? IF so, I'm not sure how to view curvature singularities topologically.

My question: is a curvature singularity, e.g. a black hole, in general relativity simply a topological hole of the spacetime manifold, or is it more topologically complicated? Or instead, is the singular structure not a part of the manifold, as suggested in @benrg's answer to this question, and thus is not a topological hole? Or is it not this simple, and there's some nuance(s) that I'm missing?

EDIT: I suppose I can phrase my confusion like this: how is it logically consistent to say that the physical singularity is not a part of the spacetime manifold (like how $$\infty$$ is not a point on the real line) AND that we can have a description of the singular structure from the metric itself (e.g., the Kerr metric has a ring singularity as can be shown from the metric)? Or do we avoid this confusion if the singularity is a topological hole?

• Thank you for the reference! Those other SE questions are helpful. benrg is correct, a singularity is a coordinate region that is not a part of the spacetime manifold. I suppose my question is then, what does it mean to "not be a part of the spacetime manifold" if geodesics terminate at that point? This is why I framed my question around topological holes. Jul 13 at 21:59
• Okay, I see. Thanks for the clarifications! I added an edited question which hopefully better approximates my confusion. Jul 14 at 0:15
• Interestingly, the hawking's area theorem also applies to a hole - that it cannot shrink? Jul 14 at 0:29
• There's some relevant info here: physics.stackexchange.com/q/144447/123208 Jul 14 at 1:34
• I posted these questions on two different posts on Math SE: math.stackexchange.com/questions/4199439/… and math.stackexchange.com/questions/4198452/… since I realized they are really math questions. Jul 15 at 19:58

In addition, if you consider an Euclidean integral of Gauss-Bonnet scalar of a BH ($$r$$ from horizon to infinity and Euclidean time from $$0$$ to $$1/T$$), you will get Euler characteristics $$\chi =2$$, which is indeed a sphere.