Given a valid classical GR spacetime manifold $M$ (i.e. 4D, Lorentzian, Hausdorff, paracompact, ?etc.), and $B\subset M$, a closed spatial subset (e.g. a closed ball at fixed $t$) to be excised,

[Aside: sorry - I try to be as precise as I can up front and this does seem to lead to unshort questions.]

The questions are:

a) Is the manifold $M' = M - B$ also a valid spacetime?

b) What is the argument/proof either way?

c) If $M'$ is not a valid spacetime, can it be repaired by some suitable surgery (other than putting back the excised subset)?

d) If $M$ is globally hyperbolic (GH), and $M'$ is either valid or can be repaired, is valid/repaired $M'$ also GH?

[NB: e.g. valid spacetime = Gödel; valid + GH = Minkowski]

My belief is that $M'$ is not a valid spacetime (or globally hyperbolic), but I can't demonstrate it*.

The objective of the question is to gain a better understanding of relevant technicalities, which I am aware of but do not comprehend; understanding the answers would be an entrée to better appreciation of theory.

Technical answers are welcome (esp. for citations and references to key theorems) even more so if they come with non-technical paraphrases that will help in understanding the technicalities.

One aspect of what I'm seeking is an insight into what you can and cannot do to a manifold in GR and why/why not. For example, in discussions of wormhole construction by surgery on Minkowksi/Schwarschild space I read the comment that after identification of surfaces (around excised open balls) "spacetime is geodesically complete" but whilst that is intuitively sensible, it is not obviously true to me.

* I believe that $M'$ is singular, having incomplete forward and backward directed timelike (and null ?+ spacelike) geodesics, but I'm also confused about completeness vs. extensibility and ? their relations to paracompactness, etc. I also don't think $M'$ is repairable because $B$ is closed and I haven't seen any reference in my searching to being able to rejoin open sets without adding "boundaries/closure". NB I am aware of extensive literature on how hard it is to define "singular" esp. w.r.t. geodesic completeness (e.g. Hawking, Penrose, Geroch, Earman, Curiel, etc.)... but I'd rather not go too deep.

[Optional subsidiary question. A potentially more complex instance might be: take two half-infinite world tubes starting/ending on $t=0$ (i.e. half-open $[0,+\infty)$ and $[0,-\infty)$) that have no spatial overlap and excise them from Minkowksi spacetime. Can the "ends" of the cylinders around $t=0$ be joined somehow to repair the spacetime?]

  • $\begingroup$ What do you mean by "valid"? A manifold with a closed ball removed will still be a manifold and, I suspect, still capable of hosting a Lorentzian metric on it (the resulting manifold won't be compact so there are no problems relating to having a continuous vector field on it). It may well be geodesically incomplete, but that usually isn't in the definition of a valid spacetime manifold (although it will make it fail to be globally hyperbolic). $\endgroup$
    – Slereah
    Apr 3, 2016 at 11:48
  • $\begingroup$ @slereah By valid I meant that which came after "i.e.". Since I'm no expert in this I thought it best not to attempt to be definitive, though given that GH typically includes the absence of naked singularities and compactness of the causal diamond IIRC, I might have said GH in place of "valid" but I didn't want to incorrectly exclude things I was ignorant of. I said suspect $M'$ is singular without being definitive about geodesics and that I am unclear about incompleteness/inextensibility... $\endgroup$ Apr 3, 2016 at 12:35
  • $\begingroup$ If you think that naked singularities are a problem for a spacetime, then yes, I'd say removing an open ball is not gonna be all that good. $\endgroup$
    – Slereah
    Apr 3, 2016 at 12:47
  • $\begingroup$ @slereah Closed ball; if I removed an open ball I could identify points on the boundary, couldn't I? (Hence the emphasis on the closed ball.) $\endgroup$ Apr 3, 2016 at 12:55
  • $\begingroup$ Oh yes sorry, I meant closed ball. $\endgroup$
    – Slereah
    Apr 3, 2016 at 13:04

1 Answer 1


If you remove a closed subset from a valid spacetime, then the result is a valid spacetime.

You can put the Lorentzian metric on the original spacetime, then simply restrict it to the part that you keep.

So it's still a 4d manifold without boundary. It's still Lorentzian and so on. You could even imagine the original manifold as some charts and transition maps with a metric on them that satisfies some compatibility conditions. Since the charts are from open sets to open sets removing a closed set doesn't change that the leftover is still open. So everything is still exactly what you require.

d) If $M$ is globally hyperbolic (GH), and $M'$ is either valid or can be repaired, is valid/repaired $M'$ also GH?

Removing a portion can leave the result as not globally hyperbolic.

An example is Minkwoski spacetime then remove a closed ball of radius $R$ at time $t=0$. It's not globally hyperbolic.

To prove it is not globally hyperbolic, basically you can start with Minkowski spacetime $M$ and use the coordinates that have spatial zero be the center of the ball. Then consider the events $p=(R/c,\vec 0)$ and $q=(-R/c,\vec 0)$.

As subsets of your manifold $M'=M-B,$ the set $J^-(p+\hat t,M')\cap J^+(q-\hat t,M')$ is not compact because the following open sets $$M-\left( J^-(p+\frac{\hat t}{n},M)\cap J^+(q-\frac{\hat t}{n}t,M)\right)$$ for $n\in\mathbb N$ cover $J^-(p+\hat t,M')\cap J^+(q-\hat t,M')$ but the open cover fails to have a finite subcover (any finite subcover is just the largest one since the open sets get strictly bigger, and that fails to cover the whole of $J^-(p+\hat t,M')\cap J^+(q-\hat t,M')$). So $J^-(p+\hat t,M')\cap J^+(q-\hat t,M')$ is just not a compact set.

But just because you have a valid spacetime doesn't make it nice. For instance these spacetimes have curves that just stop after a finite amount of time. It's not appropriate to call them singularities since the spacetime could be extended (even though the curves in the fixed spacetime cannot).

  • $\begingroup$ Thank you, and, yes, that makes sense. Minkowski - (open) ball $\rightarrow$ valid (not what I thought) but not GH (and it can't be patched up...?) which was my intuition. "Proving it is not GH is hard..." that is what I would really now like to do/see done :) Is it not GH because: (i) the curves that just end are naked singularities; (ii) because the surgery makes $J^{+}(p)\cap J^{-}(q)$ non-compact? (In which case is it because $J^{\pm}$ are no longer closed?) or (iii) because a CTC has been created (by Geroch; hasn't the topology changed?) Are (i) and (ii) the same thing here? $\endgroup$ Apr 3, 2016 at 16:36
  • $\begingroup$ [...] That curves that just stop relates to the incomplete/extensible part; could you say which category do they fall into and why? $\endgroup$ Apr 3, 2016 at 16:37
  • $\begingroup$ Erratum: Minkowski - CLOSED ball $\rightarrow$ valid etc. $\endgroup$ Apr 3, 2016 at 16:53
  • $\begingroup$ the CLOSED ball erratum was me correcting my 1st comment (out of time to re-edit); not disputing it didn't need patching to be valid. $\endgroup$ Apr 3, 2016 at 17:14
  • $\begingroup$ Thank you greatly; have to go AFK now but will be reading the edit carefully. If you are feeling generous... can this mutiliated spacetime be made GH (again the original question about joins of open sets)? [Apparently I can only +1 your answer once] $\endgroup$ Apr 3, 2016 at 17:40

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