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There is a well known fact that a compact spacetime necessarily contains a closed timelike curve (CTC). Proof can be found in several books on GR (e.g. Hawking, Ellis, Proposition 6.4.2), and in essence goes like this:

The spacetime $M$ can be covered by open sets of the form $I^+(p)$, chronological future of the point $p \in M$ (note that a priori $p$ is not an element of the set $I^+(p)$, but such situation can happen in a presence of CTCs). Now, suppose that $M$ is compact. Then there is a finite subcover, say

$$\{ I^+(p_1), \dots, I^+(p_n) \}$$

The point $p_1$ is contained in $I^+(p_{k_1})$ for some $1 \le k_1 \le n$, the point $p_{k_1}$ is contained in $I^+(p_{k_2})$, and so on. Since this subcover is finite, eventually some point $p_{k_r}$ must belong to $I^+(p_{k_s})$, with $s \le r$. Then there is a future directed timelike curve going from $p_{k_r}$ to $p_{k_s}$ (since $s \le r$) and then from $p_{k_s}$ back to $p_{k_r}$ (since $p_{k_r} \in I^+(p_{k_s})$), which gives a closed timelike curve through $p_{k_r}$ (and $p_{k_s}$) in $M$. Q.E.D.


The question that bothers me is: what is implicitly assumed about the spacetime $M$, by saying that the family of sets of the form $I^+(p)$ is indeed a covering of $M$?

Take for example a flat spacetime with compact space part and finite time direction, e.g. $M = [0,1] \times T^3$, where $T^3$ is a (spacelike) 3-torus. This is a compact manifold (since it is a product of two compact manifolds) and there are no CTCs. The loophole in the argument above seems to be in the fact that the "initial points", $\{0\} \times T^3$, are not covered by any set of the form $I^+(p)$.

One can easily modify this example by contracting the initial and final spacelike slices to points ("Big Bang" and "Big Crunch"); the resulting spacetime is still compact and contains no CTCs.

Do these manifolds fail to be "regular spacetimes" for some reason?

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$M=[0,1]XT^3$ is a manifold-with-boundary. Maybe this is the problem? –  twistor59 Nov 26 '12 at 15:46
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Yes, they fail to be "regular spacetimes" because they're not (closed - without boundaries) manifolds. In more relativistic terminology (and maths), we say that such manifolds fail to be geodesically complete, see e.g. en.wikipedia.org/wiki/Riemannian_manifold#Geodesic_completeness - Physically, it means that after the "boundaries", the spacetime doesn't say what the physics evolves into, even though it could and should, or at the beginning it doesn't say what it evolved from, even though it could and should. –  Luboš Motl Nov 26 '12 at 16:02

1 Answer 1

What is said above in the comments is correct. Note that you can violate this theorem with essential curvature singularities, too. The "Minkowski Sphere" with line element $ds^{2} = -d\theta^{2} + \sin^{2}\theta \,d\phi^{2}$ contains no CTCs (but it does contain closed null curves), but all future-pointing timelike geodesics begin at the 'south pole' and end at the 'north pole'.

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OK, so the original statement should be rephrased into "geodesically complete compact spacetime necessary contains a CTC". However, in classical GR we are typically dealing with singular (geodesically incomplete) spacetimes -- does this means that this theorem is of very little use? –  Ivica Smolić Nov 26 '12 at 16:36
    
@IvicaSmolić: I would probably say so--after all, CTC are unphysical. If they were generic features of spacetimes, we'd have some explaining to do to show why we don't observe such phenomena. Also, we typically deal with noncompact spacetimes (The closed Robertson-Walker cosmological model being the most notable exception), which would make the theorem already not apply. –  Jerry Schirmer Nov 26 '12 at 16:41
    
@IvicaSmolić, I don't think compact spacetimes are considered very often, may be because of this theorem. –  MBN Nov 28 '12 at 15:41

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