When it comes to questions of existence of bounds for PDE's and such, one must often make some assumptions regarding the topology of the space-time to use well known theorems.

My question is two-pronged:

i) I've often read, on Wikipedia (http://en.wikipedia.org/wiki/Spacetime) for example, that space-time is paracompact. I am aware of the mathematical definition and this seems counter-intuitive to me. Since the form of the stress-energy tensor entering Einstein's equations need only satisfy the conservation of energy $\nabla_{\mu} T^{\mu \nu} = 0$ condition from the Bianchi identities, how does one show this result? Can someone give me a reference for a proof?

ii) In (electro)-vacuum, what other topologies are permitted globally? I have seen some papers finding solutions that resemble black holes (in the sense they have singularities, event horizons...) but are topologically not 2-spheres at their cross section (seemingly in violation of Hawking's theorem, see e.g, http://arxiv.org/abs/hep-th/9808032, but perhaps the fact the space-time is asymptotically anti de-Sitter is why there is no violation). What space-times are compact? The spheres $S^{n}$ are all compact, but presumably even the Schwarzschild space-time is not globally? I think one can only have a compact space-time if its Euler characteristic $\chi = 0$, can this be translated into a demand on $T^{\mu \nu}$?

I am particularly interested in compact space-times, for one can then apply the Yamabe problem.


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    $\begingroup$ I can't really speak to the second point, but as for the first, a manifold is metrizable if and only if it is paracompact. So for obvious reasons, manifolds are usually defined to be paracompact to begin with. $\endgroup$
    – JohnnyMo1
    May 7, 2014 at 1:39
  • $\begingroup$ With the second point, you can think of "infinity" as a point --- generalizing the stereographic projection that embeds $\mathbb{R}^{2}$ into $S^{2}$, you can embed $\mathbb{R}^{n}$ into $S^{n}$. Then foliating Spacetimes as $\mathbb{R}\times\Sigma$ for some Cauchy surface $\Sigma$, if it's "topologically sexy" you can compactify it $\Sigma\cong S^{3}$. $\endgroup$ May 7, 2014 at 2:03
  • $\begingroup$ Thank you, JohnnyMo1; a very good point. The paracompactness is indeed an assumption then, but absolutely a necessary one! Maybe you could avoid it if you were working with some abstract non-metric quantum theory. $\endgroup$ May 7, 2014 at 2:03
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    $\begingroup$ Related: physics.stackexchange.com/q/1787/2451 and links therein. $\endgroup$
    – Qmechanic
    May 7, 2014 at 23:34
  • $\begingroup$ @JohnnyMo1: but obviously, paracompactness is only a necessary condition for compactness. Minkowski space is paracompact, but not compact, after all. $\endgroup$ May 8, 2014 at 0:00

2 Answers 2


On point i.): JohhnyMo1's comment touches the essential point, though the result he quoted holds assuming that the manifold is Hausdorff. I emphasize this because the definition of paracompactness in the literature is not uniform - sometimes it is assumed that a paracompact topological space is Hausdorff, sometimes not (M. W. Hirsch's book "Differential Topology", for instance, doesn't - neither does he assume that a manifold must be Hausdorff, by the way). More precisely, the result is a direct consequence of the Smirnov metrization theorem: a topological space is metrizable if and only if it is Hausdorff, paracompact and locally metrizable (i.e. any point has an open neighborhood whose relative topology is metrizable). Any manifold clearly satisfies the latter condition.

(EDIT: I've just got acquainted with the Smirnov metrization theorem, which allows one to do away with the connectedness hypothesis. Moreover, the counterexample I previously wrote is incorrect)

One should also add that paracompactness is equivalent to the existence of partitions of unity, which allow us to glue together locally defined objects in the manifold - for instance, this is how you prove existence of Riemannian metrics.

On point ii): if by "compact" you mean "compact without boundary" (like $S^n$), compact space-times indeed necessarily have vanishing Euler characteristic - conversely, any compact manifold with vanishing Euler characteristic admits a time oriented Lorentzian metric. However, such space-times are not physically interesting because they necessarily have closed timelike curves. The argument is simple: since any space-time $(\mathscr{M},g)$ may be covered by the chronological futures of all its points (which are open sets), using compactness one can pass to a finite subcover, say $\mathscr{M}=I^+(p_1)\cup\cdots\cup I^+(p_n)$. Therefore, $p_1$ must belong to $I^+(p_{j_1})$ for some $j_1=1,\ldots,n$, $p_{j_1}$ must belong to $I^+(p_{j_2})$ for some $j_2=1,\ldots,n$, and so on. Since we are dealing with a finite number of points, eventually one must have $p_{j_k}=p_1$ for some $k$ between $1$ and $n$, thus producing a closed timelike curve. Since such space-times are not globally hyperbolic, they are also unsuitable for the analysis of hyperbolic (i.e. wave-like) PDE's. Noncompact (Hausdorff, connected and paracompact, as in point (i)) manifolds, on the other hand, always admit a time oriented Lorentzian metric.

A reference that discusses which topological hypotheses on space-time manifolds are natural is the classic book by S. W. Hawking and G. F. R. Ellis, "The Large Scale Structure of Space-Time" (Cambridge University Press, 1973).

  • $\begingroup$ While not too physical, study of PDE on compact spacetimes does exist, for instance "Test fields on compact spacetimes : problems, some partial results and speculations". The problem being mostly that you are not guaranteed to have a well defined Cauchy problem on such spacetimes. $\endgroup$
    – Slereah
    May 7, 2014 at 5:46
  • $\begingroup$ @Slereah: I infer that you are referring to iaea.org/inis/collection/NCLCollectionStore/_Public/22/023/…. Indeed, U. Yurtsever shows that the point spectrum of the wave operator in a compact space-time is real and countable (hence, it contains no open interval), and computes it in the torus. Therefore, in such space-times (i) the Klein-Gordon equation has no nonzero global smooth (in fact, not even L2 weak) solutions for any but a countable subset of squared masses, and (ii) even in the case they do belong to this set, one cannot prescribe arbitrary smooth initial data. $\endgroup$ May 7, 2014 at 18:44

i)Since that generated a bit of discussion, let me take the point, along several authors, that the definition of a manifold includes the Hausdorff property. So I'll take this as granted. Note also that it is very sensible to define a spacetime as a connected manifold, for the reason that if this were not true we could have no causal link between the different connected parts, so there is no interesting physics in this case. As Pedro noted in the above answer, this conditions imply that the space is metrizable and existence of a riemannian metric, and as he commented below this is necessary to show existence of a lorentzian metric. Nevertheless a good deal of results in causality theory require only the existence of a conformal lorentzian metric (i.e. a metric up to a scale factor, which may differ from point to point), and this does not imply paracompactness. If you're willing to work on this low level of structure you may still question the need to include this property in the definition. Therefore I would argue that the strongest reason to think of paracompact manifolds as natural is that it implies existence of a partition of unity, which in turn allows us to make global quantities out of local ones, or to decompose global thing in terms of local objects. Therefore, even in contexts where the is no metric, e.g. gauge fields as (pullbacks) of connections in fibre bundles, you may still argue that paracompacticity is desirable.

I've found this thread in math.stackexchange useful. For the proof that every spacetime is paracompact, as well as reference for the exception for conformal lorentzian metrics, you may take a look at this paper from Geroch, "Spinor Structure of SpaceTimes in General Relativity. I".

ii)The existence of a lorentzian metric is a much more restrictive condition on a manifold than a riemannian one. In particular only compact manifolds of Euler characteristic 0 admits lorentzian metrics, as you have already said. This has nothing to do with $T_{\mu\nu}$ but with the causal structure being incompatible with other characteristics, in fact it holds independently of Einstein Equations. As Pedro also noted, compact manifolds necessarily have closed causal curves, or naked singularities, so you may argue that they are not physically relevant, although they may be interesting from the mathematical point of view. Usually we are interested in globally hyperbolical spacetimes, because the initial value problem is well-posed without further boundary conditions. In this case a well known theorem by Geroch shows that it must be $\mathbb{R}\times \Sigma$, with $\Sigma$ an arbitrary 3-manifold. One interesting case that is not globally hyperbolic but does not possess closed causal curves is the (maximal development of) anti-de Sitter spacetime. So this gives you the options in global topology. As for the case you mentioned of black holes that do not have croos sections of the event horizon $S^2$, the result I think you refer as Hawking's theorem should be this one, you are correct, the anti-de Sitter part violates one of the conditions, namely that the spacetime is asymptotically flat.

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    $\begingroup$ The proof of existence of Lorentzian metrics on noncompact manifolds or compact manifolds with vanishing Euler characteristic relies on the existence of a Riemannian metric. Namely, given such a metric $e$ and a nowhere vanishing vector field $X$ (whose existence is equivalent to the above conditions on the manifold), one can choose $X$ such that the corresponding covector $\eta$ w.r.t. $e$ has unit norm at every point. One then immediately checks that $g=e-2\eta\otimes\eta$ is a time oriented Lorentzian metric, with time orientation provided by $X$. $\endgroup$ May 8, 2014 at 0:43
  • $\begingroup$ Thanks, I was thinking in terms of more consequential results and forgot about the basics. I've edited my answer accordingly. To clarify, I was just trying to say that paracompactness is a very natural property. $\endgroup$ May 8, 2014 at 18:33

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