21
$\begingroup$

When solving the Einstein field equations,

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = 8\pi GT_{\mu\nu}$$

for a particular stress-energy tensor, we obtain the metric of the spacetime manifold, $g_{\mu\nu}$ which endows the manifold with some geometric structure. However, how can we deduce global properties of a spacetime manifold with the limited knowledge we usually have (i.e. simply the metric)? For example, how may we deduce:

  • Whether the manifold is closed or exact
  • Homology and de Rham cohomology
  • Compactness

I know if we can establish compactness, one can easily arrive at the Euler characteristic, and hence the genus of the manifold, using the Gauss-Bonnet-Chern theorem,

$$\int_M \mathrm{Pf}[\mathcal{R}] = (2\pi)^n \chi(M)$$

where $\chi$ is the Euler characteristic and $n$ half the dimension of the manifold $M$. In addition, the Chern classes of the tangent bundle computed using the metric give some information regarding the cohomology. Note this question is really not limited to spacetime manifolds. There are many scenarios in physics wherein we may only know limited information up to the metric, e.g. moduli spaces. It would be interesting to see how one can deduce global properties.


This question is inspired by brief discussions on the Physics S.E. with user Robin Ekman, and I would like to thank Danu for placing a bounty; a pleasant surprise!


Resources, especially journal papers, which focus on addressing global properties of spacetimes (or more exotic spacetimes, e.g. orbifolds) are appreciated.

$\endgroup$
  • 2
    $\begingroup$ Aren't there lots of different surfaces that can have a flat metric, e.g. the plane, torus and Klein bottle? Doesn't that mean the metric alone cannot determine the surface? $\endgroup$ – John Rennie May 7 '14 at 17:42
  • 1
    $\begingroup$ @JohnRennie Yes. I'm basically asking what we should do when all we have at our disposal is the metric in GR because it offers limited information, and no global data, e.g. homotopy. $\endgroup$ – JamalS May 7 '14 at 17:44
  • 1
    $\begingroup$ @brightmagus: That's not an answer to the question. I'm looking for constructive answers which provide an outline of an approach to determining global properties of spacetime manifolds, or a method to find out what they "are" topologically speaking, e.g. $\mathbb{R} \times T^2 \times ...$ I have seen some mention of this in Becker, Becker and Schwarz, but that's all. $\endgroup$ – JamalS May 28 '14 at 13:52
  • 1
    $\begingroup$ It is worth noting that space-time manifolds that are compact are unphysical because they admit closed time-like curves. By extension therefore, any space-time that admits $\chi(\mathcal{M}) = 0$ should be treated with caution. $\endgroup$ – Arthur Suvorov May 30 '14 at 3:30
  • 1
    $\begingroup$ There are vanishing theorems which say things to the effect of "if the curvature is sufficiently positive then some cohomology groups are trivial." Here, curvature and cohomology can mean different things. In some cases, it is the Ricci curvature and the cohomology we are talking about is the usual de Rham cohomology of the manifold. In others (en.wikipedia.org/wiki/Kodaira_vanishing_theorem) the curvature is that associated to a line bundle and the cohomology is de Rham cohomology twisted by the line bundle. $\endgroup$ – Sean Pohorence Dec 17 '16 at 14:52
19
+50
$\begingroup$

Well, the simplest case is that some topologies of spacetime may only allow a particular class of metrics. But unfortunately, it usually requires the knowledge of the metric at every point to be quite certain.

Here's a few thing we can probably assume about the spacetime manifold :

  • All the usual jazz about manifolds in general relativity (paracompactness, Hausdorff, etc). This is not necessarily the case, as some theories may allow weirder versions of it, or allow more general topologies like manifolds with boundaries and conifolds, but that is what is usually assumed to get a Lorentzian metric. The fact that spacetime is a connected manifold isn't really a physical constraint but more of a metaphysical one : you can't really say much about any disconnected piece of spacetime since it cannot affect our own. By the way, if your manifold is indeed compact, only spacetimes with Euler characteristics 0 admit a Lorentzian metric, since you need to have a line element.

  • It is also usually assumed to have some causality conditions. It may not necessarily be true, but it seems like a rather remote possibility. If the spacetime is causal (no causal loops), it cannot be compact. If you also want it to be globally hyperbolic (No loops or naked singularities), it will fix the topology as $\mathbb{R} \times \Sigma$, $\Sigma$ some 3-manifold, per Geroch's theorem.

  • To get the topology from the metric, another important constraint is geodesic completeness : no geodesic should have a finite range in its affine parameter. You can put de Sitter space in $\mathbb{R}^4$, but much like the stereographic projection of a sphere, you will reach the "edge" when a geodesic tries to go through the other pole but finds none.

  • Because of the dependance of particle physics on time reversal and space reversal, it is assumed that spacetime is both time-orientable and space-orientable. If it was not, there would be no $SO^+(3,1)$ group and as such no spin groups, but only the Pin groups, with different properties for fermions.

Those are some rather generic things you can say about spacetime from some, we hope, rather reasonable assumptions. Experimental evidence of topology is harder though.

  • The Topology Censorship Theorem is a (classical) theorem regarding the ability to gauge the topology of spacetime. From Visser :

"In any asymptotically flat, globally hyperbolic spacetime such that every inextendible null geodesic satisfies the averaged null energy condition, every causal curve from past null infinity to future null infinity is deformable to the trivial causal curve".

Which rules out, if all conditions are met, the ability to send a particle along any trajectory along a topological handle (or wormhole, in the science).

  • You can try to check the topology of the spacelike hypersurface of the spacetime by simply observing any repeating patterns, but so far this has not met with any success. The PLANCK space observatory had, among other missions, looking for any correlations in the CMB that might indicate some compactified dimension of space.

  • Edit : Oh, and by the way, space being compact in some dimension will also affect the modes of any field on it (the so called topological Casimir effect). Non-orientable compactness also has some different effects.

$\endgroup$
  • $\begingroup$ Thanks for your answer, +1. Are you able to answer the part regarding the determination of cohomologies of spacetimes? If not, are there any resources you could point me to which may be able to? $\endgroup$ – JamalS May 8 '14 at 6:48
  • $\begingroup$ I'm not that knowledgable on cohomologies unfortunately, but isn't it fixed by the topology of the manifold? As far as I'm aware, there are no conditions on forms that can influence the topology of the manifold. Although having a non-trivial topology, leading to non-closed forms, can lead you to have fields without source (charge without charge, as Wheeler said), which have not been observed. $\endgroup$ – Slereah May 8 '14 at 8:14
  • $\begingroup$ If you can deduce what the manifold is, i.e. $R^2 \times S^2$, $R \times T^2$ or whatever, then yes, that is enough information to compute the cohomology. $\endgroup$ – JamalS May 8 '14 at 8:30
3
$\begingroup$

The initial value boundary problem in general relativity only gives you the metric on a patch of the spacetime. Other methods must be used to find the true global extension of that spacetime. Therefore, Einstein's equation alone cannot tell you the topology of the spacetime.

$\endgroup$
  • $\begingroup$ See the comments to the question. The answer should outline some constructive approach to determine (as much as possible about) the global properties. Ideally, it should also point out what exactly one would need (beyond Einstein's equations) to be able to deduce even more $\endgroup$ – Danu May 28 '14 at 14:04
  • $\begingroup$ The highest-upvoted answer to the question linked by Qmechanic represents a nice first step in this direction, but I would like to hear more! $\endgroup$ – Danu May 28 '14 at 14:06
  • $\begingroup$ Thank you for your reply. What motivated this question was that I saw, in certain texts, authors present without derivation certain global properties of spacetime manifolds, so there must be some existing approach. Perhaps you could expand your answer; note you are allowed to stipulate more information than the EFE provide, but hopefully not too much. $\endgroup$ – JamalS May 29 '14 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.