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As the spacetime of the universe seems to be quite flat, a torus topology comes mind easily. How about others?

Is it issue if manifold is non-orientable? I see challenges to find 3- or 4-Klein-bottle-like combinations though postulating Cauhcy property separately in in 2+2 dimensions i.e. two 2-Klein-bottles connected with non-Cauchy connections...

In the similar question it was remainded that strong causal condition may be the way to study GR topology. Is there any reasonable doubt to set that restriction?

Maybe I'm trying impossible. I didn't find direct list of topologies tried for GR. Can anyone help?

Eusa
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  • Well, I honestly did not read the whole question, but is this what you are asking -- "why does spacetime have to be orientable and not non-orientable?" If spacetime is not (time-)orientable, there would be no distinction between causal future and past. Is this what you are asking? – VaibhavK Sep 27 '23 at 18:02
  • See also https://physics.stackexchange.com/q/3656/365939. – VaibhavK Sep 27 '23 at 18:04
  • I'm not really sure how GR and topology fit together. I asked a related question on how to solve problems in GR given periodic boundary conditions, and it's something I want to revisit some time. – user196574 Sep 27 '23 at 18:18
  • It's not obvious to me that it's OK to have an orientable closed manifold, and then add a single mass on top of that. For example, if one wants to find the electric field of some charge distribution on a closed, periodic manifold, I think the total charge must be zero. I don't work in GR, so it's likely my confusions are resolved in elementary ways, but it's something I hope to eventually learn. – user196574 Sep 27 '23 at 18:19
  • You might find Riazuelo et al., "Cosmic microwave background anisotropies in multiconnected flat spaces", Phys. Rev. D69 103518 (2004), to be of interest. That work looks at the possible imprints on the CMB of a manifold of the form $\mathbb{R} \times M_3$, where $M_3$ is a flat (but not necessarily orientable) manifold. – Michael Seifert Sep 28 '23 at 02:41

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General relativity does not care about topology of the manifold. You could for example take a flat spacetime with cubical edges and identify the edges in such a way that you have a non-orientable manifold. This satisfies the vacuum equations of GR since the metric tensor is flat everywhere.

As noted in the answers to this similar question, it might be that the physics on the manifold does not allow non-orientability since it would transform particles looping around the universe into different chirality. But GR does not care.

Anders Sandberg
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Observations suggest that our Universe is spatially homogeneous and isotropic. This means that at a manifold level, the underlying spacetime should have 6 independent spatial Killing vector fields ($J_1, J_2, J_3, P_1, P_2, P_3$) associated with it such that

$$ \left[J_{a}, J_{b}\right]=\sum_{c=1}^{3} \epsilon_{a b c} J_{c}, \quad\left[P_{a}, P_{b}\right]=0, \quad \text { and } \quad\left[J_{a}, P_{b}\right]=\sum_{c=1}^{3} \epsilon_{a b c} J_{c} $$

Here the $J$ correspond to rotation symmetry and $P$ to translational ones and $\epsilon_{abc}$ is the Levi-Civita symbol.

As the spacetime of the universe seems to be quite flat, a torus topology comes mind easily. How about others?

So one way to filter out certain spacetime models in the above question in blockquote is: does a torus as a manifold satisfy such a condition? Intuitively my guess would be "no".

Is it issue if manifold is non-orientable?

Spacetime in GR is time-orientable. This means that there exist a vector field that allows us to distinguish the past lightcone from the future light cone. So once we pick one half of the light cone, the vector field then totally determines all futur choices. This is required as although the laws of nature appear time symmetric but the Universe typically evolve only into the future. As such non-orientable manifolds will lead to issues here.

S.G
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  • I don't think that the existence of a global Killing vector field is necessary for spacetime to be locally isotropic and homogeneous, and the latter is really all the cosmological principle requires. As a 2D example, you can put a flat metric on the Klein bottle, so locally it "looks" isotropic & homogeneous, you can construct a set of four KVFs locally, and no point is "privileged." It's just that there's a topological obstruction to extending these KVFs to the full manifold. – Michael Seifert Sep 28 '23 at 02:27
  • @MichaelSeifert Yes, but if we require global hyperbolicity (which we do for our Universe) then wouldn't the homogeneity and isotropy on an initial Cauchy hypersurface eventually lead to the Killing vector fields being global? – S.G Sep 28 '23 at 04:18
  • You'll have to elaborate on that — I don't see why it would. It's not like a KVF "evolves in time"; it just exists as a vector field in 4D. – Michael Seifert Sep 28 '23 at 11:36
  • @MichaelSeifert I agree that the field would be predefined on the manifold. What I meant is that if we have access to some hypersurface of the manifold (say, the Universe today, where we are just seeing the CMB), then global hyperbolicity would dictate the structure of the rest of the manifold. So if the cosmological principle holds today and if the dynamics tells us that it will hold in the future as well, then wouldn't that result in a globally defined Killing vector field? – S.G Sep 29 '23 at 00:00