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General relativity says space-time is a $4$-dimensional manifold which may have non-zero global curvature. Now if we take a random curve or surface or $n$-fold, it may fail to be a manifold because it self-intersects (and the neigborhoods of a self-intersection point are not homeomorphic to $\mathbb R^n$). Are there any reasons this does not happen with space-time? What would be the physical meaning of the self-intersecting locus of the universe?

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Are there any reasons this does not happen with space-time?

You seem to be assuming that curves and surfaces are always embedded within a higher-dimensional space. That's not the case, and we have no evidence for any higher-dimensional space in which our universe is embedded. If it's not embedded in any higher-dimensional space, then this whole issue doesn't even arise.

It's also problematic to try to make physical theories that don't live in a nice well-behaved manifold. For instance, you could try to generalize to a manifold with boundary, but then the problem is that experiments done on the interior can never reveal to you what the laws of physics should be at the boundary. Furthermore, we prefer theories that have uniqueness and existence of solutions for Cauchy problems, but usually you will not get those properties if you tamper with the topological properties and make something that's not a manifold.

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  • $\begingroup$ I thought about this issue of being embedded in a higher dimension space. But actually, it seems to me that the fact of having a self-intersection is intrinsic to the topology of the curve or surface we consider. $\endgroup$ – Régis Jun 24 '18 at 16:09
  • $\begingroup$ I mean, if we try to locally assign to each point a curvature (that is intrinsic), odds are that in general we will end up self-intersecting, isn't it? $\endgroup$ – Régis Jun 24 '18 at 16:56
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    $\begingroup$ The self-intersection is intrinsic to the topology only if you explicitly make it so. Consider the standard immersion of a Klein bottle: it is not claimed that there are special intersection points anywhere in the Klein space, it is just that we use an awkward immersion. Self-intersections only happen if the topology states that one neighbourhood shares points with another neighbourhood. $\endgroup$ – Anders Sandberg Jun 24 '18 at 16:59
  • $\begingroup$ Ok so by Withney theorem for any $4$-dimensional manifold there is an embedding in $\mathbb R^8$. So if trying to find a $4$-fold in $\mathbb R^8$ with locally prescribed curvature we end up self-intersecting, that means there is truly no maniflod satisfying our local conditions, right? However, there might be an "intrinsically self-intersecting" $4$-fold. $\endgroup$ – Régis Jun 24 '18 at 17:49
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    $\begingroup$ Approaching a (potential) manifold as an embedding is kind of out-dated though isn't it? I mean, isn't that one of the points with the abstract definition; that it allows us to completely ignore any potential embeddings. So I think the reasoning with embedding theorem should be reversed: it states that regardless of any details of the Riemannian (note: not semi-Riemannian) manifold (such as curvature, which you mention) it can be embedded into a Euclidean space of sufficiently high dimension. Which means the problem only arises if you choose to introduce such an immersion. $\endgroup$ – Erik Jörgenfelt Jun 25 '18 at 8:19

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