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My question is soft and imprecise, as I know very little differential topology. Nevertheless, I hope it makes some $\epsilon>0$ of sense.

Assume the Universe is a 3-manifold without boundary, homeomorphic to the 3-spehere. Does this mean that it must exist as the boundary of a 4-dimensional solid, the same way 3-spehere is the boundary of the 4-dimensional solid ball. Or equivalently, must it exists as an embedding into a higher dimensional space, or it simply can exist as a 3-manifold, without the need of extra (spatial) dimensions?

Edit: To rephrase in more mathematical terms (but still soft), if we consider a 3-manifold, can we define it (understand it), and examine its mathematical properties without considering whether the 3-manifolds embeds into a higher dimensional space? Or are there properties of the manifold that require this sort of identification?

Theo
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    What does "exist as a boundary of a $4$-dimensional solid" mean? Does that mean that some $4$-dimensional solid "exists?" This isn't a good question for this site, because it borders on theological rather than mathematical. – Thomas Andrews Jan 26 '15 at 22:36
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    I said it was soft. Perhaps to rephrase in more mathematical terms (but still soft), if we consider a 3-manifold, can we define it (understand it), and examine its mathematical properties without considering whether the 3-manifolds embeds into a higher dimensional space. Or are there properties of the manifold that require this sort of identification? – Theo Jan 26 '15 at 22:49
  • Well, that seems like a very different question. – Thomas Andrews Jan 26 '15 at 23:14
  • In particular, your original question seems to be about the physical world, while your restatement is about mathematics. – Thomas Andrews Jan 26 '15 at 23:23
  • Yes, the motivation of this question was the physical world. I was trying to figure out whether the notion "outside the universe" makes any sense. Rephrased, in slightly mathematical terms, to me it meant whether a certain 3-manifold can exists on its own, or just as an embedding into a higher dimensional space. In the latter case, it seems to me that "outside the universe" makes some sense. But I agree, question is soft and tangentially mathematical. Hopefully the reformulation makes more sense. – Theo Jan 26 '15 at 23:39
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    In general, all the properties of any manifold are independent of any space we embed it in. So, for example, we can define a torus in terms of the plane, and all of its topological properties come completely from that definition. There is no need to know that the torus can be embedded in $3$-space, or that the Klein bottle cannot be. – Thomas Andrews Jan 26 '15 at 23:43
  • Thank you, this helped me realize that is impossible to give a coherent mathematical meaning to my philosophical question. – Theo Jan 26 '15 at 23:56
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    I think there's a valid math question here about whether a 3-manifold homeomorphic to 3-sphere must be the boundary of a 4-manifold. The fact that the question comes from thinking about physical space doesn't seem like a problem to me. – littleO Jan 26 '15 at 23:58

2 Answers2

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First, manifolds "exist in their own". However, every closed oriented 3-manifold does bound a compact 4-manifold (the latter is far from being unique). My suggestion is to pick up a copy of Munkres' "Topology" book and read first few chapters. This will help to clear many issues that you currently have.

Edit: Of course, Munkres does not discuss 3-dimensional manifolds (but he does explain how to classify surfaces). However, at the level of the question, before even attempting to read anything about cobordisms or characteristic classes, one should understand some basic topological concepts, in particular, that for a topological space "to exist" it does not have to be embedded in any standard ambient space (I think, this is really the thrust of the question). As for the fact that each closed oriented 3-manifold bounds, it follows from the basic cobordism theory, once you know that all the characteristic classes vanish. Vanishing is proven in detail in this MSE post.

As an alternative to learning cobordism theory, one can see that each closed oriented 3-manifold bounds by using Likorish-Wallace theorem that every such manifold is obtained via a surgery along a link in the 3-sphere.

Community
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Moishe Kohan
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  • Are manifolds discussed in the first few chapters of Munkres? – littleO Jan 26 '15 at 23:54
  • As far as I remember Munkres, he doesn't talk much about manifolds. But the point is taken, they are just topological spaces with additional properties. Whether they can be embedded into something else, or are the boundary of of another manifold, it is a separate issue. – Theo Jan 27 '15 at 00:03
  • While this answer is helpful, I do feel that a serious question has been asked here and it would be nice if someone could provide a proof that every closed oriented 3-manifold is the boundary of some compact 4-manifold. I don't think such a proof is found in Munkres, unless it's in a later edition. – littleO Jan 27 '15 at 00:56
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    Of course it is not in Munkres, for such a proof you would have to read a book on cobordism theory, or 3-topology. – Moishe Kohan Jan 27 '15 at 01:13
  • Mathematically, indeed that was the thrust of the reformulated question, to which the answer is trivial, I realize that now. However, the "existence" I had in mind in the original question was different than mathematical existence, but that made the question less mathematical and soft. Thank you. – Theo Jan 28 '15 at 05:09
  • @Theo: Yes, that much is trivially true: If a manifold $M$ homeomorphic to a manifold $M'$ and $M'=\partial W'$, for some compact manifold $W$, then also $M=\partial W$ for some $W$ homeomorphic to $W'$. I did not bother explaining this in my answer. As for physical questions, you are better off asking them at http://physics.stackexchange.com/ – Moishe Kohan Jan 28 '15 at 16:14
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As reference for the theorem in studiosus's answer, manifolds are null-cobordant (the boundary of a manifold of higher dimension) precisely when all their Stiefel-Whitney numbers vanish. The converse direction, that null-cobordant manifolds have Steifel-Whitney numbers equal to zero, is written down in Milnor and Stasheff's "Characteristic Classes" (and not terribly difficult). The forward direction is more difficult. Thom's original paper, Quelques propriétés gloables des variétés differentiables is available here, but as you might have guessed, is in French; if you know some French it is eminently readable. Milnor refers to Strong, "Notes on Cobordism Theory", as a reference for the proof. I suspect it's probably written down in any book on cobordism theory.

Using the Wu classes and some other tools given in Milnor and Stasheff's book, one can prove without much difficulty that all the Whitney classes (and hence all the Whitney numbers) of an orientable 3-manifold vanish.