Universe as a finite 3-manifold without boundary

Question

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?

Answer 1

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.

Answer 2

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.