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We know

"Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

I heard this statement is true:

  • (1) Any closed 3-manifold is a boundary of some compact 4-manifold.

See also this paper p.2's 3rd paragraph uses the fact:

  • (2) Any 3-manifold $M$ can be realized as the boundary of a 4-manifold $B$.

In particular, we know that all 3-manifolds can be triangulable. However for 4-manifolds, there are simply connected non-triangulable manifolds (such as the E$_8$ manifold). (Note: a closed 4-manifold is triangulable if and only if it's smoothable.) See this MO post: Not all manifolds can be triangulated

  • (3) For any 3-manifold $M_3$ that can be realized as the boundary of a 4-manifold $B_4$, the $M_3$ must be triangulable. So must the $M_3$ be the boundary of a triangulable 4-manifold $B_4$?

  • (4) Are there any non-triangulable 4-manifold $B_4'$ with a 3-dimensional boundary (i.e. $B_4'$ is not closed)? Then would the 3-manifold boundary $M_3'$ be triangulable (if $M_3'$ is non-triangulable, isn't that leads to a contradiction)?

Can one show these (1), (2) and explain them as intuitively as possible?

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    $\begingroup$ For your first question, all the proofs I know that 3-manifolds are the boundaries of 4-manifolds show that in fact they are the boundaries of smooth 4-manifolds (well, except for the ones that just spit out a triangulated 4-manifold immediately!), and in dimension 4 smooth and PL are the same. The second question is trivial since all 3-manifolds can be triangulated. This question is more appropriate for math.se, and I have voted to close. $\endgroup$
    – Andy Putman
    Commented Jul 21, 2018 at 2:57
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    $\begingroup$ Thanks Andy for the nice comment - I asked a more basic question at MS a week ago but only a few comments (helpful though) math.stackexchange.com/q/2850317/79069 any-closed-3-manifold-is-a-boundary-of-some-compact-4-manifold but there are no answers. $\endgroup$
    – wonderich
    Commented Jul 21, 2018 at 3:35
  • $\begingroup$ 1 follows for orientable manifolds from the fact that every 3-manifold is smoothable and every orientable smooth 3-manifold is parallelilizable since by Thom's theorem about characteristic numbers it must be null bordant. $\endgroup$
    – Connor Malin
    Commented Oct 28, 2020 at 14:19

2 Answers 2

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The first part of question (4) makes sense, and the answer is yes by a theorem of Freedman, who showed that any homology 3-sphere bounds a contractible 4-manifold. Applying this to the Poincaré homology sphere, one may show that the contractible manifold it bounds does not admit a smooth structure, since it can be embedded in the $E_8$ manifold which has no smooth structure by Rokhlin's theorem, but is smooth to one side of the Poincaré sphere by a plumbing construction.

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In this question it is shown how to see that two smooth manifolds are topologically cobordant if and only if they are smoothly cobordant, which answers some subset of the questions (the second manifold can be $S^3,$ though I am usually abusively think of it as being $\emptyset$), and in arbitrary dimension [the fact that any three-manifold bounds is more elementary and does not require the algebraic machinery].

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  • $\begingroup$ thanks +1, you mean the second manifold of what? $\endgroup$
    – wonderich
    Commented Jul 22, 2018 at 0:10
  • $\begingroup$ @wonderich The results I allude to talk about when two manifolds are cobordant, whereas you are asking about when ONE manifold bounds. Bounds, means it and the sphere (or the empty set, which is a debatable manifold) co-bound. $\endgroup$
    – Igor Rivin
    Commented Jul 22, 2018 at 1:36

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