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We know that there are non-triangulable 4-manifolds, such as the E$_8$ manifold.

  • Can E$_8$ manifold be a boundary of some 5-manifold $M_5$? Can such a $M_5$ be triangulable or non-triangulable? What are the possible $M_5$ (s)?

  • I wonder whether the non-triangulable 4-manifold can always be a boundary of some 5-manifold $N_5$? Then, can such a $N_5$ be triangulable or non-triangulable?


See also some background

Proofs of Rohlin's theorem (an oriented 4-manifold with zero signature bounds a 5-manifold),

Any 3-manifold can be realized as the boundary of a 4-manifold,

Not all manifolds can be triangulated: In which dimensions?

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    $\begingroup$ It seems like the obvious point needs to be made. If a manifold is triangulable, than so is its boundary. $\endgroup$ – Jim Conant Jul 22 '18 at 1:53
  • $\begingroup$ Good, so if such a 5-manifold exists, it must also be non-triangulable in order to have a boundary as a non-triangulable 4-manifold. $\endgroup$ – wonderich Jul 22 '18 at 1:55
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Your questions are answered by Hsu in his paper 4-Dimensional Topological Bordism.

In particular, associated to any closed oriented topological 4-manifold $X$ is a signature $\sigma(X) \in \Bbb Z$ and the Kirby-Seibenman class $\text{ks}(X) \in H^4(X;\Bbb Z/2) = \Bbb Z/2$. Hsu proves that these two invariants precisely classify 4-dimensional oriented topological manifolds up to bordism (and in particular $\text{ks}$ is an invariant of topological bordism).

A 4-manifold is triangulable if and only if it is smoothable. (This is true but not obvious; it invokes the 3D Poincare conjecture) If $X$ is smoothable then $\text{ks}(X) = 0$ (but the converse is not necessarily true).

The signature is an expected obstruction to finding a null-bordism, but Kirby-Seibenmann is less obvious. There is a manifold $F\Bbb{CP}^2$ ($F$ standing for Fake) which is homotopy equivalent to $\Bbb{CP}^2$ but not homeomorphic; it has $\text{ks} = 1$ and signature 1, and so is not smoothable. In particular, $F\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ has signature 0 but nonzero Kirby-Seibenmann invariant, and thus by Hsu's results it is not null-bordant.

There is a formula for spin 4-manifolds: $\text{ks}(X) = \sigma(X)/8 \pmod{2}$. In particular, $\text{ks}(X_{E8}) = \sigma(E8)/8 = 1$. Thus the $E8$-manifold is not null-bordant.

However, there are many that are: for instance, take $X_{E8} \# X_{E8}$. This is still not smoothable by Donaldson's theorem (the intersection form of a smooth 4-manifold, if positive/negative definite, is diagonalizable; $E8 \oplus E8$ is not). $\text{ks}$ is additive, and so $\text{ks}(X_{E8} \# X_{E8}) = 0$, and this manifold is null-bordant. As mentioned by Jim Conant in the comments above, because $X_{E8} \# X_{E8}$ is not triangulable, neither is the null-bordism.

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    $\begingroup$ I am pretty sure that PL=smooth in dimension 4 does not depend on the Poincaré conjecture (but only on the strong form of smoothability that holds in dimension 3). What you have to show is that if you attach a handle to a smooth 4-manifold, then the result is smooth. The attaching map is an embedding of one 3-manifold into another. The result then follows from the fact that every continuous embedding in dim 3 is isotopic to a smooth embedding. $\endgroup$ – Andy Putman Jul 22 '18 at 12:54
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    $\begingroup$ @Andy I am just talking about triangulations, not combinatorial triangulations - I agree PL = Smooth is much easier. I have in mind some argument showing the link of a vertex is a simply connected homology sphere, which could in principle be non-trivial. (it is not clear to me how to ensure the link of a vertex in a 4-manifold triangulation is a manifold, but I have a vague feeling I knew this long ago...) The goal in the end is to see a triangulation of a 4-manifold must be combinatorial. $\endgroup$ – Mike Miller Jul 22 '18 at 12:58
  • $\begingroup$ @Mike Miller, thanks for the nice answer +1, I assume you try to use bordant / cobordant to construct such a 5-manifold. But just to be quick, how would I construct and write down such a 5-manifold $M_5$ explicitly in my question? $\endgroup$ – wonderich Jul 22 '18 at 16:42
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    $\begingroup$ @wonderich See the first paragraph of Hsu's proof of Theorem 2.2 on page 3 of his paper for the construction. Given that any non-smoothable 4-manifold is a somewhat complicated object this puts a lower bound on the complexity of describing a 5-manifold it bounds! $\endgroup$ – Mike Miller Jul 22 '18 at 17:19

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