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?




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See also some background 


[Proofs of Rohlin's theorem (an oriented 4-manifold with zero signature bounds a 5-manifold)](https://mathoverflow.net/q/45934/27004), 


[Any 3-manifold can be realized as the boundary of a 4-manifold](https://mathoverflow.net/q/306506/27004),


[Not all manifolds can be triangulated: In which dimensions?](https://mathoverflow.net/q/264330/27004)



Non-triangulable 4-manifold as a boundary of some 5 manifold
Your questions are answered by Hsu in his paper [4-Dimensional Topological Bordism](https://core.ac.uk/download/pdf/82491711.pdf).


In particular, associated to any closed oriented topological 4-manifold $X$ is a signature $\sigma(X) \in \Bbb Z$ and the Kirby-Siebenman 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-Siebenmann 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-Siebenmann 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.


