This question is only motivated by curiosity; I don't know a lot about manifold topology.

Suppose $M$ is a compact topological manifold of dimension $n$.  I'll assume $n$ is large, say $n\geq 4$.  The question is: *Does there exist a simplicial complex which is homeomorphic to $M$?*

What I think I know is:

* If $M$ has a piecewise linear (PL) structure, then it is triangulable, i.e., homeomorphic to a simplicial complex.

* There is a well-developed technology ("[Kirby-Siebenmann invariant](https://en.wikipedia.org/wiki/Kirby-Siebenmann_invariant)") which tells you whether or not a topological manifold admits a PL-structure.

* There are exotic triangulations of manifolds which don't come from a PL structure.  I think the usual example of this is to take a [homology sphere](https://en.wikipedia.org/wiki/Homology_sphere) $S$ (a manifold with the homology of a sphere, but not maybe not homeomorphic to a sphere), triangulate it, then suspend it a bunch of times.  The resulting space $M$ is supposed to be homeomorphic to a sphere (so is a manifold).  It visibly comes equipped with a triangulation coming from that of $S$, but has simplices whose link is not homemorphic to a sphere; so this triangulation can't come from a PL structure on $M$.  

This leaves open the possibility that there are topological manifolds which do not admit *any* PL-structure but are still homeomorphic to some simplicial complex.  Is this  possible?

In other words, what's the difference (if any) between "triangulable" and "admits a PL structure"?

[This Wikipedia page on 4-manifolds](https://en.wikipedia.org/wiki/4-manifold) claims that the E8-manifold is a topological manifold which is not homeomorphic to any simplicial complex; but the only evidence given is the fact that its Kirby-Siebenmann invariant is non trivial, i.e., it doesn't admit a PL structure.