TITLE: Which manifolds are homeomorphic to simplicial complexes?
QUESTION [92 upvotes]: This question is only motivated by curiousity; 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") 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 $S$ (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 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.

REPLY [8 votes]: Regarding Charles Rezk's second question:
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?
For dimension 4, it follows from the Poincare conjecture that a 4-manifold is triangulable iff smoothable (which is also equivalent to having PL structure for dimension <8). See Problem 3 of http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf. Also see this presentation.
For dimension >4, Springer Online Reference Works claims that "the imbedding $PL \subset TRI$ is also irreversible in the same strong sense (there exist polyhedral manifolds of dimension $\geq 5$ that are homotopy inequivalent to any PL-manifold)",  but gives no examples. In this presentation it is stated that "All oriented closed 5-manifolds triangulable", so I think among them there may be some with nontrivial KS invariant and hence cannot bear PL structure.
In addition, This book (p.168, Theorem 18.4) seems to contain a result that strengthens the one mentioned in Paul's answer.
Added: This paper (22.5. Example) explicitly gives an example of "A topological manifold which is homeomorphic to a polyhedron but does not admit any PL structure".