I know there are some examples of manifolds which don't admit a PL structure (combinatorial triangulation), and that it has been recently proven that in dimension $n\geq5$ there are manifold which are not triangulable (i.e. which are not homeomorphic to a simplicial complex).


As far as I understand, in dimension $4$ the two concept (triangulable and PL) should coincide, while in dimension $n\geq 5$ they are different.


The only examples of non-combinatorial triangulations I have encountered are double suspensions of homology spheres (which are homeomorphic to spheres): so in that case there is a topological manifold which admit *also* non-combinatorial triangulations.


I was wondering if there are examples of topological manifolds which admit triangulations, but *none of them* is combinatorial. 



Example of a triangulable topological manifold which does not admit a PL structure
The answer is **yes**, see Rudyak's paper [Piecewise linear structures on topological manifolds](http://arxiv.org/abs/math/0105047), Examples 21.4: 



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> *There are topological manifolds that can be triangulated as simplicial complexes but do not admit any* PL *structure.*
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Such examples exist in fact in any dimension $n \geq 5$, and are of the form $$M\_k=V \times T^k, \quad k \geq 1,$$ where $V$ is the famous [$E\_8$-manifold](https://en.wikipedia.org/wiki/E8_manifold) constructed by Freedman. See Theorem 7.2 and Corollary 7.4 in the quoted paper.


It is worth remarking that the $4$-manifold $V$ is *not* triangulable as a simplicial complex. However, by the work of Siebenmann and others, it is known that *every orientable topological 5-dimensional closed manifold can be triangulated as a simplicial complex*, see Theorem 21.5. This implies that $M\_1= V \times S^1$ *is* triangulable, so $M\_k=M\_1 \times T^{k-1}$ is triangulable for all $k \geq 1$.


