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.


The answer is yes, see Rudyak's paper Piecewise linear structures on topological manifolds, Examples 21.4:

There are topological manifolds that can be triangulated as simplicial complexes but do not admit any PL structure.

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 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 $k \geq 1$.

  • $\begingroup$ "Friedman" or "Freedman"? $\endgroup$ – Jason Starr Aug 10 '15 at 15:13

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