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.


1 Answer 1


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"? $\endgroup$ Aug 10, 2015 at 15:13

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