W. W. Boone, W. Haken, and V. Poenaru, On Recursively Unsolvable Problems in Topology and Their Classification, Contributions to Mathematical Logic (H. Arnold Schmidt, K. Schütte, and H. J. Thiele, eds.), North-Holland, Amsterdam, 1968.

a combinatorial manifold is defined as a simplicial complex with the property that the star of every vertex is combinatorially equivalent to the standard $n$-simplex. (two simplicial complexes are combinatorially equivalent if they possess linear subdivisions, the associated abstract simplicial complexes of which are isomorphic) This is equivalent to the condition that the link of every vertex be a combinatorial $(n-1)$-sphere (=boundary of the standard $n$-simplex) if the underlying manifold has no boudnary.

However, in

A. Ranicki (ed.), The Hauptvermutung Book, K-Monographs in Mathematics, vol. 1, Kluwer Academic Publishers, Dordrecht, Boston, London, 2010.

on page 4, this condition is used to define the term

``combinatorial manifold (or PL manifold)''.

I find this very weird; a PL manifold should be defined as a topological manifold with a maximal atlas of homeomorphisms with PL coordinate changes (and I know a lot of authors who use this definition).
The obvious question now is: Is a simplicial complex, the vertices of which have $S^{n-1}$ as link, the same as a topological manifold with a maximal atlas of homeomorphisms piecewise linear coordinate changes? Of course, Ranickis nomenclature implies that it does.

Obviously, the condition on the links can be used to construct such an atlas. However, the converse puzzles me, as it seems to be equivalent to the question if every manifold with a maximal PL atlas admits a triangulation.

If anyone could point me to an article where this problem is addressed, I would be thrilled.

Best regards, Malte

  • 4
    $\begingroup$ Perhaps I'm missing something but this seems clear to me. take a locally finite PL atlas and triangulate each chart linearly on some compact polyhedral domains that still cover M. Since the transition maps are PL and the cover is locally finite you can subdivide the triangulations to make the transition maps linear on simplexes. The resulting triangulation will obviously have spheres as links of vertices. $\endgroup$ Jan 13, 2012 at 18:54
  • $\begingroup$ This looks like a fine idea. I was wondering for a moment why this wouldn't work with an arbitrary atlas, but this is required as the intersection of two simplices $f_0: \Delta_0 \rightarrow M$ and $f_1: \Delta_1 \rightarrow M$ might otherwise be non-linear'' (I guess this is what you mean by make the transition maps linear''). Anyway, you're a very sweet man, thanks a lot! $\endgroup$
    – Malte
    Jan 14, 2012 at 20:01

2 Answers 2


It is claimed here that "A PL manifold is easily shown to be PL homeomorphic to a simplicial complex that is a so-called combinatorial manifold [37]", [37] being Hudson's Piecewise Linear Topology. I think the whole thing is worked out in Chapter 3 of that book.


Take a look at MR0271956 (42 #6837) Siebenmann, L. C. Are nontriangulable manifolds triangulable? 1970 Topology of Manifolds (Proc. Inst., Univ. of Georgia, Athens, Ga., 1969) pp. 77–84 Markham, Chicago, Ill.57.01

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    $\begingroup$ Thank you for pointing out this article, I didn't know about it. However, it doesn't seem to answer the question. $\endgroup$
    – Malte
    Jan 14, 2012 at 20:01
  • $\begingroup$ I agree. It only reinforces it. I thought it might have some relevant information, but I am not surprised that it does not. In the meanwhile, Vitali's response has me fairly convinced that the two definitions are equivalent, but I have not thought it through. $\endgroup$
    – Matt Brin
    Jan 15, 2012 at 5:25

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