6
$\begingroup$

I am mainly interested in the $3$-dimensional case. It is a well-known fact, following from the work of E. E. Moise and R. H. Bing in the 1950s, that every $3$-dimensional topological manifold (with or without boundary) admits a triangulation, i.e. its homeomorphic to (the geometric realization of) an abstract simplicial complex. Furthermore, it is a well known fact that a manifold is piecewise-linear if and only if it admits a combinatorial triangulation, i.e. a triangulation in which the link of each simplex is Pl-homeomorphic to a sphere, and that in $d\leq 4$, every triangulation of a manifold is combinatorial. In other words, every $3$-manifold admits a PL-structure.

I am interested in the other way round: Is there a bunch of properties an abstract simplicial complex has to have in order to define a topological manifold? Clearly, not all $3$-dimensional simplicial complexes which one can draw give rise to a manifold. The complex should be at least pure and non-branching, I guess. Is it maybe enough to assume that a complex is combinatorial?

In the literature, I also have found the notions of ''pseudo-manifolds'', which are abstract simplicial complexes, which are pure, non-branching and strongly-connected. How is this related to my question?

Any help is appreciated. If someone could provide some reference, I would be happy too.

Improve this question
$\endgroup$
7
  • 4
    $\begingroup$ I think a 3-dim simplicial complex is a (purely 3-dimensional) topological manifold iff every every link is a 2-sphere (or 2-ball if boundary is allowed). And for a 2-dim simplicial complex, be a sphere/2-ball can be characterized (obvious local conditions + Euler characteristic). $\endgroup$
    – YCor
    Sep 7, 2021 at 17:36
  • 1
    $\begingroup$ For a simplicial complex, if all links are spheres or balls then you get a topological manifold (closed if you have only spheres). This is easy. What is not easy is to recognize a sphere, but this is not an issue in dimension 2. Also I'm not sure the converse holds in higher dimension (i.e., the links don't have to be topological spheres/balls). It might be in large dimension that checking whether a simplicial complex is a manifold is non-computable. $\endgroup$
    – YCor
    Sep 7, 2021 at 18:18
  • 1
    $\begingroup$ @YCor In dimension $n \leq 4$, a simplicial complex is a topological manifold if and only if all links of vertices are spheres if and only if all stars of vertices are balls. In dimension $n \geq 5$ this is false, because of Cannon's double suspension theorem: for every homology sphere $M$, the space $\Sigma^2 M$ is topologically a sphere. Because there is a smooth homology sphere of any dimension $n-2 \geq 3$ which has nontrivial fundamental group, triangulating it and taking the double suspension gives a triangulation of the sphere with a link homeomorphic to the non-manifold $\Sigma M$. $\endgroup$
    – mme
    Sep 7, 2021 at 18:29
  • 1
    $\begingroup$ In five dimensions we can have a triangulated topological manifold whose links are not even manifolds. $\endgroup$
    – Richard Stanley
    Sep 7, 2021 at 18:29
  • 2
    $\begingroup$ Your question has been answered in the comments. I would like to add, the combinatorial check that YCor describes is implemented in the software package "Regina". One of its functions is the generation of censuses of triangulated 3-manifolds with various properties. Basically it generates all simplicial complexes (with various constraints of your choosing) then checks to see if they are manifolds. It also implements a similar (but much slower) census algorithm to generate all triangulated 4-manifolds. $\endgroup$
    – Ryan Budney
    Sep 7, 2021 at 20:15

1 Answer 1

Reset to default
6
$\begingroup$

From the comments:

  1. Suppose that $T$ is a triangulation. If all vertex links are PL $(n-1)$-dimensional spheres then the realisation space of $T$ is a PL manifold and thus a topological manifold. (In the compact case this is equivalent to the usual definition.)

  2. There are triangulations of topological manifolds (in fact, of $S^5$) that do not have this property. Examples come from the double suspension theorem of Cannon and also Edwards.

Furthermore:

  1. There are triangulations of topological manifolds that have no PL structure. This is (essentially) the failure of the Hauptvermutung.

  2. There are closed topological manifolds that do not admit any triangulation. This is the failure of the Triangulation Conjecture, and is obtained by Casson in dimension four and to Manolescu in all higher dimensions.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.