5
$\begingroup$

I'm looking for necessary and sufficient conditions for an abstract simplicial complex so that it has a geometric realization that is homeomorphic to a manifold. I can't seem to find a single reference for this online. Any help would be greatly appreciated.

$\endgroup$
8
  • $\begingroup$ I believe there are conditions under which a manifold is a simplicial complex $\endgroup$
    – FShrike
    May 11, 2023 at 19:57
  • $\begingroup$ But not the other way around? $\endgroup$
    – the_dude
    May 11, 2023 at 20:00
  • 1
    $\begingroup$ @PeterFranek I am aware of this result, but I'm specifically interested in the other direction. $\endgroup$
    – the_dude
    May 11, 2023 at 20:09
  • 1
    $\begingroup$ This may succeed on MO $\endgroup$
    – FShrike
    May 11, 2023 at 20:17
  • 1
    $\begingroup$ A necessary condition is that links of simplices are homology spheres; a sufficient condition is that these links are topological spheres. Anything else (like the disjoint disk property, which is necessary bust not sufficient) will be well-beyond what you can handle. $\endgroup$ May 13, 2023 at 11:25

2 Answers 2

6
$\begingroup$

One very important result of geometric topology is the double suspension theorem of Bob Edwards, in which he started with a certain triangulated 3-manifold $M$ having the same homology groups as the $3$-sphere $S^3$, and proved that its double suspension $\Sigma^2 M$ is homeomorphic to $S^5$. Double suspension is a very simple construction that one can carry out on any simplicial complex $X$, producing a new simplicial complex $\Sigma^2 X$ whose dimension is $2$ plus the dimension of $X$. Roughly speaking you form one simplicial cone on $X$, and another simplicial cone on $X$, and you glue the bases of the two cones together in the obvious manner, and you get the single suspension $\Sigma X$. Then you do it again to $\Sigma X$ to get $\Sigma^2 X$.

The reason this is relevant to your question is that the hard part of Edwards' proof is to show that $\Sigma^2 M$ is a $5$-dimensional manifold. There were no techniques available to do this, Edwards proof is just shear hard work.

So if there were simple and straightforward necessary and sufficient conditions for the geometric realization of an abstract simplicial complex to be a manifold, we could apply those conditions to $\Sigma^2 M$ to get a cheapo proof of Edwards' theorem.

But no such cheapo proof has shown itself. Edwards's Theorem is a hard theorem.

Steve Ferry's lecture notes also give some historical insight into the significance of the double suspension theorem.

$\endgroup$
6
$\begingroup$

Disclaimer: I know nothing of how to prove all this, I just know the results. This is the simplicial complex recognition problem. It can be phrased so as to make it algorithmic, because a simplicial complex can just be thought of as a set of subsets of some underlying set. The set of such collections is recursive. From that perspective, it turns out that this is undecidable: no algorithm can take a simplicial complex, and tell you whether or not it's homeomorphic to a manifold. So the best that can be hoped for is a partial result.

$\endgroup$
5
  • $\begingroup$ I think wikipedia calls this the "manifold problem", which is still undecidable. $\endgroup$ May 11, 2023 at 21:20
  • $\begingroup$ @ThomasAnton, to make one point a bit clearer. When you move from a statement that such collections is recursive to undecidability, you're meaning that from this infinite set there surely be uncountable subsets? Or how do you use the fact of recursiveness here? $\endgroup$
    – A. G
    Nov 8, 2023 at 23:39
  • $\begingroup$ I'm not proving undecidability here. That's the fact I'm referring to. Simply, you can use this combinatorial description of simplicial complexes to number them, and then there is no algorithm that will take number $n$ and tell you if the corresponding simplicial complex is a manifold or not. $\endgroup$ Nov 9, 2023 at 20:27
  • $\begingroup$ @ThomasAnton, oh, you mean that this representation that you mentioned wasn't connected to some sort of inference, but just to view on what it means here to be undecidable? I could have been confused because it looks like one cannot enumerate anyway all simplicial complexes where the number of vertices is unrestricted. But I'm not sure here. $\endgroup$
    – A. G
    Nov 11, 2023 at 21:18
  • $\begingroup$ Okay, I overlooked that if there are infinitely many cells, then simplicial complexes aren't enumerable because there are uncountably many. So let's say we restrict the problem to compact simplicial complexes, which can be given finitely many cells. Then the set is recursive, and that's what the result I mention is about. So the problem can't be given a sufficiently nice solution for compact manifolds, so it definitely can't have a nice solution in general. $\endgroup$ Nov 13, 2023 at 6:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .