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We have a characterization when we want $|X|$ to be a PL-manifold, in particular that the links of all the vertices are themselves (PL) spheres. If we are in the category of PL- spaces then this is a necessary and sufficient condition. If however, we leave the PL category, then we get simplicial complexes that are topologically spheres, but not PL spheres. For instance if we take the double suspension of a homology sphere, then we get a sphere. If our homology sphere has a triangulation (and is not itself a sphere), then the double suspension also has a triangulation. The links of the suspension points are not spheres, but these links do have the homotopy type of spheres. Furthermore, the links of these suspension points are also pure simplicial complexes (pure meaning that that all top dimensional simplicies have the same dimension).

I have drawn the following conclusion: If $X$ is a simplicial complex, |X| is a topological manifold of dimension $N$, then the links of all of the vertices are pure simplicial complexes of dimension $N-1$ and have the homotopy type of the $N-1$ sphere.

What is the converse to this conclusion? Namely what extra conditions must be put on the links of the vertices other than purity and being homotopy sphere that will be equivalent to $|X|$ is a manifold?

note: This question is similar to When are (finite) simplicial complexes (smooth) manifolds? . However, the I did not find the answer there. This is also a slightly different question.

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  • $\begingroup$ A tautological answer would be "all links have to be homeomorphic, not just homotopy equivalent to spheres". But I'm not sure whether this is really what you're after. $\endgroup$
    – algori
    Dec 13, 2012 at 21:18
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    $\begingroup$ That does not seem true, since if you take the link of one of the suspension points of a double suspension of a homology three sphere (like Poincare's dodecaheadral space), you get the suspension of the homology manifold, which I do not even think is a manifold. $\endgroup$
    – Spice the Bird
    Dec 13, 2012 at 21:46
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    $\begingroup$ Spice the Brid -- this is a sufficient condition, not a necessary one. I.e., it implies that $|X|$ is a manifold, but not the other way around. $\endgroup$
    – algori
    Dec 13, 2012 at 21:58
  • $\begingroup$ OK I agree with that. I was looking for an equivalent statement. That is reflected in the title, but is not reflected properly in the grey box. Thank you. $\endgroup$
    – Spice the Bird
    Dec 13, 2012 at 22:02

2 Answers 2

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In dimensions $\geq 5$, Theorem 1.5 of Galewski and Stern implies that if the links of the vertices of a homology manifold of dimension $\geq 5$ are simply-connected, then it is a manifold.

I'm not sure if this is equivalent to your condition. Certainly your condition implies that the links of vertices are simply-connected and are homology spheres. But to be a homology $n$-manifold, the links of the barycentric subdivision must be $n-1$-homology spheres as well, and I'm not sure whether that is implied by your conditions.

In $1, 2, 3$ and (I believe) $4$ dimensions, I think it's necessary and sufficient to have links be $n-1$-spheres.

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I don't think there will be a simple necessary and sufficient condition.

For exotic triangulations, the links $L$ of vertices will not be manifolds. One thing you need (not sufficient) is for $L \times \mathbb{R}$ to be a manifold. It's kind of shocking that this is possible for non-manifolds; the first example of it is Bing's dogbone space. A weaker version of the phenomena is the Whitehead manifold $W$, which is a contractible $3$-manifold which is not homeomorphism to $\mathbb{R}^3$ but such that $W \times \mathbb{R}$ is homeomorphism to $\mathbb{R}^4$.

To get an idea of the kinds of things that are involved here, I recommend reading the introduction to Edwards's paper "Suspensions of homology spheres", available here. That paper also contains a very detailed account of the fact that the double suspension of Mazur's homology 3-sphere is a 5-sphere (with lots of pictures). Another nice account of this is in Steve Ferry's notes on geometric topology, available here.

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  • $\begingroup$ Rob Kirby's lecture here dailymotion.com/video/xuvg6y_kirby1_tech has a discussion of the examples obtained from the double suspension theorem. From what Kirby says of such examples, I would not want to meet them in a dark alley. $\endgroup$
    – Lee Mosher
    Dec 14, 2012 at 0:56
  • $\begingroup$ I met one in a dark alley once. He wasn't so mean, but sort of ugly. $\endgroup$
    – Spice the Bird
    Dec 14, 2012 at 2:27

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