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We say that a simplicial complex $K$ is acyclic if it's integral reduced simplicial homology groups are trivial in all dimensions.

For a vertex ${v} \in K$, we define the link

$$lk(v) :=\{\sigma \in K \; | \; \sigma \cup \{v\} \in K, \sigma \cap \{v\} = \emptyset\}.$$

A simplicial complex is an integral generalised homology $n$-sphere if it has the homology of $S^n$, and has the same integral local homology groups as $\mathbb{R}^n$.

There exist acyclic finite simplicial complexes such that the link of every vertex is non-acyclic. Does there exist a finite acyclic simplicial complex for which the link of every vertex is itself an integral generalised homology sphere? Alternatively, such that the link of every vertex has the homology of a sphere?

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  • $\begingroup$ Implicitly you're choosing a field with which to compute homology here, right? (Or are you talking about $\mathbb{Z}$-homology?) $\endgroup$
    – Sam Hopkins
    Commented Jun 19, 2021 at 16:56
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    $\begingroup$ Ah, sorry. Yes, coefficients in $\mathbb{Z}$. Edited the question. $\endgroup$
    – Matt
    Commented Jun 19, 2021 at 18:00
  • $\begingroup$ Note that the local homology can be recovered (after a shift) from the homology of the links. See Munkres' "Topological results in combinatorics". I haven't checked carefully, but I think the condition in the question reduces to finding a complex where the link of each vertex has nontrivial homology, and the link of each higher face has \mathbb{Z} in the top homology degree, 0 elsewhere. $\endgroup$
    – Russ Woodroofe
    Commented Jun 24, 2021 at 10:22

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If all vertex links in a finite simplicial complex $K$ are homology $n-1$-spheres (i.e., homeomorphic to $n-1$-manifolds with the same homology as an $n-1$-sphere), then the simplicial complex $K$ is a closed homology $n$-manifold. As such it has a mod-2 fundamental class: $H_n(K;\mathbb{F}_2)\cong \mathbb{F}_2$. Applying the universal coefficient theorem, it follows that $K$ cannot be integrally acyclic.

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  • $\begingroup$ Note that the OP is using a different definition of "homology $n-1$-sphere" than you. While your definition, which requires the complex to be homeomorphic to a manifold, agrees with the Wikipedia definition (en.wikipedia.org/wiki/Homology_sphere), the OP's definition of "homology manifold with the same homology as a sphere" (see en.wikipedia.org/wiki/Homology_manifold) is also reasonable and comes up often. $\endgroup$
    – Sam Hopkins
    Commented Jun 23, 2021 at 15:41
  • $\begingroup$ In particular I think the OP's definition of homology sphere is basically equivalent to being a Gorentstein* complex... $\endgroup$
    – Sam Hopkins
    Commented Jun 23, 2021 at 15:44
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    $\begingroup$ My argument applies with the OP's broader definition too; I must admit that I hadn't thought carefully about his definition - I put my own definition in to make sure that I didn't say anything false as I was worried that his definition would be even broader than it is. $\endgroup$
    – IJL
    Commented Jun 23, 2021 at 15:45
  • $\begingroup$ Thank you very much for your answer. Have you a reference for your first statement? I'll also clarify my terminology in the question - I used homology sphere to avoid using the term "homology homology sphere". $\endgroup$
    – Matt
    Commented Jun 24, 2021 at 9:09
  • $\begingroup$ To clarify, I'm asking for a proof of the statement "If all vertex links in $K$ are homology $(n-1)$-spheres, then $K$ is a closed homology $n$-manifold." where here, by "homology $(n-1)$-sphere", I mean that $K$ has the homology of an $(n-1)$ sphere and is a homology manifold. $\endgroup$
    – Matt
    Commented Jun 24, 2021 at 9:11

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