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Is it true that any compact piecewise linear homology manifold is homotopically equivalent to a (smooth?) manifold of the same dimension?

Let me say bit more since my question was wrongly understood.

  1. Any link of homology manifold has to be a homoplogy sphere.

  2. By double suspension every point on a simplex of dimension at least 1 is a manifold point (it has a neighborhood homeomorphic to an open set in $\mathbb R^n$.

  3. Therefore we have a finite discrete set of topological singularities. We can remove an $\epsilon$-neighbborhood around each, its boundary is a homological sphere so we can patch the hole by contactable manifold with the same boundary.

  4. It seems to be an answer in the topological category. Am I right?

  5. I hope that starting with dimension 5 one can do the same in smooth category.

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3 Answers 3

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On the revised question:

I am not sure what you mean by doing the same in smooth category but there are PL -manifolds that are not homotopy equivalent to smooth ones, see e.g. [M. Davis and J-C. Hausmann, Aspherical manifolds without smooth or PL structure, Springer Lecture Notes in Math. 1370, (1989), 135--142] available here. See also example 2.1 in here.

As for your question 4 you seem to be asking whether the identity map of a homology sphere $S$ extends to a homotopy equivalence between the cone $CS$ on $S$ and a contractible manifold $X$ with $\partial X=S$. This is an obstruction theory problem and (I think) there is no obstruction because all obstruction cocycles lie in relative cohomology groups with coefficients in $\pi_*(X)$ or $\pi_*(CS)$, which all vanish.

Perhaps by "doing it in the smooth category" you meant finding a smoothing on your manifold minus a star of the singular locus, and then replacing the remaining cones by smooth contractible manifolds. This might not be possible. The issue is that the smoothing must be such that all boundary homology spheres bound smooth contractible manifolds. For example, if a homotopy sphere smoothly bounds a contractible manifold, then removing a small ball from the interior of the contractible manifold gives an $h$-cobordism between the given homotopy sphere and the standard sphere. In higher dimensions the $h$-cobordism is trivial, so the homotopy sphere cannot be exotic.

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  • $\begingroup$ Thank you very much. By the way, the question was not revised — it was understood wrongly. $\endgroup$
    – ε-δ
    Commented Dec 17, 2015 at 15:07
  • $\begingroup$ You might be interested in arxiv.org/abs/1406.1735. $\endgroup$
    – Igor Belegradek
    Commented Dec 17, 2015 at 15:29
  • $\begingroup$ By the way, you say "removing a small ball from the interior of the contractible manifold gives an h-cobordism". Are you sure (I do not think it is correct). $\endgroup$
    – ε-δ
    Commented Dec 18, 2015 at 16:23
  • $\begingroup$ It is correct, if the boundary of $X$ is a homotopy sphere, which is how I apply it. Let $X_0$ be $X$ with small ball removed. The boundary of $X_0$ has two components: $S$ and $S_0$, where $S_0$ is the standard sphere. By excision the inclusion $i: S_0\to X_0$ is a homology isomorphism, and since everything is simply-connected, $i$ is a homotopy equivalence. By duality the inclusion $j: S\to X_0$ is also a homology isomorphism. If $S$ is simply-connected, then $j$ is a homotopy equivalence. $\endgroup$
    – Igor Belegradek
    Commented Dec 18, 2015 at 16:48
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    $\begingroup$ I was merely explaining that even when $S$ is simply-connected, there is no way to extend an arbitrary smoothing of $X_0$ to $X$. There is a similar phenomenon when $S$ is not simply-connected: For sufficiently large $n$ (probably $n>3$), there is a map $\phi$ from the set of homology $n$-spheres to the set of smooth structures on $S^n$ such that $\phi(\Sigma^n)$ is standard if and only if $\Sigma^n$ boudns a contractible manifolds. I suggest looking at Kervaire's paper on homology spheres. $\endgroup$
    – Igor Belegradek
    Commented Dec 22, 2015 at 20:09
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There is little hope of that, I think. The best work related to Poincare homology complexes that I am familiar with is that by Jonathan A. Hillman, which should provide many counterexamples.

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    $\begingroup$ Indeed: The paper J. Hillman, An indecomposable PD3-complex whose fundamental group has infinitely many ends, Math. Proc. Cambridge Philos. Soc. 138 (2005), 55–57, contains an example whose fundamental group is not isomorphic to a 3-manifold group. Somebody (C.B.Thomas?) also had an earlier example with finite fundamental group. $\endgroup$
    – Misha
    Commented Dec 14, 2015 at 17:42
  • $\begingroup$ @Misha, thanks for doing my work for me :-) $\endgroup$
    – Mikhail Katz
    Commented Dec 15, 2015 at 8:00
  • $\begingroup$ Sorry, I did not get it — are you trying to say that there is a counterexample in dimension 3? — all 3-dimensional homological PL-mainifolds are manifolds, aren't they? $\endgroup$
    – ε-δ
    Commented Dec 15, 2015 at 16:03
  • $\begingroup$ @ε-δ, What do you mean by a "homology manifold"? $\endgroup$
    – Mikhail Katz
    Commented Dec 15, 2015 at 16:05
  • $\begingroup$ en.wikipedia.org/wiki/Homology_manifold Please correct me if I am wrong: the link of vertex in h-manifold has to be homological sphere, so if dimension is 3 it has to be 2-sphere, so it was a manifold. $\endgroup$
    – ε-δ
    Commented Dec 15, 2015 at 16:10
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Unless I am confused, there is a counterexample in the first few lines of Bryant/Ferry/Mio/Weinberger (1996)

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    $\begingroup$ Igor: There are two issues with this example: (a) It is not a simplicial complex (actually, such homology manifolds cannot be simplicial complexes since they have to be homogeneous), (b) PI is asking about homotopy-equivalence, not a homeomorphism. The obstruction to being a manifold in BFMW-examples is $p_0$ (0-th Pontryagin class), I do not know if it is homotopy invariant or not. $\endgroup$
    – Misha
    Commented Dec 14, 2015 at 17:30
  • $\begingroup$ @Misha Re homotopy equivalence vs homeomorphism, that thought has occurred to me, but it seemed to me that it should not be hard to check that they are not homotopy equivalent, either. I could, of course, be wrong. $\endgroup$
    – Igor Rivin
    Commented Dec 14, 2015 at 19:28
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    $\begingroup$ At least some of the examples they construct are homotopy spheres. $\endgroup$
    – Misha
    Commented Dec 14, 2015 at 19:59

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