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Suppose $\Omega\subset \mathbb{S}^n$ is an open set with $\Sigma=\partial \Omega$ a smooth hypersurface. If $\Omega$ is a homology ball, in the sense that $H_i(\Omega)=H_i(\mathbb{D}^n)$ for $0\leq i \leq n$, then is it possible for $\Omega$ to be non-contractible?

Some (possibly incorrect) observations:

  • When $n=1,2$ it is clear any such $\Omega$ is contractible.
  • For all $n>1$, I believe $\Sigma$ is a homology sphere (via Poincare duality) so when $n=3$ there is no such non-contractible $\Omega$ (using the classification of surfaces and Alexander's theorem -- is there a simpler argument?)
  • By removing a small ball from a homology sphere one can produce a manifold with boundary, $X$, that is a homology ball that is not contractible. However, by the generalized Schoenflies conjecture of Brown and Mazur, such an $X$ can't embed in $\mathbb{S}^n$.

EDIT: For what it is worth, this was essentially asked (and answered) previously in this question

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  • $\begingroup$ For your last point, you can argue $X$ must be simply-connected by using $X$ in a computation of the fundamental group of $S^n$, so it is contractible. This avoids Brown-Mazur. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2021 at 21:36

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Yes, such domain exists. Let $P$ be Poincare homology 3-sphere. And $X= P\times I - D^3\times I$, then $X$ can be smoothly embedded in $S^4$(mostly the double of $X$ is $S^4$) . Let $\Omega$ be a small open neighbourhood of $X$. Then notice that $\Omega$ deformation retract onto $X$, but $\Omega$ cannot be contractible as $X$ is not.

There is a deeper reason why a homology ball that $P\# -P$ bounds cannot be contractible, and it is essentially follows from Taube's periodic end theorem.

Here is one of my attempt to draw a picture.

enter image description here

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    $\begingroup$ The embedding of the punctured Poincare dodecahedral space in S^4 comes as a Seifert surface for a spun knot. That's how I like to think about it. I believe it should be the 5-twist spin of the trefoil. maths.ed.ac.uk/~v1ranick/papers/zeemantwist.pdf $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2021 at 21:58
  • $\begingroup$ Hi Ryan, thanks for the reference. That's a good explanation. I think I am not very good in visualizing knots and (personally) try to viasualize in terms of Kirby calculus. $\endgroup$
    – Anubhav Mukherjee
    Commented Oct 26, 2021 at 22:02
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    $\begingroup$ And you have a nice answer. I think this Zeeman theorem isn't as well known as it should be. It gives a clean prescription for how to smoothly embed once-punctured 3-manifolds in $S^4$, in terms of the standard language of 3-manifold theory and their automorphism groups. There is a follow-up theorem of Litherland's that extends Zeeman's construction maximally. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2021 at 22:04

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