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Is there an $m$-dimensional simplicial complex $S$ with the following properties:

  • The cone over $S$ is homeomorphic to $\mathbb{E}^{m+1}$. Here $\mathbb{E}^{m+1}$ denoes the $(m+1)$-dimensional Euclidean space.
  • There is a vertex $v$ in $S$ such that the complement $S\backslash\{v\}$ is not simply connected.

Comments.

  • If such example exist, the cone over $v$ has to from a wild half-line in the Euclidean space. It has to be an embedding of $[0,\infty)$ which complement is not simply connected.

  • The first condition is equivalent to the following: The spherical suspension over $S$ is homeomorphic to $\mathbb{S}^{m+1}$.

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  • $\begingroup$ Btw do you define the cone of $X$ from $X\times [0,1]$ by crushing $X\times\{0\}$ to a point, or from $X\times [0,+\infty\mathclose[$? $\endgroup$
    – YCor
    Commented Jan 24, 2016 at 23:17
  • $\begingroup$ @YCor the cone is infinite. $\endgroup$
    – Anton Petrunin
    Commented Jan 24, 2016 at 23:21
  • $\begingroup$ OK... So passing to the 1-point compactification, it yields a sphere in which the complement of some segment is not simply connected. $\endgroup$
    – YCor
    Commented Jan 24, 2016 at 23:27
  • $\begingroup$ In the comment you might say that the embedding of the half-line is proper . $\endgroup$
    – YCor
    Commented Jan 25, 2016 at 11:25
  • $\begingroup$ Subquestion: does there exist a simplicial complex, homeomorphic to some sphere, and in which there exists an injective combinatorial path (= some consecutive edge with no return) whose complement is not simply connected? $\endgroup$
    – YCor
    Commented Jan 25, 2016 at 11:27

1 Answer 1

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I have got the following answer from Alexander Lytchak:

An example can be constructed the following way. Start with a nontrivial homology sphere, pass to its spherical suspension. Now shrink one of the meridians of suspension to the point, which we denote by $v$. The obtained space $S$ is the example; it admits a natural triangulation.

The first condition is easy to check. To check the second condition note that the spherical suspension $\Sigma(S)$ has a triangulation coming from $S$.

The space $\Sigma(S)$ is a homotopy equivalent to the sphere; it is homological manifold which is also manifold everywhere except maybe the poles of suspension. Each pole admits a simply connected punctured neighborhood. Therefore by disjoint disc property the poles are also manifold point. It remains to apply generalized Poincaré conjecture.

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    $\begingroup$ You start with a homology sphere of dimension $(m-1)$, so that $S$ is $m$-dimensional (and its cone is $(m+1)$-dimensional). So it gives examples for every $m\ge 4$. For $m\le 3$, do you know whether it's impossible? ($m\le 1$ is trivial, $m=2$ is probably easy but I'm not sure for $m=3$) $\endgroup$
    – YCor
    Commented Jan 25, 2016 at 20:08
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    $\begingroup$ @YCor I do not know --- I guess there are no examples for $m\le 3$. Anyway I am happy with this one --- it solves all my problems. $\endgroup$
    – Anton Petrunin
    Commented Jan 25, 2016 at 20:17

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