Let $\Delta$ be an abstract simplicial complex on finitely many vertices and $|\Delta|$ be it's geometric realization. (https://en.m.wikipedia.org/wiki/Abstract_simplicial_complex)

If $|\Delta|$ is a connected, orientable, $3$-manifold without boundary, then is $|\Delta|$ Homeomorphic to the sphere $\mathbb S^3$ ?

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    $\begingroup$ @Kevin. S: is the $3$-Torus the geometric realization of a finite simplicial complex ? $\endgroup$
    – uno
    May 21, 2020 at 8:59
  • $\begingroup$ It is known that every $3$-manifold has a triangulation (see for example Moise), and I think if the manifold is compact this triangulation is finite but I am not sure. $\endgroup$
    – William
    May 21, 2020 at 18:33
  • $\begingroup$ Yes, I believe you can prove if $|\Delta|$ is compact for an abstract simplicial complex $\Delta$ then it has finitely many simplices. For each $n$ and each $n$-simplex $\sigma$, choose an open neighbourhood of $|\sigma| \subset |\Delta|$ which does not contain the barycenter of any other $k$-simplex for $k \geq n$ (for example use the fact that $|\Delta|$ is metrizable): then I believe this open cover will not have a finite subcover unless there are finitely many simplices. Now if you have any compact $3$-manifold, not only is it triangulable but any triangulation will be finite. $\endgroup$
    – William
    May 21, 2020 at 19:43

1 Answer 1


Any $3$-manifold is triangulable and if it is compact then the triangulation is finite. In particular the $3$-torus is the geometric realization of a finite simplicial complex, so the answer to your question is "no".


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