I do not know if there are any answered question to it.
To construct the triangulation of $S^{3}$ is possible to use the fact that by taking two 3-balls and identifying their boundaries $S^{2}$, is possible to assign a tetrahedron to each sphere and then identify the two tetrahedra.
How can I think about $S^{1}\times S^{2}$?
$S^{2}$ can be decomposed into two disks, so that I can triangularize it with two triangles glued together?
What about $S^{1}\times S^{2}$: the triangulation is a tetrahedron with triangles ($S^{1}$) glued on each face or is something different?

  • $\begingroup$ How are tetrahedra $2-$balls? Tetrahedra are a $3$-simplex. $S^3$ is a $3$-dimensional manifold, so it is hard to figure how it is a union of two $2$-balls. $\endgroup$ Oct 27, 2021 at 14:36
  • $\begingroup$ Sorry I meant 3-balls. $\endgroup$
    – Pipe
    Oct 27, 2021 at 15:27
  • $\begingroup$ Can't one take the triangularizations of $S^1$ and $S^2$ separately and take their Cartesian product? The product of a $1$-ball and a $2$-ball is topologically a $3$-ball. This would result in four $3$-balls for $S^1\times S^2$. $\endgroup$ Oct 27, 2021 at 15:30
  • $\begingroup$ So the resulting triangulation will be three tetrahedra identified with each other? $\endgroup$
    – Pipe
    Oct 27, 2021 at 15:38
  • 1
    $\begingroup$ It's possible to start with triangulations of $S^1$ and $S^2$ and take a Cartesian product, but a bit more work is needed afterward, since a Cartesian product of simplices is not necessarily a simplex. $\endgroup$
    – Kajelad
    Oct 27, 2021 at 18:29


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