# Triangulation of $S^{1}\times S^{2}$

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?

• 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. Oct 27, 2021 at 14:36
• Sorry I meant 3-balls.
– Pipe
Oct 27, 2021 at 15:27
• 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 $S^1\times S^2$. Oct 27, 2021 at 15:30
• So the resulting triangulation will be three tetrahedra identified with each other?
– Pipe
Oct 27, 2021 at 15:38
• 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. Oct 27, 2021 at 18:29