Triangulation of torus $\mathbb{S}^1\times \mathbb{S}^1$ with a minimal number of triangles I want to find a minimal triangulation of torus in $\mathbb{R}^3$
To do this we need $14$
triangles How can we construct the triangulation ?
(cf. 35p. in the book From Euclid to Alexandrov a guided tour - Petrunin and Yashinski)

*

*I do not want a proof but
triangulation


*I find that minimal triangulation for $\mathbb{S}^2$ is $4$
 A: I don't think the answer above is a valid triangulation of the torus, for there would be two 1-dimensional simplices connecting the vertex of the square and the midpoint of each edge of the square. Actually it is two edges smaller than the minimum triangulation possible. Here is my solution:

A: Here this is a solution : 
(cf. https://mathoverflow.net/questions/96988/acute-triangulation)
I will sketch the reason why this is in fact a triangulation : Note that we must show that there are no two triangles $F_1,\ F_2$ sharing three vertices
Note that upper five triangles share one vertex. If $F_1$ is one of the five triangles and $F_1,\ F_2$ share three vertices, then $F_2$ is also one of the five.
Any two of six vertices on the five do not coincide.
Hence remaining last choice for $F_1$ is one of the remaining 4 triangles.
Note that $F_2$ is also one of the remaining 4 triangles. For convenience assume that $ (0,0),\ (0,1),\ (1,1),\ (1,0)  
$ are vertices for a square. If $F_1$ contains $(0,0)$, then $F_1$ contains an interior point in the square Hence since $F_2$ must contain the interior point, $F_2$ contains $(0,0)$. But mid points of side of square in $F_1,\ F_2$ do not coincide.
