Triangulation of torus $\mathbb{S}^1\times \mathbb{S}^1$ with a minimal number of triangles

Question

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)

1. I do not want a proof but triangulation

2. I find that minimal triangulation for $\mathbb{S}^2$ is $4$

Answer 1

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.