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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 $4$

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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:

enter image description here

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  • $\begingroup$ I want to know exactly why you don't think the above is valid : (1) there would be two 1-dimensional simplices connecting the vertex of the square and the midpoint of each edge of the square => I agree (2) Actually it is two edges smaller than the minimum triangulation possible => I can not understand what you mean $\endgroup$
    – HK Lee
    Apr 3 at 12:10
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Here this is a solution : enter image description here

(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.

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    $\begingroup$ I don't believe this is a triangulation. The edges of a triangulation should form a simple graph (since otherwise it would not be a simplicial complex). The answer from @Greywhite is correct in pointing out that you have double edges connecting the corners of your outer square to the midpoints of its sides. $\endgroup$
    – Brantley
    May 7 at 21:26
  • $\begingroup$ I have double edges ?? If we choose one edge in side, then it is used in different two triangles ? Am I understanding your point correctly ? $\endgroup$
    – HK Lee
    May 8 at 5:41
  • $\begingroup$ I think that the above triangulation satisfies the following : If we have two triangles ABC, DBC sharing an edge BC in the posting, then a point A is not a point D. $\endgroup$
    – HK Lee
    May 8 at 5:44
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    $\begingroup$ I agree that the triangles are uniquely determined by their vertices, but that is not enough to make it a triangulation. The edges must be uniquely determined by their vertices as well. For instance, consider the vertex in the middle of the right hand side. Both vertical edges coming out of this vertex attach to corners of the square, which are identified. That's what I mean by "double edge". $\endgroup$
    – Brantley
    May 9 at 17:52

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