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What is minimal triangulation of Klein bottle? А triangulation is a subdivision of a geometric object into simplices. Minimal in sense of vertex count.

So, I know that minimal count of vertex in the shortest triangulation must be greater then $7$, because the shortest triangulation of torus consist of $7$ vertex and Euler characteristic is equal to $0$.

I would be cool if you can show me the picture.

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I'd be impressed if there was a way to do it in less than $8$. –  Thomas Andrews Jul 27 '13 at 12:11
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them. –  achille hui Jul 27 '13 at 13:02
Thanks, it`s very good paper. –  Gleb Jul 27 '13 at 13:21

2 Answers 2

Following Daniel Robert-Nicoud's approach, if Jyrki Lahtonen is right in that the mapping must be injective for each triangle closure, then we need at least three vertices. Here is a triangulation that achieves the minimum vertex count of three. Note however that with or without Jyrki Lahtonen's restriction, we have (vertex count)$+$(face count)$-$(edge count)$=0$.

Triangulation with three vertices

Wait: Wrong. Achille Hui's reference makes the reasonable requirement that there shall be no duplicate edges, resulting in that at least seven vertices are needed. My example has lots of duplicate edges.

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achille hui gave the answer to my question. In your picture we have two vertex (upper center and right center) incidence two edges, and it isn`t triangulation. –  Gleb Jul 27 '13 at 13:42
@Gleb: Yes, I have already added why it does not work. I have considered deleting my answer, but now this is part of the discourse, so I let it stand, as shameful as it may be. –  ccorn Jul 27 '13 at 13:45

The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.

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I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle. –  Jyrki Lahtonen Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices. –  Gleb Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though. –  Daniel Robert-Nicoud Jul 27 '13 at 12:35
IOW, this is a fine structure as a CW-complex but not as a simplicial complex. –  Jyrki Lahtonen Jul 27 '13 at 12:41

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