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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

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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