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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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    $\begingroup$ 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. $\endgroup$
    – achille hui
    Jul 27, 2013 at 13:02
  • $\begingroup$ Thanks, it`s very good paper. $\endgroup$
    – Gleb
    Jul 27, 2013 at 13:21
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    $\begingroup$ These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998 $\endgroup$
    – taper
    Dec 22, 2016 at 14:22
  • $\begingroup$ Sadly, OP did not give a correct definition of a triangulation here (instead he referred to the correct definition in a comment to an answer). $\endgroup$
    – Moishe Kohan
    Nov 2 at 19:11

2 Answers 2

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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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  • $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Jul 27, 2013 at 12:32
  • $\begingroup$ Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices. $\endgroup$
    – Gleb
    Jul 27, 2013 at 12:35
  • $\begingroup$ @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. $\endgroup$
    – Daniel Robert-Nicoud
    Jul 27, 2013 at 12:35
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    $\begingroup$ IOW, this is a fine structure as a CW-complex but not as a simplicial complex. $\endgroup$
    – Jyrki Lahtonen
    Jul 27, 2013 at 12:41
  • $\begingroup$ So this 1-vertex object doesn't work, but the minimal has 8 vertices? Surely there has to be something inbetween 2, and 7 which is equal to a number of vertices of some klein bottle triangulation... This 1 vertex example is so close to working I don't see why we couldn't achieve, say, one with 7 vertices fixing the edge issue. $\endgroup$
    – Snared
    May 28 at 4:13
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Here is a obvious triangulation into two triangles:

Gluing diagram for Klein Bottle

Glue the red arrows together, and the black arrows together to get a Klein Bottle from the gray square.

The other way round, we can cut at the arrows and also cut at the yellow line to get 2 triangles from a Klein Bottle.

However, you could also glue a single triangle to a Klein Bottle, if the condition is dropped that complete edges shall be glued to complete edges.

FYI, I make the plot here (Desmos)

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