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 vertices in the shortest triangulation must be greater than $7$, because the shortest triangulation of torus consists of $7$ vertices and Euler characteristic is equal to $0$.
I would be cool if you can show me the picture.
The standard definition of a triangulation of a space $X$ consists of a simplicial complex $T$ and a homeomorphism $f: |T| \to X$. (n.b. As pointed out by M.K. in the comments, the OP gave an incorrect definition in the question, but referred to the correct definition in a reply to an answer).
The other two answers give CW decompositions of the Klein bottle $K$, but since they are not simplicial complexes, they are not triangulations of $K$ in the standard sense. A minimal CW decomposition of the Klein bottle requires one $0$-cell, two $1$-cells, and one $2$-cell. See here.
The paper Vertex-Minimal Simplicial Immersions of the Klein Bottle in Three Space linked in the first comment on OP shows that every triangulation of $K$ has at least $8$ vertices. I will summarize the argument here.
To get a lower bound they use Euler's formula for polyhedra
$$\chi(X) = V - E + F$$
where $\chi(X)$ is the Euler characteristic of $X$, and $V, E, F$ are the numbers of vertices, edges, and faces respectively. Some simple combinatorics and algebra shows that
$$V \geq \frac{7 + \sqrt{49 - 24 \chi(X)}}{2}.$$
Since $\chi(K)= 0$ (see, e.g., here) then one immediately obtains $V \geq 7$.
However, there are in fact no $7$-vertex triangulations of $K$. To see this, if $V=7$ then $E=21$ by Euler's formula, but since $21$ is the number of edges in the complete graph $K_7$ on $7$ vertices, therefore every vertex is connected to every other vertex by an edge. By a result of Franklin, every graph embeddable in the Klein bottle is $6$-colorable, but $K_7$ has chromatic number $7$, a contradiction. Therefore $V \geq 8$.
The authors go on to classify all combinatorially distinct $8$-vertex triangulations of $K$, of which there are six. Here they are:
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 induce 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.
t triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
Here is a obvious triangulation into two triangles:
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)
