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how how will the minimal triangulation of a torus look like?

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Let $\omega = e^{2 \pi i/3}$. Let $L$ be the planar lattice generated by $1$ and $\omega$. Let $M$ be the sublattice generated by $2-\omega$ and $\omega (2 - \omega)$. Tile $\mathbb{R}^2$ by equilateral triangles with vertices at $L$, and quotient by $M$. This is a triangulation of the torus with $7$ vertices (exercise!). It is minimal because any triangulation of the torus must have a vertex of degree $6$, so you can't get use fewer vertices.

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As a bonus this shows the embedding of $K_7$ in the torus, and the dual graph is the strongly regular Heawood graph (= incidence graph of the Fano plane) and tiles the torus with $7$ pairwise adjacent hexagons. – Noam D. Elkies Apr 9 '12 at 20:23
[Correction: the Heawood graph isn't quite strongly regular, but it is vertex- and edge-transitive.] – Noam D. Elkies Apr 9 '12 at 21:22
Moreover, Dewdney proved in "Wagner’s theorem for torus graphs" Discrete Math. 4 (1973), 139–149, that every triangulation of the torus can be transformed to this one by a sequence of flips and edge contractions. – Gjergji Zaimi Apr 9 '12 at 22:08
Moreover, this $7$-vertex triangulation can be realized as an embedded simplicial complex in $3$-space, so you can actually see it. (I made one out of construction paper during an idle undergraduate afternoon.) For more information about this, see, for example, . – Robert Bryant Apr 9 '12 at 23:38
Why can't I use fewer vertices? Does the definition of "triangulation" preclude an edge that joins a vertex to itself? – maproom Apr 10 '12 at 11:25

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