Triangulate a torus with 6 vertices

Question

Is it possible to triangulate a torus in such a way that each vertex has 6 triangles around it?

This is what I have so far: $0=v-e+f$

$0=2e/6-e+2e/3$

$0=0e$

this is after substituting $v=2e/6$ and $f=2e/3$ However, I am just confused since the end result is $0=0e$ so does this mean that it is possible to triangulate a torus when each vertex has $6$ triangles around it?

Answer 1

Absolutely. Take a triangulation of $\mathbb{R}^2$ with $6$ triangles meeting at each vertex, making sure that your tiling is invariant under translations in $\mathbb{Z}^2$. This will induce a tiling of $\mathbb{T}^2$ upon taking a quotient.

Answer 2

Imagine a rectangle. The diagonal will give you a 2 triangle...triangulation. the folding then bending of this rectangle into a torus doesn't change that. So in fact, you can triangluate a torus with just 2! Now just imagine 3 such rectangles and there you can easily have six or any even number for that matter. Does that make sense?