Definitions
The torus that we consider is the flat representation given by the quotient $\mathbb{C}/\mathbb{Z}[i]$, or equivalently as the set $$ \tau = \{\, (x, y) \in \mathbb{R}^{2} : 0 \leq x, y \leq 1 \,\} $$ with the points $(x, y), (x+1, y), (x, y+1)$ identified. The 3D torus is not of interest here.
Consider the torus as a subset of $\mathbb{R}^{2}$. Then a line is what we would usually consider a straight line in $\mathbb{R}^{2}$, with the identification of edges possibly splitting the line into many segments bounded by the boundary of the torus. For example, the following (left) image is the line of gradient $4/3$ on the torus that passes through $(0,0)$. The right image is twelve tilings of the flat representation to show where each segment on the left comes from.
A partitioning of the torus is an arrangement of lines on the torus. These lines must extend from boundary to boundary, as above.
A triangular grid is a partitioning of the torus by three sets of parallel lines such that every vertex has degree $6$ and every face is a triangle. We may assume the lines to be horizontal, vertical, and of a rational slope if necessary. One example of a triangular grid is given below, followed by five partitions that are allowed on the torus.
It is important that there are exactly three distinct gradients that the lines can have, and that the lines continue from boundary to boundary. For example, the flat representation of the {"Circulant", {9, {1, 2, 3}}} graph (taken from Ed Pegg's answer below) does not fit the requirements of a triangular grid as the lines do not have exactly three different gradients. This is not an allowed partition either as not all lines extend from boundary to boundary.
Useful Identities
Let every intersection of lines induce a vertex, let each line segment between connected vertices be an edge, and let the closed interior of at least two incident edges be a face. Let $V, E, F$ be the number of vertices, edges, and faces that the torus is partitioned into, respectively. It is well known that the Euler characteristic of the torus is zero, so $$ V - E + F = 0.\tag{1} $$ Let $d(v)$ denote the degree of the vertex $v$ (which must be even, by construction). It is also well known that summing the degrees of all vertices gives $$ 2E = \sum d(v).\tag{2} $$ Let $F_{s}$ denote the number of faces with $s > 1$ edges. When there are no loops in the graph, a simple counting argument shows that $$ 2E = \sum_{s \geq 2} sF_{s}.\tag{3} $$ It is also trivial that $F = \sum_{s \geq 2} F_{s}$.
Lemma -- Question
We start with a simple lemma:
Lemma. Consider a partitioning of the torus into $V$ vertices, $E$ edges, and $F$ faces that form a triangular grid. Then $$ E = 3V \qquad\text{and}\qquad F = 2V. $$
Proof. Every face has $3$ edges, so $F = F_{3}$. By $(3)$, we have $$ 2E = \sum_{s\geq 2}sF_{s} = 3F_{3} = 3F. $$ By $(1)$, we have $V - E + F = 0$ so that $3F - 3E = -3V$. We have a system of simultaneous equations, namely $$ 3F - 2E = 0 \qquad\text{and}\qquad 3F - 3E = -3V. $$ Solving this gives $E = 3V$ and $F = 2V$ as required.$\qquad\square$
My question is the following.
Is the converse of the Lemma true? That is, if some partitioning of the torus has $V$ vertices, $E$ edges, and $F$ faces such that $E = 3V$ and $F = 2V$, then is it true that the partition is necessarily a triangular grid?
I have not yet been able to find a counter-example. In the proof of the converse, I want to prove that $d(v) = 6$ for all vertices $v$, and that $F_{3} = F$ with $F_{i} = 0$ for all $i \neq 3$.
I have noticed that we cannot have any faces with more than $6$ edges so that the $s$ index only runs between $2$ and $6$, but there still seem to be too many variables to do anything (namely, each of the vertex degrees and the five $F_{s}$ for $s \in [2, 6]\cap\mathbb{Z}$).
