Does any polygon with side number $2n$ with $n \ge 2$ form a torus when all pairs of opposite sides are joined? (works for n=2, 3)

Question

Wikipedia's Eisenstein integer; Quotient of C by the Eisenstein integers says:

The quotient of the complex plane C by the lattice containing all Eisenstein integers is a complex torus of real dimension 2. This is one of two tori with maximal symmetry among all such complex tori.^{[citation needed]} This torus can be obtained by identifying each of the three pairs of opposite edges of a regular hexagon. (The other maximally symmetric torus is the quotient of the complex plane by the additive lattice of Gaussian integers, and can be obtained by identifying each of the two pairs of opposite sides of a square fundamental domain, such as [0,1] × [0,1].)

This answer explains and shows how identifying opposite sides of a hexagon produces a torus, as do answers to What surface do we get by joining the opposite edges of a hexagon?

Question: Bringing opposite sides of a quadrilateral and a hexagon both produce a torus. Does any polygon side number $2n$ with $n \ge 2$ form a torus when all pairs of opposite sides are joined?

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From the linked answer:

Image source: http://www.math.cornell.edu/~mec/Winter2009/Victor/part1.htm

Answer 1

The answer is "no" for the very next case $n=4$. In fact the answer is "no" for all cases $n \ge 4$.

I will assume that your intention is to glue opposite sides by a translation isometry of the plane; or to put it another way, to glue them using matching arrows that point in the same direction on the plane, but in opposite directions around the periphery of the polygon. It follows from this assumption that the quotient surface is orientable.

In that case, let me describe what happens in two cases, depending on whether $n$ is even or odd. The outline is: count the vertex cycles; use that to compute the Euler characteristic; and then apply the classification of surfaces.

Here's the outcome:

• If $n=2g$ is even, then the $2n$ vertices form $1$ vertex cycle (generalizing the case of the square). The quotient therefore has $1$ vertex, $n$ edges, and $1$ face, thus has Euler characteristic $1-n+1=2-2g$. From the classification of surfaces, it follows that the quotient surface is homeomorphic to the orientable surface of genus $g$.

• If $n=2g+1$ is odd, then the $2n$ vertices form two vertex cycles (generalizing the case of the hexagon). The quotient surface therefore has $2$ vertices, $n$ edges, and $1$ face, thus has Euler characteristic $2-n+1 = 2-2g$, and again is homeomorphic to the surface of genus $g$.

So to answer your question: No, not in general. The only cases where opposite side gluing gives the torus are the two cases you know, the square and the hexagon.

Generalizing from $n=2,3$ to all $n \ge 2$ is fraught with peril!

Answer 2

The simplest way to figure this out is to triangulate the polygon (add one vertex in the middle for simplicity. Then, the quotient space will have $2n$ triangles, and $3n$ edges. It will have the one central vertex, and so the gluings should somehow result in $n-1$ vertices. Whether this is true or not depends on how you do the gluing, but I leave the rest as an exercise.