The Torus Triangulated

Triangulation The most efficient way to triangulate the (2-dimensional) torus T is the following:

[figure: A; B; C; F; G; A; D; E; F; A; B; C; D; A]  (1)

This is a (distorted) rectangle with all four vertices A, and two pairs of opposite edges ABCA and ADF A identified. (In this picture, opposite edges are equal and parallel, though they are not straight lines. The pattern may be repeated over the whole plane; if you wish to tile a floor using hexagonal tiles in seven colors, this is a good way to do it, by centering each tile over a vertex of the triangulation.)

We order the seven vertices alphabetically.

Cellular chain complex As a cell decomposition, we use one 0-cell A, two 1-cells ABCA and ADF A, and the whole rectangle as the 2-cell. The two edges give obvious simplicial 1-cycles

w_{1} = [AB] + [BC] − [AC], z_{1} = [AD] + [DF] − [AF],

where we take care to keep the vertices in the correct order, and introduce signs as necessary. For the 2-cycle, we add all 14 triangles in (1), oriented anti-clockwise, with signs as necessary,

d_{2} = −[ABF] + [BF G] + …  (2)

This is a 2-cycle because all interior edges in (1) cancel, giving

∂d_{2} = [AF] − [DF] − [DA] + [AB] + [BC] − [AC]
+ [AD] + [DF] − [F A] + [AC] − [CB] − [BA]
= −z_{1} − w_{1} + z_{1} + w_{1} = 0.

Thus the cellular chain complex is

ℤ ⟶^{0} ℤ ⊕ ℤ ⟶^{0} ℤ,  (3)

and the homology groups coincide with the cellular chain groups.

Cocycles We take ℤ as the coefficient group (although the calculations are essentially the same for any coefficient group). From (3), the cellular cochain complex is

ℤ ⟵^{0} ℤ ⊕ ℤ ⟵^{0} ℤ,

and the cohomology groups coincide with the cellular cochain groups. One can check that the simplicial 1-cochains

w^{1} = [AB]^{∗} − [BF]^{∗} + [F G]^{∗} + [DG]^{∗} + [DE]^{∗} − [BD]^{∗},  (4)

where [v_{0}v_{1}]^{∗} denotes the cochain dual to the chain [v_{0}v_{1}], and

z^{1} = [AD]^{∗} + [BD]^{∗} + [BE]^{∗} + [CE]^{∗} + [CF]^{∗} + [CD]^{∗}  (5)

are in fact 1-cocycles, and it is easy to see that they are indeed dual to w_{1} and z_{1} in the sense that

⟨w^{1}, w_{1}⟩ = 1, ⟨w^{1}, z_{1}⟩ = 0, ⟨z^{1}, w_{1}⟩ = 0, ⟨z^{1}, z_{1}⟩ = 1.

You may wish to color these cocycles in the picture (1).

Let α ∈ H^{1}(T; ℤ) and β ∈ H^{1}(T; ℤ) be the classes of w^{1} and z^{1}, and let γ ∈ H^{2}(T; ℤ) be dual to d_{2}. Specifically, we could take γ to be the class of [ABD]^{∗}, since ⟨[ABD]^{∗}, d_{2}⟩ = 1. (All 2-cochains are 2-cocycles.)

Cup products The purpose of finding the representative simplicial cocycles (4) and (5) is to enable us to use the formula

⟨x ⌣ y, [v_{0}v_{1}v_{2}]⟩ = ⟨x, [v_{0}v_{1}]⟩⟨y, [v_{1}v_{2}]⟩

to compute cup products, where v_{0}, v_{1}, and v_{2} must be in order. In computing w^{1} ⌣ z^{1} for example, [v_{0}v_{1}]^{∗} must appear in w^{1} and [v_{1}v_{2}]^{∗} must appear in z^{1} in order to produce anything; only one term survives, namely

w^{1} ⌣ z^{1} = [ABD]^{∗}.

Similarly,

z^{1} ⌣ w^{1} = [BDE]^{∗},

which represents −γ (as BDE is taken clockwise),

w^{1} ⌣ w^{1} = −[ABD]^{∗} − [BDE]^{∗} − [BF G]^{∗} − [ABF]^{∗},  (6)

and z^{1} ⌣ z^{1} = 0 (exactly). One can show directly that w^{1} ⌣ w^{1} is a coboundary, but it is simpler to calculate from (6) that

⟨w^{1} ⌣ w^{1}, d_{2}⟩ = −1 + 1 − 1 + 1 = 0.

We assemble all this information.

Theorem 7 The cohomology H^{∗}(T; ℤ) is a free abelian group with basis elements 1 ∈ H^{0}(T; ℤ), α ∈ H^{1}(T; ℤ), β ∈ H^{1}(T; ℤ), and γ ∈ H^{2}(T; ℤ). The ring structure is given by α ⌣ α = 0, α ⌣ β = γ, β ⌣ α = −γ, and β ⌣ β = 0.

Remark If we take advantage of the anticommutativity of the cup product in cohomology and the fact that H^{2}(T; ℤ) = ℤ, we only need to compute α ⌣ β; the other three products are automatic.