# Thread: Homology Groups of T3

1. ## Homology Groups of T3

Find the fundamental group and the homology groups of the 3-dimensional torus, T3 = S1 X S1 X S1.

The fundamental group preserves products, so it's Z X Z X Z right? The first homology group would be the abelianization of this group... but how do you compute higher homology groups of this surface? I came across a method called the Mayor-Vietoris sequence, but I'm not sure how to apply it here.

2. Originally Posted by joeyjoejoe
Find the fundamental group and the homology groups of the 3-dimensional torus, T3 = S1 X S1 X S1.

The fundamental group preserves products, so it's Z X Z X Z right?
Right. $\pi_1(S^1 \times S^1 \times S^1) \cong \pi_1(S^1) \times \pi_1(S^1) \times \pi_1(S^1)$ and $\pi_1(S^1) = \mathbb{Z}$.

The first homology group would be the abelianization of this group... but how do you compute higher homology groups of this surface?
First of all, a 3-torus is a 3-dimensional manifold so it is not a surface.

To compute a homology groups of a 3-torus, the easiest way I think is to use a CW complex and compute a cellular homology group.

The CW structure of 3-torus consists of one 3-cell, three 2-cells, three-1 cells, and one 0-cell.

The cellular chain complexes for a 3-torus are as follows:

$0 \rightarrow \mathbb{Z} \stackrel{d_3}{\rightarrow} \mathbb{Z}^3 \stackrel{d_2}{\rightarrow} \mathbb{Z}^3 \stackrel{0}{\rightarrow} \mathbb{Z} \rightarrow 0$

Note that all the celluar boundary maps of the above are zero. I leave it to you to verify this.

Now, the (cellular) homology group of $T^3$ is

$H_0(T^3) = \mathbb{Z}$ ( a single connected component),
$H_1(T^3) = H_2(T^3) = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$,
$H_3(T^3) = \mathbb{Z}$,
$H_k(T^3) = 0$ for k > 3.