Homology of torus

This article describes the value (and the process used to compute it) of some homotopy invariant(s) for a topological space or family of topological spaces. The invariant is homology and the topological space/family is torus
Get more specific information about torus | Get more computations of homology

Statement

We denote by $T^{n}$ the $k$ -dimensional torus, which is the topological space:

$(S^{1})^{n} ≅ S^{1} \times S^{1} \times \dots S^{1}$

i.e., the product of $S^{1}$ , the circle, with itself $n$ times. It is equipped with the product topology. If we think of $S^{1}$ as a group, $T^{n}$ gets a group structure too as the external direct product.

Unreduced version over integers

The $k^{t h}$ homology group $H_{k} (T^{n})$ is a free abelian group $Z^{(\binom{n}{k})}$ of rank $(\binom{n}{k})$ , where $(\binom{n}{k})$ denotes the binomial coefficient, or the number of subsets of size $k$ in a set of size $n$ . In particular, $(\binom{n}{k})$ is positive for $k \in {0, 1, 2, \dots, n}$ and zero for other $k$ . Thus, $H_{k} (T^{n})$ is nontrivial for $0 \leq k \leq n$ and zero for $k > n$ .

Below are given the ranks of homology groups for small values of $n$ and $k$ . Each row corresponds to a value of $n$ and each column corresponds to a value of $k$ . If a cell value reads 2, for instance, that means that the corresponding homology group is $Z^{2} = Z \oplus Z$ :

$n, k$	0	1	2	3	4
0	1	0	0	0	0
1	1	1	0	0	0
2	1	2	1	0	0
3	1	3	3	1	0
4	1	4	6	4	1

Reduced version over integers

The $k^{t h}$ reduced homology group ${\tilde{H}}_{k} (T^{n})$ is a free abelian group of rank $(\binom{n}{k})$ for $k > 0$ and is trivial for $k = 0$ . In particular, it is a nontrivial group for $k \in {1, 2, \dots, n}$ and is zero for other $k$ .

Related invariants

These are all invariants that can be computed in terms of the homology groups.

Invariant	General description	Description of value for torus
Betti numbers	The $k^{t h}$ Betti number $b_{k}$ is the rank of the $k^{t h}$ homology group.	$b_{k} = (\binom{n}{k})$
Poincare polynomial	Generating polynomial for Betti numbers	$(1 + x)^{n}$
Euler characteristic	$\sum_{k = 0}^{\infty} (- 1)^{k} b_{k}$	0 -- can also be seen from the fact that we have a group.