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:

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

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 $k^{th}$ homology group $H_{k}(T^{n})$ is a free abelian group $\mathbb {Z} ^{\binom {n}{k}}$ 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 $\mathbb {Z} ^{2}=\mathbb {Z} \oplus \mathbb {Z}$:

$n,k$	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 $k^{th}$ 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 $k^{th}$ Betti number $b_{k}$ is the rank of the $k^{th}$ 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.