Homology of spheres

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 group and the topological space/family is sphere
Get more specific information about sphere | Get more computations of homology group

In this article, we briefly describe the homology groups of spheres, a proof using the Mayer-Vietoris homology sequence, and explanations in terms of cellular and simplicial homology.

Statement

Reduced version over integers

For $n$ a nonnegative integer, we have the following result for the reduced homology groups:

${\tilde {H}}_{k}(S^{n})=0,k\neq n$:

and:

${\tilde {H}}_{n}(S^{n})\cong \mathbb {Z}$

Unreduced version over integers

We need to make cases based on whether $n=0$ or $n$ is a positive integer:

• $n=0$ case: $H_{0}(S^{0})\cong \mathbb {Z} \oplus \mathbb {Z}$ and $H_{k}(S^{0})$ is trivial for $k>0$.
• $n>0$ case: $H_{0}(S^{n})\cong H_{n}(S^{n})\cong \mathbb {Z}$ and $H_{k}(S^{n})$ is trivial for $k\neq 0,n$.

Reduced version over a module $M$ over a ring $R$

For $n$ a nonnegative integer, we have the following result for the reduced homology groups:

${\tilde {H}}_{k}(S^{n})=0,k\neq n$:

and:

${\tilde {H}}_{n}(S^{n})\cong M$

Unreduced version over a module $M$ over a ring $R$

We need to make cases based on whether $n=0$ or $n$ is a positive integer:

• $n=0$ case: $H_{0}(S^{0})\cong M\oplus M$ and $H_{k}(S^{0})$ is trivial for $k>0$.
• $n>0$ case: $H_{0}(S^{n})\cong H_{n}(S^{n})\cong M$ and $H_{k}(S^{n})$ is trivial for $k\neq 0,n$.

Related invariants

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

Invariant	General description	Description of value for spheres
Betti numbers	The $k^{th}$ Betti number $b_{k}$ is the rank of the $k^{th}$ homology group.	For $n=0$, $b_{0}=2$, all other $b_{k}$ are $0$; for $n>0$, $b_{0}=b_{n}=1$, all other $b_{k}$s are $0$.
Poincare polynomial	Generating polynomial for Betti numbers	$2$ for $n=0$, $1+x^{n}$ for $n>0$
Euler characteristic	$\sum _{k=0}^{\infty }(-1)^{k}b_{k}$	$2$ for $n$ even, $0$ for $n$ odd.

Facts used

1. Homology for suspension
2. CW structure of spheres
3. Simplicial structure of spheres

Proof using singular homology

Equivalence of reduced and unreduced version

The equivalence follows from the fact that reduced and unreduced homology groups coincide for $k>0$ and for $k=0$, the unreduced homology group is obtained from the reduced one by adding a copy of $\mathbb {Z}$ (or, if working over another ring or module, the base ring or module).

Proof of reduced version

The case $n=0$ is clear: the space $S^{0}$ is a discrete two-point space, hence it has two single-point path components, so the zeroth homology group is $M^{2-1}=M^{1}=M$. Higher homology groups are trivial because the cycle and boundary groups both coincide with the group of all functions to $S^{0}$, so the homology group is trivial.

In general, we use induction, starting with the base case $n=0$. The inductive step follows from fact (1) and the fact that each $S^{n}$ is the suspension of $S^{n-1}$.

Proof using cellular homology

See (2): CW structure of spheres.

Proof using simplicial homology

See (3): Simplicial structure of spheres.