Homology of spheres

In this article, we briefly describe how to compute the homology groups of spheres using the Mayer-Vietoris homology sequence.

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:

${\tilde{H}}_{n} (S^{n}) ≅ Z$

Unreduced version over integers

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

$n = 0$ case: $H_{0} (S^{0}) ≅ Z \oplus Z$ and $H_{k} (S^{0})$ is trivial for $k > 0$ .
$n > 0$ case: $H_{0} (S^{n}) ≅ H_{n} (S^{n}) ≅ Z$ 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:

${\tilde{H}}_{n} (S^{n}) ≅ M$

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

Facts used

Homology for suspension

Proof

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 $Z$ (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 $S^{n - 1}$ .