Homology of spheres

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

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

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

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 $$k \ne 0,n$$.

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:

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

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 $$k \ne 0,n$$.

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

Facts used

 * 1) uses::Homology for suspension
 * 2) uses::CW structure of spheres
 * 3) uses::Simplicial structure of spheres

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.