Homology of spheres: Difference between revisions

Revision as of 03:43, 24 December 2010

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 $S^{n - 1}$ .

@@ Line 1: / Line 1: @@
 In this article, we briefly describe how to compute the homology groups of spheres using the [[Mayer-Vietoris homology sequence]].
-{{fillin}}
+==Statement==
+===Reduced version over integers===
+For <math>n</math> a nonnegative integer, we have the following result for the [[reduced homology group]]s:
+<math>\tilde{H}_k(S^n) = 0, k \ne n</math>:
+and:
+<math>\tilde{H}_n(S^n) \cong \mathbb{Z}</math>
+===Unreduced version over integers===
+We need to make cases based on whether <math>n = 0</math> or <math>n</math> is a positive integer:
+* <math>n = 0</math> case: <math>H_0(S^0) \cong \mathbb{Z} \oplus \mathbb{Z}</math> and <math>H_k(S^0)</math> is trivial for <math>k > 0</math>.
+* <math>n > 0</math> case: <math>H_0(S^n) \cong H_n(S^n) \cong \mathbb{Z}</math> and <math>H_k(S^n)</math> is trivial for <math>k \ne 0,n</math>.
+===Reduced version over a module <math>M</math> over a ring <math>R</math>===
+For <math>n</math> a nonnegative integer, we have the following result for the [[reduced homology group]]s:
+<math>\tilde{H}_k(S^n) = 0, k \ne n</math>:
+and:
+<math>\tilde{H}_n(S^n) \cong M</math>
+===Unreduced version over a module <math>M</math> over a ring <math>R</math>===
+We need to make cases based on whether <math>n = 0</math> or <math>n</math> is a positive integer:
+* <math>n = 0</math> case: <math>H_0(S^0) \cong M \oplus M</math> and <math>H_k(S^0)</math> is trivial for <math>k > 0</math>.
+* <math>n > 0</math> case: <math>H_0(S^n) \cong H_n(S^n) \cong M</math> and <math>H_k(S^n)</math> is trivial for <math>k \ne 0,n</math>.
+==Facts used==
+# [[uses::Homology for suspension]]
+==Proof==
+===Equivalence of reduced and unreduced version===
+The equivalence follows from the fact that reduced and unreduced homology groups coincide for <math>k > 0</math> and for <math>k = 0</math>, the unreduced homology group is obtained from the reduced one by adding a copy of <math>\mathbb{Z}</math> (or, if working over another ring or module, the base ring or module).
+===Proof of reduced version===
+The case <math>n = 0</math> is clear: the space <math>S^0</math> is a discrete two-point space, hence it has two single-point path components, so the zeroth homology group is <math>M^{2 - 1} = M^1 = M</math>. Higher homology groups are trivial because the cycle and boundary groups both coincide with the group of ''all'' functions to <math>S^0</math>, so the homology group is trivial.
+In general, we use induction, starting with the base case <math>n = 0</math>. The inductive step follows from fact (1) and the fact that each <math>S^n</math> is the suspension of <math>S^{n-1}</math>.