degree (map of spheres)

Given a non-negative integer $n$ , let $S^{n}$ denote the $n$ -dimensional sphere. Suppose $f\colon S^{n}\to S^{n}$ is a continuous map. Applying the $n^{th}$ reduced homology functor $\widetilde{H}_{n}(\_)$ , we obtain a homomorphism $f_{*}\colon\widetilde{H}_{n}(S^{n})\to\widetilde{H}_{n}(S^{n})$ . Since $\widetilde{H}_{n}(S^{n})\approx\mathbbmss{Z}$ , it follows that $f_{*}$ is a homomorphism $\mathbbmss{Z}\to\mathbbmss{Z}$ . Such a map must be multiplication by an integer $d$ . We define the degree of the map $f$ , to be this $d$ .

0.1 Basic Properties

1.

If $f,g\colon S^{n}\to S^{n}$ are continuous, then $\deg(f\circ g)=\deg(f)\cdot\deg(g)$ .
2.

If $f,g\colon S^{n}\to S^{n}$ are homotopic, then $\deg(f)=\deg(g)$ .
3.

The degree of the identity map is $+1$ .
4.

The degree of the constant map is $0$ .
5.

The degree of a reflection through an $(n+1)$ -dimensional hyperplane through the origin is $-1$ .
6.

The antipodal map, sending $x$ to $-x$ , has degree $(-1)^{n+1}$ . This follows since the map $f_{i}$ sending $(x_{1},\ldots,x_{i},\ldots,x_{n+1})\mapsto(x_{1},\ldots,-x_{i},\ldots,x_{n+1})$ has degree $-1$ by (4), and the compositon $f_{1}\circ\cdots\circ f_{n+1}$ yields the antipodal map.

0.2 Examples

If we identify $S^{1}\subset\mathbbmss{C}$ , then the map $f:S^{1}\to S^{1}$ defined by $f(z)=z^{k}$ has degree $k$ . It is also possible, for any positive integer $n$ , and any integer $k$ , to construct a map $f\colon S^{n}\to S^{n}$ of degree $k$ .

Using degree, one can prove several theorems, including the so-called ’hairy ball theorem’, which that there exists a continuous non-zero vector field on $S^{n}$ if and only if $n$ is odd.

Title	degree (map of spheres)
Canonical name	DegreemapOfSpheres
Date of creation	2013-03-22 13:22:12
Last modified on	2013-03-22 13:22:12
Owner	drini (3)
Last modified by	drini (3)
Numerical id	12
Author	drini (3)
Entry type	Definition
Classification	msc 55M25
Defines	degree