degree (map of spheres)

Given a non-negative integer $n$, let $S^n$ denote the $n$-dimensional sphere. Suppose $f:S^n\to S^n$ is a continuous map. Applying the $n^{th}$ reduced homology functor ${\tilde{H}}_n(\mathrm{\_})$, we obtain a homomorphism $f_{*}:{\tilde{H}}_n(S^n)\to {\tilde{H}}_n(S^n)$. Since ${\tilde{H}}_n(S^n)\approx \mathbb{Z}$, it follows that $f_{*}$ is a homomorphism $\mathbb{Z}\to \mathbb{Z}$. 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 $f,g:S^n\to S^n$ are continuous, then $\mathrm{deg}(f\circ g)=\mathrm{deg}(f)\cdot \mathrm{deg}(g)$.

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

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 $(x_1,\mathrm{\dots },x_i,\mathrm{\dots },x_{n+1})\mapsto (x_1,\mathrm{\dots },-x_i,\mathrm{\dots },x_{n+1})$ has degree $-1$ by (4), and the compositon $f_1\circ \mathrm{\cdots }\circ f_{n+1}$ yields the antipodal map.

0.2 Examples

If we identify $S^1\subset ℂ$, 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 $f:S^n\to S^n$ 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