AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd 2006-03-21 09:37:54 2006-03-21 10:37:25 Derivation antipodal % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem*{lemma*}{Lemma} \newtheorem*{proposition*}{Proposition} \begin{lemma*} If $X\colon S^n\to S^n$ is a unit vector field, then there is a homotopy between the antipodal map on $S^{n}$ and the identity map. \end{lemma*} \begin{proof} Regard $S^n$ as a subspace of $R^{n+1}$ and define $H\colon S^n\times[0,1]\to R^{n+1}$ by $H(v,t)=(\cos\pi t)v+(\sin\pi t)X(v)$. Since $X$ is a unit vector field, $X(v)\perp v$ for any $v\in S^n$. Hence $\|H(v,t)\|=1$, so $H$ is into $S^n$. Finally observe that $H(v,0)=v$ and $H(v,1)=-v$. Thus $H$ is a homotopy between the antipodal map and the identity map. \end{proof} \begin{proposition*} The antipodal map $A\colon S^n\to S^n$ is homotopic to the identity if and only if $n$ is odd. \end{proposition*} \begin{proof} If $n$ is even, then the antipodal map $A$ is the composition of an odd \PMlinkescapetext{number} of reflections. It therefore has degree $-1$. Since the degree of the identity map is $+1$, the two maps are not homotopic. Now suppose $n$ is odd, say $n=2k-1$. Regard $S^n$ has a subspace of $\mathbb{R}^{2k}$. So each point of $S^n$ has coordinates $(x_1,\dots,x_{2k})$ with $\sum_i x_i^2=1$. Define a map $X\colon\mathbb{R}^{2k}\to\mathbb{R}^{2k}$ by $X(x_1,x_2,\dots,x_{2k-1},x_{2k})=(-x_2,x_1,\dots,-x_{2k},x_{2k-1})$, pairwise swapping coordinates and negating the even coordinates. By construction, for any $v\in S^n$, we have that $\|X(v)\|=1$ and $X(v)\perp v$. Hence $X$ is a unit vector field. Applying the lemma, we conclude that the antipodal map is homotopic to the identity. \end{proof} \begin{thebibliography}{9} \bibitem{H} Hatcher, A. {\em Algebraic topology}, Cambridge University Press, 2002. \bibitem{M} Munkres, J. {\em Elements of algebraic topology}, Addison-Wesley, 1984. \end{thebibliography}