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Version 1 |
| \begin{lemma*} |
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| If $X\colon S^n\to S^n$ is a unit vector field, then |
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| there is a homotopy between the antipodal map on $S^{n}$ |
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| and the identity map. |
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| \end{lemma*} |
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| \begin{proof} |
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| Regard $S^n$ as a subspace of $R^{n+1}$ and define |
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| $H\colon S^n\times[0,1]\to R^{n+1}$ by |
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| $H(v,t)=(\cos\pi t)v+(\sin\pi t)X(v)$. Since $X$ is a unit |
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| vector field, $X(v)\perp v$ for any $v\in S^n$. Hence |
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| $\|H(v,t)\|=1$, so $H$ is into $S^n$. Finally observe that |
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| $H(v,0)=v$ and $H(v,1)=-v$. Thus $H$ is a homotopy between |
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| the antipodal map and the identity map. |
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| \end{proof} |
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| \begin{proposition*} |
\begin{proposition*} |
| The antipodal map $A\colon S^n\to S^n$ is homotopic |
The antipodal map $A\colon S^n\to S^n$ is homotopic |
| to the identity if and only if $n$ is odd. |
to the identity if and only if $n$ is odd. |
| \end{proposition*} |
\end{proposition*} |
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| \begin{proof} |
\begin{proof} |
| If $n$ is even, then the antipodal map $A$ is the composition |
To see the basic idea of the proof in the odd case, let us look at |
| of an odd \PMlinkescapetext{number} of reflections. It |
the case $n=1$. We need to show that the antipodal map on $S^1$ |
| therefore has degree $-1$. Since the degree of the identity |
is homotopic to the identity. The homotopy here simply rotates the |
| map is $+1$, the two maps are not homotopic. |
circle through an angle of $\pi$. We would like to use the same |
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idea for higher dimensions. |
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View $S^{2n-1}$ as a subspace of $\mathbb{R}^{2n}$. |
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So each point of $S^{2n-1}$ has coordinates $(x_1,\dots,x_{2n})$ |
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with $\sum_i x_i^2=1$. Now define a map $X\colon S^{2n-1}\to S^{2n-1}$ |
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by $X(x_1, x_2, \dots, x_{2n-1}, x_{2n})=(-x_2, x_1, \dots, -x_{2n}, x_{2n-1})$. |
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(This map happens to be a unit vector field, but we don't need that fact |
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for our proof.) The homotopy $H\colon S^{2n-1}\times[0,1]\to S^{2n-1}$ |
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we require is then given by $H(v, t)=(\cos\pi t)v+(\sin\pi t)X(v)$. For |
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any $v\in S^{2n-1}$, $v\cdot X(v)=0$, so $||H(v,t)||=1$; hence $H$ is |
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well-defined. Now observe that $H(v,0)=v$ and $H(v,1)=-v$, implying that $H$ |
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is a homotopy between the antipodal map and the identity. |
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| Now suppose $n$ is odd, say $n=2k-1$. Regard $S^n$ has a |
What happens if $n$ is even? In this case, the antipodal map is the |
| subspace of $\mathbb{R}^{2k}$. So each point of $S^n$ has |
composition of an odd number of reflections and thus has degree $-1$. |
| coordinates $(x_1,\dots,x_{2k})$ with $\sum_i x_i^2=1$. Define |
Since the identity map has degree $1$, the two maps cannot be homotopic. |
| a map $X\colon\mathbb{R}^{2k}\to\mathbb{R}^{2k}$ by |
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| $X(x_1,x_2,\dots,x_{2k-1},x_{2k})=(-x_2,x_1,\dots,-x_{2k},x_{2k-1})$, |
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| pairwise swapping coordinates and negating the even coordinates. |
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| By construction, for any $v\in S^n$, we have that $\|X(v)\|=1$ |
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| and $X(v)\perp v$. Hence $X$ is a unit vector field. Applying the |
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| lemma, we conclude that the antipodal map is homotopic to the identity. |
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| \end{proof} |
\end{proof} |
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| \begin{thebibliography}{9} |
\begin{thebibliography}{9} |
| \bibitem{H} |
\bibitem{H} |
| Hatcher, A. {\em Algebraic topology}, Cambridge University Press, 2002. |
Hatcher, A. {\em Algebraic topology}, Cambridge University Press, 2002. |
| \bibitem{M} |
\bibitem{M} |
| Munkres, J. {\em Elements of algebraic topology}, Addison-Wesley, 1984. |
Munkres, J. {\em Elements of algebraic topology}, Addison-Wesley, 1984. |
| \end{thebibliography} |
\end{thebibliography} |
