# antipodal map on $S^{n}$ is homotopic to the identity if and only if $n$ is odd

###### 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.

###### 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. ∎

###### Proposition.

The antipodal map $A\colon S^{n}\to S^{n}$ is homotopic to the identity if and only if $n$ is odd.

###### Proof.

If $n$ is even, then the antipodal map $A$ is the composition of an odd 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. ∎

## References

• 1 Hatcher, A. Algebraic topology, Cambridge University Press, 2002.
• 2 Munkres, J. Elements of algebraic topology, Addison-Wesley, 1984.
Title antipodal map on $S^{n}$ is homotopic to the identity if and only if $n$ is odd AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd 2013-03-22 15:47:33 2013-03-22 15:47:33 mps (409) mps (409) 5 mps (409) Derivation msc 51M05 msc 15-00