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

Lemma.

If $X\mathrm{:}{S}^n\mathrm{\to }{S}^n$ 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 $H:S^n\times [0,1]\to R^{n+1}$ by $H(v,t)=(\cos \pi t)v+(\sin \pi t)X(v)$. Since $X$ is a unit vector field, $X(v)⟂v$ for any $v\in S^n$. Hence $\parallel H(v,t)\parallel =1$, 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 $A\mathrm{:}{S}^n\mathrm{\to }{S}^n$ 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 ${ℝ}^{2k}$. So each point of $S^n$ has coordinates $(x_1,\mathrm{\dots },x_{2k})$ with ${\sum }_i{x}_i^2=1$. Define a map $X:{ℝ}^{2k}\to {ℝ}^{2k}$ by $X(x_1,x_2,\mathrm{\dots },x_{2k-1},x_{2k})=(-x_2,x_1,\mathrm{\dots },-x_{2k},x_{2k-1})$, pairwise swapping coordinates and negating the even coordinates. By construction, for any $v\in S^n$, we have that $\parallel X(v)\parallel =1$ and $X(v)⟂v$. Hence $X$ is a unit vector field. Applying the lemma, we conclude that the antipodal map is homotopic to the identity. ∎

Title	antipodal map on $S^n$ is homotopic to the identity if and only if $n$ is odd
Canonical name	AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd
Date of creation	2013-03-22 15:47:33
Last modified on	2013-03-22 15:47:33
Owner	mps (409)
Last modified by	mps (409)
Numerical id	5
Author	mps (409)
Entry type	Derivation
Classification	msc 51M05
Classification	msc 15-00