# Antipodal map is Smoothly homotopic to id

Problem:

Fix $m\in \mathbb{N}$. Then antipodal map $\alpha: S^m \rightarrow S^m$ is defined by $$S\in (x_1,x_2,...,x_{m+1})) \mapsto (-x_1,-x_2,...,-x_{m+1}).$$ Prove that the antipodal map $\alpha: S^m \rightarrow S^m$ is smoothly homotopic to the identity map id:$S^m \rightarrow S^m$ whenever $m$ is odd.

## My Attempt

I still haven't quite been able to figure this out. I jotted down a few things I know and can't put them together...

• The antipodal map is a diffeomorphism of compact, connected manifolds.
• $S^{2n-1} \subset \mathbb{R}^{2n}$
• If $m$ is odd let $m=2k-1$. Consider $S^m$ as a subspace of $\mathbb{R}^{2k}$. The coodinate pairs of $S^m$ is $(x_1,...,x_{2k})$ such that $\sum _{i=1}^n \vert x_i\vert ^2=1$ Define:

$$X:\mathbb{R}^{2k} \rightarrow \mathbb{R}^{2k}$$ by $$X((x_1,x_2,...,x_{2k-1},x_{2k})) = (-x_2,-x_1,...,-x_{2k},-x_{2k-1}).$$ Now,$X(v)$ is perpendicular to $v$ and $||X(v)||=1$ for all $v\in S^m$ so $X$ is a unit vector field

Hint: Let's consider the simplest case: $m = 1$.

• The identity map $(x_1, x_2) \mapsto (x_1, x_2)$ is a rotation through $0$ degrees.

• The antipodal map $(x_1, x_2 ) \mapsto (-x_1, -x_2)$ is a rotation through $180$ degrees.

How can we smoothly deform a $0$ degree rotation into a $180$ degree rotation?

• trig functions? – combo student Apr 27 '17 at 23:13
• You mean using trig functions to represent rotations? Yeah, that would work. – Kenny Wong Apr 27 '17 at 23:15
• I am not sure how to answer your question – combo student Apr 27 '17 at 23:20
• @combostudent: Remember what you want. You want a family of functions $f_{t}$ with smoothly transitions from the identity $f_{0}=\mathrm{id}$ to the antipodal map $f_{1}=\alpha.$ To do this, you'll need to come up with a parameter $t.$ What in this answer could be used as a parameter? – Will R Apr 27 '17 at 23:23
• something that gets $X(v,0)=v$ and $X(v,1)=-v$ would work to get a homotopy – combo student Apr 27 '17 at 23:28