Prove that if $S^k$ has a non vanishing vector field, then its antipodal map is homotopy to the identity if $k$ is even.
Here is what I got so far.
Suppose we have an antipodal map $x \to -x$ of $S^k \to S^k$. For $k=1$ the antipodal just rotating the point by $180^o$. So let $R(\theta)$ be the rotation map counter clock wise by angle $\theta$, meaning
$$R(\theta)= \left( \begin{array}{ccc}
Cos\theta & -sin\theta \\
sin\theta & Cos\theta \\
\end{array} \right)$$
Let $F(x,t)=R(\pi t)_x$. Know that $R$ is linear map, so it is smooth. Consider $S^k=\{x: x_1^2+x_2^2+…+x_k^2=1\}$
Now I'm stuck.
Best Answer
If S^k has a non vanishing vector field then -I will always be homotopic to I, no matter whether k is even. Since I(x) * cos(t) + v(x) * sin(t) is the desired homopoty, where v(x) denote the normalized( that is, |v(x)|=1 ) non vanishing vector field.