Prove that if $S^k$ has a non vanishing vector field, then its antipodal map is homotopy to the identity if $k$ is even.

Question

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.

Answer 1

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.

Answer 2

While @Sky's answer is excellent and standard, it may be a little difficult to see the intuition behind it. Let me try to elaborate a little (warning: It's not rigorous):

Suppose that we have a non-vanishing vector field $v(x)$. Through $x$, $v(x)$, and the origin $0$, there is exactly one plane which intersects $S^n$ in a big circle. The direction of $v(x)$ uniquely (and continuously) defines a half circle (of the said big circle). Now the homotopy between $I$ and the antipodal map of $S^n$ is the exact homotopy of those of $S^1$: Move the antipodal point $-x$ along the chosen half circle towards $x$.