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

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.

• Do you mean if $k$ is odd? Oct 5, 2014 at 14:57
• no, $k$ is even, $k$ is odd is another problem, not this one. Oct 5, 2014 at 15:18

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.

• You are right, however, I believe that $S^{2n}$ has no non-vanishing vector field. Oct 5, 2014 at 15:00
• So you want me to let $F(x,t)=I(x)cos(t)+v(x)sin(t)$ and somehow show that the sum of $F(x_i, t_i)^2 =1$? Oct 5, 2014 at 15:23
• Yes, the tangent vector v(x) is orthogonal to I(x).
– Sky
Oct 5, 2014 at 15:31
• Can you help me out a little bit more? I'm still struggling with $v(x)$ and $|v(x)|=1$ part. Oct 5, 2014 at 15:40
• Just a nitpick: Perhaps you mean $I(x)\cos(\pi t)+v(x)\sin(\pi t)$. Oct 5, 2014 at 15:46

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 $x$.