# 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? Commented Oct 5, 2014 at 14:57
• no, $k$ is even, $k$ is odd is another problem, not this one. Commented 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. Commented 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 $F(x_i, t_i)^2 =1$? Commented Oct 5, 2014 at 15:23
• Yes, the tangent vector v(x) is orthogonal to I(x).
– Sky
Commented 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. Commented Oct 5, 2014 at 15:40
• Just a nitpick: Perhaps you mean $I(x)\cos(\pi t)+v(x)\sin(\pi t)$. Commented 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 $x$.

The first thing is to try to take the path $$(1-t)x+t(-x)$$ and project it radially on the sphere, but you cannot do it since such a path intersects the origin. But with the non vanishing vector field $$v$$ you can overcome that difficulty in the following way:

First, just to have a better visualization let us do the following (this is not necessary): if $$\overline{v}$$ is the tangent vector field (with origin at $$0$$), then consider $$v(x)=\overline{v}(x)+x$$ (now with origin at $$x$$).

You just walk a little bit further from $$x$$ in the direction of $$v(x)-x$$ (you can do that once $$v(x)\neq0$$) by the path $$\dfrac{(1-t)x+t(v(x)-x)}{|(1-t)x+t(v(x)-x)|},\quad t\in[0,1].$$ And after you reached the destiny with $$t=1$$ we arrived at the point $$v(x)/|v(x)|$$, and now the path $$(1-s)v(x)/|v(x)|+s(-x)$$ never intersects the origin since $$\overline{v}(x)$$ is orthogonal to $$-x$$. Therefore you project it radially on the sphere and glue this path with the previous one, so you get the homotopy

$$H(x,t)=\begin{cases} \dfrac{(1-2t)x+2t(v(x)-x)}{|(1-2t)x+2t(v(x)-x)|}\quad\text{if}\quad t\in[0,1/2]\\ \dfrac{(2-2t)v(x)+(2t-1)(-x)}{|(2-2t)v(x)+(2t-1)(-x)|}\quad\text{if}\quad t\in[1/2,1], \end{cases}$$ which is clearly a homotopy between the identity and the antipodal map.