About a proof using complex numbers that the antipodal map on $S^n$, $n$ odd, is homotopic to the identity map

Question

This answer gives a solution to the following problem:

Prove that antipodal map on $ S^n $ homotopic to identity map if $n$ is odd.

I am wondering:

Why we had to rely on complex numbers in proving the above? And why this does not work for even spheres?"

Could anyone clarify this for me please?

Answer 1

The proof doesn't require complex numbers, but it certainly makes convenient writing down an explicit homotopy: $$H({\bf z}, t) := e^{\pi i t} {\bf z} .$$

This trick exploits the usual identification $\Bbb C^n \cong \Bbb R^{2n}$, and the unit sphere in $\Bbb R^{2n}$, $\Bbb S^{2 n - 1}$ is odd-dimensional, so it does not apply to even-dimensional spheres.

Indeed, the statement is false for even-dimensional spheres, as the antipodal map for such spheres is a composition of an odd number of reflections, each of which has degree $-1$, so the antipodal map has degree $-1$.