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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?

Travis Willse
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Idonotknow
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  • I mean by "this" the proof that I shared its link. @EricWofsey – Idonotknow Nov 04 '18 at 01:11
  • it is said in the link that we had relied on complex numbers @EricWofsey – Idonotknow Nov 04 '18 at 01:12
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    You don't "have to" rely on complex numbers if you don't want to. There are perfectly viable proofs which do not involve complex numbers. But you gotta admit, that complex number proof is very pretty. – Lee Mosher Nov 04 '18 at 01:59

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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 $-1$.

Travis Willse
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