2

I'd like some verification for the following two claims here on page four (b) and (c). That is, if we first suppose $S^{2n}$ covers $X$, then if $d:S^{2n} \rightarrow S^{2n}$ is a nontrivial deck transformation we know that $deg(d) = (-1)^{2n+1}$.

Similarly if $S^{2n+1}$ covers $X$ then nontrivial deck transformations have $deg(d) = (-1)^{2n+2}$.

Here's what I've tried thinking about. In my book (Bredon), we define the degree of a map $d:S^n \rightarrow S^n$ to be the number $a$ such that $d_*(\gamma) = a \gamma$ for all $\gamma \in \tilde{H}_n(S^n) \simeq \mathbb{Z}$. This question here showed me that only $\mathbb{Z}_2$ acts freely on $S^{2n}$. I don't know why that is, but assuming it is true this means that the action is like the antipodal identification giving us projective space, which I have shown before has degree $-1$ when doing problem related to the Hairy Ball Theorem.

user9781778
  • 31
  • 4

1 Answers1

1

Deck transformations of connected covers are determined by their value on one point which means nonidentity transformations have no fixed points. Any map $S^n \rightarrow S^n$ with no fixed points is homotopic to the antipodal map which makes its degree $-1^{n+1}$.

Connor Malin
  • 11,661
  • Thanks for your reply. Would you please explain why a continuous map of spheres with no FPs is homotopic to the antipodal map? I understand that by Hopf degree theorem, same degree implies homotopic. – user9781778 Aug 08 '19 at 14:11
  • https://math.stackexchange.com/questions/142264/what-is-the-degree-of-f-sn-to-sn-when-f-has-no-fixed-points – Connor Malin Aug 08 '19 at 14:34
  • Your question is actually the result I would use to prove the theorem about free actions on even spheres. It is really simple if you know some things about degrees. – Connor Malin Aug 08 '19 at 14:35