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Is there any short way to see that the action of $\pi_{1}(RP^{n})$ on $\pi_{n}(RP^{n}) = \mathbb{Z}$ is trivial for $n$ odd and nontrivial for $n$ even?

Maybe something without much machinery (smth related with orientability perhaps?)

Thank youuu!

Tikks
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1 Answers1

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Since $RP^n$ is the quotient of $S^n$ by the antipodal map, and $\pi_n(S^n) = \pi_n(RP^n)$, this action will coincide with the action of the antipodal map on $\pi_n(S^n)$, which is precisely as you write: trivial or not depending on whether $n$ is odd or even.

Matt E
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  • could you please give more details about the first sentence? I was thinking about something like that but... why are the two actions the same? – Tikks May 09 '13 at 05:58
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    @Tikks: Dear Tikks, One way to define the action of $\pi_1$ on higher homotopy groups is to go to the universal cover (which doesn't change the higher homotopy groups) and use the identification of $\pi_1$ with the group of deck transformations to get the action. With this definition the coincidence of the two actions is tautological. On the other hand, it shouldn't be too hard to prove that this definition coincides with whatever one you have in mind, by thinking about the explicit way that path lifting is used to identify deck tranformations of the universal cover with $\pi_1$. Regards, – Matt E May 09 '13 at 06:37