4
$\begingroup$

Let $\pi_n(\mathbb{R}P^n)$ be the $n$-th homotopy group of the $n$-dimensional projective space. Then by the long exact sequence of homotopy groups associated to the fibration $S^n\to \mathbb{R}P^n\to B\mathbb{Z}_2$, I obtain $\pi_n(\mathbb{R}P^n)=\mathbb{Z}$. Moreover, $\pi_1(\mathbb{R}P^n)=\mathbb{Z}_2$.

Question: How to prove that the action of $\pi_1(\mathbb{R}P^n)$ on $\pi_n(\mathbb{R}P^n)$ is trivial if $n$ is odd is the action is non-trivial if $n$ is even?

$\endgroup$
  • 1
    $\begingroup$ $S^n \to \mathbb RP^n$ is the universal cover (n>1), so look at the deck transformation given by the non-trivial element of $\pi_1 \mathbb RP^n$. What map does that induce on $\pi_n S^n \cong \pi_n \mathbb RP^n$? $\endgroup$ – Justin Young Apr 7 '16 at 9:31
  • $\begingroup$ @JustinYoung: But one has to justify why the action of the deck transformation group on $\pi_nS^n$ corresponds to the a priori differently defined action of $\pi_1 RP^n$ on $\pi_nRP^n$. $\endgroup$ – user39082 Apr 7 '16 at 17:00
  • $\begingroup$ Yeah, but it's nice, though. I guess start with a generator of $\pi_1$ and follow it around. $\endgroup$ – Justin Young Apr 7 '16 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.