From covering space theory we know that $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_{n+1}(\mathbb{S}^n)$.

From wikipedia I can notice that $\pi_{n+1}(\mathbb{S}^n) \cong \pi_1(\mathbb{RP}^{n-1})$.*

My question is: is there an explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$? My question is motivated by the fact that $\mathbb{RP}^n \cong \mathbb{R}^n \cup \mathbb{RP}^{n-1}$ and $I^{n+1} = I^n \times I$ with $\mathring{I^n} \cong \mathbb{R}^n$.

*I "know" the standard calculation of $\pi_{n+1}(\mathbb{S}^n)$, for example via Pontryagin construction or $J$-homomorphism. I was wondering if it's possible to compute it in the way I stated.

(I posted this originally on math.stackexchange)


I not accepted user51223's answer yet because since the groups are isomorphic for every $n$ (and not only at 2), I find artificial to distinguish the two cases. Moreover, I was looking for an isomorphism whose proof does not involve a previous knowledge of the isomorphism type of one of the groups. To be honest, I was looking for an isomorphism

  1. valid all $n$

  2. valid for the whole group (not only at 2)

  3. which does not involve the knowledge neither of $\pi_{n+1}(\mathbb{RP}^n)$ nor of $\pi_1(\mathbb{RP}^{n-1})$

  4. induced by a map (optional)

Does $\lambda_n: \mathbb{RP}^{n-1} \to \Omega^n\mathbb{RP}^n$ satisfies these conditions?

  • 1
    $\begingroup$ Where did you see that in Wikipedia? This is completely false. $\endgroup$ – abx Oct 16 '19 at 20:55
  • 7
    $\begingroup$ @abx Why? $\pi_{n+1}(\mathbb{S}^n) \cong \mathbb{Z}$ if $n = 2$, $\mathbb{Z}_2$ otherwise; $\pi_1(\mathbb{RP^{n-1}}) \cong \mathbb{Z}$ if $n-1=1$, $\mathbb{Z}_2$ otherwise. Am I wrong? $\endgroup$ – Marco Francesco Nervo Oct 16 '19 at 21:00
  • 4
    $\begingroup$ Oh, I see. You mean that they happen to be the same group, and you ask for a natural explanation. I didn't get that from your post. $\endgroup$ – abx Oct 16 '19 at 21:03

Let $n>2$. You have a map $\lambda_n:\mathbb{R}P^{n-1}\to\Omega^n S^n$ defined using reflection maps. This is the map that leads to the Kahn-Priddy theorem. This map extends to an $n$-fold loop map $$\lambda_n:\Omega^n\Sigma^n\mathbb{R}P^{n-1}\to\Omega^n S^n$$ which according to Kahn-Priddy Theorem induces an epimorphism on ${_2\pi_i}$ for $0<i<n-1$. The inclusion map $\mathbb{R}P^{n-1}\to \Omega^n\Sigma^n\mathbb{R}P^{n-1}$ induces an isomorphism on ${\pi_1}$. Now, from knowing that $\pi_1\Omega^nS^n\simeq\pi_1\mathbb{R}P^{n-1}\simeq\mathbb{Z}/2$ you can deduce that the composition $$\mathbb{R}P^{n-1}\to\Omega^n\Sigma^n\mathbb{R}P^{n-1}\to\Omega^nS^n$$ induces an isomorphism on ${_2\pi_1}$ which gives the desired isomorphism on ${\pi_1}$. Note that the geometric description of $\lambda_n$ is quit explicit.

ADDED Since the title of question is about $\mathbb{R}P^{n-1}$ and $\Omega^n\mathbb{R}P^n$, then it would suffice to compose the above composition with the $n$-loop of the covering map $S^n\to\mathbb{R}P^n$ which yields a map $$\mathbb{R}P^{n-1}\to\Omega^n\mathbb{R}P^n$$ inducing the desired isomorphism.

| cite | improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.