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, 2019 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? $\endgroup$
    – CNS709
    Oct 16, 2019 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, 2019 at 21:03

1 Answer 1


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.


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