# Explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$

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)

EDIT

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?

• Where did you see that in Wikipedia? This is completely false.
– abx
Oct 16, 2019 at 20:55
• @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? Oct 16, 2019 at 21:00
• 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.
– abx
Oct 16, 2019 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 $$0. The inclusion map $$\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.