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We know that $$ O(n+1)/O(n) \simeq SO(n+1)/SO(n) \simeq S^n, $$ based on the result of homogeneous space.

Also $$ PO(n+1)/PO(n) \simeq P^n, $$ $P^n$ is the projective space.

These are in some sense spheres.

If we embed the real projective space $\mathbb{C}P^n$ into $\mathbb{R}P^{n+1}$, we may be able to define the quotient space

$$ {\mathbb{R}P^{n+1}}/{\mathbb{R}P^{n}} \simeq ? $$

  • Do we have simpler expressions for the above "manifolds" or "quotient space"?

  • Homotopy group $\pi_i({\mathbb{R}P^{n+1}}/{\mathbb{R}P^{n}})=?$

My Attempt: Note that ${\mathbb{R}P^{n}}$ is diffeomorphic to the submanifold of $R^{(n+1)^2}$ consisting of all symmetric $(n+1) \times (n+1)$ matrices of trace 1 that are also idempotent linear transformations. , so for ${\mathbb{R}P^{2}}\simeq SO(3)$, ${\mathbb{R}P^{1}}\simeq S^1 \simeq SO(2)$ and ${\mathbb{R}P^{0}}\simeq 0$, so $$\pi_i({\mathbb{R}P^{2}}/{\mathbb{R}P^{1}})=\pi_i(SO(3)/SO(2))=?,$$ $$\pi_i({\mathbb{R}P^{1}}/{\mathbb{R}P^{0}})=\pi_i(S^1),$$ Also the $\pi_i({\mathbb{R}P^{n}})$ can be found here.

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  • $\begingroup$ Use the argument I gave in the $\mathbb{C}P^n$ case. It's almost verbatim the same. $\endgroup$ – Jason DeVito Aug 2 '18 at 2:25
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    $\begingroup$ It is not true that $\pi_1(\mathbb{R}P^n)$ is $(-1)^{n+1}$. The latter object is not even a group. $\endgroup$ – Tyrone Aug 4 '18 at 12:42
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    $\begingroup$ Also, it is $\mathbb{R}P^3\cong SO(3)$, rather than $\mathbb{R}P^2$. $\endgroup$ – Tyrone Aug 4 '18 at 12:43
  • $\begingroup$ Note that there are embeddings $\mathbb{R}P^{n-1}\hookrightarrow SO(n)$ defined by sending a line in $\mathbb{R}^n$ to a certain reflection (see James's "Topology of Stiefel Manifolds", for instance). Compare the cofibration sequence $\mathbb{R}P^{n-1}\rightarrow \mathbb{R}P^n\rightarrow \mathbb{R}P^n/\mathbb{R}P^{n-1}$ with the fibration $SO(n)\rightarrow SO(n+1)\rightarrow S^n$ using the previous embeddings, and see what you can figure out. $\endgroup$ – Tyrone Aug 4 '18 at 12:47

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