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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 complex projective space $\mathbb{C}P^n$ into $\mathbb{C}P^{n+1}$, we may be able to define the quotient space $$ {\mathbb{C}P^{n+1}}/{\mathbb{C}P^{n}} \simeq ? $$

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

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

Attempt: Note that ${\mathbb{C}P^{n}}=S^{2n+1}/U(1)$, so for ${\mathbb{C}P^{1}}\simeq S^2$ and ${\mathbb{C}P^{0}}\simeq 0$, so $$\pi_i({\mathbb{C}P^{1}}/{\mathbb{C}P^{0}})=\pi_i(S^2/0)=\pi_i(S^2),$$ which homotopy group is known.

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  • $\begingroup$ The notation $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is typically understood to be the quotient space where we identify $\mathbb{C}P^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $\mathbb{C}P^n$ into $\mathbb{C}P^{n+1}$, $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is homeomorphic to $S^{2n+2}$. That said, as a general rule, for $N\subseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold. $\endgroup$ Aug 1, 2018 at 0:43
  • $\begingroup$ thanks - if this is correct - this counts as an answer. $\endgroup$
    – wonderich
    Aug 1, 2018 at 14:24

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The notation $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is typically understood to be the quotient space where we identify $\mathbb{C}P^n$ to a point. (But not other identifications are made).

For the usual embedding of $\mathbb{C}P^n$ into $\mathbb{C}P^{n+1}$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]\mapsto [z_0:z_1:...:z_n: z_{n+1}]$, $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is homeomorphic to $S^{2n+2}$. One way to see this is as follows.

First, let $Z_i = \frac{z_i}{z_{n+1}}$ (assuming $z_{n+1}\neq 0$). Also, let $Z = \sum_{i=0}^n |Z_i|^2$.

Then, we map $\mathbb{C}P^{n+1}\rightarrow S^{2n+2}$ via $$(z_0:...:z_{n+1})\mapsto \begin{cases} \left(\frac{2Z_0}{1+Z}, \frac{2Z_1}{1+Z}, ..., \frac{2Z_n}{1+Z}, \frac{-1+Z}{1+Z} \right) & z_{n+1}\neq 0 \\ (0,0,..,0,1) & z_{n+1} = 0 \end{cases}.$$

The case that $z_{n+1} = 0$ corresponds exactly to a point being in $\mathbb{C}P^n$. As a result, $f$ is constant on $\mathbb{C}P^n$ so gives a map $\overline{f}:\mathbb{C}P^{n+1}/\mathbb{C}P^n\rightarrow S^{2n+2}$. This map is a homeomorphism.

Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2\wedge S^2$, which is not a manifold due to the wedge point.

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  • $\begingroup$ I post a new question maybe you may be also able to answer. $\endgroup$
    – wonderich
    Aug 1, 2018 at 22:51
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    $\begingroup$ I don't think it's worth bumping this answer to edit, but I just noticed my last sentence should have $S^2\vee S^2$, not $S^2\wedge S^2$. $\endgroup$ Jan 26, 2019 at 15:56

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