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Show that the complex projective space $\Bbb C P^n \cong \Bbb C^{n+1} - 0 / \Bbb C - 0.$

$\textbf {My thought} :$ We know that $\Bbb C P^n : = \Bbb S^{2n+1}/\Bbb S^1,$ where $\Bbb S^1$ is the unit circle. Now we have a surjective continuous map $f$ from $\Bbb C^{n+1} - 0$ onto $\Bbb S^{2n+1}$ defined by $$x \mapsto \frac {x} {\|x\|},\ x \in \Bbb C^{n+1} - 0.$$ Let $p : \Bbb S^{2n+1} \longrightarrow \Bbb S^{2n+1}/ \Bbb S^1 \cong \Bbb CP^n$ be the quotient map. Then the map $g : = p \circ f : \Bbb C^{n+1} - 0 \longrightarrow \Bbb C P^n$ gives us a surjective continuous map. Let $X^* = \{g^{-1} \{z\}\ |\ z \in \Bbb C P^n \}$ and endow $X^*$ with the quotient topology. Then $g$ induces a bijective continuous map $\overline {g} : X^* \longrightarrow \Bbb C P^n.$ Furthermore, $\overline {g}$ is a homeomorphism if and only if $g$ is a quotient map.

So if we can able to show that $X^* \cong \Bbb C^{n+1} - 0 / \Bbb C - 0$ and $g$ is a quotient map then we are through by universal property of quotient topology. But I am unable to show these things. Would anybody give me some suggestion in this regard?

Thanks in advance.

EDIT $:$ For any $z \in \Bbb S^{2n+1}$ there exists $z' \in C^{n+1} - 0$ such that $z = \frac {z'} {\|z'\|}.$ Now let $w \in g^{-1} \{[z]\} = g^{-1} \left \{\left [\frac {z'} {\|z'\|} \right ] \right \} ,$ where $[z]$ is the equivalence class $z \in \Bbb S^{2n+1}$ in $\Bbb C P^n : = \Bbb S^{2n+1} / \Bbb S^1.$ Then there exists $\lambda_0 \in \Bbb S^1$ such that $$\frac {w} {\|w\|} = \lambda_0 \frac {z'} {\|z'\|} \implies w = \frac {\lambda_0 \|w\|} {\|z\|} z = \lambda z$$ where $\lambda = \frac {\lambda_0 \|w\|} {\|z\|} \in \Bbb C - 0.$ Conversely, if $\lambda \in \Bbb C - 0$ then $$g(\lambda z') = \left [ \frac {\lambda} {\|\lambda\|} \frac {z'} {\|z'\|} \right ] = [\lambda' z] = [z]$$ since $\lambda' = \frac {\lambda} {\|\lambda\|} \in \Bbb S^1.$ Therefore $g^{-1} \{[z]\} = \{\lambda z'\ |\ \lambda \in \Bbb C - 0 \},$ which is nothing but the orbit of $z' \in \Bbb C^{n+1} - 0$ under the action of $\Bbb C - 0$ on $\Bbb C^{n+1} - 0$ which is given as follows $:$

For any given $z \in \Bbb C^{n+1} - 0$ and for any $z \in \Bbb C - 0$ we define $$z \cdot (z_0,z_1, \cdots, z_n) : = (zz_0, zz_1, \cdots, zz_n).$$

So we have $$X^* = (\Bbb C^{n+1} - 0) / (\Bbb C - 0).$$

Also since $f$ is a retraction it is quotient map and thus $g = p \circ f,$ is a quotient map being the composition of two quotient maps. Thus we are the done with the proof!

Anil Bagchi.
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  • Usually, the thing to be shown here is much closer to the most common definition of projective space (as space of all lines through the origin) than the quotient space of the sphere ... – Hagen von Eitzen Apr 14 '21 at 06:04
  • I think that this is the definition of $CP^n$, as quotient space of $C^{n+1} - {0}$. $CP^n$ is not a quotient space of $S^n$ (real dimension of $CP^n$ is $2n$). The $S^n/C_2$ is $RP^n$. – Peter Franek Apr 14 '21 at 06:04
  • @Hagen von Eitzen our instructor introduced the definition of the $n$-dimensional complex projective space by means of a group action where the cyclic group $C_2$ of order $2$ is acting on the $(2n+1)$-sphere $\Bbb S^{2n+1}.$ – Anil Bagchi. Apr 14 '21 at 06:06
  • (The $g$ induces a homomorphisms from $CP^n$ into $RP^{2n+1}$ but this is not an isomorphism) – Peter Franek Apr 14 '21 at 06:12
  • @Peter Franek I made a typo. Corrected it now. – Anil Bagchi. Apr 14 '21 at 06:13
  • Still the same -- the quotient space of the sphere is $RP^{2n+1}$, not $CP^n$. These are very different spaces. – Peter Franek Apr 14 '21 at 06:14
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    @Peter Franek see again carefully. This is the definition introduced to us by our instructor. I don't understand what's wrong in it as I am a new to this subject. – Anil Bagchi. Apr 14 '21 at 06:15
  • It's a wrong definition. The sphere $S^{2n+1}$ with antipodal points identified is usually the definition of $RP^{2n+1}$. It's a very different space than $CP^n := C^n-{0} / (C - {0})$. Maybe your instructor messed something up. – Peter Franek Apr 14 '21 at 06:19
  • @Peter Franek our instructor also proposed that the real projective space $\Bbb R P^n \cong \Bbb R^{n+1} - 0 / \Bbb R - 0.$ Is it correct? – Anil Bagchi. Apr 14 '21 at 06:21
  • Yes, that's correct. This is the same thing as $S^n/C_2$. – Peter Franek Apr 14 '21 at 06:21
  • So what is the correct definition of a complex projective space? How do I prove the similar statement for real projective space? Would you please help me @Peter Franek? You may post it as an answer. Also it will be of great help if you recommend me some book where I can be easily accessible. – Anil Bagchi. Apr 14 '21 at 06:24
  • The usual definition of complex projective space is $(C^{n+1} - {0}) / (C-{0})$.You can also check Wikipedia. For a "similar statement of real projective space", I would suggest to open a new clear question, as this is getting long. – Peter Franek Apr 14 '21 at 06:25
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    @Peter Franek do you want me to delete this question and ask a fresh question with the corrected version of the statement? – Anil Bagchi. Apr 14 '21 at 06:26
  • Possibly, yes -- – Peter Franek Apr 14 '21 at 06:27
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    You should define $\mathbb{RP}^n=(\Bbb R^n\setminus{0})/(\mathbb{R}\setminus{0})$ and $\mathbb{CP}^{n+1}=(\Bbb C^n\setminus{0})/(\mathbb{C}\setminus{0})$ and then you can prove $\mathbb{RP}^n=S^n/S^0$ and $\mathbb{CP}^n=S^{2n+1}/S^1$. – anon Apr 14 '21 at 06:28
  • @runway I have just contacted with our instructor and he apologizes for the unavoidable error he made. I will edit my question accordingly. – Anil Bagchi. Apr 14 '21 at 06:33
  • @Peter Franek I have edited accordingly what runway44 has suggested me to do. Can you now please have a look at the question? – Anil Bagchi. Apr 14 '21 at 06:40
  • @runway44 I have edited my question accordingly. Please have a look at it now. – Anil Bagchi. Apr 14 '21 at 06:41
  • @Phibetakappa Now the question is a bit confusing, because what does it really mean $S^{2n+1} / S^1$? You need to know what is the action of $S^1$ on $S^{2n+1}$. And this action is exactly multiplication by complex numbers in the $C^{n+1} - {0}$ space... So we are returning back to the "statement" that in fact should be a definition. – Peter Franek Apr 14 '21 at 07:22
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    @Peter Franek the action of $\Bbb S^1$ on $\Bbb S^{2n+1}$ is given as follows $:$ For any given $z \in \Bbb S^1$ and $(z_0, z_1, \cdots, z_{2n + 1} ) \in \Bbb S^{2n+1}$ we define $$z \cdot (z_0,z_1, \cdots,,z_{2n+1}) : = (zz_0, zz_1, \cdots, zz_{2n+1}).$$ – Anil Bagchi. Apr 14 '21 at 07:50
  • @Peter Franek I have written down a solution. Would you please check it? – Anil Bagchi. Apr 14 '21 at 16:43

2 Answers2

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In the literature on can find various definitions of $\mathbb CP^n$.

  1. $\mathbb CP^n = (\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$, where the group $(\mathbb C \setminus \{0\},\cdot)$ operates via scalar multiplication on $\mathbb C^{n+1} \setminus \{0\}$. This seems to be the standard (or at least most popular) definition.

  2. $\mathbb CP^n = S^{2n+1}/S^1$, where the group $(S^1,\cdot)$ operates via scalar multiplication on $S^{2n+1} \subset \mathbb C^{n+1}$.

  3. $\mathbb CP^n = \mathbf{Gr}_1(\mathbb C^{n+1})$. Here $\mathbf{Gr}_k(\mathbb C^m)$ denotes the set of all $k$-dimensional (complex) linear subspaces of $\mathbb C^m$. This is known as a Grassmann manifold. It can be endowed with a natural topology and a natural structure of a differentiable manifold. We shall not go into details here.

For the equivalence of 1. and 2. see When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map? Morever, on the level of sets it should be clear that $(\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$ and $\mathbf{Gr}_1(\mathbb C^{n+1})$ can be identified naturally: Take each equivalence class $\zeta \in (\mathbb C^{n+1} \setminus \{0\})/(\mathbb C \setminus \{0\})$ to $\zeta \cup \{0\} \in \mathbf{Gr}_1(\mathbb C^{n+1})$.

Paul Frost
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  • Can you please check my solution? – Anil Bagchi. Apr 14 '21 at 16:42
  • @Phibetakappa What you have done is correct, but too complicated. 1. You do not need to find $z'$, take $z' = z$. 2. What you have to show is that $z \sim w$ in $\mathbb C^{n+1} \setminus {0}$ iff $f(z) \sim f(w)$ in $S^{2n+1}$. If $z \sim w$, i.e. $z = \lambda w$, then $f(z) = z/\lVert z \rVert = \lambda w/\lvert \lambda \rvert \lVert w \rVert = (\lambda/\lvert \lambda \rvert)f(w)$, i.e. $f(z) \sim f(w)$. If $f(z) \sim f(w)$, i.e. $f(z) = \mu f(w)$ with $\mu \in S^1$, then $z/\lVert z \rVert = \mu w/\lVert w \rVert$, hence $z =( \mu \lVert z \rVert /\lVert w \rVert)w$, i.e. $z \sim w$. – Paul Frost Apr 14 '21 at 22:19
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    @PeterFranek You should write another answer! – Paul Frost Apr 14 '21 at 22:24
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Each complex vector $v$ in $C^{n+1}$ has a unique norm $\|v\|: = \sqrt{\sum_i |v_i|^2}$ and whenever $v\neq 0$, then $v = \|v\|\frac{v}{\|v\|}$ is a unique decomposition to a positive real number $\|v\|$ and an point on the $2n+1$ sphere.

The map $v (C\setminus\{0\}) \mapsto \frac{v}{\|v\|} S^1$ is a well defined map from $C^{n+1}\setminus\{0\} \,\, / \,\,C\setminus\{0\}$ to $S^{2n+1} / S^1$, because a different choice $v' = z v$ representing the same element is mapped to the same element $\frac{v'}{\|v'\|} S^1=\frac{v}{\|v\|} \frac{z}{|z|}S^1 = \frac{v}{\|v\|} S^1$.

Vice versa, the map $v S^1 \mapsto v (C\setminus\{0\})$ is also well defined.

You easily verify that those are inverse to each other and continuous (if you need it).

Peter Franek
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