I'm struggling to prove that $\mathbb CP^n$ is 2n-manifold.

We can defined the $\mathbb CP^n$ as the equivalence relation $(z_1,z_1,...,z_{n+1})\sim(w_1,w_1,...,w_{n+1})$ iff $z_i=\lambda w_i$, $i=1,2,...,n+1$.

In order to prove that $\mathbb CP^n$ is a $2n$-manifold, we need to define a function $f_i:U_i\to \mathbb C^n$ defined by $f_i([z_1,...,z_{n+1}])=\left(\frac{z_1}{z_i},...,\frac{z_{i-1}}{z_i},\frac{z_{i+1}}{z_i},..., \frac{z_{n+1}}{z_i}\right)$, where each $U_i$ is defined as $U_i = \{[z_0,z_1,...,z_n];z_i\neq 0\}$.

If we prove that this function is an homeomorphism, we're done.

It's easy to prove that each $f_i$ is well-defined, continuous and have this inverse $g_i:\mathbb C^n\to U_i$, defined by $g_i(z_1,...,z_n)=[z_1,...,z_{i-1},1,z_i,...,z_n]$

In order to prove that $\mathbb CP^n$ is a $2n$-manifold, it miss just the continuity of $g$, I need help in this part.


  • 2
    $\begingroup$ There is an obvious candidate for the inverse of your map, and it is very easy to show that they are in fact inverse to each other. Can you show that either of them is continuous? $\endgroup$ – Mariano Suárez-Álvarez Oct 10 '12 at 6:02
  • $\begingroup$ but that function is not continuous! It is not even well defined. $\endgroup$ – Mariano Suárez-Álvarez Oct 11 '12 at 15:56
  • $\begingroup$ @MarianoSuárez-Alvarez I've found the inverse $g$ of $f$ and I showed it's continuous, can you help me with the continuity of $g$? $\endgroup$ – user42912 Nov 23 '12 at 5:21
  • $\begingroup$ Look at $g_{1}(z_2,\dots,z_n)=[1,z_2,\dots,z_n]$. Observe we can write it as $g_{1}(z)=[1,\mathrm{id}(z)]$. You can use the definition of continuity from analysis, using $\varepsilon-\delta$ proof taking $\delta=\varepsilon$ you have it immediately... $\endgroup$ – Alex Nelson Nov 23 '12 at 5:54

Firstly, you need to prove that $\mathbb{C}P^n$ is Hausdorff. It is proved here.

Let $X = \mathbb{C}^{n+1} - \{0\}$. Let $\pi \colon X \rightarrow \mathbb{C}P^n$ be the canonical map. Let $h_0\colon \mathbb{C}^n \rightarrow X$ be the map defined by $h_0(z_1,\dots,z_n) = (1,z_1,\dots,z_n)$. Clearly $h_0$ is continuous. Since $g_0 = \pi h_0$ and $\pi$ is continuous , $g_0$ is continuous. Similarly $g_i$ is continuous for $i \neq 0$.

For the sake of completeness, I will prove that $f_i$ is continous. Since $f_0\pi(z_0,z_1,\dots,z_n) = (z_1/z_0,\dots, z_n/z_0)$, $f_0\pi\colon \pi^{-1}(U_0) \rightarrow \mathbb{C}^n$ is continous. Let $V$ be an open subset of $\mathbb{C}^n$. Since $(f_0\pi)^{-1}(V) = \pi^{-1}(f_0^{-1}(V))$ is open in $\pi^{-1}(U_0)$, it is open in $X$. Hence $f_0^{-1}(V)$ is open. Hence $f_0$ is continous. Similarly $f_i$ is continuous for $i \neq 0$.

  • $\begingroup$ But the link you provided for the proof of Hausdorffness is valid for real projective spaces. How can we do it for complex projective spaces? $\endgroup$ – Prakhar Gupta Aug 7 '18 at 14:49

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