It's not to hard to see that the quotient map $\pi\colon \mathbb{C}^{n+1}\backslash \{0\} \to \mathbb{CP}^n$ is smooth and surjective. Does that imply that it is a submersion as well?
Quotient map $\pi\colon \mathbb{C}^{n+1}\backslash \{0\} \to \mathbb{CP}^n$ is a submersion
-
0Thanks for your comment! Could you please elaborate a little bit in terms of the charts? I understand that I have to prove that the differential is surjective, so the relation between these concepts is not clear to me. – 2012-10-07
1 Answers
As Olivier Bégassat mentions in the comments, a smooth surjective map need not be a submersion. However, it is true that $\pi$ is a submersion. To see this, let $U_i = \{[x_0, \dots, x_n] \in \mathbb{CP}^n \mid x_i \neq 0\}$ and define
\begin{align*} \varphi : U_i &\to \mathbb{C}^n\\ [x_0, \dots, x_n] &\mapsto \left(\frac{x_0}{x_i}, \dots, \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \dots, \frac{x_n}{x_i}\right). \end{align*}
The pairs $(U_i, \varphi_i)$ are standard charts on $\mathbb{CP}^n$.
Let $p \in \mathbb{C}^{n+1}\setminus\{0\}$, then $\pi(p) \in U_i$ for some $i = 0, \dots, n$. Without loss of generality, suppose $\pi(p) \in U_0$. If $p = (x_0, \dots, x_n)$ then
\begin{align*} (\varphi_0\circ \pi)(x_0, \dots, x_n) &= \varphi_0(\pi(x_0, \dots, x_n))\\ &=\varphi_0([x_0, \dots, x_n])\\ &=\left(\frac{x_1}{x_0}, \dots, \frac{x_n}{x_0}\right). \end{align*}
The differential has standard matrix (of size $n\times(n+1)$) given by
$\left[\begin{array}{ccccc}-\frac{x_1}{x_0^2} & \frac{1}{x_0} & 0 & \dots & 0\\ -\frac{x_2}{x_0^2} & 0 & \frac{1}{x_0} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -\frac{x_n}{x_0^2} & 0 & 0 & \dots & \frac{1}{x_0}\end{array}\right].$
Note that the matrix has rank $n$, so the differential is surjective. Therefore, $\pi$ is a submersion.
-
0It seems like this answer is considering the spaces as complex manifolds. If we consider the spaces as ordinary smooth manifolds, the argument is not as clean. – 2018-10-03