Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's not to hard to see that the quotient map $\pi\colon \mathrm{C}^{n+1}\backslash \{0\} \to \mathbb CP^n$ is smooth and surjective. Does that imply that it is a submersion as well?

share|cite|improve this question
Generally, a surjective smooth map need not be a submersion. For instance $x\mapsto x^3$ is smooth surjective on the real line but isn't submersive over $0$. However, $\pi$ is submersive, and you can check this in local coordinates, i.e. using charts for the projective space. – Olivier Bégassat Oct 7 '12 at 19:06
Thanks 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. – user43014 Oct 7 '12 at 22:00

1 Answer 1

As Olivier Bégassat mentions in the comments, a smooth surjective map need not be a submersion. However, $\pi$ is a submersion.

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 f)(x_0, \dots, x_n) &= \varphi_0(f(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

$$\left[\begin{array}{ccccc}-\frac{x_1}{x_0^2} & \frac{1}{x_0} & 0 & \dots & 0\\ -\frac{x_1}{x_0^2} & 0 & \frac{1}{x_0} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -\frac{x_n}{x_0^2} & 0 & 0 & \dots & \frac{1}{x_0}\end{array}\right].$$

Note that the matrix has rank $n$, so the differential is surjective. Therefore, $\pi$ is a submersion.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.