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This question is motivated by this answer here.

Let $\mathbb{C}P^{n}$ be a complex projective space. Let $X\in\Gamma(T\mathbb{C}P^{n})$, be a vector field. It seems, by the answer I got in the mentioned link, that the zeros of $X$ are the points $[z]\in\mathbb{C}P^{n}$ such that $X([z])=\lambda[z]$, for some $\lambda$. Can anyone please make this clear for me ? In another way : What does it mean to be a zero of a vector field in a complex projective space ? Thanks!

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nokia
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    That is not right. A zero of $X$ is a point $p$ in $\mathbb CP^n$ such that $X(p)=0$. «What does it mean to be a zero of a vector field on a projective space?»: exactly the same as on any other manifold! – Mariano Suárez-Álvarez Dec 23 '14 at 19:14
  • In particular, in the link I mentioned, can you explain me the first few lines ? (of the answer) – nokia Dec 23 '14 at 19:17

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Let's try a different approach. We can think of $\Bbb CP^n$ as the usual quotient of $\Bbb C^{n+1}-\{0\}$ with the projection map $\pi(z) = [z]$. Let's think of a vector field $X$ on $U\subset\Bbb CP^n$ as being pushed down from an appropriate vector field $\tilde X$ on $\pi^{-1}(U)$ with the property that $\pi_{*z}\tilde X(z) = \pi_{*\lambda z}\tilde X(\lambda z)$ for all nonzero (functions) $\lambda$ on $U$. Now, we can visualize a zero $[z]$ of $X$ in terms of saying that $\pi_{*z}\tilde X(z) = 0$, which means that $\tilde X(z)$ is a multiple of $z\in\Bbb C^{n+1}$.

Along these lines, also see my answer to this question about the Euler sequence.

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Ted Shifrin
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  • Perfect. Very clear answer! Thank you. I was getting crazy, since this looks very simple. However, since I rarely work with projective spaces, I was getting very confused. Thanks once again :) Unfortunately I don't have enough reputation points to upvote your answer. – nokia Dec 23 '14 at 20:29
  • Of course. You're most welcome! – Ted Shifrin Dec 23 '14 at 20:30
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    Thanks Ted. I answered the original question but didn't make my statement as clear as this. In a word, since the projection $\pi$ maps the fiber {$\lambda z | \lambda \in \mathbb{C}$} to a single point $[z]$, its differential maps the tangent space of the fiber to zero. – Xipan Xiao Dec 23 '14 at 22:47