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Let's consider the complex projective space $\mathbb{C}P^{n}$ and let $X$ be a vector field with flow given by $X_{t}:\mathbb{C}P^{n}\rightarrow\mathbb{C}P^{n}$ such that $X_{t}([z_{0},...,z_{n}])=[z_{0},\exp(it)z_{1},...,\exp(int)z_{n}]$.

In order to apply Poincaré-Hopf Theorem and conclude that $\chi(\mathbb{C}P^{n})=n+1$, I would like to prove that $X$ has exactly $n+1$ zeros with index $1$. How can I do this ? Thanks in advance !

nokia
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1 Answers1

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At a point $z=(z_0,\cdots,z_n)\in \mathbb{C}^{n+1}$, The natural projection maps vectors like $\lambda z$ to a zero vector of $\mathbb{C}P^n$. So to find the zeros just ask $$\frac{dX_t}{dt}|_{t=0}=(0, iz_1,\cdots,inz_n)=\lambda(z_0,\cdots,z_n)$$ we get $$ikz_k=\lambda z_k, \forall k=0,\cdots, n$$ So the zeros of $X$ are the following points (setting $\lambda=0,i,2i,\cdots,ni$): $$[1,0,0,\cdots,0]$$ $$[0,1,0,\cdots,0]$$ $$\cdots$$ $$[0,0,0,\cdots,1]$$ To compute the index of a zero point, e.g. $[1,0,\cdots,0]$, use the coordinate chart $[z_0,\cdots, z_n] \mapsto (z_1/z_0,\cdots, z_n/z_0)$, the flow is $$X_t(x_1,\cdots,x_n)=(\exp(it)x_1,\cdots,\exp(int)x_n)$$ And the vector field is $$X(x_1,\cdots,x_n)=i(x_1,2x_2,\cdots,nx_n)$$ This map is obviously 1 to 1 and orientation preserving when restricting to a small sphere around 0, that is, $X$ has index 1 at 0.

Xipan Xiao
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  • Thanks for the answer. Can you explain me what do you mean with zero vector of $\mathbb{C}P^{n}$ ? The lines where you compute the zeros of $X$ are not clear for me. Why can I take $\lambda = 0$ ? I guess it is just me confused with notation and my lack of experience dealing with projective spaces, so I appreciate you patience with me and with this stupid question. Thank you again :) – nokia Dec 23 '14 at 10:54
  • Basically, my whole problem with this, was always with computing the zeros of $X$, since I really get confused with the notation. Thanks again, Xipan. – nokia Dec 23 '14 at 10:57
  • Any points $[z_0, \cdots, z_n]$ that satisfy the equation $ikz_k=\lambda z_k$ for some $\lambda$ is a zero point of X. Since $\lambda z_0=i 0 z_0=0$ this implies either $\lambda=0$ or $z_0=0$ – Xipan Xiao Dec 23 '14 at 17:17