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 !
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.