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Show that $\pi: \mathbb S^{2n+1} \to \mathbb {CP}^{n}$ is a bundle of fibre $\mathbb S^1$ ?

My attempt : Let $U_i=\{[z]\,, z_i>0\}$ and $O_i=\{ z\in \mathbb S^{2n+1}: z_i>0\}$. We have that $U_i=\pi(O_i)$ and $\mathbb{CP}=\cup_{i=1}^{n+1} U_i$ and $\mathbb S^{2n+1}=\cup_{i=1}^{n+1} O_i$.

Let $\psi_i: U_i\times \mathbb S^1\to \pi^{-1}(U_i)$ defined by : $\psi_i([z],\lambda)=([z],\lambda z)=$ $\color{red}{ (!)}$ I get stuck here and I didn't know if this application is well defined ! Any help is really appreciated !

H_K
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1 Answers1

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I think you mean: $$U_i=\{[z]\,, |z_i|>0\},\qquad\qquad O_i=\{ z\in \mathbb S^{2n+1}: |z_i|>0\}$$

Then your map $\psi_i$ should be: $$\psi_i([z],\lambda)= \lambda \frac{\overline z_i}{|z_i|}\frac z{||z||} .$$

Note this is well defined as for $\mu\in \mathbb{C}\backslash 0$ we have: $$ \lambda \frac{\overline {\mu z_i}}{|\mu z_i|}\frac{\mu z}{||\mu z||}=\lambda \frac{\overline z_i}{|z_i|}\frac z{||z||}. $$

tkf
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