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Let $S^1$ act on $S^{2n+1}$ via Hopf action and $S^1$ also acts on $\mathbb{R}^2$ via rotation about the origin. Then $S^1$ acts on $S^{2n+1}\times \mathbb{R}^2$ diagonally.

Let $M$ be the quotient of this diagonal action.

My question is why $ M$ can be viewed as the normal bundle of $\mathbb {CP}^n$ in $\mathbb {CP}^{n+1}$.

I have a feeling that it must be related to the fact that: after removing a $(2n+2)$ disk in $\mathbb {CP}^{n+1}$, the boundary $S^{2n+1}$ is fibered over the $\mathbb {CP}^n$.

But where can I find the proof of the statement.

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$S^{2n+1}$ sits in $S^{2n+3}$ with a trivial normal bundle. So the quotient map $S^{2n+3} \to \mathbb CP^{n+1}$ carries the normal bundle of $S^{2n+1}$ in $S^{2n+3}$ to the normal bundle of $\mathbb CP^n$ in $\mathbb CP^{n+1}$.

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  • $\begingroup$ Yes, we can view $S^{2n+1}\times \mathbb R^2$ as the trivial normal bundle, but the question is "the hopf action of $S^1$ on $S^{2n+3}$ is the same as the diagonal action on this normal bundle infinitesimally? $\endgroup$
    – J. GE
    Feb 10, 2010 at 5:17
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    $\begingroup$ More than infinitesimally, it's that way in a product sense of $\mathbb C^{n+1} \oplus \mathbb C = \mathbb C^{n+2}$. Just write down the simplest trivialization of the normal bundle you can imagine and it works out. $\endgroup$
    – Ryan Budney
    Feb 10, 2010 at 5:19

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