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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.

edited Feb 10 2010 at 5:02

asked Feb 10 2010 at 4:47

MG
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Ryan Budney · Accepted Answer · 2010-02-10 05:04:32Z UTC

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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 $\mathbb CP^{n+1}$.

answered Feb 10 2010 at 5:04

Ryan Budney
21.8k●1●41●102

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? – MG Feb 10 2010 at 5:17

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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. – Ryan Budney Feb 10 2010 at 5:19

Got it! Thanks! – MG Feb 10 2010 at 5:32

Normal bundle of $CP^n$ in $CP^{n+1}$

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