Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted

$S^{2n+1}$ 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 $\mathbb CP^{n+1}$.

share|improve this answer
    
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? –  John B Feb 10 '10 at 5:17
1  
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 '10 at 5:19
    
Got it! Thanks! –  John B Feb 10 '10 at 5:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.