TITLE: Normal bundle of $CP^n$ in $CP^{n+1}$
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: $S^{2n+1}$ 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 $\mathbb CP^{n+1}$.