Circle bundles over $CP^1$ and self-intersection number of $CP^1$ embeddings

Question

If $X$ is a compact oriented surface in a 4-dimensional oriented manifold $M$, then the self-intersection number $X^2$ of $X$ is given by the integral over $X$ of the Euler class of the normal bundle. In the case of $CP^1$ embedded in $CP^2$, the normal bundle is isomorphic to the Hopf bundle, therefore $X^2$ can be obtained calculating the first Chern number of the Hopf fibration (or equivalently the Euler number of its realization).

It is possible to have circle bundles on $CP^1$ with higher Chern number by taking the quotient of the total space of the Hopf fibration by the action generated by $(z^1,z^2)\mapsto(z^1 \exp(i 2 \pi/k), z^2 \exp(i 2\pi/k ))$.

Are these bundles the normal bundles of some embedding of $CP^1$ in a 4-dimensional manifold? If yes, is it possible to describe the embedding explicitly? Is there a deeper relation between the Hopf bundle and the normal bundle of $CP^1$ embedded in $CP^2$ or do they just happen to be the same?

Answer 1

You can find $\mathbb CP^1$ in a wide variety of $4$-manifolds having any Euler class you like. One really simple way is to take the connect-sum of $k$ copies of $\mathbb CP^2$. The idea is to embed $\mathbb CP^1$ in the connect sum so that you are simultaneously breaking the $\mathbb CP^1$ up as a connect sum in all the factors.

If you want a negative Euler class, you use the opposite orientation on $\mathbb CP^2$.

I'm not sure I understand your question about a "deeper relation" with the Hopf bundle. One way to interpret this is that if you remove a point from $\mathbb CP^2$ you get a complex line bundle over $\mathbb CP^1$ whose associated sphere bundle is the Hopf bundle.

Answer 2

Take your favorite complex line bundle $L$ over $\newcommand{\bCP}{\mathbb{CP}}$ $\bCP^1$. Form the rank-two complex vector bundle $E\to\bCP^1$ defined as the direct sum of $L$ with the trivial line bundle $\newcommand{\uc}{\underline{\mathbb{C}}}$ $\uc\to\bCP^1$

$$ E=L\oplus \uc. $$

Now form the projectivization $\newcommand{\bP}{\mathbb{P}}$ $\bP(E)$. This is a bundle over $\bCP^1$ whose fiver over a point $x\in\bCP^1$ is the projective space $\bP(E_x)$ consiting of all complex lines in the vector space $E_x=L_x\oplus \mathbb{C}$.

The $0$-section of $L$ defines a section $\zeta: \bCP^1\to\bP(E)$. The normal bundle of the embedding $\zeta(\bCP^1)\subset \bP(E)$ can be identified with the line bundle $L$.

Answer 3

Another approach (for any base surface $S$): take the disk bundle over $S$ with Euler number $e$ (any integer). The total space is a smooth 4-manifold with boundary with the required property (the 0-section has self-intersection $e$). If you want a closed 4-manifold, just take the double.

Regarding the second question, $\Bbb CP^2$ can be regarded as the disk bundle over $S^2$ with Euler number 1 (or $-1$ for $\overline{\Bbb CP^2}$) capped off with a 4-ball attached to the bounding 3-sphere of the total space. This is a topological description of the geometric fibration described by Ryan Budney in his answer.