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?
Best Answer
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.