Let $Q\subset\mathbb{CP}^2$ be a smooth complex conic. Its homology class is $2H$, so its <self-intersection number> is
$$
Q^2=(2H)^2=4.
$$
Rescale the <Fubini-Study form> so that $Q$ and $C$ have the same symplectic area. Their <normal bundles> have opposite Euler numbers, $+4$ and $-4$, so the <symplectic sum> can be formed along $Q$ and $C$.

Concretely, remove tubular neighborhoods $\nu(C)$ and $\nu(Q)$ and glue the boundaries by a fiber-reversing bundle map. The <symplectic neighborhood theorem> supplies the standard models needed for the gluing, and the symplectic-sum construction supplies a symplectic form on
$$
(X\setminus\nu(C))\cup_{L(4,1)}
(\mathbb{CP}^2\setminus\nu(Q)).
$$
The second piece is a rational homology ball. Thus this operation replaces the neighborhood of the $-4$ sphere by that rational ball and is the <symplectic rational blowdown of a minus-four sphere>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 4 / b / Solution

{title2=$[X]\cdot[X]$}

For an oriented submanifold of half the dimension of its ambient manifold, the self-intersection number is the intersection number with a transverse perturbation. It equals the Euler number of the <normal bundle>.

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 Incoming links: Self-intersection number