Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 146 / 1 / Solution

For a symplectic ball centered at the origin, its symplectic blowup replaces a smaller concentric ball by the disk bundle in ${\mathcal{O}}_{{\mathbb{CP}}^{n-1}}(-1)$. The zero section is the exceptional divisor $E\cong {\mathbb{CP}}^{n-1}$; the blowdown collapses $E$ to the original point and is a symplectomorphism away from $E$. The new symplectic form agrees with ${\omega }_0$ outside the surgery region, while its integral over a line in $E$ specifies the blowup size. In real dimension four, $E\cong {\mathbb{CP}}^1$ and $E^2=-1$.

A Lefschetz pencil on the compact oriented four-manifold $X$ is a mapwith finite base locus $B$, such that orientation-compatible complex coordinates give the local model $f(z_1,z_2)=[z_1:z_2]$ at a base point and $f(z_1,z_2)=z_1^2+z_2^2$ at each critical point. Blowing up all $b=|B|$ base points produces a Lefschetz fibration $\tilde{X}\to S^2$ whose regular fiber is the closed smooth fiber of the pencil.

If that fiber has genus $g$ and there are $m$ critical points, thenEvery blowup raises the Euler characteristic by one, so $\chi (\tilde{X})=\chi (X)+b$. Therefore the Euler characteristic of a Lefschetz pencil is

Now let $X={\mathbb{CP}}^2$ with its Fubini-Study form. A symplectic smooth fiber $F$ represents $dH\in H_2({\mathbb{CP}}^2;\mathbb{Z})$ for some positive integer $d$, because its symplectic area is positive. The Symplectic adjunction formula, using $K_{{\mathbb{CP}}^2}=-3H$, givesand henceThese values are $0,0,1,3,6,10,\dots$. They are all attained: two sufficiently general homogeneous polynomials of degree $d$ generate a pencil of complex plane curves with only the singularities allowed in a Lefschetz pencil, and its smooth fibers are symplectic. Thus the displayed list is exactly the list described by the Genus of a symplectic Lefschetz pencil on the complex projective plane.