Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 159 (Algebraic Surfaces)

The Picard group $\mathrm{Pic}(X)$ is the set of isomorphism classes of line bundles on $X$, with tensor product as addition, the trivial bundle as zero and the dual bundle as inverse. On a smooth projective surface, line bundles can be represented by divisors. The intersection pairing on the Picard group of a surface is the symmetric bilinear mapobtained by moving the divisors into proper position and counting their intersections with multiplicity.

The surfaceis the first Hirzebruch surface ${\mathbb{F}}_1$. If $S$ is its negative section and $F$ a fiber of the ruling, its Picard lattice of a Hirzebruch surface is

Now let $\pi :X\to {\mathbb{P}}^2$ be the given nonisomorphic birational morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let $H={\pi }^{\ast }[\text{line}]$, and take the total transform $E$ on $X$ of the exceptional curve of the first blowup. The intersection formula for blowing up a surface givesTherefore the nonzero Picard group element $H+E$ satisfiesThis is the isotropic divisor from a nontrivial birational morphism to the projective plane.

For a morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^n$, the pullback of the hyperplane bundle has the formfor an integer $d\ge 0$. If a line $\ell$ is contracted to a point, this bundle restricts trivially to $\ell$, whereasIts degree is therefore zero, so $d=0$. The homogeneous sections defining $\varphi$ are then constants, and $\varphi$ is constant. This proves the morphism from the projective plane contracting a line criterion.

Finally choose an integer $r>C^2$ and blow up $r$ distinct points of the smooth curve $C$. For the resulting morphism $\pi :X^{\prime }\to X$, let $E_1,\dots ,E_r$ be the exceptional curves. The strict transform isand the self-intersection after blowing up points on a smooth curve formula givesBlowing up a smooth point of a smooth curve does not change that curve itself, so $\pi {|}_{C^{\prime }}:C^{\prime }\to C$ is an isomorphism.

A minimal algebraic surface is a smooth projective surface containing no exceptional curve of the first kind, namely no smooth rational curve of self-intersection $-1$. An abelian surface contains no rational curve because every morphism from ${\mathbb{P}}^1$ to an abelian variety is constant. It therefore has no $(-1)$-curve and is minimal.

For every $n\ge 2$, the Minimal Hirzebruch surface ${\mathbb{F}}_n$ is a rational minimal surface. These surfaces are pairwise nonisomorphic: the negative section is the unique irreducible curve of negative self-intersection and has square $-n$, so an isomorphism would recover $n$. Thus there are infinitely many nonisomorphic minimal rational surfaces.

A K3 surface is a smooth projective surface $X$ with $K_X\cong {\mathcal{O}}_X$ and $H^1(X,{\mathcal{O}}_X)=0$. For a smooth curve $C\subset X$ of geometric genus $g$, the Adjunction formula givesand hence

An elliptic surface is a smooth projective surface with a morphism to a smooth curve whose generic fiber is a smooth genus-one curve. If $E$ is an elliptic curve, projectionis an elliptic fibration. Its canonical bundle is pulled back from $K_{{\mathbb{P}}^1}$, so every positive pluricanonical space vanishes and the Kodaira dimension is $-\infty$. This supplies the requested negative-Kodaira-dimension example.

For an elliptically fibered K3 surface, choose a smooth quartic $X\subseteq {\mathbb{P}}^3$ containing a line $L$. The canonical bundle of a smooth projective hypersurface formula makes $K_X$ trivial, and the standard cohomology sequence gives $H^1(X,{\mathcal{O}}_X)=0$, so $X$ is K3. The pencil of planes through $L$ cuts $X$ into $L$ plus a residual plane cubic. The residual linear system $|H-L|$ is basepoint-free, has square zero and defines a morphism $X\to {\mathbb{P}}^1$ whose generic fiber is a smooth plane cubic. This is the Elliptic K3 surface from a quartic containing a line.

A surface of general type is a smooth projective surface of Kodaira dimension two. Let $B\subseteq {\mathbb{P}}^2$ be a smooth plane curve of degree eight and letbe the degree-two cover branched along $B$. The branch-cover canonical-bundle formula givesThis divisor is ample, so $X$ is a double plane of general type and $\pi$ is the required finite morphism of degree two.

The irregularity of an algebraic surface and the geometric genus of an algebraic surface are respectively

If $\pi :X^{\prime }\to X$ is the blowup at a point with exceptional divisor $E$, thenA holomorphic two-form on $X$ pulls back to one on $X^{\prime }$. Conversely, a holomorphic two-form on $X^{\prime }$ descends across $E$: locally it is a form on the punctured smooth surface $X\setminus \{p\}$, and its coefficients extend over the codimension-two point $p$. Equivalently, ${\pi }_{\ast }{K}_{X^{\prime }}=K_X$. Pullback is therefore an isomorphismwhich proves the birational invariance of the geometric genus of a surface in this case.

Let $X=C\times D$, where both smooth projective curves have positive genus. The Künneth theorem gives the irregularity of a product of curvesIf a surface is a hypersurface in projective space, dimension forces it to be a smooth hypersurface $Y\subseteq {\mathbb{P}}^3$. Fromand the intermediate cohomology vanishing for line bundles on projective space, one obtains $H^1(Y,{\mathcal{O}}_Y)=0$. Thus every such hypersurface has irregularity zero, whereas $X$ has positive irregularity. Hence $C\times D$ is not isomorphic to a hypersurface. This is the product of positive-genus curves is not a projective hypersurface obstruction.

The Albanese variety $\mathrm{Alb}(X)$ is the universal abelian variety receiving a pointed morphism from $X$, andLet the smooth image curve of the Albanese morphism be $C$ of genus $g$. Pullback of holomorphic one-forms along the dominant map $X\to C$ is injective, giving $g\le q(X)$. On the other hand, the universal property of the Jacobian extends $C\to \mathrm{Alb}(X)$ to a homomorphismThe Albanese image generates the entire Albanese variety, so this homomorphism is surjective and $q(X)\le \dim \mathrm{Jac}(C)=g$. Therefore the Irregularity from a smooth Albanese curve image is