Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 159 / 3

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