Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 25 / 3

Let $D$ be an effective divisor linearly equivalent to $C$. If $C$ were not a component of $D$, all intersection multiplicities between its components and $C$ would be nonnegative, contradicting $D\cdot C=C^2<0$. Hence $D=C+D^{\prime }$ with $D^{\prime }$ effective and linearly equivalent to zero. For an ample divisor $H$, this gives $H\cdot D^{\prime }=0$. Positivity of $H$ on every irreducible component of a nonzero effective divisor forces $D^{\prime }=0$. Therefore $\boxed{D=C}$, illustrating the rigidity of a negative curve.

Use countability of negative curves. A smooth projective surface over $\mathbb{C}$ is a compact smooth manifold homotopy equivalent to a finite CW complex, so its integral cohomology group $H^2(X,\mathbb{Z})$ is finitely generated and therefore countable. Distinct irreducible curves $C,D$ with $C^2<0$ cannot have the same cohomology class: that would give $C\cdot D=C^2<0$, while distinct effective curves have nonnegative intersection multiplicities. Assigning a negative curve its cohomology class is thus injective, proving there are at most countably many negative curves.

For zero self-intersection, uncountable families do occur. On the product of two projective lines ${\mathbb{P}}^1\times {\mathbb{P}}^1$, every fiber ${\mathbb{P}}^1\times \{p\}$ has self-intersection zero, since a different fiber is disjoint and has the same numerical equivalence of divisors class. The parameter $p\in {\mathbb{P}}^1(\mathbb{C})$ varies over an uncountable set. Thus yes, there can be uncountably many such curves.

Let $X\subset {\mathbb{P}}^3$ have degree $d\ge 3$ and hyperplane class $H$. The adjunction formula gives $K_X=(d-4)H$. For a line $L\subset X$, the same formula on the smooth rational curve $L$ yields $(K_X+L)\cdot L=-2$, henceThus all lines are negative curves, and part (b) makes their set countable. The Fano scheme of lines is a closed projective scheme of $\mathrm{Gr}(2,4)$. A positive-dimensional complex algebraic component would have uncountably many points, so this scheme is zero-dimensional. A projective scheme of finite type has only finitely many zero-dimensional points. This proves the finiteness of lines on a smooth surface of degree at least three.