# V: Surfaces ## V.1: Geometry on a Surface ### V.1.1. #to-work Let $C, D$ be any two divisors on a surface $X$, and let the corresponding invertible sheaves be $\mathcal{L}, \mathcal{M}$. Show that \[ C . D=\chi\left(\mathcal{O}_X\right)-\chi\left(\mathcal{L}^{-1}\right)-\chi\left(\mathcal{M}^{-1}\right)+\chi\left(\mathcal{L}^{-1} \otimes \mathcal{M}^{-1}\right) . \] ### V.1.2. #to-work Let $H$ be a very ample divisor on the surface $X$, corresponding to a projective embedding $X \subseteq \mathbf{P}^N$. If we write the Hilbert polynomial of $X$ (III, Ex. 5.2) as \[ P(z)=\frac{1}{2} a z^2+b z+c \] show that $a=H^2, b=\frac{1}{2} H^2+1-\pi$, where $\pi$ is the genus of a nonsingular curve representing $H$, and $c=1+p_a$. Thus the degree of $X$ in $\mathbf{P}^N$, as defined in (I, §7), is just $H^2$. Show also that if $C$ is any curve in $X$, then the degree of $C$ in $\mathbf{P}^N$ is just $C . H$ ### V.1.3. #to-work Recall that the arithmetic genus of a projective scheme $D$ of dimension 1 is defined as \[ p_a=1-\chi\left(\mathcal{O}_D\right) .\] See III, Ex. 5.3. a. If $D$ is an effective divisor on the surface $X$, use (1.6) to show that \[ 2 p_a-2= D.(D+K) .\] b. $p_a(D)$ depends only on the linear equivalence class of $D$ on $X$. c. More generally, for any divisor $D$ on $X$, we define the virtual arithmetic genus (which is equal to the ordinary arithmetic genus if $D$ is effective) by the same formula: $2 p_a-2=D .(D+K)$. Show that for any two divisors $C, D$ we have \[ p_a(-D)=D^2-p_a(D)+2 \] and \[ p_a(C+D)=p_a(C)+p_a(D)+C . D-1 . \] ### V.1.4. #to-work a. If a surface $X$ of degree $d$ in $\mathbf{P}^3$ contains a straight line $C=\mathbf{P}^1$, show that $C^2=2-d$ b. Assume char $k=0$, and show for every $d \geqslant 1$, there exists a nonsingular surface $X$ of degree $d$ in $\mathbf{P}^3$ containing the line $x=y=0$. ### V.1.5. #to-work a. If $X$ is a surface of degree $d$ in $\mathbf{P}^3$, then \[ K^2=d(d-4)^2 .\] b. If $X$ is a product of two nonsingular curves $C, C^{\prime}$, of genus $g, g^{\prime}$ respectively, then \[ K^2=8(g-1)\left(g^{\prime}-1\right) .\] Cf. (II, Ex. 8.3). ### V.1.6. #to-work a. If $C$ is a curve of genus $g$, show that the diagonal $\Delta \subseteq C \times C$ has self-intersection $\Delta^2=2-2 g$. (Use the definition of $\Omega_{C / k}$ in (II, §8).) b. Let $l=C \times \mathrm{pt}$ and $m=\mathrm{pt} \times C$. If $g \geqslant 1$, show that $l, m$, and $\Delta$ are linearly independent in $\operatorname{Num}(C \times C)$. Thus $\operatorname{Num}(C \times C)$ has rank $\geqslant 3$, and in particular, \[ \operatorname{Pic}(C \times C) \neq p_1^* \operatorname{Pic} C \oplus p_2^* \Pic C .\] Cf. (III, Ex. 12.6), (V, Ex. 4.10). ### V.1.7. Algebraic Equivalence of Divisors. #to-work Let $X$ be a surface. Recall that we have defined an algebraic family of effective divisors on $X$, parametrized by a nonsingular curve $T$, to be an effective Cartier divisor $D$ on $X \times T$, flat over $T$ (III, 9.8.5). In this case, for any two closed points $0,1 \in T$, we say the corresponding divisors $D_0, D_1$ on $X$ are prealgebraically equivalent. Two arbitrary divisors are prealgebraically equivalent if they are differences of prealgebraically equivalent effective divisors. Two divisors $D, D^{\prime}$ are algebraically equivalent if there is a finite sequence $D=D_0, D_1, \ldots, D_n=D^{\prime}$ with $D_i$ and $D_{i+1}$ prealgebraically equivalent for each $i$. a. Show that the divisors algebraically equivalent to 0 form a subgroup of Div $X$. b. Show that linearly equivalent divisors are algebraically equivalent.[^hint.v.1.1.b] c. Show that algebraically equivalent divisors are numerically equivalent.[^hint.v.1.7.c][^rmk.v.1.7.c] [^hint.v.1.1.b]: Hint: If $(f)$ is a principal divisor on $X$, consider the principal divisor $(t f-u)$ on $X \times \mathbf{P}^1$, where $t, u$ are the homogeneous coordinates on $\mathbf{P}^1$. [^hint.v.1.7.c]: Hint: Use (III, 9.9) to show that for any very ample $H$, if $D$ and $D^{\prime}$ are algebraically equivalent, then $D . H=D^{\prime} . H$. [^rmk.v.1.7.c]: Note. The theorem of Néron and Severi states that the group of divisors modulo algebraic equivalence, called the Néron-Severi group, is a finitely generated abelian group. Over $\mathbf{C}$ this can be proved easily by transcendental methods (App. B, $\S 5$ ) or as in (Ex. 1.8) below. Over a field of arbitrary characteristic, see Lang and Néron [1] for a proof, and Hartshorne [6] for further discussion. Since $\Num X$ is a quotient of the Néron-Severi group, it is also finitely generated, and hence free, since it is torsion-free by construction. ### V.1.8. Cohomology Class of a Divisor. #to-work For any divisor $D$ on the surface $X$, we define its cohomology class $c(D) \in H^1\left(X, \Omega_X\right)$ by using the isomorphism Pic $X \cong$ $H^1\left(X, \mathcal{O}_X^*\right.$ ) of (III, Ex. 4.5) and the sheaf homomorphism $d \log : \mathcal{O}^* \rightarrow \Omega_X$ (III, Ex. 7.4c). Thus we obtain a group homomorphism $c: \operatorname{Pic} X \rightarrow H^1\left(X, \Omega_X\right)$. On the other hand, $H^1(X, \Omega)$ is dual to itself by Serre duality (III, 7.13), so we have a nondegenerate bilinear map \[ \langle\quad, \quad\rangle: H^1(X, \Omega) \times H^1(X, \Omega) \rightarrow k . \] a. Prove that this is compatible with the intersection pairing, in the following sense: for any two divisors $D, E$ on $X$, we have \[ \langle c(D), c(E)\rangle=(D . E) \cdot 1 \] in $k$.[^hint.v.1.8.a] [^hint.v.1.8.a]: Hint: Reduce to the case where $D$ and $E$ are nonsingular curves meeting transversally. Then consider the analogous map $c: \operatorname{Pic} D \rightarrow H^1\left(D, \Omega_D\right)$, and the fact (III, Ex. 7.4) that $c$ (point) goes to 1 under the natural isomorphism of $H^1\left(D, \Omega_D\right)$ with $k$. b. If char $k=0$, use the fact that $H^1\left(X, \Omega_X\right)$ is a finite-dimensional vector space to show that $\Num X$ is a finitely generated free abelian group. ### V.1.9. #to-work a. If $H$ is an ample divisor on the surface $X$, and if $D$ is any divisor, show that \[ \left(D^2\right)\left(H^2\right) \leqslant(D . H)^2 \text {. } \] b. Now let $X$ be a product of two curves $X=C \times C^{\prime}$. Let $l=C \times \mathrm{pt}$, and $m=$ pt $\times C^{\prime}$. For any divisor $D$ on $X$, let $a=D . l, b=D . m$. Then we say $D$ has type $(a, b)$. If $D$ has type $(a, b)$, with $a, b \in \mathbf{Z}$, show that \[ D^2 \leqslant 2 a b \text {, } \] and equality holds if and only if $D \equiv b l+a m$.[^hint.v.1.9.b] [^hint.v.1.9.b]: Hint: Show that $H=l+m$ is ample, let $E=l-m$, let $D^{\prime}=\left(H^2\right)\left(E^2\right) D-\left(E^2\right)(D \cdot H) H-\left(H^2\right)(D \cdot E) E$, and apply (1.9). This inequality is due to Castelnuovo and Severi. See Grothendieck $[2]$. ### V.1.10. Weil's Proof of the Analogue of the Riemann Hypothesis for Curves. #to-work Let $C$ be a curve of genus $g$ defined over the finite field $\mathbf{F}_q$, and let $N$ be the number of points of $C$ rational over $\mathbf{F}_q$. Then $N=1-a+q$, with $|a| \leqslant 2 g \sqrt{q}$. To prove this, we consider $C$ as a curve over the algebraic closure $k$ of $\mathbf{F}_q$. Let $f: C \rightarrow C$ be the $k$-linear Frobenius morphism obtained by taking $q$ th powers, which makes sense since $C$ is defined over $\mathbf{F}_q$, so $X_q \cong X$ (See $V, 2.4 .1)$. Let $\Gamma \subseteq C \times C$ be the graph of $f$, and let $\Delta \subseteq C \times C$ be the diagonal. Show that $\Gamma^2=q(2-2 g)$, and $\Gamma . \Delta=N$. Then apply (Ex. 1.9) to $D=r \Gamma+s \Delta$ for all $r$ and $s$ to obtain the result.[^v.1.10.seeappendix] [^v.1.10.seeappendix]: See (App. C, Ex. 5.7) for another interpretation of this result. ### V.1.11. #to-work In this problem, we assume that $X$ is a surface for which $\Num X$ is finitely generated (i.e., any surface, if you accept the Néron-Severi theorem (Ex. 1.7)). a. If $H$ is an ample divisor on $X$, and $d \in \mathbf{Z}$, show that the set of effective divisors $D$ with $D . H=d$, modulo numerical equivalence, is a finite set.[^hint.v.1.12.a.asdasdas] [^hint.v.1.12.a.asdasdas]: Hint: Use the adjunction formula, the fact that $p_a$ of an irreducible curve is $\geqslant 0$, and the fact that the intersection pairing is negative definite on $H^{\perp}$ in $\operatorname{Num} X$. b. Now let $C$ be a curve of genus $g \geqslant 2$, and use (a) to show that the group of automorphisms of $C$ is finite, as follows. Given an automorphism $\sigma$ of $C$, let $\Gamma \subseteq X=C \times C$ be its graph. First show that if $\Gamma \equiv \Delta$, then $\Gamma=\Delta$, using the fact that $\Delta^2<0$, since $g \geqslant 2$ (Ex. 1.6). Then use (a). Cf. (V, Ex. 2.5). ### V.1.12. #to-work If $D$ is an ample divisor on the surface $X$, and $D^{\prime} \equiv D$, then $D^{\prime}$ is also ample. Give an example to show, however, that if $D$ is very ample, $D^{\prime}$ need not be very ample.