Hartshorne Exercise V.1.11.
In this problem, we assume that $X$ is a surface for which $\op{Num}X$ is finitely generated (i.e., any surface, if you accept the Neron~Severi theorem (Ex. 1.7) ).
(a) If $H$ is an ample divisor on $X$, and $d\in\Z$, show that the set of effective divisors $D$ with $D.H=d$, modulo numerical equivalence, is a finite set. [Hint: Use the adjunction formula, the fact that $P_a$ of an irreducible curve is $\ge 0$, and the fact that the intersection pairing is negative definite on $H^\perp$ in Num X.
(b) Now let $C$ be a curve of genus $g\ge2$ and use (a) to show that the group of automorphisms of $C$ is finite, as follows. Given an automorphism $\sigma$ of $C$, let $\Gamma\subset X=C\times C$ be its graph.First show that if $\Gamma\equiv\Delta$, then $\Gamma=\Delta$, using the fact that $\Delta^2\lt 0$, since $g\ge 2$] (Ex. 1.6). Then use (a). Cf. (IV, Ex. 2.5).
NB: I realized I made a mistake in this part (a), so I’m working on a correct solution, hopefully coming soon.
(a) Let $D$ be an effective divisor with $D.H=d$. If $d<0$ then the described set is empty because $D.H\ge 0$. Also, if $d=0$ then $D\sim 0$ so the described set is a singleton. We may thus assume that $d>0$.
Let $\tilde D$ be the projection of $D$ onto $H^\perp$. We first note that if $D\equiv\tilde D$ then $D\sim 0$ and $d=0$. If $\tilde D\equiv 0$ then $D\in\op{Num}X$ is numerically equivalent to a multiple of $H$. In particular, we see that $\tilde D\equiv 0$ only if $H^2|d$, and in this case, $D\equiv\frac{d}{H^2}H$: the described set is a singleton. So, from now on, assume $\tilde D\not\equiv0$, in which case $\tilde D^2<0$ by the Hodge index theorem. Writing $D\equiv nH+\tilde D$ uniquely for some $n\in \Z$, we see that $nH^2=d$, so again, $H^2|d$. That is if $H^2\not|d$ then the set $\{D.H=d\}\subset\op{Num}X$ is empty. So from now on we may assume $\frac{d}{H^2}=n\in\Z_{>0}$.
Observe that $D^2=n^2H^2+\tilde D^2<n^2H^2=\frac{d^2}{H^2}$.
Let $C_1,…,C_r$ be the irreducible components of $D$ which are all possibly singular curves. Because $D.H=d$, we must have $r\le d$. For each $1\le i\le r$,
$$p_a(C_i)=\frac{C_i.(C_i+K_X)}{2}+1\ge 0.$$
Hence, $C_i^2\ge -C_i.K_X-2$. Thus,
$$D^2\ge -D.K_X-2r\ge -D.K_X-2d.$$
Now, if $X$ is fano, then it is clear that $D^2>-2d$. By Serre Duality,
$$h^0(-\omega_X)=h^2(\omega_X^{\otimes 2})
Writing $K_X$ as a union of irreducible curves, there are two cases. The first case is if
$$K_X\cap D=\{\text{some points}\}\:\text{ or }\:K_X\cap D=\varnothing,$$
and here we have $D^2\ge -2d$. The second case is if $K_X\cap D=\{\text{some curves}\}$. If $C_j$ is some curve in this intersection, then we find that $C_j^2\ge -1$ in looking at the genus formula, and so
$$D^2\ge-\displaystyle\sum_{C_j\subset K_X\cap D}1\ge -2d.$$
We conclude that in all cases,
$$-2d\le D^2\le \frac{d^2}{H^2}.$$
Under the assumption that $\op{Num}X$ is finitely generated, this bound allows $D$ to be obtained by only finitely many possible $\Z$-linear combinations of the generators, which is the desired result.
(b) Let $C$ be a curve of genus 2, $\sigma\in\op{Aut}C$, $X=C\times C$, and $\Gamma\subset X$ the graph of $\sigma$. First, suppose that $\Gamma\equiv\Delta$. By problem 13 (V, Ex. 1.6),
$$\Delta^2=\Gamma.\Delta=-2.$$
But, $\Gamma,\Delta$ are irreducible curves on $X$, hence we must have $\Gamma\sim \Delta$, and in fact $h^0(\Delta)=1$ after some thought, so actually $\Gamma=\Delta$.
Consider the map
$$\op{Aut}C\ra \op{Num}X,$$
given by
$$\sigma\mapsto\Gamma_\sigma,$$
where $\Gamma_\sigma$ is the graph of $\sigma$. Note that $\Gamma_{\op{id}}=\Delta$.
As we have seen in Problem 13 (V, Ex. 1.9), the divisor $H=\text{pt}\times C+C\times\text{pt}$ on $X$ is ample. Moreover, $\Gamma_\sigma.H=2$ for every $\sigma\in\op{Aut}C$. Hence,
$$\op{im}(\op{Aut}C)\subset \{D.H=2\}\subset\op{Num}X.$$
Lastly, I claim that this map is injective. Indeed, if $\Gamma_\sigma\equiv\Gamma_{\sigma’}$, then $\Gamma_\sigma -\Gamma_{\sigma’}+\Delta\equiv\Delta$, therefore $\Gamma_\sigma -\Gamma_{\sigma’}+\Delta=\Delta$ as divisors on $X$. It follows that $\Gamma_\sigma =\Gamma_{\sigma’}$ which occurs iff $\sigma =\sigma’$. We conclude by part (a) that
$$|\op{Aut}C|\le |\{D\in \op{Num}X:D.H=2\}|<\infty.$$
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