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)).
- If \(H\) is an ample divisor on \(X\), and \(d \in \ZZ\), 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 \(\geqslant 0\), and the fact that the intersection pairing is negative definite on \(H^{\perp}\) in \(\Num X\).
- 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).
Solution
Let \[ N=\Num X, \qquad N_\RR=N\tensor_\ZZ\RR. \] Since \(N\) is finitely generated and numerical equivalence has no torsion, \(N\) is a lattice in the finite-dimensional real vector space \(N_\RR\).
1. The intersection form is negative definite on\[ H^\perp=\{v\in N_\RR:v\cdot H=0\}, \]and\[ N_\RR=\RR H\oplus H^\perp. \]
The Hodge index theorem T-SRFHODGE says that every nonzero class in \(H^\perp\) has negative square. Thus the restriction of the intersection form to \(H^\perp\) is negative definite. Since \(H^2>0\) by the Nakai–Moishezon criterion T-SRFNAKAI, the line \(\RR H\) meets \(H^\perp\) trivially, and every \(v\in N_\RR\) decomposes uniquely as \[ v=\frac{v\cdot H}{H^2}H+ \left(v-\frac{v\cdot H}{H^2}H\right). \] The second summand is orthogonal to \(H\).
2. Put\[ K_X=\frac{K_X\cdot H}{H^2}H+z, \qquad z\in H^\perp, \]and define the positive-definite norm on \(H^\perp\) by\[ \|v\|^2=-v^2. \]If \(C\) is an irreducible curve and \(h=C\cdot H\), writing\[ C=\frac h{H^2}H+y, \qquad y\in H^\perp, \]gives\[ \|y\|^2 \leq 2+\frac{h^2+h(K_X\cdot H)}{H^2}+\|z\|\,\|y\|. \]
Adjunction gives \[ 2p_a(C)-2=C^2+K_X\cdot C. \] Since \(C\) is an integral projective curve, \(p_a(C)\geq0\), hence \[ C^2+K_X\cdot C\geq-2. \] Using the orthogonal decompositions in step 2, \[ C^2=\frac{h^2}{H^2}-\|y\|^2 \] and \[ K_X\cdot C = \frac{h(K_X\cdot H)}{H^2}+z\cdot y. \] Therefore \[ \|y\|^2 \leq 2+\frac{h^2+h(K_X\cdot H)}{H^2}+z\cdot y. \] The positive-definite inner product \[ (u,v)_+=-u\cdot v \] on \(H^\perp\) satisfies Cauchy–Schwarz, so \[ z\cdot y\leq|z\cdot y|\leq\|z\|\,\|y\|. \] Substitution proves the displayed estimate.
3. For each integer \(d\geq1\), the numerical classes of irreducible curves\(C\) satisfying\[ 1\leq C\cdot H\leq d \]form a finite set.
For such a curve, the integer \[ h=C\cdot H \] belongs to the finite set \(\{1,\ldots,d\}\). In step 2 the quantity \[ A_h=2+\frac{h^2+h(K_X\cdot H)}{H^2} \] therefore ranges over a finite set of real numbers. If \(t=\|y\|\), the estimate of step 2 is \[ t^2-\|z\|t\leq A_h. \] The left side tends to \(+\infty\) as \(t\to+\infty\). Hence there is a constant \(R_d\) such that every such curve has \[ \|y\|\leq R_d. \] Its coefficient \(h/H^2\) in the \(H\)-direction is also bounded because \(1\leq h\leq d\). Thus all these classes lie in a bounded subset of \(N_\RR\). The lattice \(N\subseteq N_\RR\) is discrete, so a bounded subset contains only finitely many lattice points. This proves the claim.
4. If \(d<0\), there is no effective divisor \(D\) with \(D\cdot H=d\); if\(d=0\), the only such effective divisor is \(D=0\).
Write a nonzero effective divisor as \[ D=\sum_i n_iC_i, \qquad n_i>0, \] with distinct irreducible curves \(C_i\). Ampleness gives \[ H\cdot C_i>0 \] for every \(i\) by Nakai–Moishezon. These intersection numbers are positive integers, so \[ D\cdot H=\sum_i n_i(H\cdot C_i)>0. \] The assertions follow.
5. For every integer \(d>0\), only finitely many numerical classes ofeffective divisors satisfy \(D\cdot H=d\).
4. Then\[ d=D\cdot H=\sum_i n_i(C_i\cdot H). \]Consequently every component satisfies\[ 1\leq C_i\cdot H\leq d, \]so by step 3 its numerical class belongs to a fixed finite set \(S_d\).Moreover,\[ \sum_i n_i\leq d \]because every \(C_i\cdot H\) is at least one. Hence \([D]\) is a sum of at most\(d\) elements of the finite set \(S_d\), counted with repetition. There are onlyfinitely many such sums in the abelian group \(N\). Together with step 4,this proves part (a) for every integer \(d\).
6. Let \(C\) have genus \(g\geq2\), let \(X=C\times C\), and let\(\Gamma_\sigma\) be the graph of an automorphism \(\sigma\) of \(C\). Then\[ \Gamma_\sigma^2=\Delta^2=2-2g<0. \]
The graph embedding has normal bundle \(\sigma^*T_C\): for a graph of a morphism \(u:C\to C\), the normal sequence identifies the quotient of \(T_C\oplus u^*T_C\) by the graph of \(du\) with \(u^*T_C\). Since \(\sigma\) is an automorphism, \[ \deg\sigma^*T_C=\deg T_C=2-2g. \] The self-intersection of a smooth curve on a smooth surface is the degree of its normal bundle. Thus \(\Gamma_\sigma^2=2-2g\). The equality \(\Delta^2=2-2g\) is Exercise V.1.6, P-AGH516DIAGONAL.
7. If\[ \Gamma_\sigma\equiv\Delta, \]then\[ \Gamma_\sigma=\Delta \]and therefore \(\sigma=\id_C\).
Numerical equivalence would give \[ \Gamma_\sigma\cdot\Delta=\Delta^2=2-2g<0. \] If the two irreducible curves \(\Gamma_\sigma\) and \(\Delta\) were distinct, their intersection number would be a sum of nonnegative local intersection multiplicities, hence would be nonnegative. This contradiction shows \[ \Gamma_\sigma=\Delta. \] Equality of the two graphs means \(\sigma(P)=P\) for every \(P\), so \(\sigma=\id_C\).
8. Distinct automorphisms of \(C\) have numerically inequivalent graphs.
Suppose \[ \Gamma_\sigma\equiv\Gamma_\tau. \] The surface automorphism \[ \id_C\times\tau^{-1}:C\times C\longrightarrow C\times C \] preserves intersection numbers and therefore numerical equivalence. It sends \(\Gamma_\tau\) to \(\Delta\) and \(\Gamma_\sigma\) to \(\Gamma_{\tau^{-1}\circ\sigma}\). Hence \[ \Gamma_{\tau^{-1}\circ\sigma}\equiv\Delta. \] Step 7 gives \(\tau^{-1}\circ\sigma=\id_C\), so \(\sigma=\tau\).
9. Every graph \(\Gamma_\sigma\) has\[ \Gamma_\sigma\cdot(l+m)=2, \]where\[ l=C\times\{P\}, \qquad m=\{P\}\times C. \]
The graph of an automorphism meets \(m\) once because fixing its first coordinate determines a unique point of the graph. It meets \(l\) once because \(\sigma^{-1}(P)\) is a single point and an automorphism has degree one. Thus \[ \Gamma_\sigma\cdot l = \Gamma_\sigma\cdot m =1. \] Exercise V.1.9, P-AGH519HODGEINDEX, proves that \(l+m\) is ample. The displayed equality follows.
10. The automorphism group of \(C\) is finite:\[ \boxed{|\Aut C|<\infty}. \]
By step 9, all graph divisors are effective and have fixed intersection degree two with the ample divisor \(l+m\). Part (a), proved in steps <1>1–<1>5, says that only finitely many numerical equivalence classes of such effective divisors exist. Step 8 says the map \[ \Aut C\longrightarrow\Num(C\times C), \qquad \sigma\longmapsto[\Gamma_\sigma] \] is injective. Hence \(\Aut C\) is finite, proving part (b).
11. Q.E.D.
Steps <1>1–<1>5 prove part (a), and steps <1>6–<1>10 prove part (b).