Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 139 (Positivity in Algebraic Geometry)

For a projective scheme $X$ of dimension of a scheme $n$ and a Cartier divisor $D$, asymptotic Riemann–Roch givesHere Euler characteristic of a coherent sheaf means $\chi ={\sum }_{i=0}^n(-1{)}^i{h}^i$, and $D^n$ is the degree of the top intersection product with the fundamental cycle of $X$, including its component multiplicities. No ampleness assumption on $D$ is needed.

The Nakai–Moishezon criterion says that a Cartier divisor $D$ on a projective scheme is ample exactly whenfor every positive-dimensional integral closed subvariety $Y$. Kleiman's criterion says that the ample cone is the interior of the nef cone; equivalently, the numerical class of $D$ is ample exactly when it is strictly positive on every nonzero element of the closed cone of curves $\overline{\mathrm{NE}}(X)$. Positivity merely on individual curves is insufficient: the closure of the cone is essential. A nef divisor has nonnegative intersection number with every integral curve, and its restriction to every closed subscheme is nef.

Choose very ample divisors $H_2$ and $H_1\sim D+H_2$, with defining sections avoiding the associated points of $X$. Such a choice is possible after taking a sufficiently high ample twist. Put $R_m=mD-H_1=(m-1)D-H_2$. The two indicated short exact sequences of sheaves and their long exact sequences in sheaf cohomology implyWe abbreviate $h^i(Z,{\mathcal{O}}_Z(L))$ by $h^i(Z,L)$. Each $H_a$ has dimension $n-1$ and the restricted divisor is nef. For positive degrees below $n-1$, the assumed induction estimate bounds the error terms by $O(m^{n-2})$. The displayed hypothesis omits top-degree cohomology; use top cohomology boundedness for nef twists for that degree, and Grothendieck vanishing above it. These give the same error bound. The auxiliary top-degree result follows from Fujita vanishing by cutting a coherent sheaf with an ample divisor, as proved in cohomology growth for nef twists.

For $n\ge 2$, summing the inequality over successive $m$ gives $h^j(X,mD)=O(m^{n-1})$ for every $j>1$. For $n\le 1$, all such groups vanish by Grothendieck vanishing.

By asymptotic Riemann–Roch and the preceding bounds for higher sheaf cohomology,A nef divisor satisfies $D^n\ge 0$: for a fixed ample divisor $A$, the divisors $D+\epsilon A$ are ample for positive rational $\epsilon$, and continuity of the intersection product gives the inequality as $\epsilon \to 0$. If $h^0(X,mD)=0$ eventually, the left side is nonpositive. Dividing by $m^n$ forces $D^n=0$, since $h^1\ge 0$ and $D^n\ge 0$. The same identity then gives $h^1(X,mD)=O(m^{n-1})$.

First suppose $X$ is integral. Induct on its dimension of a scheme. The previous two parts handle all $j>1$ and also handle $j=1$ when $h^0(X,mD)$ eventually vanishes. Otherwise choose $r>0$ with a nonzero global section of $rD$. On an integral variety it defines an effective Cartier divisor $E$, possibly empty, and multiplication by the section givesThus $h^1(X,mD)\le h^1(X,(m-r)D)+h^1(E,mD)$. For $n\ge 2$, the induction hypothesis bounds the last term by $O(m^{n-2})$; summing on each residue class modulo $r$ gives $O(m^{n-1})$. For $n=1$, $E$ is zero-dimensional and its positive-degree sheaf cohomology vanishes, so the same recurrence is bounded.

For a general projective scheme, a nonzero section can be a zero divisor, so that argument requires an additional step. The general cohomology growth for nef twists supplies it: for every coherent sheaf $\mathcal{F}$ with support dimension $d$,Its proof uses Fujita vanishing, an ample section avoiding the associated points of $\mathcal{F}$, and induction on support dimension. Taking $\mathcal{F}={\mathcal{O}}_X$ gives the required $O(m^{n-1})$ estimate for every $j\ge 1$, including nonreduced and reducible schemes; degrees above $n$ vanish.

The divisor is understood to be on $X$; the printed “on $D$” is a typo. Combining asymptotic Riemann–Roch with cohomology growth for nef twists givesSince a nef divisor has nonnegative top self-intersection number, this limit is positive exactly when $D^n>0$. This is the volume of a nef divisor criterion.

Write $F={\sum }_C{\min }_{G\in |H|}{\mathrm{mult}}_C(G)C$. This is the fixed part of a linear system. The movable part of a linear system is $|M|$, with $M=H-F$; multiplication by the defining section of $F$ identifies its sections with those of $H$. In particular $|H|=F+|M|$, and no curve occurs in every member of $|M|$. The occurrence of $|F|$ in the definition request is understood as $|H|$.

For any integral curve $C$, choose a member $G\in |M|$ not containing $C$. Their intersection number is a sum of nonnegative local intersection multiplicities, so $M\cdot C=G\cdot C\ge 0$. Therefore the movable divisor $M$ is nef. If $M=0$, the same assertion is immediate.

The relevant invariant of a divisor is its Iitaka dimension, also called its Kodaira dimension. If two linearly independent global sections $s,t$ occur in $H^0(X,nH)$, their ratio is a nonconstant rational function. The sections $s^q,s^{q-1}t,\dots ,t^q$ are linearly independent, because a nonconstant rational function over an algebraically closed field is transcendental over that field. Thus some complete linear system of a divisor has positive-dimensional image, contradicting $\kappa (H)=0$.

Consequently $h^0(X,nH)=1$ for every $n>0$; the nonemptiness of $|H|$ supplies a nonzero section in every multiple. Its unique divisor is entirely fixed. Hence every movable part $M_n$ is zero.

Because $|D|$ has no fixed components, choose two members $G_1,G_2$ with no common component. This uses the infinitude of the algebraically closed field and avoidance of finitely many proper linear subspaces of the section space. On a smooth algebraic surface, their intersection number is the sum of local intersection multiplicities. Since $G_1\cdot G_2=D^2=0$, they are disjoint. A basepoint of $|D|$ would belong to both, so $|D|$ is basepoint-free.

The resulting morphism has nonconstant image since $\dim |D|\ge 1$. Its image cannot be a surface: a generically finite morphism defined by $D$ would make $D^2$ the positive product of its degree and the degree of its image. Thus the image is a curve. Apply Stein factorization to obtain $g:X\to C$ with connected fibers and $g_{\ast }{\mathcal{O}}_X={\mathcal{O}}_C$, where $C$ is a smooth projective curve. There is a positive-degree line bundle $L$ on $C$ with ${\mathcal{O}}_X(D)=g^{\ast }L$. The projection formula for sheaves and Riemann-Roch theorem give $h^0(X,mD)=h^0(C,L^m)=m\mathrm{deg}L+O(1)$. Therefore

Each $M_n$ is a nef divisor. If $M_n^2>0$, the volume of a nef divisor formula gives quadratic growth of $h^0(X,q{M}_n)$. Multiplication by the section of $q{F}_n$ injects this space into $H^0(X,qnH)$, so the birational linear system criterion for bigness makes the Iitaka dimension of $H$ two.

Conversely, if $\kappa (H)=2$, some $|nH|$, and hence its movable part $|M_n|$, has a two-dimensional image. If $M_n^2=0$, the previous part would make $|M_n|$ basepoint-free with image a curve; this is impossible. Since $M_n$ is nef, $M_n^2\ge 0$, hence $M_n^2>0$.

If all $M_n^2=0$ and one $|M_{n^{\prime }}|$ has positive dimension, the preceding part supplies a curve image. The just-proved equivalence excludes dimension two, so $\kappa (H)=1$. Conversely $\kappa (H)=1$ excludes every positive $M_n^2$ and requires a positive-dimensional linear system. Thus

A connected integral projective variety has only constant regular functions, so $h^0(X,{\mathcal{O}}_X)=1$. By Serre duality and $K_X\sim 0$, $h^2(X,{\mathcal{O}}_X)=h^0(X,{\mathcal{O}}_X(K_X))=1$. The assumed first sheaf cohomology vanishes. Therefore the Euler characteristic of a coherent sheaf isThese hypotheses characterize the K3 surface setting used below.

Choose an ample divisor $A$. The Hodge index theorem for algebraic surfaces implies $D\cdot A>0$: if it were zero, $D^2=0$ would force the numerical class of $D$ to be zero. Consequently $-D$ cannot be effective, because every nonzero effective divisor has positive intersection with $A$, while $-D\cdot A<0$.

The Riemann–Roch theorem for algebraic surfaces and the preceding computation give $\chi (X,{\mathcal{O}}_X(D))=2+D^2/2=2$. By Serre duality, $h^2(X,D)=h^0(X,-D)=0$. Thus

First derive the intersection constraints. Since $D$ is nef and $F,M$ are effective, $D\cdot F,D\cdot M\ge 0$; their sum is $D^2=0$, so both vanish. Since the movable divisor $M$ is nef, $M^2,M\cdot F\ge 0$, and $M\cdot D=0$ forces both to vanish. In particularFor every component $E$ of $F$, nefness and $D\cdot F=0$ imply $D\cdot E=0$. The isotropic orthogonality consequence of the Hodge index theorem gives $E^2\le 0$. If $E^2=0$, Riemann–Roch theorem for algebraic surfaces and Serre duality give $h^0(X,E)\ge 2$, since $-E$ cannot be effective. Choose $E^{\prime }\in |E|$ different from $E$. It cannot contain $E$: otherwise $E^{\prime }-E$ would be a nonzero effective numerically trivial divisor, contradicting its positive intersection with an ample divisor. Choose a member of $|M|$ avoiding $E$. Replacing one copy of $E$ in $F$ by $E^{\prime }$ now produces a member of $|D|$ with smaller multiplicity along $E$, contradicting the definition of the fixed part. Therefore

The strict statement as printed needs the qualification if $F\ne 0$. Part (a) already shows $F^2=0$ in the present situation, and the next part proves $F=0$; an unconditional $F^2<0$ would contradict that conclusion.

To prove the needed conditional statement, suppose $F\ne 0$. A fixed part has $h^0(X,F)=1$. Indeed, a different $F^{\prime }\in |F|$ has smaller multiplicity along some component of $F$; otherwise $F^{\prime }-F$ would be a nonzero effective linearly trivial divisor. Adding a member of the movable part avoiding that component contradicts fixedness.

The Hodge index theorem for algebraic surfaces, together with $D\cdot F=0$ and $D^2=0$, gives $F^2\le 0$. If equality held, Riemann–Roch theorem for algebraic surfaces would give $\chi (X,F)=2$, and Serre duality would give $h^2(X,F)=h^0(X,-F)=0$. Hence $h^0(X,F)=2+h^1(X,F)\ge 2$, a contradiction. ThereforeThis is the fixed-part elimination on a K3 surface argument.

The intersection calculation in part (vi)(a) gives $F^2=0$. Part (vi)(b), with its necessary $F\ne 0$ qualification, excludes every nonzero fixed part. Thus $F=0$ and $|D|$ is movable. Part (v) gives $\dim |D|\ge 1$, and $D^2=0$ by hypothesis. Applying part (ii) provesThus a nontrivial nef divisor of square zero on a K3 surface defines a fibration over a curve; the key step is fixed-part elimination on a K3 surface.

Choose a very ample divisor $A$ such that $A-D$ has a nonzero global section; this is possible by taking a sufficiently high ample twist. Since $X$ is integral, multiplication by its $m$th power injects $H^0(X,mD)$ into $H^0(X,mA)$. By Serre vanishing and the Hilbert polynomial, $h^0(X,mA)=O(m^n)$. ConsequentlyThis is the polynomial bound for sections of a fixed divisor. The same proof works for a coherent sheaf of support dimension $d$, by choosing the multiplying section to avoid its associated points; the bound is then $O(m^d)$.

For an effective Cartier divisor $F$, the divisor restriction exact sequence givesThe scheme $F$ has dimension at most $n-1$, so the permitted scheme version of part (i) bounds $h^0(F,{\mathcal{O}}_F(kD))$ by $C{k}^{n-1}$. For the infinitely many $k$ in the hypothesis,once $k$ is sufficiently large. Thus infinitely many such $k$ have a nonzero section of $kD-F$. This dimension-drop argument is the section subtraction lemma for big divisors.

If “effective divisor” is interpreted as an effective Weil divisor on a normal variety, use its coherent divisor ideal instead. The quotient by that ideal is supported in dimension at most $n-1$, so the polynomial bound for sections of a fixed divisor gives the same conclusion. For $F=0$ the assertion is immediate.

Fix any ample divisor $A$. Choose $q\gg 0$ so that both $qA$ and $(q-1)A$ have nonzero global sections, and choose $F\in |qA|$. Condition (1) and the section subtraction lemma for big divisors yield an integer $j>0$ with $jD-qA\sim E_0\ge 0$. Adding an effective member of $|(q-1)A|$ gives $jD\sim A+E$. Thus (1) implies the stronger condition (2').

If $jD\sim A+E$ with $A$ ample and $E$ effective, multiplication by the section of $qE$ injects $H^0(X,qA)$ into $H^0(X,qjD)$. The positive leading coefficient of the ample Hilbert polynomial supplies condition (1) along the infinite sequence $m=qj$. Condition (2) implies (3). Conversely, if $jD\equiv A+E$, put $P=jD-E$. The Cartier divisor $P$ is numerically equivalent to $A$, hence ample by Kleiman's criterion; the actual equality $jD=P+E$ gives (2).

We have proved (1)$\Rightarrow$(2')$\Rightarrow$(2)$\Rightarrow$(1), and (3)$\Rightarrow$(2)$\Rightarrow$(3). Also (2') implies (3'), while (3') implies (3). Hence all five conditions are equivalent. They characterize a big divisor; the ample-plus-effective formulation is Kodaira's lemma. The inconsistent use of $n$ and $j$ in the printed multiplier is resolved by using $j$ throughout.

Choose a very ample divisor $A$. By the stronger form of Kodaira's lemma, $jD\sim A+E$ for some effective divisor $E$. The subsystem $s_E{H}^0(X,A)\subseteq H^0(X,jD)$ defines the embedding given by $A$ on $X\setminus \mathrm{Supp}E$. Ratios of its sections generate the function field $k(X)$. The map from the complete complete linear system of a divisor contains these ratios, so it induces the same function field and is birational onto its image.

The converse is true; smoothness is unnecessary. If $|jD|$ gives a birational map, choose $n$ algebraically independent ratios $s_1/s_0,\dots ,s_n/s_0$ among a generating set of its section ratios. For every $q$, the sectionsare linearly independent by algebraic independence. Therefore $h^0(X,qjD)\ge \left(\genfrac{}{}{0ex}{}{q+n}{n}\right)$, giving condition (1) of part (iii). This is the birational linear system criterion for bigness.

Condition (a) is the definition of a big real divisor. Fix an ample Cartier divisor $A$. For each big Cartier divisor $D_i$, part (iii) gives $j_i{D}_i\sim A+E_i$ with $E_i$ effective. ConsequentlyThe first coefficient is positive, proving (b).

For the converse, suppose $D{\sim }_{\mathbb{R}}tA+E$, with $t>0$ and $E$ effective. Apply rational approximation of an ample-plus-effective real divisor: in a finite-dimensional space generated by Cartier divisors and the finitely many principal divisors occurring in this relation, express $D$ as a positive convex combination of rational divisors $B_{\nu }$, each satisfying $B_{\nu }{\sim }_{\mathbb{Q}}{t}_{\nu }A+E_{\nu }$ with $t_{\nu }>0$ and $E_{\nu }$ effective. After clearing denominators, part (iii) shows that each $B_{\nu }$ is a positive rational multiple of a big Cartier divisor. This gives the actual equality required by (a), rather than merely a numerical or linear equivalence.

The approximation lemma keeps the finitely many effectivity inequalities and linear-equivalence equations simultaneously; see its proof for the rational-face argument and the reduction of nonnormal varieties by finite normalization. Therefore (a) and (b) are equivalent.

Condition (b) immediately implies (c): a positive real multiple $tA$ of an ample divisor is an ample real divisor, and real linear equivalence of divisors implies numerical equivalence of divisors.

More explicitly, the real numerical class of $D$ is the sum of a class in the ample cone and an effective real divisor class. This separates strict positivity from the possibly degenerate effective part; it does not claim that the effective part itself is ample.

Suppose $D\equiv A^{\prime }+E^{\prime }$ as in (c), and set $P=D-E^{\prime }$. Its numerical class is ample, so $P$ is an ample real divisor. Fix an ample Cartier divisor $A$. The ample cone is open by Kleiman's criterion, so choose $\epsilon >0$ with $P-\epsilon A$ still ample. Express this ample real divisor as a positive real combination $\sum b_i{B}_i$ of ample Cartier divisors. Some positive multiple of each $B_i$ has an effective representative; dividing by that multiple gives $B_i{\sim }_{\mathbb{Q}}{G}_i$ with $G_i$ effective. Hencewhich is (b). The equivalence of numerical ampleness with a positive combination of ample Cartier representatives follows by rational approximation inside the open ample cone; the rational approximation of an ample-plus-effective real divisor proof also accounts for principal-divisor directions.

Finally, if $D^{\prime }\equiv D$, the same expression $D^{\prime }\equiv A^{\prime }+E^{\prime }$ satisfies (c). By the equivalences just proved, $D^{\prime }$ satisfies all three conditions exactly when $D$ does. Thus bigness is invariant under numerical equivalence, for real as well as Cartier divisors.

The space of real numerical divisor classes is $N^1(X{)}_{\mathbb{R}}$. Write any big class as $[D]=[A]+[E]$, with $A$ an ample real divisor and $E$ effective. For a sufficiently small perturbation $\delta \in N^1(X{)}_{\mathbb{R}}$, the class $[A]+\delta$ remains in the open ample cone, so $[D]+\delta$ remains big by part (v). Thus $\mathrm{Big}(X)$ is open. Positive scaling and addition preserve the ample-plus-effective expression, so it is a convex cone.

For $n\ge 1$, fix a very ample divisor $H$. The linear functional $\ell ([D])=D\cdot H^{n-1}$ is strictly positive on every big class: the ample part contributes positively and the effective part nonnegatively. No nonzero linear subspace can be contained in this cone, since it would contain both $v$ and $-v$. In particular the zero class is not big in positive dimension. When $n=0$, $N^1(X{)}_{\mathbb{R}}=0$, so there is no positive-dimensional subspace to consider. Therefore