19.9 Flags, convex bodies and section growth926A00196D5E

Proof. Sections with distinct valuations are linearly independent: in a nontrivial finite linear combination of them, the smallest valuation occurs in exactly one summand and cannot cancel. Hence there are at most $\dim V$ values.

For the reverse inequality, we argue by induction on $\dim V$, the case $\dim V=0$ being immediate. The nonempty set of values has a smallest element $a$ in the lexicographic order. The Taylor coefficient of the monomial $z^ a$ is a nonzero linear functional on $V$. Its kernel has codimension one, and every nonzero element of the kernel has valuation strictly larger than $a$. By the induction hypothesis the kernel contributes $\dim V-1$ distinct values, all different from $a$. $\square$

Proof. For each $i$, equip $Y_{i-1}$ with a Gauduchon metric $\gamma _ i$; when $Y_{i-1}$ is a curve, any Hermitian metric will do. Put $d_ i=\dim Y_{i-1}=n-i+1$ and

For $i=n$ the integral defining $a_ i$ is the integral of $1$ over the point $p$, so $a_ n=1$.

For $j<i$, also put

All these numbers depend only on $L$, the flag and the chosen metrics.

At the $i$th stage of the construction of $\nu (s)$, the residual section is a section of the line bundle

Its zero divisor contains $\nu _ i(s)Y_ i$. The Poincaré–Lelong formula, tested against $\gamma _ i^{d_ i-1}$ as in the proof of theorem 19.8.5, therefore gives

Define recursively

Induction on $i$ in (19.9.4) gives $\nu _ i(s)\leq C_ i m$. $\square$

Proof. The empty case is immediate. Otherwise, let $S_ m=\{ \nu (s):0\neq s\in W_ m\} $. By lemma 19.9.2, $S_ m$ has $\dim W_ m$ elements, and $m^{-1}S_ m\subseteq \Delta (W_\bullet )$. Attach to each $a\in S_ m$ the half-open cube $m^{-1}(a+[0,1)^ n)$. These cubes are disjoint, their total volume is $m^{-n}\dim W_ m$, and their union is contained in

These compact neighborhoods decrease to $\Delta (W_\bullet )$ as $m\to \infty $, so their volumes converge to its volume. This proves the inequality. A compact convex set has positive $n$-dimensional volume exactly when it has nonempty interior, which gives the last statement. $\square$

Proof. The semigroup $S$ satisfies the cone conditions established above and, by hypothesis, generates $\mathbb Z^{n+1}$. The semigroup counting theorem therefore applies to it. By lemma 19.9.2, the number of points of $S$ at height $m$ is $\dim W_ m$, and the height-one slice of the closed convex cone generated by $S$ is $\Delta (W_\bullet )$. $\square$

Proof. Choose $g_0=(b,e)\in G$ and an integral basis $\lambda _1,\ldots ,\lambda _ d$ of $\Lambda $. Every element of $G$ has a unique expression

These coordinates identify $G$ with $\mathbb Z^{d+1}$. Let $S'$ be the image of $S$ under this identification. It generates $\mathbb Z^{d+1}$, and the height of the image of $(x,m)\in S$ is $q=m/e\geq 0$.

The points $(m^{-1}x,1)$, for $(x,m)\in S$ with $m>0$, span the $(d+1)$-dimensional real vector space $\operatorname {span}_{\mathbb R}G$. Since they all have height one, their affine span has dimension $d$. Taking the convex hull and the closure does not change this affine span, which proves the dimension assertion.

Intersect the cone containing $S$ with $\operatorname {span}_{\mathbb R}G$. The linear change of coordinates carries this intersection to a closed cone with compact height-one slice whose only point at height zero is the origin. Indeed, a vector $(x,m)$ with $|x|\leq Cm$ has horizontal coordinates bounded by a constant times $m$, hence by a constant times $q$. The height-one body of $S'$ is

where the right-hand side is expressed in the basis $\lambda _1,\ldots ,\lambda _ d$. It is full-dimensional in $\mathbb R^ d$, and therefore

Apply the semigroup counting theorem of subsection 19.9.5 to $S'$, which generates its full lattice. The points of $S'$ at height $q$ correspond bijectively to the valuations of the nonzero elements of $W_{eq}$, and by lemma 19.9.2 there are $\dim W_{eq}$ of them. Hence

Dividing by $e^ d$ gives the asserted limit, and its positivity gives eventual nonvanishing. For $d=0$ the body is a point of volume one, so the integer dimensions converge to one and are eventually equal to one. Finally, every degree with a nonzero section lies in the projection $e\mathbb Z$ of $G$, which proves the vanishing outside these multiples. $\square$

Proof. Suppose first that $d>0$. Choose sections $s_0,\ldots ,s_ d$ of positive degrees $m_0,\ldots ,m_ d$ whose normalized valuations $\nu (s_ i)/m_ i$ are affinely independent. Let $M$ be a common multiple of their degrees and put $t_ i=s_ i^{M/m_ i}\in W_ M$. The ratios $f_ i=t_ i/t_0$ belong to $K(W_\bullet )$. The flag valuation extends to quotients by subtraction, giving

These vectors are linearly independent. Distinct monomials in the $f_ i$ therefore have distinct valuations, so the term of smallest valuation in a nonzero polynomial expression cannot cancel, and the ratios are algebraically independent. Thus the transcendence degree is at least $d$; this is trivial when $d=0$.

Conversely, let $f_1,\ldots ,f_ r$, with $r\geq 1$, be algebraically independent elements of the field. Using the product of their denominators, write $f_ i=A_ i/B$ with $A_ i,B\in W_ M$ and $B\neq 0$ for a common positive degree $M$. For every integer $q\geq 1$ and every multi-index $a=(a_1,\ldots ,a_ r)\in \mathbb N^ r$ with $|a|=a_1+\cdots +a_ r\leq q$, the sections

are linearly independent, because division by $B^ q$ turns a linear relation among them into a polynomial relation among the $f_ i$. There are $\binom {q+r}{r}$ such sections. Since $W_ M\neq 0$, the degree $M$ is a multiple of $e$, and theorem 19.9.9 gives $\dim W_{Mq}=O(q^ d)$. Comparing with the binomial lower bound forces $r\leq d$, which proves equality of the transcendence degree with $d$. A body with nonempty interior has $d=n$, and this gives the stated consequence for $X$.

The field and the set of degrees with nonzero sections do not depend on the flag, so neither $d$ nor $e$ does. By theorem 19.9.9, the lattice volume is, for either flag, the limit of the same dimensions with the same normalization. It is therefore independent of the flag as well, although the bodies and their lattices may differ. $\square$

Proof. Fix a Hermitian form $\omega $ on $X$ and a Kähler form $\kappa $ on $Z$, and choose $C>0$ with $\pi ^*\omega \leq C\kappa $. By proposition 13.14.15, it suffices to bound the masses of smooth functions $u\in \mathrm{PSH}(X,\omega )$. Their pullbacks satisfy

Taking top powers and integrating gives

The first equality follows by change of variables on the set where $\pi $ is biholomorphic, since smooth top-degree forms give the exceptional analytic sets zero mass. The inequality follows by expanding in the smooth semipositive difference $C\kappa -\pi ^*\omega $, and the last equality follows from Stokes' theorem because $\kappa $ is closed. The bound does not depend on $u$, which proves the bounded mass property. $\square$

Proof. If $\Delta (W_\bullet )$ had nonempty interior, $X$ would have $n$ algebraically independent meromorphic functions by proposition 19.9.10. Moishezon's theorem would then give a modification of $X$ from a smooth projective, hence Kähler, manifold, contradicting lemma 19.9.12. The body therefore has empty interior, and proposition 19.9.6 gives both conclusions. $\square$

Proof. If $n=1$, take $\pi $ to be the identity and any point as the last member of the flag; the remaining assertions are then immediate. If $n\geq 2$, blow up a point $p\in X$. The local description of the blowup of $\mathbb C^ n$ at the origin shows that $\widetilde X$ is smooth and that its exceptional divisor is $E\simeq \mathbb P^{n-1}$. A nested sequence of linear subspaces of $E$, together with $\widetilde X$, gives a flag

Each member is smooth, compact and connected, and is a prime divisor in the preceding member. The modification is an isomorphism off $E$.

Let $s$ be a section of $\pi ^*L^ m$. Off $E$, it gives a section of $L^ m$ on $X\setminus \{ p\} $. In a trivialization of $L^ m$ on a coordinate ball centered at $p$, its coefficient is holomorphic on the punctured ball, and by Hartogs' extension theorem, valid since $n\geq 2$, it extends holomorphically across $p$. The extension glues with the section away from $p$ to a section of $L^ m$ on $X$. Its pullback agrees with $s$ off $E$, hence everywhere by the identity theorem. This proves surjectivity; injectivity and compatibility with products follow because pullback is an isomorphism on the dense open set $\widetilde X\setminus E$. $\square$

Proof. Pullback preserves the graded dimensions and the degrees with nonzero sections. Applying theorem 19.9.3, theorem 19.9.9 on $\widetilde X$ gives the displayed limit and the positivity of the lattice volume. Pullback also identifies the section-ratio fields, since their elements are the corresponding ratios and pullback is injective on the dense set where $\pi $ is an isomorphism. Thus proposition 19.9.10 identifies $d$ with the transcendence degree of the original field. The unchanged degrees with nonzero sections determine $e$, and the limit of the unchanged dimensions determines $V(W_\bullet )$. None of the three depends on the auxiliary choices. $\square$

Proof. If all positive-degree section spaces vanish, there is nothing to prove. Otherwise, apply the theorem to the full section ring. Since $W_ m=0$ for $e\nmid m$, the limit gives $h^0(X,L^ m)=O(m^ d)$ for all degrees, with $d\leq n$. If $d=n$, the section-ratio field contains $n$ algebraically independent meromorphic functions on $X$, so $X$ is Moishezon. Thus $d<n$ when $X$ is not Moishezon. Finally, by Moishezon's theorem a Moishezon manifold admits a modification from a projective, hence Kähler, manifold, so it has the bounded mass property by lemma 19.9.12. The contrapositive gives the last assertion. $\square$

Proof. Choose a basis $s_0,\ldots ,s_ N$ of $W$. Principalizing the ideal of their local coefficients gives a smooth compact modification $\pi :Y\to X$, isomorphic over $X\setminus B_ W$, and an effective divisor $D$ with

where the $v_ j$ have no common zero. The resolved map $f=[v_0:\cdots :v_ N]:Y\to \mathbb P^ N$ has maximal differential rank $d$, since it agrees with $\Phi _ W\circ \pi $ on a dense open set.

The critical set $C=\{ y:\operatorname {rank}\mathrm df_ y<d\} $ is analytic. Proper mapping and the image-dimension theorem imply that $Z=f(Y)$ and $T=f(C)$ are compact analytic sets, with $\dim Z=d$ and $\dim T\leq d-1$ when $T$ is nonempty. For the last inequality apply the rank bound to the smooth locus of every irreducible component of $C$. By Chow's theorem $T$ is algebraic. Choose a homogeneous polynomial $P$ of degree $a\geq 1$ that vanishes on $T$ and does not vanish identically on $Z$. If $T$ is empty, take $P=1$ and $a=0$, which also covers $d=0$.

Then $\sigma =P(s_0,\ldots ,s_ N)$ is a nonzero section of $L^{pa}$. Indeed, the set where $P(v_0,\ldots ,v_ N)\ne 0$ meets the dense locus where $\pi $ is an isomorphism. The smooth function

has a positive maximum $x_*$. At that point $\sigma \ne 0$ and some $s_ j\ne 0$, so $x_*\notin B_ W$ and its lift is outside $C$. The tuples $(\sigma s_ j)$ and $(s_ j)$ define the same projective map near $x_*$. For $\omega _{\mathrm{FS}}=\mathrm{dd}^{\mathrm{c}}\log (1+|w|^2)$ in an affine chart, the Hessian test gives

The first form on the right is semipositive of rank $d$. Thus $\beta _{x_*}$ is semipositive and its kernel is contained in the kernel of that form, proving the assertion. The polynomial degree may depend on $p$ and $W$; no bound on it is needed. $\square$

Proof. The proposition applied to a single section excludes every positive power if $S_ h$ is empty. If all positive-degree section spaces vanish, the remaining section bounds are immediate and $\kappa (L)=-\infty $. Otherwise, suppose a graded algebra has a nonzero positive-degree piece. Write any finitely many algebraically independent elements of its ratio field with a common denominator in one positive degree, as in proposition 19.9.10. They are coordinate ratios of a section map. Its resolved image is projective algebraic by proper mapping and Chow's theorem; its dimension is therefore at least their number. The proposition bounds this dimension by $r_ h^+$, so the field's transcendence degree is at most $r_ h^+$.

Apply theorem 19.9.16 to this algebra and to the full section ring. The degree-period formula gives the growth bound in every degree, including the zero spaces outside the period. The earlier flag identification gives the body dimensions. Finally, $\mathcal H_ p(u)\subseteq H^0(X,L^ p)$, so the same upper bound holds even though the integrable spaces need not form a graded algebra. $\square$

Proof. Induct on the dimension, simultaneously for all compact complex manifolds and all integer upper bounds on curvature rank. Dimension zero is immediate. If all these section spaces vanish, there is nothing to prove. Otherwise fix a nonzero section in degree $a\geq 1$ and resolve its zero divisor on a smooth compact modification $\mu :Y\to X$. With $M=\mu ^*L$, the pulled-back twist is $M^{-a}(D)$ for a fixed effective divisor $D$ whose components are smooth. Pullback is injective on sections, and the pulled-back curvature has rank at most $r$.

Write $D=H_1+\cdots +H_ s$, allowing repeated components, and put $D_ j=H_1+\cdots +H_ j$. For $m=p-a>0$, the divisor sequences give

The first term is $O(m^ r)$ by the preceding corollary; every other term is $O(m^ r)$ by induction, since curvature rank cannot increase on restriction. The number of terms and their twists are fixed. Absorb the finitely many $p\leq a$ in the constant. On disconnected manifolds apply this argument componentwise; this also makes the induction valid for disconnected divisor components. $\square$

Proof. For $E$ of positive rank, let $\pi :Y=\mathbb P(E^*)\to X$ parametrize lines, and let $\mathcal O_ Y(1)$ be dual tautological. Fiberwise linear forms give $\pi _*\mathcal O_ Y(1)=E$, hence

The pulled-back curvature has rank at most $r$, so the fixed-line-twist lemma, proved in all dimensions, gives the first bound. Serre duality identifies the dual of the top cohomology with

The dual curvature is $-\beta $ and has the same rank; apply the degree-zero bound to $L^*$. A bundle of rank zero has zero cohomology and satisfies both bounds. $\square$

Proof. The analytic Hirzebruch–Riemann–Roch theorem gives

Here the integral selects real degree $2n$, and the Chern character and Todd forms are computed with fixed Hermitian metrics. The curvature normalization represents $c_1(L)$, and $\beta ^{r+1}=0$ pointwise by its rank bound. Every coefficient is independent of $p$. $\square$

Proof. The endpoints have already been bounded. Since $\chi =h^0-h^1+h^2$, the middle dimension is at most $h^0+h^2+|\chi |=O(p^ r)$ by the Euler polynomial. $\square$

Proof. A section, expressed in a line-bundle frame on the projective factor, is a holomorphic function of $Y$ and hence is constant there. Restriction to any slice $\mathbb P^ r\times \{ y_0\} $ is therefore inverse to pullback on sections. Counting degree-$p$ homogeneous monomials in $r+1$ variables gives the formula; for $r=0$ it equals one. $\square$

Proof. By proposition 19.8.13, sections of $L_ k^ m$ are homogeneous polynomials of degree $km$. Dividing successively by the highest powers of $z_1,\ldots ,z_{n-1}$ that divide the section, and restricting at each step, leaves a nonzero multiple of $z_ n^{km-a_1-\cdots -a_{n-1}}$, where $a_ i$ is the $i$th valuation. This residual section has no zeros on $Y_{n-1}$, so the last valuation is zero, and the preceding ones are nonnegative with sum at most $km$.

Conversely, every such tuple is the valuation of the monomial

The normalized values lie in the displayed simplex and, for $m=1$, include all its vertices, so their closed convex hull is the simplex. For $k>0$, its coordinate-lattice volume is $k^{n-1}/(n-1)!$, the leading coefficient of the binomial count in proposition 19.8.13. For $k=0$, only the constant polynomials remain. $\square$