19.9.18 Curvature rank and section growth1936F96341B0

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. $\square$