Proposition 19.9.19.C8B2CB1B6819 For a nonzero subspace $W\subset H^0(X,L^ p)$, $p\geq 1$, let $B_ W$ be its common zero set and let $\Phi _ W:X\setminus B_ W\to \mathbb P(W^*)$ be its section map. If $d$ is the maximal rank of its differential, there is $x_*\notin B_ W$ such that
Proposition 19.9.19C8B2CB1B6819
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. $\square$
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