PomeLemma 19.9.2.2B103C3B777A For every finite-dimensional vector subspace $V\subset H^0(X,L^ m)$,
\[ \# \{ \nu (s):0\neq s\in V\} =\dim V. \]
19.9.1 Flag valuations and degree bounds954DDD998705
Successive vanishing orders along a flag compute the lexicographically smallest exponent in a local Taylor expansion. The coefficient of that exponent is a linear functional on sections, and this shows that the number of distinct values on a space of sections equals its dimension. At each step, the effective divisor that remains gives, through a Gauduchon degree, a linear bound for the next order.
Until subsection 19.9.14, suppose that $X$ carries a complete flag, that is, a chain
\[ X=Y_0\supset Y_1\supset \cdots \supset Y_ n=\{ p\} , \]
where $p$ is a point and each $Y_ i$, $1\leq i\leq n$, is a smooth connected compact prime divisor in $Y_{i-1}$. For a nonnegative integer $m$ and a nonzero section $s\in H^0(X,L^ m)$, let $\nu _1(s)$ be its vanishing order along $Y_1$. Dividing $s$ by the corresponding power of the canonical section of $\mathcal O_ X(Y_1)$ and restricting to $Y_1$ gives a nonzero section of $L^ m(-\nu _1(s)Y_1)|_{Y_1}$. Repeating this procedure along the flag defines $\nu _2(s),\ldots ,\nu _ n(s)$, and the vector $\nu (s)=(\nu _1(s),\ldots ,\nu _ n(s))\in \mathbb N^ n$ is the flag valuation of $s$.
Near $p$, choose holomorphic coordinates in which $Y_ i=\{ z_1=\cdots =z_ i=0\} $, and a nonvanishing local frame of $L$. Then $\nu (s)$ is the lexicographically smallest exponent occurring in the Taylor expansion of the local coefficient of $s$. Consequently, with respect to the lexicographic order on $\mathbb N^ n$,
\[ \nu (st)=\nu (s)+\nu (t), \qquad \nu (s+t)\geq \min \{ \nu (s),\nu (t)\} , \]
where the first identity holds for nonzero sections $s,t$ of arbitrary degrees, and the second when $s,t$ have the same degree and $s+t\neq 0$. If $\nu (s)\neq \nu (t)$, equality holds in the second formula.
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 $a$. $\square$
The valuations of sections of $L^ m$ grow at most linearly in $m$. To see this, pair the residual divisor at each stage of the construction with a Gauduchon metric on the corresponding member of the flag.
Theorem 19.9.3.EA8340BC8D54 There are constants $C_1,\ldots ,C_ n\geq 0$, depending only on $L$ and the chosen flag, such that
\[ 0\leq \nu _ i(s)\leq C_ i m \quad (1\leq i\leq n) \]
for every $m\geq 1$ and every nonzero $s\in H^0(X,L^ m)$.
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
\[ a_ i=\int _{Y_ i}\gamma _ i^{d_ i-1}>0, \qquad A_ i=\deg _{\gamma _ i}(L|_{Y_{i-1}}). \]
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
\[ B_{ij}=\deg _{\gamma _ i} \left(\mathcal O_{Y_{j-1}}(Y_ j)|_{Y_{i-1}}\right). \]
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
\[ M_{i-1}=L^ m|_{Y_{i-1}}\otimes \bigotimes _{j<i}\mathcal O_{Y_{j-1}}(-\nu _ j(s)Y_ j)|_{Y_{i-1}}. \]
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
E150336DE571
19.9.4
\begin{equation} a_ i\nu _ i(s)\leq mA_ i-\sum _{j<i}B_{ij}\nu _ j(s). \end{equation}
Define recursively
\[ C_ i=\frac{|A_ i|+\sum _{j<i}|B_{ij}|C_ j}{a_ i}. \]
Induction on $i$ in (19.9.4) gives $\nu _ i(s)\leq C_ i m$. $\square$
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