PomeE9A3FA804369The valuations of the nonzero homogeneous elements of a graded algebra of sections form a semigroup whose part in degree $m$ has exactly as many elements as the dimension of the degree-$m$ piece. The linear bound of theorem 19.9.3 places this semigroup in a cone with compact height-one slice, and a semigroup counting theorem relates the asymptotic count to the volume of this slice.
Let $W_\bullet =\bigoplus _{m\geq 0}W_ m$ be a graded subalgebra of the section ring $\bigoplus _{m\geq 0}H^0(X,L^ m)$, with $W_0=\mathbb C$. For a weight $u\in \mathrm{QPSH}(X)$, the bounded weighted section algebra $\mathcal W_\bullet (u)$ of subsection 19.8.8 is an example.
Set
This is an additive semigroup, because the product of two nonzero sections on the connected manifold $X$ is nonzero and valuations add. Choose constants $C_ i$ as in theorem 19.9.3. Then $S$ lies in the closed cone
whose only point at height zero is the origin. Define
where $\overline{\operatorname {conv}}$ denotes the closed convex hull. If some positive-degree piece of $W_\bullet $ is nonzero, this is a compact convex subset of $\mathbb R^ n$; equivalently, it is the height-one slice of the closed convex cone generated by $S$. If every positive-degree piece is zero, $\Delta (W_\bullet )=\varnothing $. Write $\operatorname {vol}_{\mathbb R^ n}$ for Euclidean volume.
Proposition 19.9.6.E6B4502ECB3B For every such algebra,
In particular, if $\Delta (W_\bullet )$ has empty interior, then $\dim W_ m=o(m^ n)$.
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. $\square$
We use the following semigroup counting theorem, recalled in Theorem 19.7.20 (Kaveh–Khovanskii). Let $r\geq 0$, and let $\Sigma \subset \mathbb Z^{r+1}$ be an additive semigroup that generates $\mathbb Z^{r+1}$ as a group and lies in a closed convex cone in $\mathbb R^ r\times \mathbb R_{\geq 0}$ whose height-one slice is compact and whose only point at height zero is the origin. Then the number of points of $\Sigma $ at height $m$, divided by $m^ r$, converges to the Euclidean volume of the height-one slice of the closed convex cone generated by $\Sigma $.
Corollary 19.9.7.071721B08FC2 If the subgroup generated by $S$ is all of $\mathbb Z^{n+1}$, then
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 $\Delta (W_\bullet )$. $\square$
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