PomeProposition 18.1.48.A51A6DE92117 Under the assumptions of theorem 18.1.46, assume furthermore that $\varphi $ has analytic singularities such that $\theta _{\varphi }$ is a Kähler current, then (18.1.47) holds.
Proposition 18.1.48A51A6DE92117
Proof. Step 1. Reduce to the case of log singularities.
Let $\pi \colon Y\rightarrow X$ be a log resolution of $\varphi $. In this case, for each $k\in \mathbb {Z}_{>0}$, we have
\[ h^0\Bigl (X,T\otimes L^ k\otimes \mathcal{I}(k\varphi )\Bigr )=h^0\Bigl (Y,K_{Y/X}\otimes \pi ^*T\otimes \pi ^*L^ k\otimes \mathcal{I}(k\pi ^*\varphi )\Bigr ). \]
By proposition 16.6.10, we have
\[ \operatorname{vol}(\mathrm{dd}^{\mathrm{c}}h)=\operatorname{vol}(\mathrm{dd}^{\mathrm{c}}\pi ^*h). \]
Therefore, it suffices to argue (18.1.47) with $K_{Y/X}\otimes \pi ^*T$, $\pi ^*L$ and $\pi ^*h$ in place of $T$, $L$ and $h$.
Note that $\theta _{\varphi }$ is no longer a Kähler current, but it still has positive mass.
Step 2. Assume that $\varphi $ has log singularities along an effective $\mathbb {Q}$-divisor $D$. We may assume that $D$ is simple normal crossing by replacing $X$ by a modification and $D$ by the pull-back. We decompose $D$ into irreducible components, say
\[ D=\sum _{i=1}^ N a_ iD_ i. \]
In this case, we can easily compute
\[ \mathcal{I}(k\varphi )=\mathcal{O}_ X\left(-\sum _{i=1}^ N\lfloor k a_ i \rfloor D_ i\right) \]
for each $k\in \mathbb {Z}_{>0}$. Observe that $L-D$ is nef (see 5.18.5). Choose an integer $q>0$ so that $qD$ is integral. For each residue class $s=0,\ldots ,q-1$, we have
\[ \left\lfloor (mq+s)D\right\rfloor =mqD+\lfloor sD\rfloor . \]
Hence, on the residue class $k=mq+s$, the line bundle
\[ T\otimes L^ k\otimes \mathcal{O}_ X\left(-\sum _{i=1}^ N\lfloor k a_ i \rfloor D_ i\right) \]
is a fixed twist of $\left(q(L-D)\right)^ m$. Applying the asymptotic Riemann–Roch theorem [Corollary 1.4.41, PAG1] to the nef line bundle $q(L-D)$ and then letting $m\to \infty $ in each residue class, we conclude that
\[ \lim _{k\to \infty }\frac{n!}{k^ n}h^0\left(X,T\otimes L^ k\otimes \mathcal{O}_ X\left(-\sum _{i=1}^ N\lfloor k a_ i \rfloor D_ i\right)\right)= (L-D)^ n. \]
Observe that by proposition 5.20.4,
\[ \theta _{\varphi }=[D]+S, \]
where $S$ is a closed positive $(1,1)$-current with bounded potential. Therefore,
\[ (L-D)^ n=\int _ X S^ n=\int _ X \theta _{\varphi }^ n. \]
By example 18.1.10, we know that the right-hand side is exactly $\operatorname{vol}\theta _{\varphi }$. $\square$
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