Let $T$ be a holomorphic line bundle on $X$. The goal of this section is to prove the following:
. Note that it is defined when $k$ is large enough.
Although we are only interested in the untwisted case, the proof given below requires the twisted case.
We first observe that the sequence $\Delta _{k,T}(\theta ,\varphi )$ is uniformly bounded: This follows easily from proposition 19.4.12. So Blaschke's selection theorem theorem 19.7.4 is applicable. We will apply this observation without further comments.
Proof.
Up to replacing $X$ by a modification and twisting $T$ accordingly, we may assume that $\varphi $ has log singularities along an effective $\mathbb {Q}$-divisor $D$, see proposition 19.4.32 and theorem 5.18.8.
Take $\epsilon \in \mathbb {Q}\cap (0,1)$. In this case, for large enough $k\in \mathbb {Z}_{>0}$ we have
\[ \begin{split} \mathrm{H}^0\left(X,T\otimes L^ k\otimes \mathcal{I}_{\infty }(k\varphi )\right) \subseteq & \mathrm{H}^0\left(X,T\otimes L^ k\otimes \mathcal{I}(k\varphi )\right) \\ \subseteq & \mathrm{H}^0\left(X,T\otimes L^ k\otimes \mathcal{I}_{\infty }(k(1-\epsilon )\varphi )\right). \end{split} \]
Take an integer $N\in \mathbb {Z}_{>0}$ so that $ND$ is a divisor and $N\epsilon $ is an integer.
Let $\Delta '$ be the limit of a subsequence of $(\Delta _{k,T}(\theta ,\varphi ))_ k$, say the sequence defined by the indices $k_1,k_2,\ldots $. Thanks to theorem 19.7.4, it suffices to show that $\Delta '=\Delta _{\nu }(\theta ,\varphi )$.
There exists $t\in \{ 0,1,\ldots ,N-1\} $ such that $k_ i\equiv t$ modulo $N$ for infinitely many $i$, up to replacing $(k_ i)_ i$ by a subsequence, we may assume that $k_ i\equiv t$ modulo $N$ for all $i$. Write $k_ i=N g_ i+t$. Then for large enough $i$, we have
\[ \begin{split} \mathrm{H}^0\left(X,T\otimes L^{-N+t} \otimes L^{N(g_ i+1)}\otimes \mathcal{I}_{\infty }\left(N(g_ i+1)\varphi \right)\right) \subseteq \mathrm{H}^0\left(X,T\otimes L^{k_ i}\otimes \mathcal{I}(k_ i\varphi )\right) \\ \subseteq \mathrm{H}^0\left(X,T\otimes L^ t \otimes L^{Ng_ i}\otimes \mathcal{I}_{\infty }\left(g_ i N(1-\epsilon )\varphi \right)\right). \end{split} \]
So
\[ \begin{split} (g_ i+1) \Delta _{g_ i+1,T\otimes L^{-N+t}}(NL-ND)+N(g_ i+1)\nu (D)\subseteq (Ng_ i+t)\Delta _{k_ i,T}(\theta ,\varphi )\\ \subseteq g_ i\Delta _{g_ i,T\otimes L^ t}\left(NL-N(1-\epsilon )D\right)+Ng_ i(1-\epsilon )\nu (D). \end{split} \]
Letting $i\to \infty $, by proposition 19.4.12,
\[ \Delta _{\nu }(L-D)+\nu (D)\subseteq \Delta '\subseteq \Delta _{\nu }(L-(1-\epsilon )D)+(1-\epsilon )\nu (D). \]
Letting $\epsilon \to 0+$, using the continuity of classical Okounkov bodies in the big cone recalled in theorem 19.5.4, we find that
\[ \Delta '=\Delta _{\nu }(L-D)+\nu (D)=\Delta _{\nu }(\theta ,\varphi ), \]
where we applied 19.4.23 as well. Our assertion follows.
$\square$
Here and in the sequel, $\Delta _{\nu }((1-\beta )\theta ,\varphi )=\Delta _{\nu }((1-\beta )\theta +\mathrm{dd}^{\mathrm{c}}\varphi )$ is defined in (
).
Proof of theorem 19.4.41.
Fix a very ample line bundle $H$ on $X$ and a Kähler form $\omega \in c_1(H)$ such that $\omega \geq \theta $.
Step 1. We first handle the case where $\theta _{\varphi }$ is a Kähler current, say $\theta _{\varphi }\geq 2\delta \omega $ for some $\delta \in (0,1)$. Take a quasi-equisingular approximation $(\varphi _ j)_ j$ of $\varphi $ in $\mathrm{PSH}(X,\theta )$. We may assume that $\theta _{\varphi _ j}\geq \delta \omega $ for all $j\geq 1$. After adding constants, we also assume that $\varphi \leq 0$ and $\varphi _ j\leq 0$ for all $j$.
Let $\Delta '$ be a limit of a subsequence of $(\Delta _{k,T}(\theta ,\varphi ))_ k$. Let us say the indices of the subsequence are $k_1<k_2<\cdots $. By theorem 19.7.4, it suffices to show that $\Delta '=\Delta _{\nu }(\theta ,\varphi )$.
Observe that for each $j\geq 1$, we have $\Delta '\subseteq \Delta _{\nu }(\theta ,\varphi _ j)$ by 19.4.43. Letting $j\to \infty $, we find $\Delta '\subseteq \Delta _{\nu }(\theta ,\varphi )$ as a consequence of theorem 19.4.30. Therefore, it suffices to prove that
\[ \operatorname{vol}\Delta '\geq \operatorname{vol}\Delta _{\nu }(\theta ,\varphi ). \]
Fix an integer $N>\delta ^{-1}$. Observe that for any $j\geq 1$, we have $\varphi _ j\in \mathrm{PSH}(X,(1-N^{-1})\theta )$. Similarly, $\varphi \in \mathrm{PSH}(X,(1-N^{-1})\theta )$. By 19.4.44, it suffices to argue that
6A18B89932A8
19.4.45
\begin{equation} \operatorname{vol}\Delta '\geq \operatorname{vol}\Delta _{\nu }\left((1-N^{-1})\theta ,\varphi \right). \end{equation}
Step 1.1. We first reduce to the case where $N|k_ i$ for all $i$.
We are free to replace $(k_ i)_ i$ by a subsequence, so we may assume that $k_ i\equiv a$ modulo $N$ for all $i\geq 1$, where $a\in \{ 0,1,\ldots ,N-1\} $. We write $k_ i=g_ i N+a$. Observe that for each $i\geq 1$,
\[ \mathrm{H}^0\left(X,T\otimes L^{k_ i}\otimes \mathcal{I}(k_ i\varphi )\right)\supseteq \mathrm{H}^0\left(X,T\otimes L^{-N+a}\otimes L^{g_ i N+ N}\otimes \mathcal{I}\left((g_ iN+N)\varphi \right)\right). \]
Up to replacing $T$ by $T\otimes L^{-N+a}$, we may therefore assume that $a=0$, so that $k_ i=g_ i N$.
Step 1.2. Write $k_ i=g_ i N$ for all $i$. We prove (19.4.45).
Since
\[ \int _ X\theta _{\varphi _ j}^ n\to \int _ X\theta _{P_\theta [\varphi ]_{\mathcal I}}^ n, \]
we can apply 13.13.30 to find $j'\in \mathbb {Z}_{>0}$ such that for all $j\geq j'$, there is $\psi _ j\in \mathrm{PSH}(X,\theta )_{>0}$ satisfying
8D5878FDF4FD
19.4.46
\begin{equation} P_{\theta }[\varphi ]_{\mathcal{I}}\geq \left(1-N^{-1}\right)\varphi _ j+N^{-1} \psi _ j. \end{equation}
Fix $j\geq j'$. It suffices to show that
DA123199D17C
19.4.47
\begin{equation} \Delta _{\nu }\left((1-N^{-1})\theta ,\varphi _{j}\right)+v'\subseteq \Delta ' \end{equation}
for some $v'\in \mathbb {R}^ n$. In fact, if this is true, we have
\[ \operatorname{vol}\Delta '\geq \operatorname{vol}\Delta _{\nu }\left((1-N^{-1})\theta ,\varphi _ j\right). \]
Letting $j\to \infty $ and applying corollary 16.9.36 and theorem 19.4.30, we conclude (19.4.45).
It remains to prove (19.4.47).
By 13.13.48, there is $g'>0$ such that for any $g\geq g'$, we can find a non-zero section
\[ s_ g\in \mathrm{H}^0\left(X,L^ g\otimes \mathcal I(g\psi _ j)\right). \]
By 5.18.15, after possibly increasing $g'$, for all $g\geq g'$,
\[ \mathcal I\left(gN\varphi _ j\right)\subseteq \mathcal I_\infty \left(g(N-1)\varphi _ j\right). \]
Moreover, the inequality (19.4.46) gives that for $g\geq g'$,
\[ \begin{aligned} \mathcal I(gN\varphi _ j)\cdot \mathcal I(g\psi _ j)\subseteq & \mathcal I_\infty \left(g(N-1)\varphi _ j\right)\cdot \mathcal I(g\psi _ j)\\ \subseteq & \mathcal I\Bigl (g(N-1)\varphi _ j+g\psi _ j\Bigr )\\ \subseteq & \mathcal I\Bigl (gN P_\theta [\varphi ]_{\mathcal I}\Bigr )\\ =& \mathcal I(gN\varphi ), \end{aligned} \]
where the second inclusion follows directly from the definition of $\mathcal I_\infty $. Thus multiplication by $s_ g$ gives an injective linear map
\[ \mathrm{H}^0\left(X,T\otimes L^{g(N-1)}\otimes \mathcal I(gN\varphi _ j)\right) \xrightarrow {\times s_ g} \mathrm{H}^0\left(X,T\otimes L^{gN}\otimes \mathcal I(gN\varphi )\right). \]
In particular, when $g=g_ i$ for some $i$ large enough, we then find
\[ \Delta _{g_ i,T}\left((N-1)\theta ,N\varphi _{j}\right)+(g_ i)^{-1}\nu (s_{g_ i})\subseteq N\Delta _{k_ i,T}(\theta ,\varphi ). \]
We observe that the $(g_ i)^{-1}\nu (s_{g_ i})$'s are bounded, since both convex bodies appearing in this equation remain bounded as $i$ varies. After passing to a subsequence, we may assume that $(g_ i)^{-1}\nu (s_{g_ i})$ converges. Then 19.4.43, followed by division by $N$, gives a vector $v'\in \mathbb {R}^ n$ such that (19.4.47) holds.
Step 2. Next we handle the general case.
Let $\Delta '$ be the Hausdorff limit of a subsequence of $(\Delta _{k,T}(\theta ,\varphi ))_ k$, say the subsequence with indices $k_1<k_2<\cdots $. By theorem 19.7.4, it suffices to prove that $\Delta '=\Delta _{\nu }(\theta ,\varphi )$.
Take $\psi \in \mathrm{PSH}(X,\theta )$ such that $\theta _{\psi }$ is a Kähler current and $\psi \leq \varphi $. The existence of $\psi $ follows from 13.13.47.
Then for any $\epsilon \in \mathbb {Q}\cap (0,1)$,
\[ \Delta _{k,T}(\theta ,\varphi )\supseteq \Delta _{k,T}\Bigl (\theta ,(1-\epsilon )\varphi +\epsilon \psi \Bigr ) \]
for all $k\geq 1$. It follows from Step 1 that
\[ \Delta ' \supseteq \Delta _{\nu }\Bigl (\theta ,(1-\epsilon )\varphi +\epsilon \psi \Bigr ). \]
As $\epsilon \to 0+$, we have
\[ (1-\epsilon )\varphi +\epsilon \psi \xrightarrow {d_ S}\varphi \]
by 16.9.25. Hence theorem 19.4.30 gives
\[ \Delta _{\nu }\Bigl (\theta ,(1-\epsilon )\varphi +\epsilon \psi \Bigr )\to \Delta _{\nu }(\theta ,\varphi ) \]
in the Hausdorff metric. It follows that $\Delta ' \supseteq \Delta _{\nu }(\theta ,\varphi )$.
It remains to establish that
CFA7D4F4C2EB
19.4.48
\begin{equation} \operatorname{vol}\Delta '\leq \operatorname{vol}\Delta _{\nu }(\theta ,\varphi ). \end{equation}
Fix an integer $N>0$, it suffices to argue that
05403B615855
19.4.49
\begin{equation} \operatorname{vol}\Delta '\leq \frac{1}{n!} \int _ X \left(N^{-1}\omega +\theta +\mathrm{dd}^{\mathrm{c}}P_{N^{-1}\omega +\theta }[\varphi ]_{\mathcal{I}}\right)^ n. \end{equation}
Assuming this, letting $N\to \infty $ and using 18.1.31 together with proposition 18.1.27 6, we conclude (19.4.48).
After adding a constant to $\varphi $, which does not change the multiplier ideals or the current $\theta +\mathrm{dd}^{\mathrm{c}}\varphi $, we may assume that $\varphi \leq 0$.
Step 2.1. We first reduce to the case $N|k_ i$ for all $i$.
For this purpose, we are free to replace $k_1<k_2<\cdots $ by a subsequence. We may then assume that $k_ i\equiv a$ modulo $N$ for all $i\geq 1$ for some $a\in \{ 0,1,\ldots ,N-1\} $. We write $k_ i=g_ iN +a$. Observe that
\[ \mathrm{H}^0\left(X,T\otimes L^{k_ i}\otimes \mathcal{I}(k_ i \varphi )\right)\subseteq \mathrm{H}^0\left(X, T\otimes L^ a \otimes L^{g_ i N}\otimes \mathcal{I}(g_ i N \varphi ) \right). \]
Up to replacing $T$ by $T\otimes L^ a$, we may assume that $a=0$.
Step 2.2. We write $k_ i=g_ i N$ for all $i$.
Take a non-zero section $s\in \mathrm{H}^0(X,H)$.
We have an injective linear map
\[ \mathrm{H}^0\left(X,T\otimes L^{gN}\otimes \mathcal{I}(gN\varphi )\right)\xrightarrow {\times s^ g}\mathrm{H}^0\left(X,T\otimes H^ g\otimes L^{gN}\otimes \mathcal{I}(gN\varphi )\right) \]
for each $g\geq 1$. In particular, for each $i\geq 1$,
\[ k_ i\Delta _{k_ i,T}(\theta ,\varphi )+g_ i\nu (s)\subseteq g_ i\Delta _{g_ i,T}\left(\omega +N\theta ,N\varphi \right). \]
Letting $i\to \infty $, by Step 1, we have
\[ N\Delta ' + \nu (s) \subseteq \Delta _{\nu }(\omega +N\theta ,N\varphi ). \]
So
\[ \operatorname{vol}\Delta '\leq \operatorname{vol}\Delta _{\nu }\left(N^{-1}\omega +\theta ,\varphi \right)=\frac{1}{n!}\int _ X \left(N^{-1}\omega +\theta +\mathrm{dd}^{\mathrm{c}}P_{N^{-1}\omega +\theta }[\varphi ]_{\mathcal{I}}\right)^ n. \]
This completes the proof.
$\square$
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