Theorem 19.4.41.76132CC6940C As $k\to \infty $, we have
958BCE448264
19.4.42
B6A67594E2A0Theorem 19.4.41.76132CC6940C As $k\to \infty $, we have
958BCE448264
19.4.42
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.
Lemma 19.4.43.FD338F03BE48 Assume that $\varphi $ has analytic singularities and $\theta _{\varphi }$ is a Kähler current. Then (19.4.42) holds.
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
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
So
Letting $i\to \infty $, by proposition 19.4.12,
Letting $\epsilon \to 0+$, using the continuity of classical Okounkov bodies in the big cone recalled in theorem 19.5.4, we find that
where we applied 19.4.23 as well. Our assertion follows. $\square$
Lemma 19.4.44.F3D0645BDF93 Assume that $\theta _{\varphi }$ is a Kähler current. Then as $\mathbb {Q}\ni \beta \to 0+$, we have
Proof. By proposition 19.4.33 and proposition 19.4.28, we have
for any small enough $\beta \in \mathbb {Q}_{>0}$. In particular, if $\Delta '$ is the Hausdorff limit of a subnet of $(\Delta ((1-\beta )\theta ,\varphi ))_{\beta }$, then $\Delta '\subseteq \Delta _{\nu }(\theta ,\varphi )$. But
Since we have not developed the theory of nef b-divisors yet, we give a direct proof as well. Take a Kähler form $\omega $ so that $\theta _{\varphi }\geq \omega $. Let $\psi =P_{\theta -\omega }[\varphi ]_{\mathcal{I}}$. Then $\varphi \sim _{\mathcal{I}}\psi $. In order to establish the last equality, we may replace $\varphi $ by $\psi $ and hence assume that $\varphi $ is $\mathcal{I}$-good. For small enough $\beta >0$, the form $(1-\beta )\theta -(\theta -\omega )=\omega -\beta \theta $ is Kähler, so this remains true in the class $(1-\beta )\{ \theta \} $ by 18.1.8. Thus the equality of masses in an $\mathcal{I}$-equivalence class reduces the desired equality to
which is obvious.
It follows that $\Delta '=\Delta _{\nu }(\theta ,\varphi )$. We conclude by theorem 19.7.4. $\square$
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
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
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 $i\geq 1$,
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
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
Fix $j\geq j'$. It suffices to show that
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19.4.47
for some $v'\in \mathbb {R}^ n$. In fact, if this is true, we have
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
By 5.18.15, after possibly increasing $g'$, for all $g\geq g'$,
Moreover, the inequality (19.4.46) gives that for $g\geq g'$,
where the second inclusion follows directly from the definition of $\mathcal I_\infty $. Thus multiplication by $s_ g$ gives an injective linear map
In particular, when $g=g_ i$ for some $i$ large enough, we then find
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 $\epsilon \in \mathbb {Q}\cap (0,1)$,
for all $k\geq 1$. It follows from Step 1 that
As $\epsilon \to 0+$, we have
by 16.9.25. Hence theorem 19.4.30 gives
in the Hausdorff metric. It follows that $\Delta ' \supseteq \Delta _{\nu }(\theta ,\varphi )$.
It remains to establish that
CFA7D4F4C2EB
19.4.48
Fix an integer $N>0$, it suffices to argue that
05403B615855
19.4.49
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
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
for each $g\geq 1$. In particular, for each $i\geq 1$,
Letting $i\to \infty $, by Step 1, we have
So
This completes the proof. $\square$
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