Theorem 19.4.4176132CC6940C

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

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

Fix $j\geq j'$. It suffices to show that

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

Fix an integer $N>0$, it suffices to argue that

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. $\square$