19.5.31 The valuative characterizationBB37B398D505

Proof. Step 1. We first reduce to the case where $\beta $ is a Kähler class.

Take a Kähler class $\alpha $ on $X$. It follows from the volume formula in theorem 19.5.4 that

Since $\beta |_{Y_1}$ is nef, hence pseudo-effective, definition 19.5.24 similarly gives

So it suffices to prove (19.5.33) with $\beta +\epsilon \alpha $ in place of $\beta $.

Step 2. Assume that $\beta $ is a Kähler class. The $\supseteq $ direction in (19.5.33) follows from the extension theorem theorem 5.18.18. To prove the other direction, recall that by theorem 19.5.4, for $t>0$ small enough, we have

As $t\to 0+$, the right-hand side converges to $\Delta _{Y_1\supseteq \cdots \supseteq Y_ n}(\beta |_{Y_1})$ with respect to the Hausdorff metric as a consequence of theorem 19.5.4, while the left-hand side converges to

by 19.7.9. We conclude our assertion. $\square$

Proof. Let $\omega $ be a Kähler form on $X$. The last assertion follows from the first by perturbing $\theta $ to $\theta +\epsilon \omega $.

Step 1. We first handle the case where $T$ has analytic singularities. Let $\pi \colon Z\rightarrow X$ be a modification such that

1. 90254E23C2B2

   $Y_{\bullet }$ admits a lifting $(W_{\bullet },g)$, and

2. CA8486D14D75

   $\pi ^*T=[D]+R$, where $D$ is an effective $\mathbb {Q}$-divisor on $Z$ and $R$ is a closed positive $(1,1)$-current with bounded potential.

This is possible by theorem 5.18.8 and theorem 19.3.19.

By 20.2.24,

where $\Pi \colon W_1\rightarrow Y_1$ is the restriction of $\pi $. It follows from theorem 19.5.26 that

Taking (19.3.13) into account, we find that it suffices to show that

We may assume that $\pi $ is the identity map. Since $\nu (T,Y_1)=0$, the divisor $D$ does not contain $Y_1$. Hence we have

Note that $[D]|_{Y_1}$ is the current of integration along an effective $\mathbb {Q}$-divisor on $Y_1$.

In particular,

The first coordinate of $\nu _{Y_{\bullet }}([D])$ is $0$, and the remaining coordinates are exactly $\nu _{Y_1\supseteq \cdots \supseteq Y_ n}([D]|_{Y_1})$. Thus it suffices to show that

Since $R$ has bounded potential, its cohomology class is nef; it is big because $T$ is a Kähler current. Thus the last identity is exactly 19.5.32.

Step 2. Next we consider the case where $T$ is a Kähler current. Take a quasi-equisingular approximation $(T_ j)_ j$ of $T$ in $\mathcal{Z}_+(X,\alpha )$. From Step 1, we know that for large $j\geq 1$,

Letting $j\to \infty $ and applying theorem 19.5.26 and proposition 20.2.22, we conclude (19.5.35). $\square$

Proof. We argue by induction on $n\geq 0$. The case $n=0$ is of course trivial. Let us assume that $n>0$ and the case $n-1$ has been proved.

We first observe that by theorem 19.5.26,

By proposition 18.1.27 5, removing the divisorial part does not change the volume. Comparing the volumes of both sides using theorem 19.5.26 and proposition 18.1.27, we find that equality holds:

Replacing $T$ by $T-\nu (T,Y_1)[Y_1]$, we may therefore assume that $\nu (T,Y_1)=0$. It suffices to apply 19.5.34 and the inductive hypothesis. $\square$

Proof. When $T$ is a Kähler current, this follows from theorem 19.5.37.

In general, by definition, $\nu _{Y_{\bullet }}(T)=\nu _{Y_{\bullet }}(T+\omega )$ for any Kähler form $\omega $ on $X$. It follows that

for any Kähler form $\omega $. It follows that $\nu _{Y_{\bullet }}(T)\in \Delta _{Y_{\bullet }}(T)$. $\square$

Proof. The $\supseteq $ direction in (19.5.41) follows from corollary 19.5.39 together with theorem 19.5.26 2.

Let us write

for the time being.

Step 1. Assume that $T$ has analytic singularities. We have

It follows easily from theorem 19.5.4 that the volume of the right-hand side is equal to the volume of $\Delta _{Y_{\bullet }}(T)$, so (19.5.41) holds.

Step 2. Assume that $T$ is a Kähler current. Take a quasi-equisingular approximation $T_ j\in \mathcal{Z}_+(X,\alpha )$ of $T$. Next we use the language of psh functions. Let $\varphi _ j,\varphi \in \mathrm{PSH}(X,\theta )$ be the potentials corresponding to $T_ j,T$ for each $j\geq 1$.

Fix an integer $N>0$. For large enough $j\geq 1$, we can find $\psi _ j\in \mathrm{PSH}(X,\theta )_{>0}$ such that

The existence of $\psi _ j$ follows from 13.13.30. It follows that

We claim that

Let $x$ be a point in the right-hand side. Fix $N>0$. For all large $j$, choose

with $y_ j\to x$. By the preceding inclusion,

The points $x,y_ j,z_ j$ all lie in the compact set $\Delta _{Y_{\bullet }}(\alpha )$ after taking closures, by corollary 19.5.39. Hence, letting $j\to \infty $ and then $N\to \infty $, we get $x\in \overline{D_{Y_{\bullet }}(T)}$, proving the claim. Therefore, by Step 1, we conclude that

The reverse direction is already known.

Step 3. Finally, consider the general case. Take a Kähler current $T'\in \mathcal{Z}_+(X,\alpha )$ more singular than $T$ (the existence of which is given by 13.13.47). For each $\epsilon \in (0,1)$, we know that

Letting $\epsilon \to 0+$ and using the $d_ S$-continuity in theorem 19.5.26 4, we find that

As the other inclusion is already known, we conclude. $\square$

Proof. By theorem 19.5.40, it is clear that

On the other hand, we clearly have

for any Kähler form $\omega $ on $X$. It follows that

By theorem 19.5.37, the right-hand side is just $\nu _{Y_{\bullet }}(T+\omega )=\nu _{Y_{\bullet }}(T)$. We conclude the proof. $\square$