20.1.38 Properties of the intersection product066B0B2E68CA

Proof. 1, 2 and 3 follow immediately from the corresponding results for mixed volumes, as proved in proposition 18.1.27.

4 Take $\mathcal{I}$-good non-divisorial closed positive $(1,1)$-currents $T_1,\ldots ,T_ n$ and $T'$ so that $\mathbb {D}(T_ i)=\mathbb {D}_ i$ for all $i=1,\ldots ,n$ and $\mathbb {D}(T')=\mathbb {D}'$. Furthermore, we may assume that the $T_ i$'s and $T'$ are Kähler currents by the perturbation argument.

Let $(T_{i}^ j)_ j$ be a quasi-equisingular approximation of $T_ i$ for $i=2,\ldots ,n$. It follows from theorem 16.9.30 that

It suffices to show that for all $j\geq 1$,

Therefore, we have reduced to the case where $T_2,\ldots ,T_ n$ have analytic singularities. After a resolution, we may assume that they have log singularities along $\mathbb {Q}$-divisors. By proposition 18.1.27 5, we can further reduce to the case where $T_2,\ldots ,T_ n$ have bounded local potentials. Perturbing $T_2,\ldots ,T_ n$ by a Kähler form, we may further assume that $\{ T_2\} ,\ldots , \{ T_ n\} $ are Kähler classes. By proposition 18.1.27 4, we finally reduce to the case where $T_2,\ldots ,T_ n$ are Kähler forms. In this case, the assertion follows from the pseudo-effectivity of $\{ T'\} -\{ T_1\} $. $\square$

Proof. 1, 2, 3 and 4 These are immediate consequences of the corresponding results in the big and nef case established in 20.1.40.

5 By (18.1.26) and (20.1.42), we may assume that $T_1,\ldots ,T_ n$ are Kähler currents. In this case, the assertion follows from the definitions. $\square$

Proof. By 20.1.40 4, the left-hand side is non-negative. For the upper bound, by the multi-linearity of the product proved in proposition 20.1.43, we can rewrite the difference as a sum of $2^ n-1$ terms like

where $J\subseteq \{ 1,\ldots ,n\} $ is a non-empty subset of cardinality $m$, and $\{ i_1,\ldots ,i_{n-m}\} $ is the complement of $J$. We need to show that these coefficients are uniformly bounded by a constant depending only on the given data. For this purpose, we add $\mathbb {D}(\omega )$ to each $\mathbb {D}_ i$ and then replace $K$ by $K+\{ \omega \} $. Therefore, we reduce to the case where each $\mathbb {D}_ i$ is nef and big, and hence $\mathbb {D}_ i=\mathbb {D}(T_ i)$ for some $\mathcal{I}$-good closed positive $(1,1)$-currents on $X$ with positive volumes. Then we need to bound

For this purpose, let $D\geq 0$ be the minimal non-negative number so that $D\{ \omega \} -\alpha $ is nef for any $\alpha \in K$. Then the integral is bounded by the cohomological intersection $\{ \omega \} ^ m\cap (D\{ \omega \} )^{n-m}$. Our assertion follows. $\square$

Proof. Using 20.1.44 and 20.1.7, we may easily reduce to the case where $\mathbb {D}$ is nef and big. In this case, take a non-divisorial closed positive $(1,1)$-current $T$ in $\mathbb {D}_ X$ such that $\mathbb {D}(T)=\mathbb {D}$. Then we need to show that

which is proved in theorem 20.1.15. $\square$

Proof. We may assume that $\operatorname{vol}\mathbb {D}_ i>0$ for each $i=1,\ldots ,n$ since there is nothing to prove otherwise. In this case, our assertion follows from proposition 18.1.36. $\square$

Proof. Step 1. We first assume that the $\mathbb D_ i^ j$'s and the $\mathbb D_ i$'s are all big.

Choose $\mathcal I$-good non-divisorial currents $T_ i^ j$ and $T_ i$ such that $\mathbb D(T_ i^ j)=\mathbb D_ i^ j$ and $\mathbb D(T_ i)=\mathbb D_ i$. Fix a Kähler form $\omega $ on $X$ and fix $\epsilon >0$. Since $\mathbb D_{i,X}^ j\to \mathbb D_{i,X}$, for all sufficiently large $j$ the class

is Kähler. Choose a Kähler form $\chi _ i^ j$ in this class and set $S_ i^ j=T_ i^ j+\chi _ i^ j$. Then $S_ i^ j$ lies in the fixed class $\mathbb D_{i,X}+\epsilon \{ \omega \} $ and has the same divisorial Lelong numbers as $T_ i^ j$. Hence, by theorem 18.1.42 and proposition 18.1.27 2,

Therefore,

Letting $\epsilon \to 0+$ gives (20.1.48).

Step 2. Next we handle the general case.

Take a Kähler form $\omega $ on $X$. Then by 20.1.44, for any $\epsilon \in (0,1]$, we have

Since $\epsilon $ is arbitrary, our assertion follows. $\square$

Proof. This is a straightforward consequence of 20.1.40 4 and proposition 20.1.47. $\square$