13.9 Maximal mass and comparisonCCB85E469493

Proof. Big classes. Any two potentials with minimal singularities in a fixed big class differ by a bounded function, and a change of smooth representative changes each potential by a smooth function. Hence corollary 13.8.8 shows that $I_\omega $ does not depend on these choices. The same corollary and minimality give the upper bound for closed positive currents in big classes. If $\omega '=\omega +\mathrm{dd}^{\mathrm{c}}f$ is another Kähler form in the class of $\omega $, with $f$ smooth, then $\omega '^{n-p}-\omega ^{n-p}=\mathrm d\bigl (\mathrm d^{\mathrm c}f\wedge \sum _{i=0}^{n-p-1}\omega '^ i\wedge \omega ^{n-p-1-i}\bigr )$ is a smooth exact form. Its integral against the closed current $\langle T_1\wedge \cdots \wedge T_ p\rangle $ vanishes, because $S(\mathrm d\psi )=\pm (\mathrm dS)(\psi )=0$ for a closed current $S$ and a smooth form $\psi $ on the compact manifold $X$. Hence the numbers for big classes do not depend on $\omega $ within its class, and neither do their limits, since the perturbing classes $\varepsilon \{ \omega \} $ are the same.

Let $\alpha $ be big and $\gamma $ pseudo-effective, represented by a closed positive current $S$. Adding $S$ to a current with minimal singularities in $\alpha $ gives a current in the big class $\alpha +\gamma $, which by the upper bound has mixed mass at most the number with $\alpha $ replaced by $\alpha +\gamma $. By multilinearity and positivity, its mixed mass is at least the number for $\alpha $. Hence, when all arguments are big, replacing an argument $\alpha $ by $\alpha +\gamma $ increases the number. Homogeneity for big classes follows by scaling a minimal potential together with its background: if $c>0$, then $cV_\theta $ has minimal singularities in $\mathrm{PSH}(X,c\theta )$, and the product is homogeneous in each factor by proposition 13.2.6.

Pseudo-effective classes. By this monotonicity for big classes, the perturbed numbers decrease as $\varepsilon \downarrow 0$; they are nonnegative, so the limit exists. Every Kähler form $\kappa $ satisfies $c^{-1}\omega \leq \kappa \leq c\omega $ for some $c\geq 1$ (lemma 6.14.1), so monotonicity places the perturbation by $\varepsilon \kappa $ between the perturbations by $c^{-1}\varepsilon \omega $ and $c\varepsilon \omega $, with the same fixed test form $\omega ^{n-p}$; this proves independence of the perturbing class. The same comparison shows that perturbing the arguments by different fixed multiples of $\varepsilon \omega $ gives the same limit, which extends homogeneity to pseudo-effective classes. Monotonicity also extends: if $\gamma $ is pseudo-effective, then for every $\varepsilon >0$ the number with an argument $\alpha +\varepsilon \{ \omega \} $ replaced by $\alpha +\gamma +\varepsilon \{ \omega \} $ is at least the original one, by the big case, and we let $\varepsilon \downarrow 0$.

If all $\alpha _ j$ are big, choose $\delta >0$ such that each $\alpha _ j$ contains a Kähler current $R_ j\geq \delta \omega $. Then $(1+\varepsilon /\delta )\alpha _ j-(\alpha _ j+\varepsilon \{ \omega \} )=(\varepsilon /\delta )\{ R_ j-\delta \omega \} $ is pseudo-effective, so monotonicity and homogeneity give

and the limit is $I_\omega (\alpha _1,\ldots ,\alpha _ p)$. For arbitrary closed positive currents $T_ j\in \alpha _ j$, the upper bound in the big classes $\alpha _ j+\varepsilon \{ \omega \} $, applied to $T_ j+\varepsilon \omega $, and multilinearity give $\int _ X\langle \bigwedge _ jT_ j\rangle \wedge \omega ^{n-p}\leq I_\omega (\alpha _1+\varepsilon \{ \omega \} ,\ldots )$, and letting $\varepsilon \downarrow 0$ proves the upper bound.

Nef and big classes. If the $\alpha _ j$ are nef, their perturbations are Kähler classes: $\alpha _ j+\varepsilon \{ \omega \} $ contains $\theta _{j,\varepsilon /2}+\varepsilon \omega \geq \varepsilon \omega /2$, where $\theta _{j,\varepsilon /2}\geq -(\varepsilon /2)\omega $ is a smooth representative of $\alpha _ j$ given by definition 10.4.1. If $\theta _ j+\varepsilon \omega +\mathrm{dd}^{\mathrm{c}}f$ is a Kähler form with $f$ smooth, then $f\in \mathrm{PSH}(X,\theta _ j+\varepsilon \omega )$, so $V_{\theta _ j+\varepsilon \omega }\geq f-\sup _ Xf$ by lemma 10.3.8; since also $V_{\theta _ j+\varepsilon \omega }\leq 0$ by the same lemma, the minimal potentials are bounded. By proposition 13.5.1, the perturbed numbers are the integrals $\int _ X\bigwedge _ j(\theta _ j+\varepsilon \omega )\wedge \omega ^{n-p}$, and their limits are the cohomological intersection numbers. Finally, let $\alpha $ be big and $T_0\in \alpha $ a Kähler current with $T_0\geq \delta \omega $. Expanding $T_0=\delta \omega +(T_0-\delta \omega )$ by multilinearity gives $\langle T_0^ n\rangle \geq \delta ^ n\omega ^ n$, so $\operatorname {vol}(\alpha )\geq \delta ^ n\int _ X\omega ^ n>0$ by the upper bound. $\square$

Proof. By theorem 13.8.7, the mass of each $u_ k$ is at most that of $\phi $, which equals the mass of $u$. Hence theorem 13.7.5, with $p=n$, gives weak convergence, and its lower inequality together with this bound gives convergence of the masses. If $u\in \mathcal E(X,\theta )$, then by proposition 13.9.2 the mass of each $u_ k$ is at most $\operatorname {vol}(\{ \theta \} )$, which is the mass of $u$, and the same argument applies. Finally, decreasing sequences converge in extended capacity by lemma 13.7.3. $\square$

Proof. Let $\varepsilon >0$. The function $w=\max \{ u,v-\varepsilon \} $ is $\theta $-plurisubharmonic. The sets $\{ u<v-\varepsilon \} $ and $\{ u>v-\varepsilon \} $ are disjoint and plurifine open. Indeed, near each point we have $u-v=(u+h)-(v+h)$ for a smooth local potential $h$ of $\theta $, and $u+h$, $v+h$ are plurisubharmonic, so by proposition 5.47.25 the function $u-v$ is plurifinely continuous off the common pole set of $u$ and $v$, which is disjoint from both sets. Moreover $w$ equals $v-\varepsilon $ on the first set and $u$ on the second. By plurifine locality (proposition 13.2.6),

The left side is at most $\operatorname {vol}(\{ \theta \} )$ by proposition 13.9.2. The complement of $\{ u>v-\varepsilon \} $ consists of $\{ u\leq v-\varepsilon \} $, which is contained in $\{ u<v\} $ up to the common pole set of $u$ and $v$, and that pluripolar set carries no mass. Hence

As $\varepsilon \downarrow 0$, the sets $\{ u<v-\varepsilon \} $ increase to $\{ u<v\} $, which proves the inequality. If $u$ has full mass, the last two terms cancel. If $u\geq v-O(1)$, then $w-u$ is bounded, so $\mu _ w(X)=\mu _ u(X)$ by corollary 13.8.8, and the same computation with $\mu _ u(X)$ in place of the volume gives the inequality without these terms. $\square$

Proof. After adding a constant to $\rho $ we may assume $\rho \leq u$. For $0<t<1$ put $q_ t=(1-t)v+t\rho $. Then $u\geq q_ t-O(1)$, and $\{ u<q_ t\} \subset \{ u<v\} $, because $u<(1-t)v+t\rho \leq (1-t)v+tu$ forces $u<v$. Multilinearity and proposition 13.9.5 therefore give

As $t\downarrow 0$, every point where $u<v$ and $\rho >-\infty $ eventually lies in $\{ u<q_ t\} $, so Fatou's lemma, applied along a sequence $t\downarrow 0$, gives $\mu _\rho (\{ u<v\} \cap \{ \rho >-\infty \} )=0$. Since $\mu _\rho $ gives no mass to the pole set of $\rho $, this proves the first assertion.

Now let the class be big and $u$ have minimal singularities. The class contains a Kähler current $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega $ with $\delta >0$. Then $u\geq \rho -O(1)$ by minimality, and $\mu _\rho \geq \delta ^ n\omega ^ n$, because multilinearity expands $\langle (\theta +\mathrm{dd}^{\mathrm{c}}\rho )^ n\rangle =\langle (\delta \omega +(\theta +\mathrm{dd}^{\mathrm{c}}\rho -\delta \omega ))^ n\rangle $ into positive terms, one of which is $\delta ^ n\omega ^ n$. The first assertion gives $u\geq v$ almost everywhere with respect to $\omega ^ n$, that is, almost everywhere for Lebesgue measure in charts. Near each point, $u$ and $v$ become plurisubharmonic after adding a common smooth local potential $h$ of $\theta $, and then $\max \{ u,v\} +h$ and $u+h$ are plurisubharmonic functions that agree almost everywhere; by theorem 5.9.1, theorem 3.4.12 they agree everywhere, that is, $u\geq v$ everywhere. $\square$