1BFD41225C11
Fix a smooth closed real $(1,1)$-form $\theta $ on $X$.
See Definition 16.11.5.
When we want to be more specific, we may also write $\varphi \land _{\theta }\psi $. Note that it is possible that the set in (
16.3.1) is empty, in which case $\varphi \land \psi =-\infty $.
Proposition 16.3.2.C1971A5F3DFD The operator $\land $ is a well-defined commutative, associative binary operator
\[ \bigl (\mathrm{PSH}(X,\theta )\cup \{ -\infty \} \bigr )\times \bigl (\mathrm{PSH}(X,\theta )\cup \{ -\infty \} \bigr )\rightarrow \mathrm{PSH}(X,\theta )\cup \{ -\infty \} . \]
Proof.
We first show that the map is well-defined. For this purpose, take $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. When the set in (16.3.1) is empty, there is nothing to prove. So let us assume that the set is not empty.
Define
\[ \gamma =\operatorname *{sup}_{}{}\! {}^{*}\left\{ \eta \in \mathrm{PSH}(X,\theta ):\eta \leq \varphi ,\eta \leq \psi \right\} . \]
Then by proposition 5.14.2, we find that $\gamma \in \mathrm{PSH}(X,\theta )$ and hence $\gamma $ is a candidate for the supremum in (16.3.1). Therefore, $\gamma \leq \varphi \land \psi $. The reverse inequality follows from the definition of $\gamma $, so
\[ \varphi \land \psi =\gamma \in \mathrm{PSH}(X,\theta ). \]
The commutativity and the associativity of $\land $ follow directly from the defining maximality property.
$\square$
Note that from the proof, we have
\[ \varphi \land \psi =P_{\theta }\Bigl ( \min \{ \varphi ,\psi \} \Bigr ). \]
See definition 13.13.18 for the notation $P_{\theta }(\bullet )$.
Lemma 16.3.3.4CF7246E6E5A Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. Assume that $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$. Then
E08D87E98114
16.3.4
\begin{equation} \theta _{\varphi \land \psi }^ n\leq \mathds {1}_{\{ \varphi \land \psi =\varphi \} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ \varphi \land \psi =\psi \} } \, \theta _{\psi }^ n. \end{equation}
In particular, $\theta _{\varphi \land \psi }^ n$ is carried by the union $\{ \varphi \land \psi =\varphi \} \cup \{ \varphi \land \psi =\psi \} $.
Proof.
We first observe that as a consequence of 13.13.14, we have
\[ \mathds {1}_{\{ \varphi \land \psi =\varphi \} } \, \theta _{\varphi \land \psi }^ n\leq \mathds {1}_{\{ \varphi \land \psi =\varphi \} } \, \theta _{\varphi }^ n,\quad \mathds {1}_{\{ \varphi \land \psi =\psi \} } \, \theta _{\varphi \land \psi }^ n\leq \mathds {1}_{\{ \varphi \land \psi =\psi \} } \, \theta _{\psi }^ n. \]
Applying theorem 13.13.24 to $\min \{ \varphi ,\psi \} $, we conclude that
\[ \theta _{\varphi \land \psi }^ n\leq \mathds {1}_{\{ \varphi \land \psi =\varphi \} } \, \theta _{\varphi \land \psi }^ n+\mathds {1}_{\{ \varphi \land \psi =\psi \} }\, \theta _{\varphi \land \psi }^ n\leq \mathds {1}_{\{ \varphi \land \psi =\varphi \} } \, \theta _{\varphi }^ n+\mathds {1}_{\{ \varphi \land \psi =\psi \} } \, \theta _{\psi }^ n, \]
and (16.3.4) is established.
$\square$
We recall that the relations $\preceq $ and $\sim $ are introduced in definition
16.7.1.
Definition 16.3.5.395DB7860949 Given $\varphi \in \mathrm{PSH}(X,\theta )$, we define its $P$-envelope as follows:
7E001B03F24A
16.3.6
\begin{equation} P_{\theta }[\varphi ]\coloneqq \operatorname *{sup}_{}{}\! {}^{*}\left\{ \psi \in \mathrm{PSH}(X,\theta ): \psi \leq 0,\psi \preceq \varphi \right\} . \end{equation}
Note that $\varphi -\sup _ X \varphi $ is always a candidate for the supremum. By proposition
5.14.2, we have $P_{\theta }[\varphi ]\in \mathrm{PSH}(X,\theta )$ and $P_{\theta }[\varphi ]\leq 0$. Moreover, the definition can be equivalently described as
FA68BB354359
16.3.7
\begin{equation} P_{\theta }[\varphi ]=\operatorname *{sup}_{C\in \mathbb {Z}_{>0}}{}\! {}^{*}(\varphi +C)\land V_{\theta }. \end{equation}
Recall that $V_{\theta }$ is introduced in (13.13.7). For any $C\in \mathbb {R}$, we have $(\varphi +C)\land V_{\theta }\in \mathrm{PSH}(X,\theta )$ and
\[ (\varphi +C)\land V_{\theta }\sim \varphi . \]
In other words, in (16.3.6), we may replace the condition $\psi \preceq \varphi $ by $\psi \sim \varphi $. Also note that as a consequence,
268F8CAEF7E2
16.3.8
\begin{equation} \varphi \preceq P_{\theta }[\varphi ]. \end{equation}
The idea lying behind the definition of $P_{\theta }[\varphi ]$ is that we choose the least singular element out of all potentials with the same singularity type as $\varphi $. As we shall see in example 16.3.22 below, $P_{\theta }[\varphi ]$ does not necessarily have the same singularity type as $\varphi $. This forces us to define a rougher equivalence relation in definition 16.8.2.
The envelope depends on the choice of $\theta $, but the dependence is easy to understand:
Proposition 16.3.9.EF5AC4B19C4D Let $\theta '=\theta +\mathrm{dd}^{\mathrm{c}}g$ for some $g\in C^{\infty }(X)$. Then for any $\varphi \in \mathrm{PSH}(X,\theta )$, we have $\varphi '\coloneqq \varphi -g\in \mathrm{PSH}(X,\theta ')$ and
\[ P_{\theta }[\varphi ]\sim P_{\theta '}[\varphi ']. \]
Proof.
By symmetry, it suffices to show that
\[ P_{\theta }[\varphi ]\preceq P_{\theta '}[\varphi ']. \]
We may assume that $g\geq 0$ (after replacing $g$ by $g+C$ with $C\geq -\inf _ X g$, which leaves $\theta '$ unchanged and shifts $\varphi '$ by an additive constant). Then for any $\psi \in \mathrm{PSH}(X,\theta )$ with $\psi \preceq \varphi $ and $\psi \leq 0$, we set $\psi '\coloneqq \psi -g\in \mathrm{PSH}(X,\theta ')$. Then $\psi '\preceq \varphi '$ and $\psi '\leq 0$, so $\psi '\leq P_{\theta '}[\varphi ']$. Since $\psi $ is arbitrary, it follows that
\[ P_{\theta }[\varphi ]-\sup _ X g\leq P_{\theta }[\varphi ]-g\leq P_{\theta '}[\varphi '], \]
which gives $P_{\theta }[\varphi ]\preceq P_{\theta '}[\varphi ']$.
$\square$
The $P$-envelope preserves the non-pluripolar masses:
Proposition 16.3.10.E58386609911 Suppose that $\theta _1,\ldots ,\theta _ n$ are smooth closed real $(1,1)$-forms on $X$. Let $\varphi _ i\in \mathrm{PSH}(X,\theta _ i)$ for each $i=1,\ldots ,n$. Then
0F7C5A9FEE2D
16.3.11
\begin{equation} \int _ X \theta _{1,P_{\theta _1}[\varphi _1]}\wedge \cdots \wedge \theta _{n,P_{\theta _ n}[\varphi _ n]}=\int _ X \theta _{1,\varphi _1}\wedge \cdots \wedge \theta _{n,\varphi _ n}. \end{equation}
This proposition together with theorem
13.13.11 will be combined to a common generalization in proposition
16.8.22 after introducing the $P$-preorder.
Proof.
For each $C\in \mathbb {Z}_{>0}$ and each $i=1,\ldots ,n$, we have
\[ (\varphi _ i+C)\land V_{\theta _ i}\sim \varphi _ i. \]
It follows from the monotonicity theorem theorem 13.13.11 that
\[ \int _ X \theta _{1,(\varphi _1+C)\land V_{\theta _1}}\wedge \cdots \wedge \theta _{n,(\varphi _ n+C)\land V_{\theta _ n}}=\int _ X \theta _{1,\varphi _1}\wedge \cdots \wedge \theta _{n,\varphi _ n}. \]
So (16.3.11) follows from (16.3.7) and corollary 13.13.12.
$\square$
Conversely, proposition
16.3.10 characterizes the $P$-envelope, as we will see in a moment in theorem
16.3.15. We need some preparation for the proof. We first show that the $P$-envelope can be regarded as a concentration of the non-pluripolar mass:
Theorem 16.3.12.553335F185CE Let $\varphi \in \mathrm{PSH}(X,\theta )$. Then
5CEAE61A4277
16.3.13
\begin{equation} \theta _{P_{\theta }[\varphi ]}^ n\leq \mathds {1}_{\{ P_{\theta }[\varphi ]=0\} }\, \theta ^ n. \end{equation}
In fact, equality holds as we will see in proposition
16.3.41 below.
Proof.
Thanks to 16.3.3, for each $C>0$, we have
\[ \begin{aligned} \theta _{(\varphi +C)\land V_{\theta }}^ n\leq & \mathds {1}_{\{ (\varphi +C)\land V_{\theta }=\varphi +C\} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ (\varphi +C)\land V_{\theta }=V_{\theta }\} }\, \theta _{V_{\theta }}^ n\\ \leq & \mathds {1}_{\{ \varphi +C\leq V_{\theta }\} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ P_{\theta }[\varphi ]=V_{\theta }\} } \, \theta _{V_{\theta }}^ n. \end{aligned} \]
We wish to let $C\to \infty $. The dominated convergence theorem and the semicontinuity theorem corollary 13.13.12 give, as $C\to \infty $,
\[ \mathds {1}_{\{ \varphi +C\leq V_{\theta }\} } \theta _{\varphi }^ n\rightharpoonup 0,\quad \theta _{(\varphi +C)\land V_{\theta }}^ n\rightharpoonup \theta _{P_{\theta }[\varphi ]}^ n. \]
So we conclude that
\[ \theta _{P_{\theta }[\varphi ]}^ n\leq \mathds {1}_{\{ P_{\theta }[\varphi ]=V_{\theta }\} } \, \theta _{V_{\theta }}^ n. \]
Taking corollary 13.13.61 into consideration, we conclude (16.3.13).
$\square$
Using essentially the same proof, we arrive at the following conclusion:
Corollary 16.3.14.B6EF8456606F Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. Let
\[ \eta \coloneqq \operatorname *{sup}_{C>0}{}\! {}^{*} (\varphi +C)\land \psi .{}^{\href{#sgpt-a0000019653}{1}} \]
Assume that $\eta \not\equiv -\infty $ (for example when $\varphi \preceq \psi $). Then
\[ \theta _{\eta }^ n\leq \mathds {1}_{\{ \eta =\psi \} }\, \theta _{\psi }^ n. \]
Proof.
Thanks to 16.3.3, for each $C>0$, we have
\[ \begin{aligned} \theta _{(\varphi +C)\land \psi }^ n\leq & \mathds {1}_{\{ (\varphi +C)\land \psi =\varphi +C\} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ (\varphi +C)\land \psi =\psi \} }\, \theta _{\psi }^ n\\ \leq & \mathds {1}_{\{ \varphi +C\leq \psi \} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ \eta =\psi \} } \, \theta _{\psi }^ n. \end{aligned} \]
We wish to let $C\to \infty $. The dominated convergence theorem and the semicontinuity theorem corollary 13.13.12 give, as $C\to \infty $,
\[ \mathds {1}_{\{ \varphi +C\leq \psi \} } \theta _{\varphi }^ n\rightharpoonup 0,\quad \theta _{(\varphi +C)\land \psi }^ n\rightharpoonup \theta _{\eta }^ n. \]
So we conclude that
\[ \theta _{\eta }^ n\leq \mathds {1}_{\{ \eta =\psi \} } \, \theta _{\psi }^ n. \]
This completes the proof.
$\square$
Theorem 16.3.15.7ADF30509E4E Assume that $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$. Then
2B585EFD6781
16.3.16
\begin{equation} P_{\theta }[\varphi ]=\sup \left\{ \psi \in \mathrm{PSH}(X,\theta ): \psi \leq 0,\varphi \preceq \psi , \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n\right\} . \end{equation}
In particular, in this case,
A43EDA3894B0
16.3.17
\begin{equation} P_{\theta }\Bigl [P_{\theta }[\varphi ]\Bigr ]=P_{\theta }[\varphi ]. \end{equation}
Note that in (
16.3.16) and (
16.3.6), the auxiliary function $\psi $ lies on different sides of $\varphi $.
Proof.
Let $\psi $ be a candidate of the right-hand side of (16.3.16). It follows from theorem 16.3.12 that
\[ \int _{\{ P_{\theta }[\varphi ]<\psi \} } \theta _{P_{\theta }[\varphi ]}^ n\leq \int _{\{ P_{\theta }[\varphi ]<\psi \} \cap \{ P_{\theta }[\varphi ]=0\} } \theta ^ n=0. \]
On the other hand, we have
\[ P_{\theta }[\varphi ]\preceq P_{\theta }[\psi ],\quad \psi \preceq P_{\theta }[\psi ] \]
and all these potentials have the same mass due to proposition 16.3.10. So the domination principle theorem 13.13.43 is applicable and gives
\[ P_{\theta }[\varphi ]\geq \psi . \]
Hence we get the $\geq $ direction in (16.3.16).
On the other hand, $P_{\theta }[\varphi ]$ itself is a candidate of the right-hand side of (16.3.16), as a consequence of proposition 16.3.10. Therefore, (16.3.16) follows.
As for (16.3.17), the $\geq $ direction is trivial. For the other direction, let $\psi \in \mathrm{PSH}(X,\theta )$ be a potential satisfying
\[ \psi \leq 0,\quad P_{\theta }[\varphi ]\preceq \psi ,\quad \int _ X \theta _{P_{\theta }[\varphi ]}^ n=\int _ X \theta _{\psi }^ n. \]
Then it follows from proposition 16.3.10 that
\[ \psi \leq 0,\quad \varphi \preceq \psi ,\quad \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n. \]
In view of (16.3.16), we conclude the $\leq $ direction of (16.3.17).
$\square$
Definition 16.3.18.6623E38E5FC2 If $\varphi =P_{\theta }[\varphi ]$ and $\int _ X \theta _{\varphi }^ n>0$, we say $\varphi $ is a model potential.
Recall that the notion of model potentials depends heavily on the choice of $\theta $. When there is a risk of confusion, we also say $\varphi $ is a model potential in $\mathrm{PSH}(X,\theta )$.
There are plenty of model potentials:
Corollary 16.3.20.E10CA97A11EF Let $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$. Then $P_{\theta }[\varphi ]$ is a model potential in $\mathrm{PSH}(X,\theta )$. Moreover,
\[ \int _ X \theta _{P_{\theta }[\varphi ]}^ n=\int _ X \theta _{\varphi }^ n. \]
Proof.
This follows immediately from theorem 16.3.15 and proposition 16.3.10.
$\square$
As we have already seen in the proof of
13.13.30, we have the following interesting property:
Proposition 16.3.21.4C5C62135747 Let $\varphi \in \mathrm{PSH}(X,\theta )$. Consider a model potential $\phi \in \mathrm{PSH}(X,\theta )_{>0}$. Assume that $\varphi \preceq \phi $. Then
\[ \sup _{\{ \phi \neq -\infty \} } (\varphi -\phi )=\sup _ X \varphi . \]
In particular, for any $\varphi \in \mathrm{PSH}(X,\theta )$, $\varphi \preceq \phi $ and $\varphi \leq 0$ implies $\varphi \leq \phi $.
Proof.
Observe that $\phi \leq 0$, so on the set $\{ \phi \neq -\infty \} $, we have
\[ \varphi -\phi \geq \varphi . \]
So the $\geq $ direction follows from corollary 5.15.21. Conversely, by assumption, we can find a constant $C>0$ so that
\[ \varphi -\sup _ X \varphi \leq \min \{ \phi +C,0 \} . \]
It follows that $\varphi -\sup _ X \varphi \leq P_{\theta }[\phi ]=\phi $. Therefore, the reverse inequality follows.
$\square$
Example 16.3.22.E548BFB3BF04 We continue our favorite example example 5.20.6. Let $X=\mathbb {P}^1$ and $\omega $ be the Fubini–Study metric. We define $\varphi \in \mathrm{PSH}(X,\omega )$ as follows: For $z\in \mathbb {C}$, we let
\[ \varphi (z)= \begin{cases} -\log \left(|z|^2+1\right)+\Bigl (-\log \left(-\log |z|^2\right)\Bigr ) \lor \Bigl (2+\log |z|^2\Bigr ),& \textup{if }|z|<1/\sqrt{2},\\ 2+\log \frac{|z|^2}{|z|^2+1},& \textup{otherwise}, \end{cases} \]
while $\varphi (\infty )=2$. Note that $\varphi $ is $\omega $-psh on $X=\mathbb {P}^1$ by the gluing lemma 5.14.16. The singularity of $\varphi $ only occurs at $z=0$, close to which, $\varphi $ is approximately $-\log \left(-\log |z|^2\right)$. This type of singularity is therefore called the log-log type singularity.
We claim that
B2DFFDD6A0F3
16.3.23
\begin{equation} P_{\omega }[\varphi ]=0. \end{equation}
In particular, we find that $\varphi $ and $P_{\omega }[\varphi ]$ have different singularity types.
By theorem 16.3.15, in order to verify (16.3.23), it suffices to verify that
C825DA5CEC8C
16.3.24
\begin{equation} \int _ X \omega _{\varphi }^1=1. \end{equation}
Here $\omega _{\varphi }^1$ is taken in the non-pluripolar sense. Since $\{ 0,\infty \} \subseteq \mathbb {P}^1$ is pluripolar, this reduces to showing that
\[ \int _{\mathbb {C}^*} \mathrm{dd}^{\mathrm{c}}\psi =\frac{1}{4\pi }\int _{\mathbb {C}^*} (\Delta \psi )\, \mathrm{d}\mu =1, \]
where $\psi (z)=\varphi (z)+\log (|z|^2+1)$ and $\mu $ is the standard Lebesgue measure on $\mathbb {C}$.
Note that the Laplacian vanishes outside $\overline{B(0,0.7)}$ since $\psi (z)=2+\log |z|^2$ there, which is harmonic. Therefore,
\[ \int _{\mathbb {C}^*} \mathrm{dd}^{\mathrm{c}}\psi =\frac{1}{4\pi }\int _{|z|<1/\sqrt{2}} (\Delta \psi )(z)\, \mathrm{d}\mu . \]
It is an elementary exercise to see that the right-hand side is exactly equal to $1$. If you are familiar with toric geometry, this is more or less trivial since
\[ \nabla _ r \Bigl ( \left(-\log (-r)\right)\lor (2+r) \Bigr )\bigl ((-\infty ,-\log 2)\bigr )= (0,1]. \]
Otherwise, just try to evaluate the integral using Green's identities. Therefore, (16.3.24) is proved and our assertion (16.3.23) follows.
Next we give a criterion on when the rooftop operator is not identically $-\infty $.
Proposition 16.3.25.F7C0BACE770D Assume that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ and
E6B68D406648
16.3.26
\begin{equation} \int _ X \theta _{\varphi }^ n+\int _ X \theta _{\psi }^ n>\int _ X \theta _{\varphi \lor \psi }^ n. \end{equation}
Then $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$.
Thanks to the monotonicity theorem theorem
13.13.11, we may also replace $\varphi \lor \psi $ on the right-hand side of (
16.3.26) by any $\gamma \in \mathrm{PSH}(X,\theta )$ such that $\varphi \lor \psi \preceq \gamma $.
Proof.
Without loss of generality, we may assume that $\varphi ,\psi \leq 0$. For simplicity, we write
\[ \eta =P_{\theta }[\varphi \lor \psi ]. \]
Note that due to our assumption (16.3.26), $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$, and hence so is $\varphi \lor \psi $ thanks to the monotonicity theorem theorem 13.13.11. By corollary 16.3.20, $\eta $ is a model potential, and $\int _ X\theta _\eta ^ n=\int _ X\theta _{\varphi \lor \psi }^ n>0$.
Take $C>0$ large enough, so that
E47535691487
16.3.27
\begin{equation} \int _{\{ \varphi >\eta -C\} } \theta _{\varphi }^ n+\int _{\{ \psi >\eta -C\} } \theta _{\psi }^ n>\int _ X \theta _{\eta }^ n. \end{equation}
This is possible thanks to proposition 13.11.4 4. Fix $C'>C$. Write
\[ \gamma _{C'}\coloneqq \Bigl (\varphi \lor (\eta -C')\Bigr )\land \Bigl (\psi \lor (\eta -C')\Bigr ). \]
Then observe that $\gamma _{C'}\sim \eta $, and
\[ \inf _{C'>C}\gamma _{C'}=\varphi \land \psi . \]
Suppose, for contradiction, that $\varphi \land \psi \equiv -\infty $. Then
\[ \adjustlimits \lim _{C'\to \infty } \sup _ X\gamma _{C'}=-\infty . \]
Thanks to proposition 16.3.21, for each $C'>C$,
\[ \sup _{X} \gamma _{C'}=\sup _{\{ \eta \neq -\infty \} } (\gamma _{C'}-\eta ), \]
since $\eta $ is a model potential. It follows that
74C044AB05CB
16.3.28
\begin{equation} \adjustlimits \lim _{C'\to \infty } \sup _{\{ \eta \neq -\infty \} } (\gamma _{C'}-\eta )=-\infty . \end{equation}
In particular, we can take $C'$ large enough so that
\[ \gamma _{C'}\leq \eta -C. \]
For such $C'$, we have
\[ \begin{aligned} \int _ X \theta _{\gamma _{C'}}^ n=& \int _{\{ \gamma _{C'}\leq \eta -C\} } \theta _{\gamma _{C'}}^ n \\ \leq & \int _{\left\{ \varphi \lor (\eta -C')\leq \eta -C\right\} } \theta _{\varphi \lor (\eta -C')}^ n +\int _{\left\{ \psi \lor (\eta -C')\leq \eta -C \right\} } \theta _{\psi \lor (\eta -C')}^ n\\ =& 2\int _ X \theta _{\eta }^ n-\int _{\{ \varphi >\eta -C\} }\theta _{\varphi }^ n-\int _{\{ \psi >\eta -C\} }\theta _{\psi }^ n\\ <& \int _ X \theta _{\eta }^ n, \end{aligned} \]
where the second line follows from 16.3.3, the fourth line follows from (16.3.27). This contradicts the fact that $\gamma _{C'}\sim \eta $ in view of theorem 13.13.11.
$\square$
Proposition 16.3.29.B9461C380E5C Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. Assume that
8997CB842099
16.3.30
\begin{equation} \varphi =P_{\theta }[\varphi ],\quad \psi =P_{\theta }[\psi ],\quad \varphi \land \psi \not\equiv -\infty . \end{equation}
Then
7E981960ACBC
16.3.31
\begin{equation} P_{\theta }[\varphi \land \psi ]=\varphi \land \psi . \end{equation}
Proof.
Observe that by proposition 16.4.2 3,
\[ P_{\theta }[\varphi \land \psi ]\leq P_{\theta }[\varphi ]=\varphi ,\quad P_{\theta }[\varphi \land \psi ]\leq P_{\theta }[\psi ]=\psi . \]
So the $\leq $ direction in (16.3.31) holds. The reverse direction is trivial.
$\square$
Lemma 16.3.32.BCC1B8FD1BED Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. Assume that (16.3.30) holds and
\[ \int _ X \theta _{\varphi \land \psi }^ n>0. \]
Suppose that $\gamma ,\eta \in \mathrm{PSH}(X,\theta )$ and satisfy
\[ \gamma \preceq \varphi ,\quad \eta \preceq \psi ,\quad \int _ X \theta _{\gamma }^ n=\int _ X \theta _{\varphi }^ n,\quad \int _ X \theta _{\eta }^ n=\int _ X \theta _{\psi }^ n. \]
Then $\gamma \land \eta \not\equiv -\infty $ and
725175FB4F3C
16.3.33
\begin{equation} \int _ X \theta _{\gamma \land \eta }^ n=\int _ X \theta _{\varphi \land \psi }^ n. \end{equation}
Later on in section
16.7, we shall reformulate and generalize this lemma in terms of the $P$-preorder, which is a more natural language to express the result. See proposition
16.8.29.
Proof.
Choose constants $A,B>0$ such that
\[ \gamma -A\leq \varphi ,\quad \eta -B\leq \psi . \]
Since
\[ (\gamma -A)\land (\eta -B)\sim \gamma \land \eta , \]
the non-pluripolar masses of these two rooftop potentials are equal by the monotonicity theorem theorem 13.13.11. Hence it suffices to prove the lemma after replacing $\gamma ,\eta $ by $\gamma -A,\eta -B$. We can therefore assume that
\[ \gamma \leq \varphi ,\quad \eta \leq \psi . \]
Step 1. We first show that $\gamma \land \psi \not\equiv -\infty $ and
500FC88C1939
16.3.34
\begin{equation} \int _ X \theta _{\gamma \land \psi }^ n=\int _ X \theta _{\varphi \land \psi }^ n. \end{equation}
For any $a>1$, we define
\[ \gamma _ a\coloneqq P_{\theta }\Bigl ( a\gamma +(1-a)\varphi \Bigr ). \]
When $a=1$, we simply write $\gamma _1=\gamma $.
Thanks to corollary 13.13.42, for any $a\geq 1$, we have $\gamma _ a\leq \varphi $ and
\[ \int _ X \theta _{\gamma _ a}^ n=\int _ X \theta _{\varphi }^ n. \]
By our assumption, for any $a\geq 1$, we have
\[ \int _ X \theta _{\gamma _ a}^ n+\int _ X \theta _{\varphi \land \psi }^ n> \int _ X \theta _{\varphi }^ n,\quad \gamma _ a\preceq \varphi ,\quad \varphi \land \psi \preceq \varphi . \]
So proposition 16.3.25 gives
\[ \gamma _ a\land \varphi \land \psi =\gamma _ a\land \psi \in \mathrm{PSH}(X,\theta ). \]
In particular, for $a=1$, we find $\gamma \land \psi \not\equiv -\infty $. It remains to verify (16.3.34).
For $a>1$, we have
\[ \gamma \geq a^{-1}\gamma _ a+\left(1-a^{-1}\right)\varphi , \]
thanks to the definition of $\gamma _ a$ (cf. (13.13.34)). Hence
\[ \gamma \land \psi \geq a^{-1}\left( \gamma _ a\land \psi \right)+\left(1-a^{-1}\right)\left(\varphi \land \psi \right). \]
Therefore, by the monotonicity theorem theorem 13.13.11,
\[ \int _ X \theta _{\gamma \land \psi }^ n\geq \left(1-a^{-1}\right)^ n \int _ X \theta _{\varphi \land \psi }^ n. \]
Letting $a\to \infty $ and taking theorem 13.13.11 into account, (16.3.34) follows.
Step 2. We complete the proof.
For any $a>1$, we let
\[ \eta _ a\coloneqq P_{\theta }\Bigl ( a\eta +(1-a)\psi \Bigr ). \]
When $a=1$, we set $\eta _1=\eta $. Thanks to corollary 13.13.42, for each $a\geq 1$, we have $\eta _ a\leq \psi $ and
\[ \int _ X \theta _{\eta _ a}^ n=\int _ X \theta _{\psi }^ n. \]
From Step 1, we have
\[ \int _ X \theta _{\gamma \land \psi }^ n+\int _ X \theta _{\eta _ a}^ n=\int _ X \theta _{\varphi \land \psi }^ n+\int _ X \theta _{\psi }^ n>\int _ X \theta _{\psi }^ n. \]
Applying proposition 16.3.25 again, we find
\[ \gamma \land \psi \land \eta _ a =\gamma \land \eta _ a\in \mathrm{PSH}(X,\theta ). \]
When $a=1$, we find $\gamma \land \eta \in \mathrm{PSH}(X,\theta )$. It remains to prove (16.3.33).
By definition of $\eta _ a$, we have
\[ \eta \geq a^{-1}\eta _ a+\left(1-a^{-1}\right)\psi . \]
Therefore,
\[ \gamma \land \eta \geq a^{-1} (\gamma \land \eta _ a)+\left(1-a^{-1}\right) \left( \gamma \land \psi \right). \]
By the monotonicity theorem theorem 13.13.11,
\[ \int _ X \theta _{\gamma \land \eta }^ n\geq \left(1-a^{-1}\right)^ n\int _ X \theta _{\gamma \land \psi }^ n. \]
Letting $a\to \infty $ and applying theorem 13.13.11 again, we arrive at (16.3.33).
$\square$
There is an interesting diamond-like inequality regarding the non-pluripolar masses:
Theorem 16.3.35.5B751E9DFBC1 Assume that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$ and $\varphi \land \psi \in \mathrm{PSH}(X,\theta )$. Then
020126C445C3
16.3.36
\begin{equation} \int _ X \theta _{\varphi }^ n+\int _ X \theta _{\psi }^ n\leq \int _ X \theta _{\varphi \lor \psi }^ n+\int _ X \theta _{\varphi \land \psi }^ n. \end{equation}
Proof.
We may assume that $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$, as otherwise (16.3.36) follows immediately from the monotonicity theorem theorem 13.13.11.
Step 1. We claim that it suffices to prove (16.3.36) with $P_{\theta }[\varphi ]$ and $P_{\theta }[\psi ]$ in place of $\varphi $ and $\psi $.
In fact, as $\varphi \land \psi \not\equiv -\infty $, we also have $P_{\theta }[\varphi ]\land P_{\theta }[\psi ]\not\equiv -\infty $. Moreover,
\[ \int _ X \theta _{P_{\theta }[\varphi ]}^ n=\int _ X \theta _{\varphi }^ n,\quad \int _ X \theta _{P_{\theta }[\psi ]}^ n=\int _ X \theta _{\psi }^ n \]
by proposition 16.3.10. On the other hand, as $C\to \infty $, the potentials
\[ \Bigl ( (\varphi +C)\land V_{\theta } \Bigr ) \lor \Bigl ( (\psi +C)\land V_{\theta } \Bigr ) \]
converge to $P_{\theta }[\varphi ]\lor P_{\theta }[\psi ]$ almost everywhere. Therefore, thanks to corollary 13.13.12 and theorem 13.13.11,
\[ \int _ X \theta _{P_{\theta }[\varphi ]\lor P_{\theta }[\psi ]}^ n=\int _ X \theta _{\varphi \lor \psi }^ n. \]
Finally, we have
\[ \int _ X \theta _{P_{\theta }[\varphi ]\land P_{\theta }[\psi ]}^ n= \int _ X \theta _{\varphi \land \psi }^ n \]
as well. When the left-hand side vanishes, this follows from theorem 13.13.11. Otherwise, it follows from 16.3.32.
In particular, (16.3.36) is equivalent to the corresponding result with $P_{\theta }[\varphi ]$ and $P_{\theta }[\psi ]$ in place of $\varphi $ and $\psi $.
Step 2. We shall assume that $\varphi $ and $\psi $ are model potentials in the sequel.
We write $\gamma =\varphi \lor \psi $ for simplicity. Note that $\gamma \leq 0$.
For each $C>0$, we introduce
\[ \varphi _ C=\varphi \lor (\gamma -C),\quad \psi _ C=\psi \lor (\gamma -C). \]
Note that $\varphi _ C\sim \psi _ C\sim \gamma $. Observe that
\[ \inf _{C>0} \varphi _ C\land \psi _ C=\varphi \land \psi . \]
The $\geq $ direction is trivial, and the $\leq $ direction follows from the observation that the left-hand side is dominated by both $\varphi _ C$ and $\psi _ C$ for any $C>0$, and therefore by both $\varphi $ and $\psi $.
For $C>0$, we compute
35F5A4FBAC68
16.3.37
\begin{equation} \begin{aligned} \theta _{\varphi _ C}^ n=& \mathds {1}_{\{ \varphi >\gamma -C\} } \, \theta _{\varphi _ C}^ n+ \mathds {1}_{\{ \varphi \leq \gamma -C\} } \, \theta _{\varphi _ C}^ n & \\ =& \mathds {1}_{\{ \varphi >\gamma -C\} } \, \theta _{\varphi }^ n+ \mathds {1}_{\{ \varphi \leq \gamma -C\} } \, \theta _{\varphi _ C}^ n & \textup{by proposition \href{../../tag/9E81D36B1290/}{13.11.4} \href{../../tag/9E81D36B1290/#sgpt-1C036AD6}{1}}\\ =& \theta _{\varphi }^ n+ \mathds {1}_{\{ \varphi \leq \gamma -C\} } \, \theta _{\varphi _ C}^ n & \textup{by theorem \href{../../tag/553335F185CE/}{16.3.12}}. \end{aligned} \end{equation}
In the last line we use that $\varphi $ is model, so $\theta _\varphi ^ n$ is carried by $\{ \varphi =0\} \subseteq \{ \varphi >\gamma -C\} $. We also get a similar formula for $\theta _{\psi _ C}^ n$.
Therefore, taking the monotonicity theorem theorem 13.13.11 into consideration, we find
BCB4335E3D6E
16.3.38
\begin{equation} \int _ X \theta _{\gamma }^ n-\int _ X \theta _{\varphi }^ n=\int _{\{ \varphi \leq \gamma -C\} } \theta _{\varphi _ C}^ n,\quad \int _ X \theta _{\gamma }^ n-\int _ X \theta _{\psi }^ n=\int _{\{ \psi \leq \gamma -C\} } \theta _{\psi _ C}^ n. \end{equation}
On the other hand, using 16.3.3 and (16.3.37), we find
\[ \begin{aligned} \theta _{\varphi _ C\land \psi _ C}^ n \leq & \mathds {1}_{\{ \varphi _ C\land \psi _ C=\varphi _ C\} } \, \theta _{\varphi _ C}^ n+\mathds {1}_{\{ \varphi _ C\land \psi _ C=\psi _ C\} } \, \theta _{\psi _ C}^ n \\ \leq & \mathds {1}_{\{ \varphi _ C\land \psi _ C=\varphi _ C=0\} }\, \theta _{\varphi }^ n+\mathds {1}_{\{ \varphi _ C\land \psi _ C=\psi _ C=0\} }\, \theta _{\psi }^ n+\mathds {1}_{\{ \varphi \leq \gamma -C\} }\, \theta _{\varphi _ C}^ n\\ & +\mathds {1}_{\{ \psi \leq \gamma -C\} }\, \theta _{\psi _ C}^ n. \end{aligned} \]
In particular,
\[ \mathds {1}_{\{ \varphi _ C\land \psi _ C<0 \} }\, \theta _{\varphi _ C\land \psi _ C}^ n \leq \mathds {1}_{\{ \varphi \leq \gamma -C\} }\, \theta _{\varphi _ C}^ n+\mathds {1}_{\{ \psi \leq \gamma -C\} }\, \theta _{\psi _ C}^ n. \]
Integrating, we find
\[ \begin{aligned} \int _{\{ \varphi _ C\land \psi _ C=0 \} } \theta _{\varphi _ C\land \psi _ C}^ n \geq & \int _ X \theta _{\gamma }^ n-\int _{\{ \varphi \leq \gamma -C\} } \theta _{\varphi _ C}^ n-\int _{\{ \psi \leq \gamma -C\} } \theta _{\psi _ C}^ n\\ =& \int _ X \theta _{\varphi }^ n+\int _ X \theta _{\psi }^ n-\int _ X \theta _{\gamma }^ n, \end{aligned} \]
where on the first line we used the fact that $\varphi _ C\land \psi _ C\sim \gamma $, while on the second line, we used (16.3.38).
Letting $C\to \infty $, and using corollary 13.13.16, we conclude (16.3.36).
$\square$
For later use, we also introduce a generalized envelope operator:
Definition 16.3.39.B791A3D8205D Given $\varphi \in \mathrm{PSH}(X,\theta )$ and a function $f\colon X\to [-\infty ,\infty ]$, we define
\[ P_\theta [\varphi ](f)\coloneqq \operatorname *{sup}_{}{}\! {}^{*}\bigl \{ \eta \in \mathrm{PSH}(X,\theta ): \eta \leq f\textup{ quasi-everywhere},\ \eta \preceq \varphi \bigr \} . \]
In general, we cannot replace $\eta \leq f$ quasi-everywhere by everywhere.
Example 16.3.40.17FC69E7E25E When $f=0$, we have $P_\theta [\varphi ](f)=P_{\theta }[\varphi ]$.
When $\varphi =V_{\theta }$, we have $P_\theta [V_{\theta }](f)=P_{\theta }(f)$.
So definition 16.3.39 is a common generalization of definition 16.3.5 and definition 13.13.18.
Proposition 16.3.41.BE692352C556 Let $\varphi \in \mathrm{PSH}(X,\theta )$ and $f\in C^{1,\overline{1}}(X)$ with $\varphi \leq f$. Then
\[ \theta ^ n_{P_\theta [\varphi ](f)} = \mathds {1}_{\left\{ P_\theta [\varphi ](f)=f\right\} }\, \theta ^ n_ f. \]
In particular,
\[ \theta ^ n_{P_\theta [\varphi ]} = \mathds {1}_{\left\{ P_\theta [\varphi ]=0\right\} }\, \theta ^ n. \]
This is a vast generalization of theorem
16.3.12.
Proof.
By definition, we have
\[ P_\theta [\varphi ](f)=P_\theta [\varphi ]\Bigl (P_\theta (f)\Bigr ) \]
and $\varphi \leq P_\theta (f)$. By corollary 16.3.14 and corollary 13.13.61,
\[ \theta ^ n_{P_\theta [\varphi ](f)} \leq \mathds {1}_{\left\{ P_\theta [\varphi ](f)=P_\theta (f)\right\} }\, \theta ^ n_{P_\theta (f)} = \mathds {1}_{\left\{ P_\theta [\varphi ](f)=f\right\} }\, \theta ^ n_ f. \]
The reverse inequality follows immediately from theorem 13.13.57.
$\square$
Proposition 16.3.42.1F5A8A58FB4D Let $\varphi \in \mathrm{PSH}(X,\theta )$ and $f\in C^{1,\overline{1}}(X)$ with $\varphi \leq f$. Then
\[ \begin{aligned} \mathds {1}_{\left\{ \varphi =P_\theta (f)\right\} }\, \theta ^ n_{P_\theta (f)} =& \mathds {1}_{\left\{ \varphi =P_\theta (f)\right\} }\, \theta ^ n_\varphi , \\ \mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)\right\} }\, \theta ^ n_{P_\theta [\varphi ](f)} =& \mathds {1}_{\{ \varphi =P_\theta [\varphi ](f)\} }\, \theta ^ n_\varphi . \end{aligned} \]
Proof.
By corollary 13.13.61 and theorem 13.13.57, we have
\[ \mathds {1}_{\{ \varphi =P_\theta (f)\} }\, \theta ^ n_{P_\theta (f)}=\mathds {1}_{\{ \varphi =P_\theta (f)=f\} }\, \theta ^ n_{f}=\mathds {1}_{\{ \varphi =P_\theta (f)=f\} }\, \theta ^ n_{\varphi }\leq \mathds {1}_{\{ \varphi =P_\theta (f)\} }\, \theta ^ n_{\varphi }. \]
The reverse inequality follows from 13.13.14. The first equality follows.
As for the second, we compute using proposition 16.3.41 and theorem 13.13.57 that
\[ \begin{split} \mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)\right\} }\, \theta ^ n_{P_\theta [\varphi ](f)}=\mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)=f\right\} }\, \theta ^ n_{f}\\ =\mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)=f\right\} }\, \theta _{\varphi }^ n\leq \mathds {1}_{\{ \varphi =P_\theta [\varphi ](f)\} }\, \theta ^ n_\varphi . \end{split} \]
Since $\varphi \leq P_{\theta }[\varphi ](f)$, 13.13.14 gives the opposite inequality. The second equality follows.
$\square$
Pome
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