Theorem 17.6.5

Theorem 17.6.5.C6C3A19AAB0D Let $\theta $ be a smooth closed real $(1,1)$-form on $X$ with big class, and let $u\in \mathrm{PSH}(X,\theta )$. Then the singularity ceiling of $u$ is its singularity envelope, that is,

\[ \Big(\sup _{v\in \mathcal F_ u}v\Big)^*=P_\theta [u], \]

and $P_\theta [P_\theta [u]]=P_\theta [u]$. In particular, both identities hold when $\int _ X\langle \theta _ u^ n\rangle =0$.

Proof. By the remark following definition 17.6.4, both identities are unchanged when a constant is added to $u$, so we may assume $u\leq 0$. Put $\phi =P_\theta [u]$.

We first show that $\phi $ lies below the singularity ceiling of $u$. By definition 16.21.1, $\phi \leq 0$, and $u\leq \phi $ because $u$ is itself a competitor. Applying the mixed mass preservation of proposition 16.21.2(3) to the tuple consisting of $k$ copies of $u$ and $n-k$ copies of $V$, and using $P_\theta [V]=V$ (proposition 16.21.2(4)), we see that $\phi $ has the same mixed masses against $V$ as $u$. Hence $\phi \in \mathcal F_ u$, and therefore $\phi \leq (\sup _{v\in \mathcal F_ u}v)^*$.

Conversely, let $v\in \mathcal F_ u$, and for $0<\varepsilon <1$ put

\[ w_\varepsilon =(1-\varepsilon )u+\varepsilon V, \qquad z_\varepsilon =(1-\varepsilon )v+\varepsilon V. \]

Expanding by multilinearity and using the mixed mass identities that define $\mathcal F_ u$, we get

\[ \int _ X\langle \theta _{w_\varepsilon }^ n\rangle =\int _ X\langle \theta _{z_\varepsilon }^ n\rangle \geq \varepsilon ^ n\int _ X\langle \theta _ V^ n\rangle >0. \]

Moreover $w_\varepsilon \preceq z_\varepsilon $, since $u\preceq v$. Let $\zeta _\varepsilon =P_\theta [z_\varepsilon ]$. By theorem 16.22.11, $\zeta _\varepsilon $ is a model potential with the same mass as $z_\varepsilon $, hence as $w_\varepsilon $. As $w_\varepsilon \preceq z_\varepsilon \leq \zeta _\varepsilon $, the potential $w_\varepsilon $ belongs to $\mathcal E(X,\theta ,\zeta _\varepsilon )$, and proposition 16.22.14 gives

\[ P_\theta [w_\varepsilon ]=P_\theta [\zeta _\varepsilon ]=\zeta _\varepsilon =P_\theta [z_\varepsilon ]. \]

Now let $\varepsilon \downarrow 0$. Proposition 17.6.1, applied to $u$ and to $v$, gives $P_\theta [u]=P_\theta [v]$. Since $v\leq 0$, the potential $v$ is itself a competitor for $P_\theta [v]$, so $v\leq P_\theta [v]=\phi $. Taking the supremum over $v\in \mathcal F_ u$ and its upper semicontinuous regularization, we obtain $(\sup _{v\in \mathcal F_ u}v)^*\leq \phi $.

For idempotence, let $\phi _\varepsilon =P_\theta [w_\varepsilon ]$ be as in proposition 17.6.1. Since $\phi \leq \phi _\varepsilon $, the monotonicity of $P_\theta [\cdot ]$ and theorem 16.22.11, which applies to $w_\varepsilon $ because $\int _ X\langle \theta _{w_\varepsilon }^ n\rangle >0$, give

\[ P_\theta [\phi ]\leq P_\theta [\phi _\varepsilon ] =\phi _\varepsilon . \]

Letting $\varepsilon \downarrow 0$ and using the proposition, we obtain $P_\theta [\phi ]\leq \phi $. Conversely, $\phi \leq 0$ is a competitor for $P_\theta [\phi ]$, so $\phi \leq P_\theta [\phi ]$. This proves idempotence. $\square$

Theorem 17.6.5`C6C3A19AAB0D`

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Theorem 17.6.5C6C3A19AAB0D

Comments

Theorem 17.6.5`C6C3A19AAB0D`