18.12 All smooth obstaclesB61E129D450F

Proof. For either envelope operator, the obstacle comparison gives

Hence the two envelope distances for a pair $[u],[v]$ differ by at most $2\| F-G\| _\infty $, uniformly in the pair. The space $C^\infty (X,\mathbb R)$ contains a countable subset that is dense for the uniform norm: approximate each member of a countable dense subset of $C(X)$ by smooth functions with errors $1/m$, $m\geq 1$. Given an entourage, we approximate each of its finitely many obstacles by a member of the family and shrink the radii accordingly; this proves that the family generates the same uniformity. The divisorial coordinates still range over all fixed compact modifications.

The transport in proposition 18.11.1 relabels the entire family of smooth obstacles by $F\mapsto F-\chi $, or by $F\mapsto sF$. These are bijections of the family, which proves both isomorphisms.

For nonseparation, use the unbounded function $u=1-\sqrt{1-w}$ of proposition 18.10.1. It has trivial multiplier ideals and zero divisorial Lelong numbers, and $\delta w\preceq u$ for $0<\delta \leq 1$. Fix $F$ and put $q=P_\omega (F)$, the envelope of the obstacle $F$ in the sense of section 5.40. Since $\omega >0$, the constant $\min _ XF$ is a competitor, so $q$ is bounded and $q\leq F$. The function

is $\omega $-plurisubharmonic, lies below $F$, and satisfies $h_\delta \preceq u$. Letting $\delta \downarrow 0$ off the pole of $w$ gives $P_{\omega ,F}[u]=q=P_{\omega ,F}[0]$. Since $u$ and $0$ have the same multiplier ideals, their ideal envelopes below $F$ also agree. Every generating entourage therefore has identical sections at $[u]$ and $[0]$, although these ordinary classes differ. $\square$

Proof. Set $q_ F=P_{\theta ,F}[u]_{\mathcal I}$. Translation of competitors gives

so $q_ F\sim q$. The identity $q_ F-F=P_{\theta +\mathrm{dd}^{\mathrm{c}}F}[u-F]_{\mathcal I}$ and theorem 18.4.2 give $q_ F\sim _{\mathcal I}u$. Since $q_ F\leq F$ and $q_ F\sim q$, it is a competitor for $P_{\theta ,F}[q]$. Conversely, the maximum of a competitor for $P_{\theta ,F}[q]$ with a sufficiently low translate of $q$ is a competitor for $P_{\theta ,F}[q]_{\mathcal I}$, and this ideal envelope equals $q_ F$ because $q\sim _{\mathcal I}u$. These bounds give

The last assertion follows from proposition 16.28.65, which gives $P_{\theta ,F}[u]=P_{\theta ,F}[q]$. $\square$

Proof. Fix a smooth obstacle $F$. As in the preceding proof, transport of the obstacle and theorem 18.4.2 give $P_{\omega ,F}[u]_{\mathcal I}\sim _{\mathcal I}u$. For $a$ with analytic singularities, lemma 5.38.11 therefore makes this ideal envelope a competitor for the ordinary envelope, so $P_{\omega ,F}[a]=P_{\omega ,F}[a]_{\mathcal I}$.

Write $p=P_{\omega ,F}[u]_{\mathcal I}$ and $q_ j=P_{\omega ,F}[b_ j]_{\mathcal I}$. The dyadic construction gives $u\preceq _{\mathcal I}b_ j$, decreasing ordinary types, and convergence of every local divisorial Lelong number of $b_ j$ to that of $u$. The ideal envelope preserves the ideal order, as one sees by taking maxima with a translate of $b_ j$ below $F$, so $p\leq q_ j\leq F$. Since $q_ j=P_{\omega ,F}[b_ j]$ and the ordinary types decrease, the sequence $q_ j$ decreases. Its limit $q$ is bounded below by $p$, so it is $\omega $-plurisubharmonic, and $p\leq q\leq F$; the same bounds give convergence in $L^1$.

Convergence after pullback and semicontinuity of Lelong numbers, as in proposition 18.7.1, give $\nu _ E(q)\geq \nu _ E(u)$ for every local divisor $E$, and $p\leq q$ gives the reverse inequality. The Lelong-number criterion of theorem 5.37.3 yields $q\sim _{\mathcal I}u$. Since $q\leq F$, it is a competitor for $p$, which proves

The controlled ideal relations and all divisorial distances on fixed modifications converge by the proof of theorem 18.9.1; they do not depend on $F$. Consequently this one dyadic sequence converges in $\tau ^{\mathrm{all}}_\omega $ exactly when the stated envelope equalities hold.

For each $F$, equality of the two $L^1$ envelope coordinates is a closed condition. Analytic classes satisfy all these equalities, so the closure of the analytic classes under nets lies in $\mathcal G^{\mathrm{all}}_\omega $, and the dyadic sequence proves the converse. Finally, the uniform obstacle bound in the proof of proposition 18.12.1 extends equality from a dense family to every smooth obstacle. $\square$

Proof. For $v\in \mathrm{PSH}(Y,\theta ')$, define

Over a sufficiently small ball in $X$, this is the finite maximum of the functions obtained by composing $v$ with the local inverse branches of $r$. Each of them is $\theta $-plurisubharmonic, so $v_{\max }\in \mathrm{PSH}(X,\theta )$ by lemma 5.24.1. If $v\preceq u'$, then $v_{\max }\preceq u$. If $v\sim _{\mathcal I}u'$, a local biholomorphic change of variables makes each branch multiplier-ideal equivalent to $u$, and the maximum rule in lemma 5.37.4 gives $v_{\max }\sim _{\mathcal I}u$. The obstacle bound $v\leq F\circ r$ gives $v_{\max }\leq F$.

Thus every competitor upstairs is dominated by the pullback of a corresponding competitor downstairs. Conversely, pullback of a competitor downstairs preserves the background inequality, the ordinary order or multiplier-ideal equivalence, and the obstacle bound. These two correspondences between the competitor families give the envelope identities; upper regularization commutes with $r$, since $r$ is locally biholomorphic.

Since $r$ is surjective, it reflects the ordinary order: $u'\preceq \varphi \circ r$ if and only if $u\preceq \varphi $. Fix $a\in (0,1)$. Pullback takes any stretching competitor $h$ with $a\varphi +(1-a)h\leq u$ to a stretching competitor upstairs. Conversely, if $v\in \mathrm{PSH}(Y,\theta ')$ satisfies

then taking the maximum over the fibers gives $a\varphi +(1-a)v_{\max }\leq u$; this also holds at poles, because the convex sums are defined there. Applying the definition of the relative class for every $a$ proves the last assertion. $\square$