13.24 Volumes of multiplier-ideal envelopes146F3DD22A85

Proof. By theorem 18.7.6, the analytic ordinary models $q_ j=P_{\theta _ j}[a_ j]$ decrease to $P_\theta [u]_{\mathcal I}$, and the backgrounds $\theta _ j$ decrease to $\theta $. Thus theorem 13.21.2 applies to the $q_ j$. Since $q_ j\sim a_ j$ by proposition 18.4.8, the volumes of $q_ j$ and $a_ j$ agree. $\square$

Proof. Recall that $q$ is an ordinary model and $u\preceq P_\theta [u]\leq q$. If (i) holds, then $P_\theta [u]=q$ has positive lower volume, and theorem 13.19.5 shows that both volumes of $u$ equal those of $q$; in particular (ii) holds, together with the final assertion.

Conversely, (ii) and $u\preceq q$ give $u\in \mathcal E_{\mathrm{vol}}(X,\theta ,q)$, and corollary 13.19.8 gives $P_\theta [u]=q$, which is (i). The same corollary shows that (i) is equivalent to (iii). By proposition 18.12.2, (iii) implies (iv), and (iv) with $F=0$ is (i). $\square$

Proof. Step 1: the approximants. Fix a Hermitian form $h$. Bigness and theorem 5.36.1 give an analytic $\rho \in \mathrm{PSH}(X,\theta )$ with $\rho \leq 0$ and a rational $\kappa >0$ such that $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \kappa h$. Choose the dyadic regularizations $a_ j$ as in (18.7.5), enlarging their errors to positive rational numbers $\epsilon _ j\downarrow 0$. Put

Then $\theta +\mathrm{dd}^{\mathrm{c}}b_ j\geq ((1-c_ j)\kappa -c_ j\epsilon _ j)h=0$. The coefficients are rational, so $b_ j$ has analytic singularities by proposition 5.34.1. Since $a_ j\to u$ in $L^1$ and $c_ j\to 1$, also $b_ j\to u$ in $L^1$.

Step 2: two-sided bounds. Choose $A>0$ with $\theta \leq Ah$, and set

These backgrounds decrease to $\theta $ and dominate $\theta +\epsilon _ jh$, so the $a_ j$ also satisfy (18.7.5) with the larger errors $(1+A/\kappa )\epsilon _ j$. Theorem 18.7.6 therefore gives $t_ j\searrow q$. Moreover

By proposition 18.4.8, $r_ j\sim b_ j$. Hence $r_ j/c_ j$ is a nonpositive $\widehat\theta _ j$-plurisubharmonic function that is more singular than $a_ j$, because $b_ j\leq c_ ja_ j$; it is therefore a competitor for $t_ j$, and $r_ j\leq c_ jt_ j$. Also $q\leq t_ j\leq c_ jt_ j$.

For a lower bound, recall that $q\sim _{\mathcal I}u$, and $u\preceq _{\mathcal I}a_ j$ because $a_ j\geq u$. Addition and scaling preserve this order by theorem 5.37.3, lemma 5.37.4. Thus $c_ jq+(1-c_ j)\rho \preceq _{\mathcal I}b_ j$. This potential lies below zero, so envelope order and proposition 18.4.8 give

Step 3: volumes. For either singularity volume, homogeneity, addition and monotonicity now yield

The last comparison uses both $\theta \leq c_ j\widehat\theta _ j$ and $\max (r_ j,q)\leq c_ jt_ j$. By theorem 13.21.2, the outer terms tend to $\operatorname {vol}_\theta (q)$. This proves the convergence of both volumes of $r_ j$ and the first limit in (13.24.4). In particular, after discarding finitely many indices, the ordinary models $r_ j$ and $q$ have a common positive lower-volume bound, so $r_ j\to q$ in $L^1$ and in capacity by proposition 13.22.4, theorem 13.23.3.

For the second limit, translate $u$ so that $u\leq q$; this does not change $d_\theta (r_ j,u)$. Then

Both bounds tend to $\overline{\operatorname {vol}}_\theta (q)$, so $d_\theta (r_ j,u)=2U(\max (r_ j,u))-U(r_ j)-U(u)$ tends to $U(q)-U(u)$, as claimed. The last equivalence follows from proposition 13.24.2. $\square$

Proof. If $p=P_\theta [p]_{\mathcal I}$, apply theorem 13.24.3 with $u=p$. For the converse it suffices to prove the more general assertion, because analytic ordinary models are ideal models by proposition 18.4.8.

Put $q=P_\theta [p]_{\mathcal I}$ and $\delta =\underline{\operatorname {vol}}_\theta (p)>0$. Choose $s_ j\in (0,1)$ tending to zero with $d_\theta (p_ j,p)<\delta s_ j^ n$, after discarding finitely many indices. For $m_ j=\max (p_ j,p)$ one has

By lemma 13.18.1 with $b=s_ j^{-1}$, there is $h_ j\in \mathrm{PSH}(X,\theta )$ with $(1-s_ j)m_ j+s_ jh_ j\leq p_ j$. Set $H_ j=P_\theta [h_ j]_{\mathcal I}$. Since $m_ j\geq p$, monotonicity, addition and scaling of ideal envelopes (proposition 18.4.6) give

Consequently $\underline{\operatorname {vol}}_\theta (p_ j)\geq (1-s_ j)^ n\underline{\operatorname {vol}}_\theta (q)\geq (1-s_ j)^ n\delta $ by addition and volume monotonicity, since $q\geq p$. The models $p_ j,p$ therefore lie in $\mathcal M_{\delta /2}(\theta )$ for large $j$, so proposition 13.22.4 gives $p_ j\to p$ in $L^1$.

The $H_ j$ have supremum zero, so normalized compactness bounds $\int _ X(-H_ j)\, \mathrm{d}V$ by a constant $C_\theta $. Since $q\leq 0$, the convex inequality above gives $q-p_ j\leq s_ j(q-H_ j)\leq -s_ jH_ j$, and hence

Passing to the $L^1$ limit gives $q\leq p$ almost everywhere, and hence everywhere by lemma 5.24.3. The opposite inequality $p=P_\theta [p]\leq P_\theta [p]_{\mathcal I}=q$ holds by proposition 18.4.4, so $p=q$. $\square$

Proof. The Poincaré–Lelong formula gives $\mathrm{dd}^{\mathrm{c}}\ell _ j=[H_ j]-\kappa $. Unitary invariance reduces its integral to the distribution of $x=|Z_0|^2/\| Z\| ^2$, whose density on $[0,1]$ is $2(1-x)$. Thus

The decreasing series has finite $L^1$ norm and a common lower curvature bound $-A\kappa $. Its canonical limit is quasi-plurisubharmonic. Distributional convergence gives the current identities and strict positivity.

Fix $\pi $ and $E$, and choose a Gauduchon form $\gamma $ on the compact surface $Z$. For the tail $r_ N=\sum _{j>N}a_ j\ell _ j$ of total coefficient $A_ N=\sum _{j>N}a_ j$, the closed positive current

has trace mass $A_ N\int _ Z\pi ^*\kappa \wedge \gamma $, by distributional Stokes. The pullback of $r_ N$ is quasi-plurisubharmonic, since $\pi $ is an isomorphism on a dense open set where $r_ N$ is finite almost everywhere. Divisorial removal gives

The denominator is positive and the ratio is finite. Finite additivity of Lelong numbers now identifies the partial profile with $\nu (u\circ \pi ,E)$ up to an error tending to zero. This proves convergence and the asserted equality. $\square$

Proof. Ideal profile and envelope. Choose an integer $C>A$ with $C\kappa \geq \theta $. Both $u$ and $v$ are $C\kappa $-plurisubharmonic. The projective valuative characterization of all-weight ideal types applies in this closed background: generic orders on all smooth projective modifications recover point Lelong numbers by further point blowups, and equality of those numbers is equivalent to ideal equivalence. The stated profile therefore implies $v\sim _{\mathcal I}u$. Conversely, ideal equivalence pulls back under every compact modification by lemma 5.43.23; the local valuative criterion theorem 5.37.3 then gives equality of generic coefficients on every such modification. The common factor between growth against $\log |z|$ and the book's $\log |z|^2$ convention does not affect these equalities.

The potential $u$ has complex Hessian $S-A\kappa $. By corollary 18.4.9, its ordinary and ideal envelopes agree and are ordinarily equivalent to $u$. Ideal equivalence identifies the defining families for the two ideal envelopes in the statement, proving the assertion about $q$. The inequality $\theta +\mathrm{dd}^{\mathrm{c}}u\geq R>0$ proves bigness. For any Hermitian form $h$ on $X$, choose $D$ with $D\kappa \geq h$. Bounded $h$-plurisubharmonic potentials are $D\kappa $-plurisubharmonic, and positivity of bounded mixed products bounds their masses by $\int _ X(D\kappa )^2=D^2$. This proves the bounded mass property.

The bounded residual. Fix $w\sim u$ in $\mathrm{PSH}(X,\theta )$ and put $T=C\kappa +\mathrm{dd}^{\mathrm{c}}w$. Its base divisorial coefficients are $a_ j$, so divisorial removal and passage through increasing finite sums give

The distributional difference has a canonical quasi-plurisubharmonic representative, which is bounded by ordinary equivalence and almost-everywhere order. Thus $K$ has bounded local potentials, and its trace charges no pluripolar set. The current $S$ is carried by the actual pole union $\bigcup _ jH_ j\subset \{ w=-\infty \} $. Thus $\langle T\rangle =K$. For every smooth real $(1,1)$-form $L$ with $\mathrm{dd}^{\mathrm{c}}L=0$, distributional Stokes gives

Closed top mass and smooth correction. Put $T_ u=C\kappa +\mathrm{dd}^{\mathrm{c}}u=S+(C-A)\kappa $. For the finite partial sums $u_ N=\sum _{j\leq N}a_ j\ell _ j$, the closed non-pluripolar mass is $(C-\sum _{j\leq N}a_ j)^2$: outside the finite divisor union the current is the smooth residual, and the measure gives that union zero mass. Since $u_0\leq u_ N$, closed-background mass monotonicity, the closed case of theorem 13.17.1, gives

For the reverse inequality, locally write $S=\mathrm{dd}^{\mathrm{c}}b$ with $b$ plurisubharmonic. The currents $(C-A)\kappa +\mathrm{dd}^{\mathrm{c}}\max (b,-k)$ have bounded local potentials and dominate $(C-A)\kappa $. Their squared products, restricted to $\{ b>-k\} $, therefore dominate $(C-A)^2\kappa ^2$ on that set. Letting $k$ increase and using the local truncation definition gives $\langle T_ u^2\rangle \geq (C-A)^2\kappa ^2$. The two bounds prove equality of total masses. Closed mass invariance under bounded changes then gives

Set $L=C\kappa -\theta \geq 0$, so $\mathrm{dd}^{\mathrm{c}}L=0$. Expanding the bounded products of $\max (w,-k)$ in the two backgrounds and restricting to $\{ w>-k\} $ gives, in the limit,

All correction measures are finite: the closed top measure has the mass just computed, the degree-one limit is $K$, and the smooth measure $L^2$ gives the pole set zero mass. The same truncations are used in both backgrounds, so this identity is an equality of Borel measures. Integrating yields

This holds throughout the ordinary type, proving both singularity-volume identities and their strict positivity. $\square$

Proof. In a coordinate ball choose a smooth real cutoff $\chi $ equal to one near the origin, and extend $\alpha =\chi |z_2|^2\mathrm dz_1$ by zero. Put

It is real and smooth, with $\mathrm{dd}^{\mathrm{c}}\tau =0$. Where $\chi =1$,

Choose $\epsilon >0$ small enough that $\theta =\kappa +\epsilon \tau \geq \kappa /2$. Set $A=1/4$, $a_ j=A2^{-j}$, $t_ j=2^{-j}$ and

The Vandermonde determinant shows that any three defining linear forms are independent; hence the hyperplanes are not in one pencil. Here $R=\theta -A\kappa \geq \kappa /4$, so $\int _ XR^2\geq 1/16$. The preceding proposition gives the mass throughout the ordinary type and both singularity volumes. There are infinitely many positive base divisor coefficients. An analytic singularity on a compact manifold has only finitely many prime divisorial components, so no analytic potential has this ideal type. $\square$