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The maximum of two potentials is a common upper bound for both, and the sum of the two upper-volume deficits from this maximum measures how far apart their ordinary envelopes are. A uniform positive lower-volume bound controls how this quantity changes when a third potential is introduced, and this replaces the triangle inequality in the Cauchy argument below. From now on we write $U(u)=\overline{\operatorname {vol}}_\theta (u)$ and $M=U(V_\theta )$; by monotonicity, $U\leq M$ on $\mathrm{PSH}(X,\theta )$. We also fix a smooth positive probability volume form $\mathrm{d}V$ for $L^1$ norms.
Definition 13.22.1.A220D62C8412 For $u,v\in \mathrm{PSH}(X,\theta )$, define
\[ d_\theta (u,v)=2U(\max (u,v))-U(u)-U(v). \]
For $\delta >0$, put
\[ \mathcal M_\delta (\theta )= \{ u\in \mathrm{PSH}(X,\theta ):P_\theta [u]=u, \ \underline{\operatorname {vol}}_\theta (u)\geq \delta \} . \]
We call $d_\theta $ the volume comparison. A sequence is $d_\theta $-Cauchy if $d_\theta (u_ j,u_ k)\to 0$ as $j,k\to \infty $, and it converges to $u$ for $d_\theta $ if $d_\theta (u_ j,u)\to 0$.
The volume comparison is finite, nonnegative and symmetric. It need not satisfy the triangle inequality (example 13.27.1), which is why we do not call it a distance. It depends only on the ordinary singularity types, since replacing either input by an equivalent one changes the maximum by a bounded amount. Monotonicity also gives
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13.22.2
\begin{equation} |U(u)-U(v)|\leq d_\theta (u,v),\qquad u\leq v\ \Longrightarrow \ d_\theta (u,v)=U(v)-U(u). \end{equation}
In the separation result below, the restriction to positive lower volume is necessary, even for a closed positive background.
Example 13.22.3.4B14FA731306 For every $n\geq 2$ there are a closed positive $(1,1)$-form $\theta $ on the projective manifold $\mathbb P^1\times \mathbb P^{n-1}$ and two ordinary models $a\not\sim b$ in $\mathrm{PSH}(\mathbb P^1\times \mathbb P^{n-1},\theta )$ such that
\[ P_\theta [a]_{\mathcal I}=b,\qquad \underline{\operatorname {vol}}_\theta (a) =\underline{\operatorname {vol}}_\theta (b)=0,\qquad d_\theta (a,b)=0. \]
For every $n\geq 3$, the same assertions hold for a suitable nonclosed positive form $\theta $ on a projective manifold of dimension $n$ with the bounded mass property.
Proof.
Take the closed positive product background of example 18.15.4 on $\mathbb P^1\times \mathbb P^{n-1}$. Its models satisfy $a\leq b$ and $U(a)=U(b)=0$. Thus $\max (a,b)=b$, and the defining formula gives
\[ d_\theta (a,b)=2U(b)-U(a)-U(b)=0. \]
The other assertions are part of that example. For the nonclosed case use example 18.15.5; its models are again ordered and have zero upper volumes, so the same calculation applies.
$\square$
Proposition 13.22.4.38359A611D4C For potentials $u,v$ of positive lower volume,
\[ d_\theta (u,v)=d_\theta (P_\theta [u],P_\theta [v]),\qquad d_\theta (u,v)=0\ \Longleftrightarrow \ P_\theta [u]=P_\theta [v]. \]
There is $C_\theta >0$, depending on $\theta $ and $\mathrm{d}V$, such that, for $u,v\in \mathcal M_\delta (\theta )$,
\[ \| u-v\| _{L^1}\leq 2C_\theta \min \left\{ 1, \left(\frac{2d_\theta (u,v)}\delta \right)^{1/n}\right\} . \]
Proof.
Translate the inputs below zero, and put $p=P_\theta [u]$, $q=P_\theta [v]$, $m=\max (u,v)$ and $r=P_\theta [m]$. Since $u\leq p$, $v\leq q$ and the envelope is monotone,
\[ m\leq \max (p,q)\leq r. \]
All these potentials have positive lower volume. Theorem 13.19.5 and monotonicity therefore give $U(m)=U(\max (p,q))=U(r)$, while $U(u)=U(p)$ and $U(v)=U(q)$. This proves the first identity. If $d_\theta (u,v)=0$, then $U(u)=U(v)=U(m)$, since both are at most $U(m)$. Thus $u,v\in \mathcal E_{\mathrm{vol}}(X,\theta ,m)$, and theorem 13.18.5, proposition 16.28.65 give $P_\theta [u]=P_\theta [m]=P_\theta [v]$. Conversely, if $p=q$, the first identity gives $d_\theta (u,v)=d_\theta (p,p)=0$.
Now let $u,v\in \mathcal M_\delta (\theta )$ and set $s=d_\theta (u,v)$. By theorem 5.25.2, there is $C_\theta $ bounding the $L^1$ norm of every $\theta $-plurisubharmonic potential with supremum zero. Since ordinary models have supremum zero, the estimate holds if $s\geq \delta /2$. Suppose $0<s<\delta /2$ and put $t=(2s/\delta )^{1/n}$. The upper-volume deficit from $u$ to $m$ is at most $s$, and the lower volume of $m$ is at least $\delta $. Thus lemma 13.18.1, with $b=t^{-1}$, gives $h\in \mathrm{PSH}(X,\theta )$ such that
\[ (1-t)m+th\leq u. \]
For each competitor $g$ for $P_\theta [h]$ (so $g\leq 0$ and $g\preceq h$), the potential $(1-t)m+tg$ lies below zero and is more singular than $u$. Since $u=P_\theta [u]$, it lies below $u$, and taking the envelope gives
\[ (1-t)m+tP_\theta [h]\leq u. \]
Hence $0\leq m-u\leq t(m-P_\theta [h])\leq -tP_\theta [h]$, and $\int _ X(m-u)\, \mathrm{d}V\leq tC_\theta $ because $P_\theta [h]$ has supremum zero. The same argument applies to $v$, and adding the two estimates gives the assertion, since $|u-v|\leq (m-u)+(m-v)$. If $s=0$, the argument works for every $t\in (0,1)$; let $t\downarrow 0$.
$\square$
Although the triangle inequality fails, the volume comparison satisfies it up to an error that is controlled uniformly on $\mathcal M_\delta (\theta )$.
Proposition 13.22.5.412DBAD823B9 For $u,v,w\in \mathcal M_\delta (\theta )$,
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13.22.6
\begin{equation} d_\theta (u,w)\leq d_\theta (u,v)+d_\theta (v,w) +2nM\min \left\{ 1, \left(\frac{2d_\theta (v,w)}\delta \right)^{1/n}\right\} . \end{equation}
Consequently every $d_\theta $-convergent sequence in $\mathcal M_\delta (\theta )$ is $d_\theta $-Cauchy, and its limit is unique.
Proof.
Put $s=d_\theta (v,w)$. If $s\geq \delta /2$, the inequality follows from $d_\theta (u,w)\leq 2M$. If $0<s<\delta /2$, set $t=(2s/\delta )^{1/n}$. Lemma 13.18.1, applied to $v\preceq \max (v,w)$, gives $h\in \mathrm{PSH}(X,\theta )$ with
\[ (1-t)\max (v,w)+th\leq v. \]
This inequality also gives $h\leq v$. Set $r=\max (u,v)$ and $z=\max (u,v,w)$. Where $z=\max (v,w)$, the inequality above bounds $(1-t)z+th$ by $v\leq r$; where $z=u$, both $u$ and $h$ lie below $r$. Thus
\[ (1-t)z+th\leq r. \]
Addition, homogeneity and volume monotonicity give
\[ U(z)-U(r)\leq \bigl (1-(1-t)^ n\bigr )U(z)\leq nMt. \]
Since $\max (u,w)\leq z$, it follows that
\[ \begin{split} d_\theta (u,w) & \leq 2U(r)+2nMt-U(u)-U(w)\\ & =d_\theta (u,v)+U(v)-U(w)+2nMt\\ & \leq d_\theta (u,v)+s+2nMt. \end{split} \]
Here the last step uses (13.22.2). For $s=0$, the same proof works with every $t\in (0,1)$; let $t\downarrow 0$. This proves (13.22.6). Applying it with the limit as the middle point shows that a convergent sequence is Cauchy. If a sequence converges to two limits, applying it with a member of the sequence as the middle point shows that the two limits have volume comparison zero; by proposition 13.22.4, they are equal.
$\square$
Theorem 13.22.7.62ECCE0A41E6 Every $d_\theta $-Cauchy sequence in $\mathcal M_\delta (\theta )$ converges for $d_\theta $ to a unique element of $\mathcal M_\delta (\theta )$. It also converges to that element in $L^1(X,\mathrm{d}V)$.
Proof.
The plan is to dominate a fast subsequence by the ordinary envelopes of its tail suprema; these decrease to the limit, and the stretching lemma controls the volume they add. Choose a subsequence $a_ j$ of the given sequence such that
\[ d_\theta (a_ j,a_{j+1})<\delta \, 2^{-(j+1)n} \qquad (j\geq 1). \]
For $k\geq 0$, set
\[ v_ j^ k=\max (a_ j,\ldots ,a_{j+k}),\qquad v_ j=\left(\sup _{l\geq j}a_ l\right)^*. \]
For every $l\geq j$, lemma 13.18.1 with $b=2^ l$, applied to $a_ l\preceq \max (a_ l,a_{l+1})$, gives $h_ l\leq a_ l$ satisfying
\[ (1-2^{-l})\max (a_ l,a_{l+1})+2^{-l}h_ l\leq a_ l. \]
The argument by cases in the proof of proposition 13.22.5 gives
\[ (1-2^{-l})v_ j^{l+1-j}+2^{-l}h_ l\leq v_ j^{l-j}. \]
Consequently
\[ U(v_ j^{l+1-j})-U(v_ j^{l-j})\leq nM2^{-l}. \]
Summing from $l=j$ to $j+k-1$ and then letting $k\to \infty $, using theorem 13.21.1, yields
\[ 0\leq U(v_ j)-U(a_ j)\leq nM2^{1-j}. \]
Here $v_ j\in \mathrm{PSH}(X,\theta )$ has lower volume at least $\delta $, because $a_ j\leq v_ j\leq 0$; this also justifies the use of that theorem.
Put $q_ j=P_\theta [v_ j]$. By theorem 13.19.5, $q_ j$ is an ordinary model with the same two volumes as $v_ j$. The sequence $q_ j$ decreases, and each $q_ j$ has supremum zero. By normalized compactness, its decreasing limit $q$ is not identically $-\infty $, so $q\in \mathrm{PSH}(X,\theta )$. By theorem 13.21.2,
\[ q=P_\theta [q],\qquad \underline{\operatorname {vol}}_\theta (q) =\lim _ j\underline{\operatorname {vol}}_\theta (q_ j)\geq \delta , \qquad U(q_ j)\longrightarrow U(q). \]
Both $a_ j$ and $q$ lie below $q_ j$. Hence
\[ \begin{split} d_\theta (a_ j,q) & \leq \bigl (U(q_ j)-U(a_ j)\bigr )+\bigl (U(q_ j)-U(q)\bigr )\\ & \leq nM2^{1-j}+U(q_ j)-U(q)\longrightarrow 0. \end{split} \]
To treat the whole sequence, apply (13.22.6) with a late member $a_ j$ of the subsequence as the middle point. The Cauchy property makes the first term small for all sufficiently late members of the original sequence, while $d_\theta (a_ j,q)$ makes the other two terms small. Thus the original sequence converges to $q$. Uniqueness follows from proposition 13.22.5, and the $L^1$ estimate in proposition 13.22.4 proves the last assertion.
$\square$
The non-strict inequality $\underline{\operatorname {vol}}_\theta \geq \delta $ in the definition of $\mathcal M_\delta (\theta )$ is essential for completeness.
Example 13.22.8.AB2E8992474C Let $\theta $ be a smooth positive $(1,1)$-form of integral one on a compact Riemann surface, and let $0<\delta <1$. The ordinary models of mass strictly greater than $\delta $ do not form a complete space for $d_\theta $.
Proof.
Fix a point $x$, and let $g_ x\leq 0$ be the Green function used in theorem 20.3.10, with $\mathrm{dd}^{\mathrm{c}}g_ x=\delta _ x-\theta $, where $\delta _ x$ is the Dirac measure at $x$. For $0\leq a<1$, put $q_ a=P_\theta [ag_ x]$. The current $\theta +\mathrm{dd}^{\mathrm{c}}(ag_ x)$ equals $a\delta _ x+(1-a)\theta $, with non-pluripolar mass $1-a$. Since the background is closed, both singularity volumes of $ag_ x$ equal this mass, and theorem 13.19.5 gives
\[ q_ a=P_\theta [q_ a],\qquad \underline{\operatorname {vol}}_\theta (q_ a)=U(q_ a)=1-a. \]
Since $g_ x\leq 0$, the potentials $q_ a$ decrease as $a$ increases, and (13.22.2) gives $d_\theta (q_ a,q_ b)=|a-b|$. Choose $a_ j<1-\delta $ with $a_ j\uparrow 1-\delta $. The sequence $q_{a_ j}$ is $d_\theta $-Cauchy, and every term has mass $1-a_ j>\delta $. By (13.22.2), any $d_\theta $-limit has upper volume $\lim _ j(1-a_ j)=\delta $. In this closed background the two volumes coincide, so no such limit has mass strictly greater than $\delta $.
$\square$
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