13.27 A complete metric on positive-volume models7EAA1D7446C4

Proof. Write $|Z|^2=|Z_0|^2+|Z_1|^2+|Z_2|^2$ in homogeneous coordinates. Fix a rational $a\in (0,1/2)$ and put

With $H_ i=\{ Z_ i=0\} $, the corresponding currents are $a[H_0]+(1-a)\theta $, $a[H_1]+(1-a)\theta $ and $a[H_0]+a[H_1]+(1-2a)\theta $. The background is closed, so each singularity volume equals the non-pluripolar mass, which is the mass of the smooth part:

The maximum $\max (u,v)$ has the ordinary type of

To compute its volume, blow up the common point $[0:0:1]$ of $H_0$ and $H_1$. Denote the blowup by $\pi :Y\to \mathbb P^2$, its exceptional divisor by $E$, and the pencil map $[Z_0:Z_1]$ by $f:Y\to \mathbb P^1$. For the Fubini–Study form $\omega _1$ of mass one on $\mathbb P^1$, the Poincaré–Lelong formula gives

Here $\beta $ is smooth and semipositive. Also

The mixed integral is one because every fiber of the pencil maps onto a line. By lemma 10.17.3 and change of variables off $E$,

Since $w\leq u$ and $w\leq v$, (13.22.2) and the definition of $d_\theta $ give

Set $p=P_\theta [u]$, $q=P_\theta [v]$ and $r=P_\theta [w]$. By proposition 18.4.8, these models have the same analytic singularity types as $u$, $v$ and $w$, and their lower volumes are at least $(1-2a)^2$. By projection invariance (proposition 13.22.4), they have the same comparisons, and the difference between the two sides is $2a^2>0$. $\square$

Proof. Put $s=d_\theta (p,q)$ and $m=\max (p,q)$. If $s\geq \delta /2$, the bound follows from $|L(p)-L(q)|\leq M$, and if $s=0$, then $p=q$ by proposition 13.22.4. Suppose $0<s<\delta /2$ and take $t=(2s/\delta )^{1/n}$. The upper-volume deficit from either $p$ or $q$ to $m$ is at most $s$, while $L(m)\geq \delta $. Thus lemma 13.18.1 gives a potential $h\in \mathrm{PSH}(X,\theta )$ with

Volume addition and monotonicity imply $L(p)\geq (1-t)^ nL(m)\geq (1-t)^ nL(q)$, so $L(q)-L(p)\leq (1-(1-t)^ n)M\leq nMt$. Interchanging $p$ and $q$ proves the bound.

For the last assertion, put $s_ j=d_\theta (p_ j,p)$. Whenever $0<s_ j<L(p)/2$, the same stretching argument applied to $\max (p_ j,p)$ gives

If $s_ j=0$, then $p_ j=p$. Hence $L(p_ j)\geq L(p)/2$ for all sufficiently large $j$, and (13.27.3) with $\delta =L(p)/2$ gives $L(p_ j)\to L(p)$. $\square$

Proof. Step 1: a decreasing family of symmetric relations. Set $C=1+M$ and $m(p)=L(p)/C$, so that $0<m(p)<1$. Put $a_1=1$, $a_{k+1}=na_ k+1$, and choose $0<\varepsilon _1<1$. Recursively choose $\varepsilon _{k+1}>0$ so small that, with $r_ k=4^{a_{k+1}}\varepsilon _{k+1}$,

Let $\mathcal U_0=\mathcal M_+(\theta )^2$ and, for $k\geq 1$, set

These symmetric sets contain the diagonal and decrease with $k$, and their intersection is the diagonal by proposition 13.22.4. We claim that

where composition of relations means joining successive pairs. For $k=0$ this is immediate. For $k\geq 1$, suppose $(p_{i-1},p_ i)\in \mathcal U_{k+1}$ for $i=1,2,3$, and put $m_*=\min _{0\leq i\leq 3}m(p_ i)$. The reciprocal differences along this chain have total less than $3\varepsilon _{k+1}<1$, so $1/m(p_ i)>1/m_*-1$. Hence $m(p_ i)\leq 4m_*$ for every $i$: if $m_*\leq 3/4$, this follows from $1/m_*-1\geq 1/(4m_*)$, and otherwise from $m(p_ i)<1$. Each successive comparison is therefore less than

All four models have lower volume at least $Cm_*$. Applying (13.22.6) with $\delta =Cm_*$ twice, first with middle point $p_2$ and then with middle point $p_1$, yields

Here $(a_{k+1}-1)/n=a_ k$, $a_{k+1}\geq a_ k$ and $m_*<1$. The endpoint reciprocal difference is less than $3\varepsilon _{k+1}<\varepsilon _ k$, proving (13.27.6).

Step 2: the chain metric. Define

where the infimum runs over finite chains $p=p_0,p_1,\ldots ,p_ N=q$ and integers $k_ i\geq 0$ with $(p_{i-1},p_ i)\in \mathcal U_{k_ i}$; we call $\sum _ i2^{-k_ i}$ the cost of the chain. A single edge in $\mathcal U_0$ gives $\mathfrak d_\theta \leq 1$. Reversing and concatenating chains gives symmetry and the triangle inequality, while diagonal edges with arbitrarily large indices give $\mathfrak d_\theta (p,p)=0$.

We next prove the inclusions

The second inclusion follows by using a single edge. For the first, we prove by induction on the number of edges that every chain of cost less than $2^{-k}$ has its endpoints in $\mathcal U_{k-1}$, simultaneously for all $k\geq 1$. For one edge this follows because the $\mathcal U_ k$ decrease. For a chain of cost $A<2^{-k}$, choose $l\geq k$ with

Split the chain at the first edge at which the cumulative cost reaches $2^{-(l+1)}$. The subchains before and after this edge both have cost less than $2^{-(l+1)}$, so by induction their endpoint pairs lie in $\mathcal U_ l$; an empty subchain is interpreted as a diagonal pair. Every edge cost is a power of two smaller than $2^{-l}$, so the middle edge lies in $\mathcal U_{l+1}\subset \mathcal U_ l$. By (13.27.6), the endpoints of the whole chain form a pair in $\mathcal U_{l-1}\subset \mathcal U_{k-1}$. This completes the induction and proves (13.27.8). Since the $\mathcal U_ k$ intersect in the diagonal, $\mathfrak d_\theta $ separates points, so it is a metric.

Step 3: convergence and completeness. By (13.27.8), $\mathfrak d_\theta (p_ j,p)\to 0$ if and only if $(p_ j,p)$ eventually lies in every $\mathcal U_ k$. This implies $d_\theta (p_ j,p)\to 0$. Conversely, if $d_\theta (p_ j,p)\to 0$, then proposition 13.27.2 gives $L(p_ j)\to L(p)>0$, so for each fixed $k$ both inequalities defining $\mathcal U_ k$ hold eventually. This proves (13.27.5).

The same inclusions show that a $\mathfrak d_\theta $-Cauchy sequence is $d_\theta $-Cauchy and has Cauchy reciprocal lower volumes. These reciprocals are bounded, so a tail of the sequence lies in some $\mathcal M_\delta (\theta )$. By theorem 13.22.7, the sequence converges there for $d_\theta $, and hence for $\mathfrak d_\theta $. This proves completeness. Conversely, a $d_\theta $-Cauchy sequence with Cauchy reciprocal lower volumes has a common positive lower-volume bound on a tail, and then both inequalities defining each $\mathcal U_ k$ hold for all pairs of sufficiently late terms. This proves the Cauchy criterion.

On $\mathcal M_\delta (\theta )$, (13.27.3) controls the reciprocal differences uniformly in terms of $d_\theta $, while $m(p)\geq \delta /C$ controls the second threshold in $\mathcal U_ k$. Thus $\mathfrak d_\theta $ and $d_\theta $ define the same uniform structure there. Continuity of the lower volume makes $\mathcal M_\delta (\theta )$ closed for $\mathfrak d_\theta $, so the restriction of $\mathfrak d_\theta $ to it is complete. $\square$

Proof. By theorem 13.24.5, an ordinary model of positive lower volume that is a $d_\theta $-limit of ideal models is again an ideal model. The same theorem, which relies on theorem 13.24.3, gives the density of analytic models. Now apply (13.27.5) and the completeness of $\mathfrak d_\theta $. $\square$