18.26 Complete model topologies987BD480177F

Proof. Choose the cutoff potentials $u_ a$ of lemma 5.37.2 for the single point $x$, with $a_0$ smaller than the reciprocal of the constant in that lemma, and put $p_ a=P_\omega [u_ a]$. By proposition 18.4.8, $p_ a$ is an ideal model with analytic singularities, ordinarily equivalent to $u_ a$. It is locally bounded away from $x$, and its only pole has the stated logarithmic coefficient. Distinct coefficients give unbounded differences near $x$. Formula (18.26.3) follows from lemma 5.37.1 with codimension $n$, and its exponents distinguish any two coefficients at sufficiently large weights.

It remains to show that the current has no coefficient mass at $x$. In coordinates centered at $x$, let $\beta =\mathrm{dd}^{\mathrm{c}}|z|^2$ and let

This is a positive measure. Choose smooth cutoff functions $0\leq \chi _\varepsilon \leq 1$, equal to one on the ball of radius $\varepsilon $ and supported in the ball of radius $2\varepsilon $, with $-C\varepsilon ^{-2}\beta \leq \mathrm{dd}^{\mathrm{c}}\chi _\varepsilon \leq C\varepsilon ^{-2}\beta $; their complex Hessians are supported in the annulus between the two balls. Since $\beta $ is closed, distributional integration by parts gives

The logarithmic bound for $p_ a$ on the annulus and the Euclidean volume of the annulus therefore give

This is where $n\geq 2$ is used. By positivity, the coefficient measures of $\omega +\mathrm{dd}^{\mathrm{c}}p_ a$ are bounded by its trace measure $\nu $, so they have no mass at $x$. Hence the restriction to the pole set is zero, as asserted. The zero potential is an ordinary and an ideal model because $\omega >0$, and its pole set is empty. $\square$

Proof. On relatively compact blowup charts, with $z=(s,sw)$, one has $p_ a\circ \pi =a\log |s|^2+O(1)$ along $E=\{ s=0\} $. Thus the pole set is $E$ and the generic Lelong number along it is $a$. Add a local smooth function $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \pi ^*\omega $. By the argument of lemma 5.38.1, the restriction of the closed positive current $\mathrm{dd}^{\mathrm{c}}(p_ a\circ \pi +\rho )$ to $E$ is $a[E]$, and subtracting the smooth correction does not change the coefficient mass on $E$. Consequently $\mu _ a=a[E]\wedge \eta ^{n-1}$. The positive measure $[E]\wedge \eta ^{n-1}$ has mass $\int _ E\eta ^{n-1}>0$, which proves the formula for the total variation. $\square$

Proof. For each $j$, choose a proper closed analytic subset $S_ j\subset X$ over whose complement $\pi _ j$ is an isomorphism. Each $S_ j$ is nowhere dense, so the Baire category theorem gives a point $x\notin \bigcup _ jS_ j$. Apply proposition 18.26.2 at this point.

Each $\pi _ j$ is biholomorphic near the unique pole $y_ j$ of $p_ a\circ \pi _ j$, and the pulled-back potential has the same logarithmic coefficient there. Its smooth background and positive curvature current satisfy the trace estimate in the proof of proposition 18.26.2, which does not use strict positivity of the background. Thus the coefficient measures have no mass at $y_ j$, and their restriction to the pole set is zero. The same proposition gives the ordinary inequivalence and the distinct multiplier ideals. $\square$

Proof. The form $\pi ^*\omega $ is semipositive, and it is positive on the nonempty open set where $\pi $ is an isomorphism, so $A_\pi >0$. By proposition 5.43.18, $u\circ \pi \in \mathrm{PSH}(Y_\pi ,\pi ^*\omega )$. Since $\mathrm{dd}^{\mathrm{c}}(\eta _\pi ^{n-1})=0$, distributional Stokes gives

Thus $\mu _\pi (u)$ is a positive measure of mass at most $A_\pi $, and the normalized total variation distance of two such measures is at most two. Symmetry and the triangle inequality persist under the supremum, which proves the pseudometric assertion.

Let $p,q$ be ordinary models with analytic singularities, choose a common log resolution $\pi $ in $\mathscr R$, and suppose that $\delta _\omega ^{\mathrm{pol}}(p,q)=0$. Then their measures in this coordinate agree. Let $E_ i$ be the prime components of the resolved ideals on $Y_\pi $, and $a_ i,b_ i$ the coefficients of $p$ and $q$ along them. The divisorial restriction argument of lemma 5.38.1 gives

Indeed, a smooth local correction makes the complex Hessian closed and positive without changing its coefficient mass on a divisor. Distinct components do not share trace mass on their intersections, and each component has strictly positive trace mass. Equality of these measures therefore implies $a_ i=b_ i$ for every $i$. The criterion of lemma 5.38.11, which uses only the finitely many divisors $E_ i$, applied in both directions gives $p\sim q$. An ordinary model is the ordinary envelope of every potential of its type, so ordinary models with the same type are equal. This proves separation.

On a curve, every smooth modification is an isomorphism and $\eta _\pi ^{n-1}=1$. Hence $A_\pi =\int _ X\omega $, and pullback preserves the total variation of polar curvature. The formula on curves and the assertions following it are therefore consequences of corollary 13.28.7, proposition 13.28.11.

Finally, $u-f$ has, in background $\omega '$, the same pulled-back curvature current and the same pole set as $u$ in background $\omega $, and Stokes' theorem gives $A_{\pi ,\omega '}=A_{\pi ,\omega }$. The gauge formula follows. It holds for the potentials themselves, before any ordinary projection. $\square$

Proof. The metric is finite, bounded by two, and separates points. A Cauchy sequence is Cauchy in total variation in each coordinate, and completeness of the space of finite signed measures gives a limit there; positivity and the mass bound pass to the limit. Given $\varepsilon >0$, choose $N$ such that every normalized coordinate distance is at most $\varepsilon $ for all $j,k\geq N$. Letting $k$ tend to infinity separately in each coordinate preserves this inequality, and taking the supremum proves convergence for $\widehat\delta $. The last assertion follows from the definition and the mass bound proved above. $\square$

Proof. Boundedly equivalent potentials have boundedly equivalent pullbacks on every modification, with the same pole set. Proposition 5.46.3 therefore gives equality of their polar currents and of their measures $\mu _\pi $ in every coordinate, which proves descent to ordinary classes.

If $p=P_\omega [p]$, the version of proposition 18.5.1 for ordinary envelopes gives

Thus $G_ f(p)\sim p-f$. Since ordinary envelopes depend only on the ordinary type, $P_{\omega '}[G_ f(p)]=P_{\omega '}[p-f]=G_ f(p)$, and similarly $P_\omega [G_ f(p)+f]=P_\omega [p]=p$. The same argument applied to models in background $\omega '$ proves surjectivity and the formula for the inverse. Invariance of the polar coordinates and the gauge identity of proposition 18.26.8 now give

Subtraction of a smooth function and bounded equivalence preserve analytic singularities. If $p=P_\omega [p]_{\mathcal I}$, the gauge estimate for ideal envelopes gives $P_{\omega '}[p-f]_{\mathcal I}\sim p-f\sim G_ f(p)$. By corollary 18.4.4, this ideal envelope is an ordinary model, and two ordinary models of the same ordinary type coincide. Hence it equals $G_ f(p)$, which shows that $G_ f$ maps ideal models to ideal models; the inverse map gives the converse. $\square$

Proof. By proposition 5.43.18, the pullbacks belong to $\mathrm{PSH}(Y_\pi ,\pi ^*\omega )$, and proposition 5.46.7, applied with background $\pi ^*\omega $, shows that their polar currents are closed and positive. Both currents are carried by the union of the two Borel pole sets, which has Lebesgue measure zero. Equality of their $\eta _\pi $-traces therefore gives equality of the currents by lemma 5.46.4, and the converse is immediate. Since every $A_\pi $ is positive, the last assertion follows. $\square$

Proof. Fix $\pi $. By proposition 18.26.9, the measures $\mu _\pi (p_ j)$ converge in total variation to a positive measure $\lambda _\pi $. Put

This is a Borel set of Lebesgue measure zero. Every measure $\mu _\pi (p_ j)$ vanishes on its complement, so $\lambda _\pi (Y_\pi \setminus B_\pi )=0$. The polar currents are closed and positive by corollary 18.26.11, and their $\eta _\pi $-trace masses are at most $A_\pi $. By compactness of positive currents of bounded mass, a subsequence converges weakly. Its limit $R$ is closed and positive, and $R\wedge \eta _\pi ^{n-1}=\lambda _\pi $. Since positivity bounds all coefficient measures by the Hermitian trace, $R$ is carried by $B_\pi $. Any two subsequential limits have the same null carrier and the same trace, so they agree by lemma 5.46.4. As every subsequence has a further convergent subsequence, the whole sequence of currents converges weakly to this unique limit $R_\pi $.

Apply proposition 5.46.5 to $g$, with background $\pi ^*\omega $ and potential $p_ j\circ \pi $. It gives $g_*\operatorname {Pol}_{\pi '}(p_ j)=\operatorname {Pol}_\pi (p_ j)$. Proper pushforward is weakly continuous, since a smooth test form downstairs pulls back to a smooth test form upstairs, and passing to the limit proves compatibility. The limits are constructed separately in each coordinate, so no diagonal subsequence over $\mathscr R$ is needed.

To obtain one pole carrier, put $h_ m=\sum _{j=1}^ m2^{-j}p_ j$. These functions decrease, and they are $\omega $-plurisubharmonic because the coefficients have sum at most one and $\omega >0$. Normalized compactness, theorem 5.25.2, bounds their $L^1$ norms uniformly, so corollary 5.25.3 gives a limit $h\in \mathrm{PSH}(X,\omega )$. Since all summands are nonpositive, each pole set of a $p_ j$ is contained in the pole set of $h$. Hence $B_\pi \subset \{ h\circ \pi =-\infty \} $, which gives the last assertion. $\square$

Proof. At $x=[1:0:0]$, put

The homogeneous quotient defines a global nonpositive potential. Its curvature is positive, since $\log (|Z_1|^2+|Z_2|^2)$ is plurisubharmonic in a local trivialization and the remaining background is $(1-a)\omega $. Its only pole is $x$, where it is $a\log |z|^2+O(1)$. By proposition 18.4.8, $p_ x=P_\omega [u_ x]$ is an ideal model with analytic singularities, ordinarily equivalent to $u_ x$. Unitary projective automorphisms preserve $\omega $ and commute with its envelopes for the zero obstacle. Transporting this construction to pairwise distinct points $x_ j$, we obtain models $p_ j$.

Fix $\pi :Y_\pi \to X$ and some Gauduchon form $\gamma _\pi $ on $Y_\pi $. The closed polar current of $p_ j\circ \pi $ is supported on the analytic fiber over $x_ j$. The divisorial restriction in lemma 5.38.1 and the support theorem, [Chapter III, Corollaries 2.11 and 2.14, pp. 141–143, DemCADG], give

where the sum runs over the irreducible curves $E\subset Y_\pi $ with $\pi (E)=x_ j$. Indeed, corollary 18.26.11 shows that the polar current is closed, and after the curve components are removed, the remaining closed current of order zero is supported at finitely many points and therefore vanishes. The sum is finite. Every such curve is contracted by $\pi $, so the restriction of $\pi ^*\omega $ to it is zero. Put

A modification of a surface has finitely many exceptional curves, so there are only finitely many points over which $\pi $ has exceptional curves. Since the points $x_ j$ are distinct, $J_\pi $ is finite. Choose $\varepsilon _\pi >0$ with $\varepsilon _\pi C_{\pi ,j}\leq 1/j$ for every $j\in J_\pi $; if $J_\pi $ is empty, take $\varepsilon _\pi =1$. Now fix

This form is positive and Gauduchon, since on a surface the Gauduchon equation is $\mathrm{dd}^{\mathrm{c}}\eta _\pi =0$. Moreover

These forms are fixed for the whole sequence, and the estimate holds for every modification. Since $\mu _\pi (0)=0$, taking the supremum gives the asserted convergence.

To compute the singularity volumes, let $\rho :Y\to X$ be the blowup at $x$, with exceptional curve $E$, and let $F:Y\to \mathbb P^1$ be the morphism resolving the pencil $[Z_1:Z_2]$. Write $\omega _1$ for the normalized Fubini–Study form on $\mathbb P^1$. Then

For every $v\sim u_ x$, divisorial removal gives $H_ a+\mathrm{dd}^{\mathrm{c}}w\geq 0$, where $w=v\circ \rho -u_ x\circ \rho $ is bounded after extension across $E$. The bounded product of proposition 5.28.2 assigns no mass to $E$, so change of variables off $E$ identifies the non-pluripolar mass of $v$ with $\int _ Y(H_ a+\mathrm{dd}^{\mathrm{c}}w)^2$. Since $H_ a$ is closed, integration by parts for bounded potentials gives

For the last equality, note that $H_ a$ represents the class $H-aE$, where $H=\rho ^*[\omega ]$, and that $H^2=1$, $H\cdot E=0$ and $E^2=-1$. Hence both singularity volumes of $p_ x$, and by unitary transport those of every $p_ j$, are equal to $1-a^2$. The zero model has both volumes equal to one. Since $p_ j\leq 0$, definition 13.22.1 gives $d_\omega (p_ j,0)=1-(1-a^2)=a^2$. This also proves the common positive lower bound for the volumes and the incompatibility of the topologies. $\square$

Proof. Use the binary Cantor set and the nonatomic probability measure $\sigma $ of example 13.28.14, placed in a small real interval in $\Delta $. Its intervals of level $k$ have length $r_ k=\exp (-4^{k+1})$ and mass $2^{-k}$. For every $s>0$,

Covering $K$ by its $2^ k$ intervals of level $k$, and $S$ by the $2^{nk}$ products of such intervals, which have diameter at most $\sqrt nr_ k$, shows that $K$ and $S$ have Hausdorff dimension zero. Define the Green potential of the disk

Then $\mathrm{dd}^{\mathrm{c}}g=\sigma $, with the normalization of $\mathrm{dd}^{\mathrm{c}}$ fixed in the conventions. The Green kernel is nonpositive, vanishes on the boundary and is smooth away from its pole. On a fixed smaller disk it is at most $\log |z-y|^2+C$. For $z\in K$, integration over the interval of level $k$ containing $z$, together with nonpositivity of the kernel elsewhere, gives

Outside $K$ the potential is finite, smooth and harmonic. Its Lelong number at $a$ is $\sigma (\{ a\} )=0$. Since $K$ stays away from the boundary, $g$ is smooth near the boundary and tends uniformly to zero there.

Near a point off $S$, some coordinate $z_ j$ lies outside $K$, and the finite value $g(z_ j)$ locally exceeds all branches $g(z_ i)$ with $z_ i$ near $K$. Hence $h$ is locally the maximum of at most $n$ pluriharmonic functions, and in particular it is locally Lipschitz. For $m\leq n$ pluriharmonic functions $a_ j$, use the smooth approximations

which decrease to $\max _ ja_ j$ as $\varepsilon \downarrow 0$. Their complex Hessian is a positive multiple of the weighted covariance of the forms $\partial a_ j$, with probability weights proportional to $e^{a_ j/\varepsilon }$. The centered forms satisfy one weighted linear relation, so this Hessian has rank at most $m-1<n$, and its $n$th power vanishes. Bounded decreasing convergence in theorem 5.27.1 gives the same conclusion for the maximum.

The bounded truncation $h_ L$ is continuous, since near a pole of any curve potential the corresponding branch lies below the constant $-L$. The same argument, applied to the $n+1$ branches $g(z_1),\ldots ,g(z_ n),-L$, shows that $(\mathrm{dd}^{\mathrm{c}}h_ L)^ n$ is carried by

Indeed, off $Q_ L$ one branch lies strictly below the maximum. Discarding it, together with any branches near poles, leaves at most $n$ pluriharmonic branches locally, so the measure vanishes there. Since $Q_ L$ is compact, the Chern–Levine–Nirenberg estimate (5.27.3) gives finite mass.

For $L,L'>0$, the functions $h_ L$ and $h_{L'}$ agree near $\partial D$, since $h$ tends uniformly to zero there. Integration by parts for bounded potentials gives

The current on the right has compact support, so its integral is zero. To see that the common mass $M$ is positive, fix $L$ and choose $\varepsilon >0$ so small that

is nonempty. This set is relatively compact in $D$, because $h_ L$ tends to zero at the boundary while the quadratic function is strictly negative on $\overline D$. The comparison (5.27.7) for bounded potentials gives

Finally, $g$ is bounded on compact subsets of $\overline\Delta \setminus K$. Its sublevel sets $\{ g\leq -L\} $ therefore converge to $K$ as $L\to \infty $, and $Q_ L$ converges to $S$. $\square$

Proof. The truncation $\max (g,-s)$ agrees with $g$ near the boundary of $\Delta $. Their difference is a compactly supported integrable function, so its complex Hessian has integral zero, and $\mu _ s(\Delta )=\sigma (\Delta )=1$. The truncation is constant near $K$ and harmonic off $K$ wherever $g\ne -s$, which proves the support assertion. Since the truncation is bounded, $\mu _ s$ has no point masses, by theorem 5.27.1. Decreasing local $L^1$ convergence to $g$ gives $\mu _ s\to \mathrm{dd}^{\mathrm{c}}g=\sigma $; for large $s$, the supports lie in one fixed compact subset of $\Delta $.

The rank argument of lemma 18.26.15 applies with the weights $a_ j$. Away from $S$, the finite branches dominate the branches near poles, so $h_ a$ is locally a Lipschitz maximum of at most $n$ pluriharmonic functions, with zero top-degree product. For the bounded maximum with the constant branch $-L$, the same argument shows that its top-degree product is supported on

Indeed, off this set one branch lies strictly below the maximum, leaving at most $n$ active pluriharmonic branches; this also applies near branches with poles, which lie below the constant branch.

Near a point of $Q_{a,L}$ at which all the differentials $\mathrm dg(z_ j)$ are nonzero, set $x_ j=a_ jg(z_ j)+L$. Each $x_ j$ is the real part of a local holomorphic coordinate in its own factor. For $\varepsilon >0$, smooth the maximum locally by

with weights

Then $-L+f_\varepsilon $ decreases to $h_{a,L}$. The Hessian matrix of $f_\varepsilon $ in the variables $x_ j$ is $\varepsilon ^{-1}(\operatorname {diag}(w_ j)-(w_ jw_ k)_{j,k})$, whose determinant is $\varepsilon ^{-n}w_0\prod _ jw_ j$. Hence

The scalar kernel on $\mathbb R^ n$ is an approximate identity of mass $1$. To check the constant, write $y=x/\varepsilon $. The map

is a diffeomorphism onto the open simplex $\{ w_ j>0,\ \sum _ jw_ j<1\} $, with Jacobian $w_0\prod _ jw_ j$. The simplex has volume $1/n!$, so the kernel above has integral $1$. Integrability of the fixed kernel in $y$ gives concentration at $x=0$ as $\varepsilon \downarrow 0$.

Bounded decreasing convergence of products therefore identifies the local limit with the tensor product of the one-variable level measures. In the $j$th factor, the chain rule for $\max (x_ j,0)$ gives

where the Dirac notation denotes the level measure in the real normal coordinate $x_ j$. The critical points of the nonconstant harmonic function $g$ off $K$ are isolated. In each coordinate neighborhood, their inverse images are analytic hypersurfaces, hence pluripolar. The bounded top-degree product gives them zero mass by theorem 5.27.1, and so does the candidate tensor measure, because each $\mu _{L/a_ j}$ has no point masses. Thus the formula extends across these points and, together with the support assertion, holds on all of $D$.

Tensor products of the curve probabilities converge to $\sigma ^{\otimes n}$: this follows first for products of continuous functions of one variable, and then for all continuous test functions by uniform approximation on a common compact product. Finally, $h_{a,L}\downarrow h_ a$, whose unbounded locus is the closed set $S$ of Hausdorff dimension zero. By theorem 5.27.18, the weak limit is the full top-degree product of $h_ a$. Taking $a_ j=1$ proves $M=1$. $\square$

Proof. Place the polydisk of lemma 18.26.15 in a holomorphic coordinate chart of $X$. Choose a smooth cutoff function $0\leq \chi \leq 1$, compactly supported in that polydisk and equal to one on a neighborhood of $S$, and extend $H=\chi h$ by zero. Then $H$ is quasi-plurisubharmonic and not identically $-\infty $. Indeed, outside $S$ the gradients of $h$ are locally bounded, so the terms containing derivatives of $\chi $ in

have bounded coefficients on their compact support, while the first term is positive. Hence $\mathrm{dd}^{\mathrm{c}}H\geq -C\omega $ for some $C>0$. Take $0<t<1/C$ and put $u=tH$. Then $u\leq 0$, $\sup _ Xu=0$, the pole set of $u$ is $S$, and

Set $u_ k=\max (u,-k)$. By Stokes' theorem in the closed background, the bounded currents $\omega +\mathrm{dd}^{\mathrm{c}}u_ k$ have total Monge–Ampère mass $A=\int _ X\omega ^ n$. Near $S$, where $\chi =1$,

For large $k$, the whole support of the measure on the right lies in this neighborhood; these measures have mass $t^ nM$ and their supports converge to $S$. Take a subsequence along which both these local measures and the global measures $(\omega +\mathrm{dd}^{\mathrm{c}}u_ k)^ n$ converge weakly. The global limit has mass at least $t^ nM$ on $S$. Off $S$, the function $u$ is locally bounded, and $u_ k=u$ eventually on every compact set, so there the global limit is the non-pluripolar measure of $u$. It follows that

The lower bound follows from the current inequality above and bounded local products off the Lebesgue-null set $S$.

In a closed Kähler background, bounded changes of a potential do not change its non-pluripolar mass. Both singularity volumes of $u$ therefore equal this positive number. By theorem 13.19.5, $p=P_\omega [u]$ is a normalized ordinary model with the same volumes. In particular $p\ne 0$, and $d_\omega (p,0)=A-\overline{\operatorname {vol}}_\omega (p)>0$. Taking for $\delta $ half the lower volume of $p$ puts both models in $\mathcal M_\delta (\omega )$. Also $u\leq p\leq 0$, so the pole set of $p$ is contained in $S$.

Fix a smooth modification $\pi $, and let $E_ i$ be the irreducible divisorial components of its exceptional locus, away from which $\pi $ is an isomorphism. The polar current $R$ of $p\circ \pi $ is closed and positive by proposition 5.46.7. On the set where $\pi $ is an isomorphism, $R$ is carried by the inverse image of $S$, which has Hausdorff dimension zero in each relatively compact chart; by corollary 5.34.3, $R$ vanishes there. The support theorem for closed positive currents therefore gives

see [Chapter III, Corollaries 2.11 and 2.14, pp. 141–143, DemCADG]. Components of codimension at least two carry no closed $(1,1)$-current of order zero, and the sum is finite by compactness. Since $R\leq \pi ^*\omega +\mathrm{dd}^{\mathrm{c}}(p\circ \pi )$, its coefficients satisfy $b_ i\leq \nu (p\circ \pi ,E_ i)$.

If $\pi (E_ i)$ is not contained in $S$, some regular point of $E_ i$ maps outside $S$. Near such a point $u\circ \pi $, and hence $p\circ \pi $, is bounded below, so the generic Lelong number of $p\circ \pi $ along $E_ i$ vanishes and $b_ i=0$. Every remaining image $\pi (E_ i)$ is an irreducible analytic set contained in the set $S$ of Hausdorff dimension zero, hence a point $a=(a_1,\ldots ,a_ n)$. Near that point, $u=t\max _ jg(z_ j)$. At a general point of $E_ i$, with a local equation $\zeta $ of $E_ i$, the first coordinate satisfies

A local holomorphic $m$th root of the unit $f$ turns this expression into $\zeta '^ m$ for another local equation $\zeta '$ of $E_ i$. Taking circular means gives

Since $p\circ \pi \geq u\circ \pi \geq t\, g(z_1\circ \pi )$ there, again $b_ i=0$. Thus $R=0$ for every modification.

The same local comparison gives zero Lelong numbers for $u$, and hence for $p\geq u$, at every point. Skoda's local integrability theorem, recalled in [SGPT], then makes $\mathrm e^{-sp}$ locally integrable for every $s>0$, so all multiplier ideals of $p$ are trivial. The ideal envelope of $p$ is therefore the zero model, and since $p\ne 0$, $p$ is not an ideal model. The zero model has analytic singularities, and $\delta _\omega ^{\mathrm{pol}}(p,0)=0$ puts $p$ in the closure of the analytic models for the polar comparison. This proves the final assertions. $\square$

Proof. Fix $t>0$. Let $u_{t,k}=\max (tH,-k)$ and $\Theta _ k=\omega +\mathrm{dd}^{\mathrm{c}}u_{t,k}$. These bounded closed positive currents satisfy, for $1\leq j\leq n$,

Indeed, subtract $\omega ^ j$, write the difference as a telescoping sum of mixed products, and integrate $\mathrm{dd}^{\mathrm{c}}\bigl (u_{t,k}\sum _{i=0}^{j-1}\Theta _ k^ i\wedge \omega ^{j-1-i}\bigr )\wedge \omega ^{n-j}$ by parts for bounded potentials.

Choose an open neighborhood $U$ of $S$ on which $\chi =1$. There $\Theta _ k=\omega +t\mathrm{dd}^{\mathrm{c}}h_{k/t}$, so positivity gives

For the finitely many $1\leq j<n$, take a common subsequence along which the local closed positive currents $(\mathrm{dd}^{\mathrm{c}}h_{k/t})^ j$ converge weakly on $U$, with limits $R_ j$; the inequalities above give the required local bounds on their traces. Along a further subsequence, the global measures $\Theta _ k^ n$ and the local top-degree measures also converge weakly. By lemma 18.26.15, the latter have mass $M$ and supports eventually contained in a fixed compact subset of $U$ and converging to $S$. Their extensions by zero therefore converge, along a further subsequence, to a measure $\rho $ supported on $S$ with $\rho (S)=M$.

Since $S$ has Hausdorff dimension zero, corollary 5.34.3 shows that every $R_ j$, $j<n$, has coefficient measures vanishing on $S$. The smooth measure $\omega ^ n$ also gives $S$ zero mass. On $U$ we expand

If $\eta $ is the chosen global weak limit, this expansion gives $\eta (S)=t^ n\rho (S)=Mt^ n$, while $\eta (X)=A$. Off $S$, the function $u_ t$ is locally bounded and its truncations eventually agree with it on each compact set, so there $\eta $ agrees with $\mathrm{MA}_\omega (u_ t)$. The non-pluripolar measure gives $S$ zero mass, which proves

This argument uses only a weak subsequential limit in top degree; it does not define an unbounded Bedford–Taylor product on $S$.

The current inequality $\omega +\mathrm{dd}^{\mathrm{c}}u_ t\geq (1-Ct)\omega $ gives $A-Mt^ n\geq (1-Ct_0)^ nA>0$. Invariance of non-pluripolar mass under bounded changes in the closed background and theorem 13.19.5 now give the formula for both singularity volumes of the normalized ordinary model $p_ t$. Since $H\leq 0$, monotonicity of the ordinary envelope gives $p_ s\geq p_ t$ for $s\leq t$, and the zero endpoint satisfies the same order. Thus $\max (p_ s,p_ t)=p_ s$, and the formula for ordered pairs in (13.22.2) gives $d_\omega (p_ s,p_ t)=M(t^ n-s^ n)$. Hence $p_ t\mapsto Mt^ n$ is an isometry onto $[0,Mt_0^ n]$, and the interval is complete.

The proof of proposition 18.26.17 applies to every $0<t\leq t_0$: all first polar currents vanish on all modifications, and all multiplier ideals are trivial. The polar currents also vanish for $p_0=0$, so the polar comparison vanishes on the family. For $t>0$, the strict volume deficit makes $p_ t\ne 0$, and its trivial ideals then show that it is not an ideal model. Finally, choose $t_0<1/C$ so small that $(1-Ct_0)^ nA>\delta $; the uniform lower bound for the volumes puts every $p_ t$ in $\mathcal M_\delta (\omega )$. $\square$

Proof. Outside a fixed neighborhood of $S$, the truncations $h_ L$ eventually agree with $h$ on every compact set, in particular on the supports of the derivatives of $\chi $. The cutoff calculation in proposition 18.26.17 therefore gives a lower bound $-C\omega $ for $\mathrm{dd}^{\mathrm{c}}H_ L$ and $\mathrm{dd}^{\mathrm{c}}H$, uniform in $L$.

Let $\kappa =\mathrm{dd}^{\mathrm{c}}|z|^2$ on the coordinate polydisk $D$, and choose $B>0$ so that $T_ L=B\kappa +\mathrm{dd}^{\mathrm{c}}H_ L$ is positive and closed on $D$. The potentials of $T_ L$ are locally bounded. Since $H_ L$ has compact support, integration by parts for bounded potentials gives, for $1\leq j\leq n$,

Indeed, every nonconstant term of the binomial expansion is the complex Hessian of a compactly supported current built from bounded products, wedged with a closed power of $\kappa $. Thus the traces in all degrees are bounded independently of $L$.

For $j<n$, every subsequential weak limit of $T_ L^ j$ has coefficient measures vanishing on $S$, by corollary 5.34.3. Off $S$, local boundedness and eventual equality identify it with $(B\kappa +\mathrm{dd}^{\mathrm{c}}H)^ j$. These two properties determine the limit uniquely. Expanding $(T_ L-B\kappa )^ j$ therefore shows that $(\mathrm{dd}^{\mathrm{c}}H_ L)^ j$ converges weakly to a finite signed current $W_ j$ with no coefficient mass on $S$. It vanishes near $\partial D$ and extends by zero to $X$. Integration by parts for bounded potentials gives

The left side converges to $\int _ XW_ j\wedge \omega ^{n-j}$, which proves that $c_ j$ exists and is finite.

Fix $0<t<1/C$. The positive measures $(\omega +t\mathrm{dd}^{\mathrm{c}}H_ L)^ n$ have uniformly bounded mass, by their local expansion in powers of $T_ L$ and smooth forms. Their total masses satisfy

since the top power $(\mathrm{dd}^{\mathrm{c}}H_ L)^ n$ integrates to zero by Stokes' theorem for bounded potentials. Let $\eta $ be a weak subsequential limit. Its total mass is $A+\sum _{j=1}^{n-1}\binom nj c_ jt^ j$.

Near $S$, where $\chi =1$, expand $(\omega +t\mathrm{dd}^{\mathrm{c}}h_ L)^ n$. All the limits of lower degree give $S$ zero mass, while the top-degree measures have mass $M$ and supports converging to $S$, by lemma 18.26.15. Hence $\eta (S)=Mt^ n$. Off $S$, eventual equality with the locally bounded function $tH$ identifies $\eta $ with the non-pluripolar measure of $tH$. Subtracting the mass on $S$ proves the formula.

If $\mathrm{dd}^{\mathrm{c}}(\omega ^ q)=0$ for $1\leq q<n$, every $c_ j$ is zero. For $n=2$, dominated $L^1$ convergence $H_ L\to H$ gives $c_1=\int _ XH\mathrm{dd}^{\mathrm{c}}\omega $, which proves the formula on surfaces. These formulas give the masses of the potentials $tH$ themselves. Passing to singularity volumes in a nonclosed background requires control of the masses of all potentials ordinarily equivalent to $tH$. $\square$

Proof. By theorem 11.1.7, every bounded $\omega $-plurisubharmonic potential has mass $A$. Thus $v_+(\omega )=A$ for the upper mass of definition 13.14.14, and $X$ has the bounded mass property by proposition 13.14.15.

Fix $v\sim tH$. Its pole set is $S$, and it is locally bounded off $S$. In a coordinate neighborhood of $S$, choose a smooth local correction $q$ such that $a=v+q$ is plurisubharmonic. Since $\chi =1$ there, $a-th$ is bounded near $S$. By theorem 5.27.18, the full product of $h$ is the weak limit of the measures $(\mathrm{dd}^{\mathrm{c}}h_ L)^ n$, and the support and mass assertions of lemma 18.26.15 show that this limit is supported on $S$ with total mass $M$. The invariance of residual measures in lemma 5.27.19 therefore gives

Set $v_ k=\max (v,-k)$. After each fixed local smooth correction, these functions decrease to a plurisubharmonic function whose unbounded locus lies in $S$. The theorem on full local products and the expansion (5.28.1) identify the weak limit of $(\omega +\mathrm{dd}^{\mathrm{c}}v_ k)^ n$ with a global positive measure $T_ v$ of total mass $A$; the local expressions agree on overlaps because they are limits of the same bounded global measures. By corollary 5.34.3, every closed positive power $(\mathrm{dd}^{\mathrm{c}}a)^ j_{\mathrm{full}}$ with $j<n$ has coefficient measures vanishing on $S$, and smooth forms also give $S$ zero mass. Hence only the top-degree power contributes on $S$, and $T_ v(S)=Mt^ n$. Off $S$, the truncations eventually agree with $v$ on each compact set, so there $T_ v$ equals the non-pluripolar measure of $v$. Removing the mass on $S$ proves the formula for every representative $v\sim tH$.

Both singularity volumes of $tH$ thus equal $A-Mt^ n$. The current inequality $\omega +\mathrm{dd}^{\mathrm{c}}(tH)\geq (1-Ct)\omega $ shows that this volume is positive, uniformly for $0<t\leq t_0<1/C$. By theorem 13.19.5, ordinary projection preserves both volumes and gives the normalized models. Their order and distances follow as in corollary 18.26.18, with the zero endpoint defined separately.

The support and Lelong-number argument in the proof of proposition 18.26.17 uses that the background is smooth and that its polar restriction is closed, but not that the background is closed. It therefore gives vanishing first polar currents on all modifications and trivial multiplier ideals at all weights in the present setting as well. The strict volume deficit shows that every model with positive parameter is not an ideal model. Choosing $t_0$ small enough puts the interval in $\mathcal M_\delta (\omega )$. Finally, on a surface the only condition is $\mathrm{dd}^{\mathrm{c}}\omega =0$, which is the Gauduchon condition, and Gauduchon forms exist by theorem 11.4.2. $\square$

Proof. The pole set of $v$ is $S$, and $v$ is locally bounded off $S$. On each relatively compact coordinate neighborhood, choose a fixed smooth local correction $q$ such that $v+q$ and every $v_ k+q$ are plurisubharmonic. By theorem 5.27.18, all closed powers of $\mathrm{dd}^{\mathrm{c}}(v_ k+q)$ converge weakly to finite currents of order zero. Expanding $\mathrm{dd}^{\mathrm{c}}v_ k=\mathrm{dd}^{\mathrm{c}}(v_ k+q)-\mathrm{dd}^{\mathrm{c}}q$ gives weak convergence of $(\mathrm{dd}^{\mathrm{c}}v_ k)^ j$ in every degree $j$. The local limits patch together, because the bounded currents $(\mathrm{dd}^{\mathrm{c}}v_ k)^ j$ are global; write $W_ j$ for the limits.

Integration by parts for bounded potentials gives

this follows by uniformly bounded decreasing regularization, as in theorem 11.1.5, theorem 5.32.1. Pairing $W_ j$ with the smooth form $\omega ^{n-j}$ shows that $c_ j(v)$ exists and is finite. The top power $(\mathrm{dd}^{\mathrm{c}}v_ k)^ n$ has zero integral for every $k$, by Stokes' theorem for bounded potentials. Expanding the positive bounded measures $(\omega +\mathrm{dd}^{\mathrm{c}}v_ k)^ n$ therefore gives a global positive weak limit $T_ v$ of finite total mass $A+\sum _{j=1}^{n-1}\binom nj c_ j(v)$.

Near $S$, compare $a=v+q$ with $th$; their difference is bounded there. The full top-degree product of $h$ has mass $M$ and is supported on $S$, by theorem 5.27.18, lemma 18.26.15. Hence lemma 5.27.19 gives $(\mathrm{dd}^{\mathrm{c}}a)^ n_{\mathrm{full}}(S)=Mt^ n$. Every closed power of lower degree has zero coefficient mass on $S$, by corollary 5.34.3. In the expansion with the smooth form $\omega -\mathrm{dd}^{\mathrm{c}}q$, only the top power therefore contributes to $T_ v(S)$, and $T_ v(S)=Mt^ n$. Off $S$, the truncations eventually agree with $v$, so $T_ v$ is the non-pluripolar measure of $v$ there. Subtracting the mass on $S$ proves the formula.

Each measure $(\omega +\mathrm{dd}^{\mathrm{c}}v_ k)^ n$ has total mass between $v_-(\omega )$ and $v_+(\omega )$. Passing to the weak limit and subtracting $Mt^ n$ gives the two bounds, the upper one in the extended real sense. In dimension two, $L^1$ convergence of $v_ k$ to $v$ gives $c_1(v)=\int _ Xv\mathrm{dd}^{\mathrm{c}}\omega $, which proves the formula on surfaces. $\square$

Proof. By the current inequality $\omega +\mathrm{dd}^{\mathrm{c}}u\geq (1-Ct)\omega $, the potential $u+\varepsilon f$ is $\omega $-plurisubharmonic for both signs of $\varepsilon $, provided $|\varepsilon |$ is small: choose $c_ f>0$ with $-c_ f\omega \leq \mathrm{dd}^{\mathrm{c}}f\leq c_ f\omega $ and take $|\varepsilon |<(1-Ct)/c_ f$. These potentials differ boundedly from $u$, so proposition 18.26.21 gives the stated difference.

If $\mathrm{dd}^{\mathrm{c}}\omega =0$, the same proposition gives mass $A-Mt^2$ for every potential ordinarily equivalent to $u$. Conversely, write $\mathrm{dd}^{\mathrm{c}}\omega =\rho \omega ^2$ with $\rho $ smooth and real. If this form is not zero, choose $f=\rho $. Then $\int _ Xf\mathrm{dd}^{\mathrm{c}}\omega =\int _ X\rho ^2\omega ^2>0$, so the masses of the smooth perturbations above vary. Thus constancy of the mass forces $\mathrm{dd}^{\mathrm{c}}\omega =0$, which is the Gauduchon condition on a surface. $\square$

Proof. Taking the infimum and the supremum in proposition 18.26.21 gives the bounds for the singularity volumes of $tH$. The lower volume is positive, so by theorem 13.19.5 ordinary projection preserves both volumes. Since $tH\leq p_ t\leq 0$ and $\sup _ XtH=0$, the model is normalized.

The type of zero consists of the bounded potentials, so its upper volume is $v_+(\omega )$. Since $p_ t\leq 0$, (13.22.2) gives $d_\omega (p_ t,0)=v_+(\omega )-\overline{\operatorname {vol}}_\omega (p_ t)\geq Mt^ n$. In particular $p_ t\ne 0$.

The part of the proof of proposition 18.26.17 that uses supports and Lelong numbers applies to this model: its pole set lies in $S$, its polar restriction is closed and positive, and every exceptional divisor whose image lies in $S$ maps to a point. Since the curve potential $g$ has zero Lelong numbers, every divisorial coefficient vanishes. The lower bound $p_ t\geq tH$ and Skoda's integrability theorem give trivial multiplier ideals at all weights. Hence the ideal envelope of $p_ t$ is the zero model, while $p_ t\ne 0$. All first polar comparisons with zero vanish, which proves that $p_ t$ is not an ideal model and lies in the closure of the analytic models.

Given $0<\delta <v_-(\omega )$, choose $t_0<1/C$ with $Mt_0^ n<v_-(\omega )-\delta $. The lower bound for the volume then puts zero and every $p_ t$ with $0<t\leq t_0$ in $\mathcal M_\delta (\omega )$.

On a compact complex surface, a Gauduchon form has finite upper mass by theorem 11.4.2, theorem 11.1.7. By the metric independence in proposition 13.14.15, every Hermitian form then has finite upper mass, and corollary 12.19.15 gives positive lower mass. $\square$

Proof. Put $u_ t=tH$ for $t>0$ and $u_0=0$. For $0<t<1/C$, the potential $u_ t$ is $Ct\omega $-plurisubharmonic. Adding it to the zero potential in the background $(1-Ct)\omega $, proposition 13.17.2 and homogeneity show that either singularity volume of $u_ t$ is at least $(1-Ct)^ n$ times the corresponding volume of zero. In particular, the lower volume of $u_ t$ is positive on the whole range $0<t<1/C$, and ordinary projection preserves both volumes by theorem 13.19.5. The upper bound follows from proposition 18.26.21, and the zero endpoint has $L(0)=v_-(\omega )$ and $U(0)=v_+(\omega )$.

Near $S$, one has $u_ t=th$. After a fixed smooth local correction, invariance of residual measures and lemma 18.26.15 show that $u_ t$ has residual measure of mass $Mt^ n$, and by theorem 13.19.6 so does $p_ t$. Zero has residual measure zero, since its corrected top-degree measure is smooth. Thus two models of the family coincide only if their parameters are equal.

For $s<t$, set $c=(1-Ct)/(1-Cs)$. Then $0<c<1$ and

The second summand is $(1-c)\omega $-plurisubharmonic. Addition and homogeneity give $V(t)\geq c^ nV(s)$, whereas monotonicity gives $V(t)\leq V(s)$. The same computation holds for $s=0$, with the separately defined zero potential. Monotonicity of the envelope gives $p_ s\geq p_ t$, so (13.22.2) gives $d_\omega (p_ s,p_ t)=U(s)-U(t)$. Since $U(s)\leq v_+(\omega )$, the Lipschitz bound follows from

By this bound, $U$ is continuous. Distinct ordinary models of positive lower volume are separated by proposition 13.22.4, so $U$ is strictly decreasing. Thus $t\mapsto v_+(\omega )-U(t)$ identifies the family isometrically with the closed interval in the statement, which proves completeness.

The support and Lelong-number argument in the proof of proposition 18.26.17 applies on the whole range $0<t<1/C$: all first polar currents vanish and all multiplier ideals are trivial. The deficit $Mt^ n$ of the upper volume separates every positive parameter from zero, which proves that the models with positive parameter are not ideal models and that their polar comparisons vanish. Finally, choose $t_0<1/C$ with $(1-Ct_0)^ nv_-(\omega )>\delta $ to place the whole interval in $\mathcal M_\delta (\omega )$. $\square$

Proof. Choose two disjoint coordinate polydisks, and in each apply the construction of proposition 18.26.19, obtaining a cutoff $H_ i$ with pole set $S_ i$, curvature constant $C_ i>0$ and truncation mass $M_ i>0$. For small $r>0$ put

Choose $r$ so small that $C_ it_ i<1$ and $(1-C_ it_ i)^ nv_-(\omega )>\delta $ for both $i$. By corollary 18.26.24, the $p_ i$ are normalized ordinary models in $\mathcal M_\delta (\omega )$; they are not ideal models, their multiplier ideals are trivial and their first polar currents vanish. Their upper volumes $U_ i$ satisfy $U_ i\leq v_+(\omega )-r$.

The full top-degree product of $h$ in the $i$th chart is a positive measure $\rho _ i$ supported on $S_ i$, of mass $M_ i$. Near $S_ i$ one has $u_ i=t_ ih$, so invariance of residual measures under bounded changes and theorem 13.19.6 give $\mathbf1_{S_ i}\Phi _{p_ i}=t_ i^ n\rho _ i$, extended by zero outside that chart. Outside the compact support of $H_ i$, the bounds $u_ i\leq p_ i\leq 0$ force $p_ i=0$. In particular $p_ i$ is smooth near $S_ j$ for $j\ne i$, and its full measure there is $\omega ^ n$, which gives $S_ j$ zero mass. Hence $R_ i=t_ i^ n\rho _ i$. These measures have the same total mass $r>0$ and disjoint supports, so they differ.

Since the cutoffs have disjoint supports, at every point at least one $p_ i$ vanishes, and both are nonpositive; this proves $\max (p_1,p_2)=0$. By definition 13.22.1,

All first polar coordinates of both models are zero, so their polar comparison vanishes. When $\mathrm{dd}^{\mathrm{c}}(\omega ^ q)=0$ for $1\leq q<n$, corollary 18.26.20 gives both volumes equal to $A-M_ it_ i^ n=A-r$, and $v_+(\omega )=A$. Substituting these values gives the exact distance $2r$ and the final assertion. $\square$

Proof. Place the polydisk $D$ of lemma 18.26.16 in a coordinate chart, and use one cutoff function $0\leq \chi \leq 1$, compactly supported there and equal to one near $S=K^ n$. For $a\in B$, extend $H_ a=\chi h_ a$ by zero and put $u_ a=tH_ a$. We first prove a curvature bound that is uniform in $a$.

Choose a compact neighborhood $\Sigma $ of the supports of the derivatives of $\chi $, disjoint from $S$. The isotropic maximum $\max _ j g(z_ j)$ is bounded below on $\Sigma $ by $-B_0$, for some $B_0>0$. Since $g\leq 0$ and $a_ j\leq \beta $, it follows that $-\beta B_0\leq h_ a\leq 0$ there. If the $j$th branch $a_ jg(z_ j)$ attains the maximum, then $a_ jg(z_ j)\geq -\beta B_0$, so $g(z_ j)\geq -\beta B_0/\alpha $. Within the fixed compact range of the coordinates, this superlevel set stays away from $K$, and $g$ and its gradient are bounded on a neighborhood of it. Thus all maximizing branches have uniformly bounded gradients, and the maxima $h_ a$ satisfy one local Lipschitz bound on $\Sigma $, independent of $a\in B$.

In the distributional product rule for $\mathrm{dd}^{\mathrm{c}}(\chi h_ a)$, the term $\chi \mathrm{dd}^{\mathrm{c}}h_ a$ is positive. The remaining terms contain derivatives of $\chi $ together with $h_ a$ or its first derivatives, so the preceding bounds control their coefficients uniformly. The cutoff argument of proposition 18.26.19 therefore gives one $C>0$ with $\mathrm{dd}^{\mathrm{c}}H_ a\geq -C\omega $ for every $a\in B$. Choose $t_0<1/C$ with $t_0^ n\beta ^ n<A-\delta $. Then all $u_ a$, $0<t\leq t_0$, are normalized and satisfy $\omega +\mathrm{dd}^{\mathrm{c}}u_ a\geq (1-Ct)\omega $.

By theorem 11.1.7, all bounded $\omega $-plurisubharmonic functions have mass $A$, so $v_-(\omega )=v_+(\omega )=A$. For any $v\sim u_ a$, the global limit of the measures of the truncations has total mass $A$, as in the proof of corollary 18.26.20. Near $S$, the difference between $v$ plus a smooth local correction and $th_ a$ is bounded. By lemma 5.27.19, lemma 18.26.16, the residual measure of $v$ is

extended by zero from the chart. Every corrected closed power of lower degree gives $S$ zero coefficient mass, so this is also the restriction to $S$ of the global limit of the truncations. Off $S$ that limit equals the non-pluripolar measure, which proves that every such representative has mass

Hence, by theorem 13.19.5, ordinary projection preserves both singularity volumes, and by theorem 13.19.6 it preserves the residual measure. Setting $p_ a=P_\omega [u_ a]$ proves the formulas for the residual measure and the volumes simultaneously for all $a\in B$.

For $a,b\in B$ and $c_ j=\min (a_ j,b_ j)$, positivity of the weights and $g\leq 0$ give, pointwise and also at the poles,

The invariance of the comparison under ordinary projection in proposition 13.22.4 and the definition of the comparison give

For the boxes $R_ a=\prod _ j[0,a_ j]\subset \mathbb R^ n$ and Lebesgue measure $\lambda $, the last expression is $t^ n\lambda (R_ a\mathbin {\triangle }R_ b)$, since $R_ a\cap R_ b=R_ c$; equivalently, it is $t^ n\| \mathbf1_{R_ a}-\mathbf1_{R_ b}\| _{L^1(\lambda )}$. This proves the triangle inequality within the family. A slab in the coordinate where $|a_ j-b_ j|$ is largest, with every other coordinate in $[0,\alpha ]$, lies in the symmetric difference up to its boundary, and its volume gives the lower bound. The symmetric difference is covered by the $n$ coordinate slabs between $a_ j$ and $b_ j$, with every other coordinate in $[0,\beta ]$, and their total volume gives the upper bound. In particular, the distance separates the parameters, and the two estimates show that it induces the Euclidean topology of $B$. A sequence that is Cauchy for the distance has Cauchy weights in the Euclidean sense, whose limit lies in $B$, and the upper bound gives convergence for the distance. Compactness follows from the same bounds.

The support and Lelong-number argument in the proof of proposition 18.26.17 applies for every positive weight vector. Every exceptional divisor mapping into $S$ maps to a point, and comparison with $ta_1g(z_1\circ \pi )$ shows that the generic Lelong number along it is zero. Thus every first polar current is zero. The same comparison gives zero Lelong numbers at all points and trivial multiplier ideals at all weights. The strict volume deficit $t^ n\prod _ j a_ j$ shows that $p_ a\ne 0$, whereas the ideal envelope of $p_ a$ is zero, so $p_ a$ is not an ideal model.

Finally, the constraint $\prod _ ja_ j=P$ defines a closed subset of $B$, so the subfamily is complete. Taking logarithms coordinatewise identifies it with

For $\alpha ^ n<P<\beta ^ n$, this convex polytope meets the interior of the cube at the point with $x_ j=(\log P)/n$ for all $j$, and it has dimension $n-1$. The formulas for the volume and the residual measure are constant on it, while the distance separates its members. For example, taking $\alpha =1/2$, $\beta =2$, $P=1$, $a=(2,1/2,1,\ldots ,1)$ and $b=(1,1,1,\ldots ,1)$ gives two models with common residual measure $t^ n\rho $, both volumes equal to $A-t^ n$, and distance $t^ n$. $\square$