Proposition 18.26.25.085AE6785656 Let $\omega $ be a Hermitian form on a connected compact complex manifold $X$ of dimension $n\geq 2$, with $0<v_-(\omega )\leq v_+(\omega )<+\infty $. Fix $0<\delta <v_-(\omega )$. For every sufficiently small $r>0$, there are distinct normalized ordinary models $p_1,p_2\in \mathcal M_\delta (\omega )$ and disjoint compact sets $S_1,S_2$ of Hausdorff dimension zero such that the full Monge–Ampère measures $\Phi _{p_ i}$ of theorem 13.19.6, the weak limits of the Monge–Ampère measures of the bounded truncations, satisfy
\[ R_ i:=\mathbf1_{S_1\cup S_2}\Phi _{p_ i},\qquad R_ i(X)=r,\qquad \operatorname {supp}R_ i\subset S_ i. \]
Both models are not ideal models, all their multiplier ideals are trivial, and their first polar currents vanish on every smooth modification. Moreover,
\[ \max (p_1,p_2)=0,\qquad d_\omega (p_1,p_2)\geq 2r, \qquad \delta _\omega ^{\mathrm{pol}}(p_1,p_2)=0. \]
If $\mathrm{dd}^{\mathrm{c}}(\omega ^ q)=0$ for $1\leq q<n$, both singularity volumes of both models equal $A-r$, where $A=\int _ X\omega ^ n$, and their distance is exactly $2r$. Thus the first polar currents together with the total residual mass do not separate ordinary models, and when $\mathrm{dd}^{\mathrm{c}}(\omega ^ q)=0$ for $1\leq q<n$, they still do not separate them after both singularity volumes are added.
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
\[ t_ i=(r/M_ i)^{1/n},\qquad u_ i=t_ iH_ i,\qquad p_ i=P_\omega [u_ i]. \]
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,
\[ d_\omega (p_1,p_2)=2v_+(\omega )-U_1-U_2\geq 2r. \]
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.
$\square$
Pome
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