Proposition 18.26.2.088A7D40B1D1 For every $x\in X$, there is $a_0>0$ such that, for each rational $a\in (0,a_0)$, there is an ideal model $p_ a$ with analytic singularities and pole set $\{ x\} $ satisfying
\[ p_ a=a\log |z|^2+O(1)\quad \text{near }x,\qquad \operatorname {Pol}_1(p_ a)=0. \]
The models $p_ a$ are pairwise ordinarily inequivalent and have pairwise distinct families of multiplier ideals $\mathcal I(tp_ a)$, $t>0$, whereas also $\operatorname {Pol}_1(0)=0$. More precisely, with $\mathfrak m_ x$ the maximal ideal at $x$,
55D42653A018
18.26.3
\begin{equation} \mathcal I(tp_ a)_ x =\mathfrak m_ x^{\max \{ \lfloor ta\rfloor -n+1,0\} } \qquad (t>0). \end{equation}
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
\[ \nu =(\omega +\mathrm{dd}^{\mathrm{c}}p_ a)\wedge \beta ^{n-1}. \]
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
\[ \int \chi _\varepsilon \, \mathrm d\nu =\int \chi _\varepsilon \omega \wedge \beta ^{n-1} +\int p_ a\, \mathrm{dd}^{\mathrm{c}}\chi _\varepsilon \wedge \beta ^{n-1}. \]
The logarithmic bound for $p_ a$ on the annulus and the Euclidean volume of the annulus therefore give
\[ 0\leq \nu (\{ x\} )\leq \int \chi _\varepsilon \, \mathrm d\nu \leq C_ a\bigl (\varepsilon ^{2n} +\varepsilon ^{2n-2}(1+|\log \varepsilon |)\bigr ) \longrightarrow 0. \]
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.
$\square$
Pome
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