PomeLemma 18.27.2.96F4F5465740 If (18.27.1) holds, then $q$ has zero point Lelong numbers and zero polar curvature. The pullback of $q$ to every smooth connected compact modification has the same small-curvature property, and its point Lelong numbers and polar curvature also vanish.
18.27 Small curvature and strict ordinary types0A2CF48796FA
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Multiplier ideals can miss an unbounded residual. To decide whether an ordinary type containing such a residual admits a strict current, we need representatives in that same ordinary type with a controlled curvature bound. Fix a Hermitian form $h$ on $X$ and a quasi-plurisubharmonic function $q$ such that
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18.27.1
\begin{equation} \text{for every }\delta >0\text{ there is }q_\delta \sim q \quad \text{with}\quad \mathrm{dd}^{\mathrm{c}}q_\delta \geq -\delta h. \end{equation}
This property does not depend on the choice of $h$, since any two Hermitian forms are uniformly comparable on $X$.
Proof. The function $q_\delta /\delta $ belongs to $\mathrm{PSH}(X,h)$. The uniform Lelong bound of lemma 5.38.8, and invariance under bounded changes, give
\[ 0\leq \nu (q,x)=\nu (q_\delta ,x) =\delta \nu (q_\delta /\delta ,x)\leq C_ h\delta . \]
Letting $\delta $ tend to zero proves the first assertion.
For polar curvature, put $S=\{ q=-\infty \} $ and let $R=\mathbf1_ S\mathrm{dd}^{\mathrm{c}}q$, a closed positive current by proposition 5.46.7. Choose a Gauduchon metric $\gamma $ when $n\geq 2$, using theorem 11.4.2; when $n=1$ take any Hermitian form and use $\gamma ^0=1$. Bounded-type invariance in proposition 5.46.3 gives $R=\mathbf1_ S(\delta h+\mathrm{dd}^{\mathrm{c}}q_\delta )$. The positive current $\delta h+\mathrm{dd}^{\mathrm{c}}q_\delta $ has total $\gamma $-trace $\delta \int _ Xh\wedge \gamma ^{n-1}$ by distributional Stokes. Consequently
\[ 0\leq \int _ XR\wedge \gamma ^{n-1} \leq \delta \int _ Xh\wedge \gamma ^{n-1}. \]
Thus $R=0$ by positivity and $\delta \downarrow 0$.
Now let $\pi :X'\to X$ be a smooth connected compact modification, and fix $h'>0$ on $X'$. Genuine pullbacks are quasi-plurisubharmonic by proposition 5.43.18. Their ordinary inequalities hold pointwise, including at poles, and $\pi ^*h\leq Ch'$ for some fixed $C$. Hence $\pi ^*q_\delta \sim \pi ^*q$ and $\mathrm{dd}^{\mathrm{c}}\pi ^*q_\delta \geq -C\delta h'$. Since $C\delta $ can be arbitrarily small, the preceding argument on $X'$ proves both vanishing assertions. $\square$
The point Lelong statement and the polar statement use different parts of the current: a pole of codimension at least two need not carry any coefficient mass of the first Hessian. Small curvature controls both.
Lemma 18.27.3.E5C4A193363D Let $B$ be a smooth real $(1,1)$-form. If $q'$ is quasi-plurisubharmonic, all its point Lelong numbers vanish, and $B+\mathrm{dd}^{\mathrm{c}}q'\geq \epsilon h$ with $\epsilon >0$, then there is a smooth real $f$ such that $B+\mathrm{dd}^{\mathrm{c}}f\geq \epsilon h/2$.
Proof. Apply the classical closed-current regularization of 5.36.3 to $T=\mathrm{dd}^{\mathrm{c}}q'$ with smooth representative zero and lower bound $\gamma =\epsilon h-B$. The lower bound is allowed to be nonclosed. More explicitly, fix any positive attenuation level. Its Lelong level set is empty, so the regularization gives global smooth $f_ k$ and
\[ \mathrm{dd}^{\mathrm{c}}f_ k\geq \gamma -C\lambda _ kh-\epsilon _ kh, \]
where the continuous nonnegative $\lambda _ k$ decrease pointwise to zero and $\epsilon _ k\downarrow 0$. A fixed smooth metric on $TX$ has bounded curvature, so a sufficiently large multiple $Ch$ supplies the curvature bound in that theorem. Dini's theorem gives uniform convergence of $\lambda _ k$ to zero. Choose $k$ with $C\| \lambda _ k\| _\infty +\epsilon _ k\leq \epsilon /2$ and set $f=f_ k$. The smooth approximants need not remain in the ordinary type of $q'$. $\square$
Let $\Delta =\sum _{j=1}^ Na_ jD_ j$ be a finite effective real divisor, where the $D_ j$ are distinct prime divisors and $a_ j>0$. Choose smooth Hermitian metrics on their divisor bundles and put
\[ \sigma _\Delta =\sum _ ja_ j\log \| s_ j\| ^2,\qquad \mathrm{dd}^{\mathrm{c}}\sigma _\Delta =[\Delta ]-\chi _\Delta , \]
where $\chi _\Delta $ is smooth and closed. The case $\Delta =0$ means $\sigma _\Delta =\chi _\Delta =0$; zero coefficients are omitted from the logarithmic sum.
Theorem 18.27.4.0E6CABEB995A Suppose (18.27.1) holds and put $B=\theta -\chi _\Delta $. The ordinary type of $\sigma _\Delta +q$ contains a strict $\theta $-current if and only if $B$ has a smooth strictly positive representative: $B+\mathrm{dd}^{\mathrm{c}}f>0$ for some smooth real $f$.
Proof. Suppose first that $v\sim \sigma _\Delta +q$ and $\theta +\mathrm{dd}^{\mathrm{c}}v\geq \epsilon h$ with $\epsilon >0$. By lemma 18.27.2, $\nu (v,D_ j)=a_ j$. Apply lemma 5.38.1 with background $\theta -\epsilon h$ to obtain
\[ \theta +\mathrm{dd}^{\mathrm{c}}v-[\Delta ]\geq \epsilon h. \]
For $\Delta =0$ this is the original strict inequality. The $L^1$ difference $v-\sigma _\Delta $ consequently has a global quasi-plurisubharmonic representative $q'$ with $B+\mathrm{dd}^{\mathrm{c}}q'\geq \epsilon h$. Off the divisor it satisfies $q'\sim q$; these inequalities extend everywhere by lemma 5.24.3. Thus $q'$ has zero point Lelong numbers, and lemma 18.27.3 gives $B+\mathrm{dd}^{\mathrm{c}}f>0$ for a smooth $f$.
Conversely, compactness gives $B+\mathrm{dd}^{\mathrm{c}}f\geq \epsilon h$ for some $\epsilon >0$. Choose $q_\delta $ in (18.27.1) with $\delta <\epsilon /2$. Then $v=\sigma _\Delta +q_\delta +f$ is in the specified ordinary type, and Poincaré–Lelong gives
\[ \theta +\mathrm{dd}^{\mathrm{c}}v=[\Delta ]+B+\mathrm{dd}^{\mathrm{c}}f+\mathrm{dd}^{\mathrm{c}}q_\delta \geq \epsilon h/2. \]
$\square$
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