Pome8C7192C73C68The pullback of a positive form from a singular space loses positivity along the contracted curves. In the standard small resolution of a node, the logarithm of the norm in the normal bundle of the exceptional curve provides the missing curvature along the curve. Cutting this logarithm off where the pullback is already positive gives a global strict barrier, so the results above apply.
Assume now $n=3$. Let $Y$ be a connected compact normal complex threefold, smooth outside a finite set $\Sigma $ of nodes. A node is locally the hypersurface
Let $\pi :X\to Y$ be a proper holomorphic map that is biholomorphic over $Y\setminus \Sigma $. Assume each local resolution is the standard small resolution: a neighborhood of its exceptional curve is identified with a neighborhood of the zero section of
and the contraction, in bundle coordinates $(z,y_1,y_2)$, is
Interchanging the two rulings gives the other standard model. Write $E=\pi ^{-1}(\Sigma )$; its components $E_ j$ are disjoint copies of $\mathbb P^1$.
Fix a positive form $\omega _ Y$ on $Y$ in the local embedding sense: in each ambient chart it is induced by a smooth strictly positive real $(1,1)$-form. Set $\theta =\pi ^*\omega _ Y$, and let $\eta $ be an arbitrary Hermitian form on $X$.
Lemma 15.19.10.C6428C798D5F The form $\theta $ is smooth and semipositive, positive on $X\setminus E$. There are $\delta >0$ and $\rho \in \mathrm{PSH}(X,\theta )$ with analytic poles precisely on $E$, smooth on $X\setminus E$, such that
Proof. Smoothness and positivity off $E$ follow from the pullback and the biholomorphism. In one standard model put
The function $r^2$ is the sum of the squared absolute values of the four contracted coordinates. It is therefore globally defined on the bundle; $b$ is the pullback of the Fubini–Study form from its base. In the chart (15.19.9),
AC4599B20814
15.19.11
This is an inequality of currents across the zero section as well as a smooth inequality off it.
Ambient positivity gives $\theta \geq c\mathrm{dd}^{\mathrm{c}}r^2$ on a sufficiently small tube, for some $c>0$. The smooth form $\mathrm{dd}^{\mathrm{c}}r^2+b$ is positive definite there. Indeed the contraction is an immersion off the zero section, and on that section $\mathrm{dd}^{\mathrm{c}}r^2$ is positive in both fiber directions while $b$ is positive along the base. Compactness of the zero section and a smaller closed tube give a uniform lower bound by a positive multiple of $\eta $. Thus, for every fixed $a>0$,
near the zero section, with $\delta _ a>0$.
Choose disjoint model tubes with nonempty complement and numbers $0<t_1<t_2<1$ in each. Let $0\leq \chi \leq 1$ be a smooth cutoff of $r^2$, equal to one for $r^2\leq t_1$ and zero for $r^2\geq t_2$. Extend $\chi (r^2)\log r^2$ by zero outside its tube. On the compact transition annuli, the pullback $\theta $ is uniformly positive and the second derivatives of this cutoff potential are bounded. A sufficiently small positive rational $a$ therefore makes
satisfy $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \eta $ there and near every $E_ j$. On the remaining compact set, $\rho =0$ and $\theta $ is positive, so decreasing $\delta $ proves the global inequality. Near $E_ j$ the function is $a\log (|y_1|^2+|y_2|^2)+a\log (1+|z|^2)$, hence has analytic singularities with rational coefficient and pole set exactly $E_ j$. It is smooth elsewhere and nonpositive; its zero extension gives supremum zero. If $E$ is empty, take $\rho =0$ and use the positivity of $\theta $ on compact $X$. $\square$
Corollary 15.19.12.A1536B3C6F39 For the preceding geometric data and every smooth positive probability volume form $\mu $ on $X$, the solutions of (15.19.2) have the bounds and subsequential convergence of theorem 15.19.7, with $Z=E$. Moreover, for each exceptional curve,
Proof. Apply the theorem using lemma 15.19.10. Since $\pi $ contracts $E_ j$, the restriction of $\theta $ to it is zero. Stokes on the compact smooth curve gives $\int _{E_ j}\mathrm{dd}^{\mathrm{c}}u_\varepsilon =0$, proving the area identity. $\square$
When $K_ X$ is torsion, the density can be chosen so that these metrics are Chern–Ricci flat. Recall that the Chern–Ricci form $\operatorname {Ric}^{\mathrm{Ch}}(\omega )$ of a Hermitian metric $\omega $ is $-\mathrm{i}\partial \bar\partial $ of the logarithm of the density of $\omega ^ n$ with respect to a coordinate volume form. Choose a nowhere vanishing holomorphic section of $K_ X^{\otimes m}$ for some $m\geq 1$. If its local expression is $f(\mathrm{d}z_1\wedge \mathrm{d}z_2\wedge \mathrm{d}z_3)^{\otimes m}$, use $|f|^{2/m}$ as the density relative to the standard coordinate volume form. The coordinate Jacobians cancel, so this defines a smooth positive volume form on $X$; normalize it to total mass one and take it as $\mu $. Its logarithmic density is pluriharmonic. Applying $-\mathrm{i}\partial \bar\partial \log $ to $\omega _\varepsilon ^3=c_\varepsilon \mu $ therefore gives $\operatorname {Ric}^{\mathrm{Ch}}(\omega _\varepsilon )=0$, and the same holds for the smooth limiting metric on $X\setminus E$.
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