11.6 The Gauduchon criterion for positivity6029C37CE10F

of theorem 11.6.1. Necessity is proposition 11.5.1. For sufficiency, let $F$ be the real vector space of smooth real $(n-1,n-1)$-forms, and let

Fix a strictly positive $\Gamma _0\in E$, which exists by theorem 11.4.2. By the root calculation above, the hypothesis says that $\ell $ is nonnegative on the strictly positive elements of $E$. Adding $\epsilon \Gamma _0$ and letting $\epsilon \downarrow 0$, we see that $\ell $ is also nonnegative on the semipositive elements of $E$.

Step 1: a positive extension of $\ell $. For $\Phi \in F$ put

This is a finite sublinear function. To see finiteness, choose $C$ with $-C\Gamma _0\leq \Phi \leq C\Gamma _0$. The admissible element $C\Gamma _0$ gives an upper bound, while every admissible $\Gamma $ satisfies $\Gamma +C\Gamma _0\geq 0$ and hence $\ell (\Gamma )\geq -C\ell (\Gamma _0)$. Sublinearity follows by adding and rescaling competitors. For $\Phi \in E$, the nonnegativity of $\ell (\Gamma -\Phi )$ for admissible $\Gamma $ shows that $p(\Phi )=\ell (\Phi )$.

The Hahn–Banach theorem gives a real linear extension $T$ of $\ell $ to $F$ with $T(\Phi )\leq p(\Phi )$ for all $\Phi \in F$. If $\Phi \geq 0$, the zero form is admissible for $p(-\Phi )$, so $T(-\Phi )\leq 0$, that is, $T(\Phi )\geq 0$. Moreover, the bounds by $C\Gamma _0$ show that $T$ is continuous for the uniform norm. It is therefore a positive $(1,1)$-current, and the current $S=T-\theta $ annihilates $E$.

Step 2: $S$ is $\mathrm{dd}^{\mathrm{c}}$-exact. To recover a potential from $S$, consider the scalar operator

where $\mathrm dV$ is the smooth positive volume form fixed above. Since $\mathrm{dd}^{\mathrm{c}}\Gamma _0=0$, this is an elliptic scalar operator with no zeroth-order term. Its kernel consists of the constants and its index is zero, while Stokes' theorem gives $\int _ XDv\, \mathrm{d}V=0$. Consequently its range is exactly the space of smooth functions of integral zero, and by elliptic estimates its inverse on functions normalized by $\int _ Xv\, \mathrm{d}V=0$ is continuous in the smooth topology.

For a smooth top-degree form $\zeta $ with $\int _ X\zeta =0$, solve $Dv=\zeta /\mathrm{d}V$ with this normalization and set $R\zeta =v\Gamma _0$. Then $\mathrm{dd}^{\mathrm{c}}R\zeta =\zeta $, and $R$ is continuous. Define a distribution $u$ on such top-degree forms by

and extend it to all top-degree forms by assigning an arbitrary value on one fixed volume form of integral one. For every $\Phi \in F$, the form $\Phi -R(\mathrm{dd}^{\mathrm{c}}\Phi )$ belongs to $E$, and $S$ annihilates it. It follows that

Thus $\theta +\mathrm{dd}^{\mathrm{c}}u=T\geq 0$, and proposition 5.22.2 gives a representative of $u$ in $\mathrm{PSH}(X,\theta )$. $\square$

Proof. Since $\theta $ is smooth, $-C\omega \leq \theta \leq C\omega $ for some $C$, so the infimum lies in $[-C,C]$. If $\theta +\mathrm{dd}^{\mathrm{c}}u\geq a\omega $, the trace identity applied to the background $\theta -a\omega $ shows that $a$ is at most each of the ratios in the definition, hence $a\leq \kappa _\omega (\theta )$. Conversely, for $a=\kappa _\omega (\theta )$, all Gauduchon degrees of $\theta -a\omega $ are nonnegative, and theorem 11.6.1 applied to this smooth form gives the required $u$. The last assertion follows, since bigness means the existence of a $\theta $-plurisubharmonic $u$ with $\theta +\mathrm{dd}^{\mathrm{c}}u\geq \delta \omega $ for some $\delta >0$. $\square$

Proof. Choose $u\in \mathrm{PSH}(X,\theta )$ and put $S=\theta +\mathrm{dd}^{\mathrm{c}}u\geq 0$. By proposition 11.5.1, $\int _ XS\wedge \gamma ^{n-1}=0$. Since the trace measure $S\wedge \gamma ^{n-1}$ of the positive current $S$ controls all its coefficient measures, $S=0$. Taking the trace of the equation $\mathrm{dd}^{\mathrm{c}}u=-\theta $ with respect to $\omega $ gives an elliptic equation with smooth right-hand side, so $u$ is smooth by elliptic regularity, and $f=-u$ has the required property. The converse is Stokes' theorem. For $n=1$, the trace measure is $S$ itself and the same argument applies. $\square$