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Proposition 23.2.59EBFA309383E5

Proposition 23.2.59.EBFA309383E5 The cone $\operatorname {LG}(X)$ is nonempty, convex and open in $W_ X$. Its linear span is $W_ X$.

Proof. Let $\mathcal Z_ X$ be the real vector space of smooth $(n-1,n-1)$-forms $\Gamma $ satisfying $\mathrm{dd}^{\mathrm{c}}\Gamma =0$, and set

\[ q_ X(\Gamma )=[\mathrm d^{\mathrm c}\Gamma ]. \]

We first show that $q_ X(\mathcal Z_ X)=W_ X$. The inclusion into $W_ X$ follows from (23.2.58). If $[\lambda ]$ annihilates the image of $q_ X$, then by (23.2.58) the class $\{ \kappa _\lambda \} $ has zero pairing with every Gauduchon form. The generalized Lamari criterion, theorem 11.6.1, implies that both $\{ \kappa _\lambda \} $ and $-\{ \kappa _\lambda \} $ are pseudoeffective, so $[\lambda ]\in K_ X$ by lemma 23.2.56. Thus the image of $q_ X$ and $W_ X$ have the same annihilator in $V_ X$, and since the Poincaré pairing between $H^1_{\mathrm{dR}}$ and $H^{2n-1}_{\mathrm{dR}}$ is nondegenerate, they are equal.

Let $\mathcal Z_ X^+$ be the set of strictly positive forms in $\mathcal Z_ X$. By the linear-algebra construction used in the proof of theorem 11.6.1, every such form is the $(n-1)$st power of a unique Hermitian form, which is then Gauduchon. Hence

\[ \operatorname {LG}(X)=q_ X(\mathcal Z_ X^+). \]

By Gauduchon's theorem, $\mathcal Z_ X^+$ is nonempty, and it is a convex cone; hence so is $\operatorname {LG}(X)$.

For openness, choose forms $\Gamma _1,\ldots ,\Gamma _ r\in \mathcal Z_ X$ whose images form a basis of $W_ X$. If $\Gamma \in \mathcal Z_ X^+$, then $\Gamma +\sum _ i t_ i\Gamma _ i$ remains strictly positive for all sufficiently small real $t_ i$, and the images of these forms cover a neighborhood of $q_ X(\Gamma )$ in $W_ X$. This includes the case $W_ X=\{ 0\} $, with the empty basis. Since a nonempty open subset spans the ambient vector space, the assertion about the span follows. $\square$

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