23.2.55 The Lee–Gauduchon cone25E90D86853A

Proof. If $j_ X([\lambda ])=0$, write $\mathrm d^{\mathrm c}\lambda =\mathrm{dd}^{\mathrm{c}}f$ for a smooth real function $f$. Since $\mathrm d^{\mathrm c}\mathrm{d}=-\mathrm{d}\mathrm d^{\mathrm c}$, the form $\lambda +\mathrm{d}f$ is both $\mathrm{d}$-closed and $\mathrm d^{\mathrm c}$-closed. Since the two equations separate the $\partial $- and $\bar\partial $-derivatives, its $(1,0)$-part is holomorphic and closed. Thus $\lambda +\mathrm{d}f$ is the real part of the closed holomorphic form $2(\lambda +\mathrm{d}f)^{1,0}$. Conversely, the real part of a closed holomorphic one-form is closed for both operators, so its image under $j_ X$ is zero.

By proposition 10.11.2, the pseudoeffective Bott–Chern cone contains no line. Thus $j_ X(\xi )$ and $-j_ X(\xi )$ are both pseudoeffective exactly when $j_ X(\xi )=0$, which is the identity $\mathcal C_ X\cap (-\mathcal C_ X)=K_ X$. $\square$

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

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

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. $\square$

Proof. The first equality is theorem 11.6.1 applied to the smooth closed form $\kappa _\lambda $, together with (23.2.58). It expresses $\mathcal C_ X$ as an intersection of closed half-spaces, so this cone is closed.

If $a=[\mathrm d^{\mathrm c}(\gamma ^{n-1})]\in \operatorname {LG}(X)$ and $\xi =[\lambda ]\in \mathcal C_ X\setminus K_ X$, choose a closed positive current $T=\kappa _\lambda +\mathrm{dd}^{\mathrm{c}}u$. It is nonzero because it represents the nonzero class $j_ X(\xi )$. Therefore

Conversely, suppose that $a\in W_ X\setminus \operatorname {LG}(X)$. Separating the point $a$ from the nonempty open convex cone $\operatorname {LG}(X)$ in the finite-dimensional space $W_ X$ gives a nonzero linear functional $\ell $ on $W_ X$ such that

Indeed, in the usual separating inequalities the constant can be taken to be zero, by scaling the elements of the cone. By Poincaré duality, $\ell =\langle \xi ,\cdot \rangle $ for a class $\xi \in V_ X$ that is unique modulo $K_ X$, and $\xi \notin K_ X$ because $\ell \neq 0$. The first equality of the theorem gives $\xi \in \mathcal C_ X$, and then $\langle \xi ,a\rangle \leq 0$ shows that $a$ does not belong to the right side of the second equality. This proves the second equality; if $\operatorname {LG}(X)=W_ X$, there is nothing to prove. $\square$