Proposition 23.2.694CCB765ECCA2

Proof. Put $\Gamma =\gamma ^{n-1}$ and use $\beta $ as above. Then

Thus $[\mathrm d^{\mathrm c}\Gamma ]=0$.

Next, let $S$ be a real current of degree one such that $T=\mathrm{d}S$ is a nonzero positive $(1,1)$-current, and let $\Theta $ be a closed real form with strictly positive $(n-1,n-1)$-part $\Gamma $. By type considerations and Stokes' theorem for currents,

But the first integral is strictly positive because $T\ne 0$ and $\Gamma >0$, a contradiction. Thus a strongly Gauduchon metric excludes nonzero positive exact currents.

Conversely, suppose that no strongly Gauduchon metric exists. In the real Fréchet space of smooth $(2n-2)$-forms, let $\mathcal Z$ be the closed subspace of closed forms, and let $\mathcal P$ be the open convex cone of forms with strictly positive $(n-1,n-1)$-part. This cone is nonempty, since it contains $h^{n-1}$ for any Hermitian metric $h$. The two sets are disjoint, because the positive $(n-1)$st root of the $(n-1,n-1)$-part of a form in their intersection would be a strongly Gauduchon metric. The Hahn–Banach separation theorem therefore gives a nonzero real current $T$ of degree two such that

Fix $\Theta _0\in \mathcal P$. If a real form $\Psi $ has zero $(n-1,n-1)$-part, then $\Theta _0+t\Psi \in \mathcal P$ for every $t\in \mathbb R$, so $T(\Psi )=0$. Thus $T$ is of type $(1,1)$. For any semipositive $(n-1,n-1)$-form $\Gamma $, we have $\Gamma +\varepsilon \Theta _0\in \mathcal P$ for $\varepsilon >0$, and letting $\varepsilon \downarrow 0$ gives $T(\Gamma )\geq 0$; thus $T$ is positive.

Since $T$ annihilates every exact test form, it is closed. By the de Rham theorem for currents and Poincaré duality, its cohomology class is determined by its pairings with closed forms of complementary degree. All these pairings vanish, so $T=\mathrm{d}S$ for a real current $S$. Thus $T$ is a nonzero positive exact current, which proves the converse. $\square$