23.2.67 Strongly Gauduchon metrics5A635830FDEB

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. $\square$

Proof. Suppose that $\gamma $ is strongly Gauduchon, and choose a smooth closed real form $\Theta $ whose $(n-1,n-1)$-part is $\gamma ^{n-1}$. By compactness, $h^{n-1}\leq A\gamma ^{n-1}$ for some $A>0$. Positivity and type considerations then give

The last pairing is a fixed linear functional on the finite-dimensional de Rham group, which proves (23.2.71). Conversely, this bound forces every closed positive de Rham-exact current to vanish, and proposition 23.2.69 gives a strongly Gauduchon metric.

For the compactness assertion, the bound gives uniformly bounded trace masses. A weak limit of such currents is positive and closed, and its de Rham class is the limit of their classes, as one sees by pairing with a finite basis of smooth closed forms of complementary degree and using Poincaré duality. Thus the family is weakly closed, and the usual weak compactness of positive currents with bounded mass proves the assertion.

Let $c_ j$ be a convergent sequence in the image cone, and choose closed positive currents $T_ j$ with $[T_ j]_{\mathrm{dR}}=c_ j$. By the compactness assertion, a subsequence converges to a closed positive current representing the limit of the classes. Its Bott–Chern class is pseudoeffective, so the image cone is closed. Convexity follows from addition and nonnegative scaling. Finally, if both $c$ and $-c$ lie in the image, choose closed positive currents $T$ and $S$ representing them. Their sum is de Rham exact, so (23.2.71) gives $T+S=0$, and positivity gives $T=S=0$, hence $c=0$. $\square$

Proof. Consider the holomorphic action of the additive group $\mathbb C$ on

The action is free, since $\mathrm{e}^ t=\mathrm{e}^{\mathrm i t}=1$ forces $t=0$. It is also proper. Indeed, if both $(z,w)$ and $t\cdot (z,w)$ range in compact subsets of $\widetilde X$, then the norms of the four vectors $z$, $w$, $\mathrm e^ tz$ and $\mathrm e^{\mathrm it}w$ in $\mathbb C^2$ are bounded above and away from zero. Writing $t=a+\mathrm i b$, the ratios of these norms bound $\mathrm{e}^ a$ and $\mathrm{e}^{-b}$ above and away from zero, so $a$ and $b$ stay bounded. A free proper holomorphic action has local holomorphic slices, and its quotient $X$ is a Hausdorff complex manifold of dimension three.

Every orbit meets $S^3\times S^3$ exactly once, at the image of $(z,w)$ under the group element with

These expressions depend smoothly on $(z,w)$, so the slice $S^3\times S^3$ identifies the quotient with $S^3\times S^3$ as a smooth manifold. In particular, $X$ is compact and connected, and

It follows that $V_ X=K_ X=W_ X=0$ and $\operatorname {LG}(X)=\{ 0\} $; the latter cone is nonempty by Gauduchon's theorem and lies in the zero group $H^5_{\mathrm{dR}}(X,\mathbb R)$.

The map $(z,w)\mapsto ([z],[w])$ is invariant under the action and descends to a holomorphic map

It is a submersion, because the map on $\widetilde X$ is a submersion and the orbits of the action lie in its fibers. Let $\eta $ be the sum of the Fubini–Study forms on the two factors. The smooth form

is closed, semipositive and of rank two everywhere. It is nonzero and de Rham exact, since $H^2_{\mathrm{dR}}(X,\mathbb R)=0$. By proposition 23.2.69, $X$ admits no strongly Gauduchon metric.

The Bott–Chern class of $\alpha $ is nonzero: otherwise $\alpha =\mathrm{dd}^{\mathrm{c}}f$ for a smooth real function $f$, and the integral of $\alpha $ against any Gauduchon form would vanish, contradicting $\alpha \geq 0$ and $\alpha \ne 0$. The class is pseudoeffective, since $\alpha $ is a positive representative. Finally, $j_ X$ has zero domain, so this class is not in its image. $\square$