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Corollary 23.2.7086E2247D209D

Corollary 23.2.70.86E2247D209D Fix a Hermitian metric $h$ on $X$ and a norm on $H^2_{\mathrm{dR}}(X,\mathbb R)$. Then $X$ admits a strongly Gauduchon metric if and only if there is a constant $C>0$ such that every closed positive $(1,1)$-current $T$ satisfies

24F2C25D4A8F 23.2.71
\begin{equation} \int _ XT\wedge h^{n-1}\leq C\| [T]_{\mathrm{dR}}\| . \end{equation}

Under this condition, the family of closed positive $(1,1)$-currents whose de Rham classes belong to a fixed compact set is weakly compact, and the image of the pseudoeffective Bott–Chern cone in $H^2_{\mathrm{dR}}(X,\mathbb R)$ is a closed convex cone containing no line.

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

\[ \int _ XT\wedge h^{n-1} \leq A\int _ XT\wedge \gamma ^{n-1} =A\langle [T]_{\mathrm{dR}},[\Theta ]_{\mathrm{dR}}\rangle . \]

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 $c=0$. $\square$

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