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11.5 Conservation of the trace massA21D56C40154

A21D56C40154

The power $\gamma ^{n-1}$ of a Gauduchon metric annihilates every complex Hessian under integration. It therefore assigns to a positive current $\theta +\mathrm{dd}^{\mathrm{c}}u$ a mass depending only on its smooth background $\theta $, also when the potential $u$ is unbounded.

Proposition 11.5.1.828FAD1A35BF Let $\theta $ be a smooth real $(1,1)$-form, let $u$ be a $\theta $-plurisubharmonic function, and let $\gamma $ be a Gauduchon metric. Then

97DCB5703B01 11.5.2
\begin{equation} \int _ X(\theta +\mathrm{dd}^{\mathrm{c}}u)\wedge \gamma ^{n-1} =\int _ X\theta \wedge \gamma ^{n-1}. \end{equation}

Proof. Since $\mathrm{dd}^{\mathrm{c}}(\gamma ^{n-1})=0$, distributional integration by parts gives

\[ \langle \mathrm{dd}^{\mathrm{c}}u,\gamma ^{n-1}\rangle =\langle u,\mathrm{dd}^{\mathrm{c}}(\gamma ^{n-1})\rangle =0. \]

The product on the left of (11.5.2) is a positive measure, whose total mass is therefore the finite number on the right. $\square$

We call $\int _ X\theta \wedge \gamma ^{n-1}$ the Gauduchon degree of $\theta $ with respect to $\gamma $. Here $\theta +\mathrm{dd}^{\mathrm{c}}u$ is the full distributional current, including any mass carried by the pole set of $u$. The non-pluripolar Monge–Ampère measure of section 13.14 discards this mass; example 13.15.17 illustrates the difference on a curve.

Corollary 11.5.3.A9609E0AE7B9 There are constants $0<a\leq b<+\infty $, depending only on $\omega $ and a chosen Gauduchon metric, such that every $u\in \mathrm{PSH}(X,\omega )$ satisfies

\[ a\leq \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge \omega ^{n-1}\leq b. \]

Proof. Fix a Gauduchon form $\gamma $ on $X$ and choose positive constants $c,C$ with $c\gamma ^{n-1}\leq \omega ^{n-1}\leq C\gamma ^{n-1}$. Wedging these inequalities with the positive current $\omega +\mathrm{dd}^{\mathrm{c}}u$ and integrating, we obtain, by proposition 11.5.1, the bounds $cA$ and $CA$, where $A=\int _ X\omega \wedge \gamma ^{n-1}>0$. $\square$

The trace bound controls the mass of positive currents and hence their weak compactness. By itself it gives no bound for the Monge–Ampère mass, since a bound on the integral of a trace does not bound the integral of its $n$th power.

The pointwise arithmetic–geometric mean inequality relates the conserved trace to the density in a smooth Monge–Ampère equation. Fix a smooth positive volume form $\mathrm{d}V$ on $X$.

Proposition 11.5.4.BFC1330527CA For any two Hermitian forms $\eta ,\gamma $,

1F99A7154650 11.5.5
\begin{equation} \eta \wedge \gamma ^{n-1} \geq \left(\frac{\eta ^ n}{\gamma ^ n}\right)^{1/n}\gamma ^ n \end{equation}

pointwise. Consequently, let $\theta $ be a smooth real $(1,1)$-form and $f$ a smooth strictly positive function. If $u\in C^\infty (X,\mathbb R)$ and $c>0$ satisfy $(\theta +\mathrm{dd}^{\mathrm{c}}u)^ n=c f\, \mathrm{d}V$ with $\theta +\mathrm{dd}^{\mathrm{c}}u>0$, then

B78F87EF7411 11.5.6
\begin{equation} c^{1/n} \int _ X\left(\frac{f\, \mathrm{d}V}{\gamma ^ n}\right)^{1/n}\gamma ^ n \leq \int _ X\theta \wedge \gamma ^{n-1} \end{equation}

for every Gauduchon metric $\gamma $.

Proof. In a frame where $\gamma $ is the identity, let $\lambda _1,\ldots ,\lambda _ n>0$ be the eigenvalues of $\eta $. The ratios of the two sides of (11.5.5) to $\gamma ^ n$ are $n^{-1}\sum _ j\lambda _ j$ and $(\prod _ j\lambda _ j)^{1/n}$, respectively, and their comparison is the arithmetic–geometric mean inequality.

For the second assertion, apply (11.5.5) with $\eta =\theta +\mathrm{dd}^{\mathrm{c}}u$, integrate, and use (11.5.2). $\square$

The corresponding density inequality for unbounded potentials in a Hermitian background is proved in lemma 13.14.12.

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