Proposition 10.29.10E080ADB5BEF1

Proof. (1) Since $\mathrm{dd}^{\mathrm{c}}|z|^2$ has constant coefficients and $y_ j$ is pluriharmonic,

where

The Hermitian matrix of $\mathrm{dd}^{\mathrm{c}}(y_1y_2)$ in the basis $\mathrm dw_1,\mathrm dw_2$ is a positive multiple of $\left(\begin{smallmatrix}0& 1\\ 1& 0\end{smallmatrix}\right)$, which is indefinite. Accordingly, for $\eta _\pm =\mathrm dw_1\pm \mathrm dw_2$ the $(3,3)$-forms $\mathrm{dd}^{\mathrm{c}}\omega \wedge \mathrm i\eta _\pm \wedge \overline\eta _\pm $ are respectively negative and positive multiples of $\mathrm dV_0$, so $\mathrm{dd}^{\mathrm{c}}\omega =\mathrm{dd}^{\mathrm{c}}\beta $ is neither semipositive nor seminegative. Since $\theta $ is closed and $\beta ^2=0$, we have $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=2\theta \wedge \mathrm{dd}^{\mathrm{c}}\beta $, and this vanishes because every term of $\theta \wedge (\mathrm dw_1\wedge \mathrm d\overline w_2+\mathrm dw_2\wedge \mathrm d\overline w_1)$ repeats a differential.

(2) By theorem 11.1.5 with $n=3$ and $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$, $\mathrm M_\omega (u)-\int _ X\omega ^3=3\int _ Xu\, \mathrm{dd}^{\mathrm{c}}u\wedge \mathrm{dd}^{\mathrm{c}}\omega $. The form $\mathrm{dd}^{\mathrm{c}}\omega $ is closed, so Stokes' theorem gives $\int _ Xu\, \mathrm{dd}^{\mathrm{c}}u\wedge \mathrm{dd}^{\mathrm{c}}\omega =-\int _ X\mathrm du\wedge \mathrm d^{\mathrm c}u\wedge \mathrm{dd}^{\mathrm{c}}\omega $. Now $\mathrm du\wedge \mathrm d^{\mathrm c}u=(\mathrm i/2\pi )\, \partial u\wedge \overline\partial u$, and a direct computation in the coordinates $w_1,w_2,z$ gives

Collecting the constants $(1/2\pi )(1/8\pi )(1/2\pi )$ gives the formula.

(3) Since $\mathrm{dd}^{\mathrm{c}}u$ has no components in the $z$-direction, $\omega +\mathrm{dd}^{\mathrm{c}}u=(\theta +\mathrm{dd}^{\mathrm{c}}u)+\beta $ splits into blocks, and $u\in \mathrm{PSH}(X,\omega )$ if and only if $u\in \mathrm{PSH}(X,\theta )$. By theorem 10.29.4, $u=f(y)+\log (y_1y_2)$ with $f$ convex, so $g$ is locally Lipschitz, hence Lipschitz by periodicity. In the variables $\varsigma _ j=\log y_ j$, convexity of $f$ is equivalent to positivity of the symmetric matrix

as one checks from $y_ iy_ j\, \partial ^2f/\partial y_ i\partial y_ j=\partial ^2\Phi /\partial \varsigma _ i\partial \varsigma _ j-\delta _{ij}\, \partial \Phi /\partial \varsigma _ i$ for $\Phi (\varsigma )=f(\mathrm e^{\varsigma })=g(\varsigma )-\varsigma _1-\varsigma _2$. This condition is linear in $g$ with constant coefficients, so it is preserved by convolution with a smooth kernel on $\mathbb R^2$; the regularized functions are periodic, and they converge uniformly to $g$ with gradients converging almost everywhere. By the continuity of bounded Monge–Ampère products under uniform convergence and dominated convergence, it suffices to prove all assertions for smooth $g$ with $H\geq 0$.

For such $g$, the function $u$ depends only on $y$, so $\partial u/\partial w_ j=-\frac{\mathrm i}2\partial u/\partial y_ j$ and the integrand in (2) is $\frac14\partial _{y_1}u\, \partial _{y_2}u=\partial _1g\, \partial _2g/(4y_1y_2)$. A direct computation gives $\omega ^3=3\theta ^2\wedge \beta =\frac{3}{64\pi ^3}\, \mathrm dV_0/(y_1y_2)$. By (10.29.9) and (2),

where the second equality uses $\int _{T_ U}\partial _ jg\, \mathrm d\varsigma =0$.

The diagonal entry $H_{11}=P-\partial _1P\geq 0$ says that $\mathrm e^{-\varsigma _1}P$ is nonincreasing in $\varsigma _1$, that is,

If $P(\varsigma )<0$, the right side tends to $-\infty $ as $\tau \to \infty $, which is impossible for the bounded function $P$; hence $P\geq 0$, and likewise $Q\geq 0$. Thus $\partial _ jg\leq 1$. The lattice $\Lambda _ U$ contains two independent vectors $\lambda ,\lambda '$ with positive coordinates, and every point of $T_ U$ has the form $\varsigma +\tau \lambda +\tau '\lambda '$ with $\tau ,\tau '\in [0,1)$. Along $\tau \mapsto \varsigma +\tau \lambda $, the derivative of $g$ is at most $|\lambda |_1=\lambda _1+\lambda _2$, and its integral over $[0,1]$ vanishes by periodicity, so $|g(\varsigma +\tau \lambda )-g(\varsigma )|\leq |\lambda |_1$. Hence $\operatorname {osc}g\leq L$, with $L=|\lambda |_1+|\lambda '|_1$ depending only on $\Lambda _ U$.

Averaging (10.29.11) in the form $P(\varsigma -\tau e_1)\geq \mathrm e^{-\tau }P(\varsigma )$ over $\tau \in [0,1]$ gives

On the other hand, averaging $P(\varsigma )\geq \mathrm e^{-\tau }P(\varsigma +\tau e_1)$ over $\tau \in [0,R]$ with $R=\max (2L,1)$ gives

The same bounds hold for $Q$. Therefore the average of $PQ$ over $T_ U$ lies between $\mathrm e^{-2R}/4$ and $\bigl ((1+L)/(1-\mathrm e^{-1})\bigr )^2$, which proves (3).

(4) Let $k$ belong to the dual lattice $\Lambda _ U^*=\{ k\in \mathbb R^2:\langle k,\lambda \rangle \in \mathbb Z\text{ for all }\lambda \in \Lambda _ U\} $, and put $g_\epsilon (\varsigma )=\epsilon \cos (2\pi \langle k,\varsigma \rangle )$. For small $\epsilon >0$, the matrix $H$ of $g_\epsilon $ is positive definite, so $u_\epsilon =g_\epsilon (\varsigma )$ is a smooth function with $\omega +\mathrm{dd}^{\mathrm{c}}u_\epsilon >0$. By (3),

Since the lattice $\Lambda _ U^*$ has vectors in every open quadrant, we may choose $k$ with $k_1k_2>0$, respectively $k_1k_2<0$, which gives the two inequalities. $\square$