10.29 Oeljeklaus–Toma manifolds9F59E018A291

Proof. A subgroup $G$ of $\mathbb R^ s$ is dense unless there is a nonzero linear form $\xi $ on $\mathbb R^ s$ with $\xi (G)\subseteq \mathbb Z$: the closure of $G$ is the sum of a linear subspace $V$ and a discrete subgroup of a complement of $V$, and if $V\neq \mathbb R^ s$ one can take $\xi $ vanishing on $V$ and integral on that discrete subgroup. Suppose $\xi (\sigma _{\mathbb R}(a))=\sum _ j\xi _ j\sigma _ j(a)\in \mathbb Z$ for every $a\in \mathcal O_ K$. Since $\mathcal O_ K$ spans $K$ over $\mathbb Q$, the map $a\mapsto \sum _ j\xi _ j\sigma _ j(a)$ is a $\mathbb Q$-linear map $K\to \mathbb Q$. The trace form of $K$ is nondegenerate, so there is $b\in K$ with

where $\sigma $ runs over all $s+2t$ embeddings of $K$ into $\mathbb C$. Distinct field embeddings are linearly independent over $\mathbb C$, by Dedekind's lemma on characters. Comparing the coefficients of a complex embedding $\sigma _{s+1}$, which occurs only on the right, gives $\sigma _{s+1}(b)=0$. Hence $b=0$, and then $\xi _ j=\sigma _ j(b)=0$ for every $j$. $\square$

Proof. We first note that a function $g$ on $\mathbb H^ s$ that depends only on $y$ is plurisubharmonic if and only if, as a function on $(0,\infty )^ s$, it is convex (with values in $[-\infty ,+\infty )$ and not identically $-\infty $). For smooth $g$ this follows from $\partial ^2g/\partial w_ j\partial \overline w_ k=\frac14\partial ^2g/\partial y_ j\partial y_ k$. In general, convolution with a smooth radial kernel in $w$ produces smooth functions of $y$ alone; they decrease to $g$ if $g$ is plurisubharmonic, and they converge locally uniformly to $g$ if $g$ is convex, because a convex function that is not identically $-\infty $ on the open convex set $(0,\infty )^ s$ is finite and continuous there. Since decreasing limits of convex functions are convex and locally uniform limits of plurisubharmonic functions are plurisubharmonic, the claim follows.

The identity (10.29.5) says exactly that $u_ f$ is invariant under $U$, and $u_ f$ is clearly invariant under the translations. On the universal cover, $\theta +\mathrm{dd}^{\mathrm{c}}u_ f=\mathrm{dd}^{\mathrm{c}}f(y)\geq 0$. Hence $u_ f\in \mathrm{PSH}(X,\theta )$, and $f$ is recovered from $u_ f$, so the map is injective.

Conversely, let $u\in \mathrm{PSH}(X,\theta )$, and let $\tilde u$ be its lift to $\mathbb H^ s\times \mathbb C^ t$. Since $\theta +\mathrm{dd}^{\mathrm{c}}\tilde u=\mathrm{dd}^{\mathrm{c}}(\tilde u-\sum _ j\log y_ j)$ there, the function $q=\tilde u-\sum _ j\log y_ j$ is plurisubharmonic on $\mathbb H^ s\times \mathbb C^ t$.

Step 1: constancy on the leaves. Fix $w^0\in \mathbb H^ s$. The restriction of $q$ to the slice $\{ w^0\} \times \mathbb C^ t$ is plurisubharmonic or identically $-\infty $, and it is bounded above because $u$ is upper semicontinuous on the compact manifold $X$. A plurisubharmonic function on $\mathbb C^ t$ that is bounded above is constant, since its restriction to every complex line through two given points is a subharmonic function on $\mathbb C$ bounded above. Hence $\tilde u(w,z)=\tilde u(w,0)$ for all $(w,z)$. In particular, the function $F(w)=q(w,0)$ is not identically $-\infty $, because $\tilde u$ is not, and it is plurisubharmonic on $\mathbb H^ s$.

Step 2: invariance under real translations. The translations by $\mathcal O_ K$ and Step 1 give $F(w+\sigma _{\mathbb R}(a))=F(w)$ for every $a\in \mathcal O_ K$, since $\log y_ j$ does not change under real translations. For $r>0$, the averages $F_ r$ of $F$ over the Euclidean balls of radius $r$ are defined on $\{ w:\min _ jy_ j>r\} $ and continuous there, because $F$ is locally integrable, and they are invariant under the same translations. By lemma 10.29.3 and continuity, each $F_ r$ is invariant under all translations by $\mathbb R^ s$. Since $F_ r$ decreases to $F$ as $r\downarrow 0$, the function $F$ is invariant under all real translations as well, so $F(w)=f(y)$ for a function $f$ on $(0,\infty )^ s$.

Step 3: convexity and automorphy. Since $F=f(y)$ is plurisubharmonic, the first paragraph of the proof shows that $f$ is convex, finite and continuous. Finally, $u=u_ f$, and invariance of $u$ under $U$ is (10.29.5). $\square$

Proof. Continuous functions on a compact manifold are bounded, so any two of them are equivalent, and their Lelong numbers and multiplier ideals are those of a bounded function. Since $\theta \geq 0$, the constant $0$ belongs to $\mathrm{PSH}(X,\theta )$, so $V_\theta =0$, and $P_\theta [u]=P_\theta [0]=V_\theta $. By theorem 10.29.4, locally $\theta +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}f(y)$ with $f$ continuous and depending on the $s$ real variables $y_ j$; the Bedford–Taylor products of such a current vanish in degree $s+1\leq n$, because they are limits of the corresponding products for smooth convex approximants of $f$, whose complex Hessians have rank at most $s$. Thus $(\theta +\mathrm{dd}^{\mathrm{c}}u)^ n=0$. The form $\theta $ is smooth and semipositive, hence nef. Suppose that $\rho \in \mathrm{PSH}(X,\theta )$ satisfied $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega $ with $\delta >0$. The function $\rho $ is continuous, and on a small coordinate ball with coordinates $\zeta $ we may write $\theta =\mathrm{dd}^{\mathrm{c}}a$ with $a$ smooth and $\delta \omega \geq \delta '\mathrm{dd}^{\mathrm{c}}|\zeta |^2$ with $\delta '>0$. Then $a+\rho -\delta '|\zeta |^2$ is plurisubharmonic, and expanding $(\mathrm{dd}^{\mathrm{c}}(a+\rho ))^ n$ as a sum of positive mixed Bedford–Taylor products gives $\mathrm{MA}_\theta (\rho )\geq \delta '^ n(\mathrm{dd}^{\mathrm{c}}|\zeta |^2)^ n\neq 0$ on the ball, a contradiction. Hence $\theta $ is not big. $\square$

Proof. By Siu's theorem (5.34.2), the set $E_ c=\{ \nu (u,\cdot )\geq c\} $ is a compact analytic subset of $X$. It is a proper subset, because $\nu (u,x)=0$ outside the pluripolar set $\{ u=-\infty \} $, which has Lebesgue measure zero. Hence $E_ c$ is finite by the theorem of Ornea and Verbitsky. By Skoda's integrability criterion, recalled in [SGPT], $\mathcal I(ku)_ x=\mathcal O_{X,x}$ whenever $k\nu (u,x)<1$, so $\mathcal I(ku)$ is trivial outside the finite set $E_{1/k}$. A coherent ideal sheaf that is trivial outside a finite set has finite colength. $\square$

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. $\square$

Proof. Suppose $\widetilde\omega $ is such a form, with $\mathrm{dd}^{\mathrm{c}}\widetilde\omega \geq 0$ on every complex $2$-plane; the other case is analogous. Average the lift of $\widetilde\omega $ over the translations $(w,z)\mapsto (w+r,z+\zeta )$, $(r,\zeta )\in (\mathbb R^ s\times \mathbb C)/(\sigma _{\mathbb R},\sigma _{\mathbb C})(\mathcal O_ K)$, as in the definition of $Au$ above. The averaged form $\overline\omega $ is again $\Gamma $-invariant and Hermitian, its coefficients in the frame $\mathrm dw_1,\ldots ,\mathrm dw_ s,\mathrm dz$ depend only on $y$, and $\mathrm{dd}^{\mathrm{c}}\overline\omega \geq 0$ on complex $2$-planes, because translations are holomorphic and positivity on $2$-planes is preserved by averaging.

Let $h(y)>0$ be the coefficient of $\mathrm i\, \mathrm dz\wedge \mathrm d\overline z$ in $\overline\omega $. Invariance under a unit $\varepsilon $ gives $h(\sigma (\varepsilon )y)|\sigma _{s+1}(\varepsilon )|^2=h(y)$, that is, $h(\sigma (\varepsilon )y)=\sigma _1(\varepsilon )\cdots \sigma _ s(\varepsilon )\, h(y)$, since $\varepsilon $ has norm one. Hence $h/(y_1\cdots y_ s)$ is invariant under $U$, and since $\Lambda _ U$ is a lattice it is bounded between two positive constants:

Let $v$ be a nonzero vector in the span of $\partial /\partial w_1,\ldots ,\partial /\partial w_ s$, and let $P$ be the complex $2$-plane spanned by $v$ and $\partial /\partial z$. Write $\overline\omega =\sum _{a,b}g_{ab}(y)\, \mathrm i\, \xi _ a\wedge \overline\xi _ b$ with $\xi =(\mathrm dw_1,\ldots ,\mathrm dw_ s,\mathrm dz)$. Since the forms $\xi _ a$ are closed, $\mathrm{dd}^{\mathrm{c}}\overline\omega =\sum _{a,b}\mathrm{dd}^{\mathrm{c}}g_{ab}\wedge \mathrm i\, \xi _ a\wedge \overline\xi _ b$, and each $\mathrm{dd}^{\mathrm{c}}g_{ab}$ has only $w$-components. A term restricted to $P$ is nonzero only if it contains both $\mathrm dz$ and $\mathrm d\overline z$, which happens only for $\xi _ a=\xi _ b=\mathrm dz$. Hence the restriction of $\mathrm{dd}^{\mathrm{c}}\overline\omega $ to $P$ equals the restriction of $\mathrm{dd}^{\mathrm{c}}h\wedge \mathrm i\, \mathrm dz\wedge \mathrm d\overline z$, and its sign is that of the value of $\mathrm{dd}^{\mathrm{c}}h$ on $v$. Thus $\mathrm{dd}^{\mathrm{c}}h\geq 0$ in the $w$-variables, and $h$ is convex in $y$.

Let $R>1$ and put $A=(R,R^{-1},1,\ldots ,1)$, $B=(R^{-1},R,1,\ldots ,1)$ and $M=(A+B)/2$. By (10.29.15), $h(A),h(B)\leq C$, whereas $h(M)\geq c(R+R^{-1})^2/4$, which contradicts convexity for large $R$. In the seminegative case, $h$ is concave, so $t\mapsto h(t,\ldots ,t)$ grows at most linearly as $t\to \infty $; this contradicts the lower bound $h(t,\ldots ,t)\geq ct^ s$ with $s\geq 2$. $\square$