10.29.2 Potentials of the transverse form992E9A24F805

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. $\square$