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Theorem 10.29.4244BCBFB750A

Theorem 10.29.4.244BCBFB750A Let $\mathcal F_ U$ be the set of convex functions $f:(0,\infty )^ s\to \mathbb R$ such that

8DB42C059FB9 10.29.5
\begin{equation} f(\sigma (\varepsilon )y)=f(y)-\sum _{j=1}^ s\log \sigma _ j(\varepsilon ) \qquad (\varepsilon \in U). \end{equation}

For $f\in \mathcal F_ U$, the function $u_ f(w,z)=f(y)+\sum _ j\log y_ j$ is $\Gamma $-invariant, and $f\mapsto u_ f$ is a bijection from $\mathcal F_ U$ onto $\mathrm{PSH}(X,\theta )$. In particular, every $\theta $-plurisubharmonic function on $X$ is continuous and depends only on $y=\operatorname {Im}w$.

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). $\square$

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