9.9 Plurisubharmonic functions on reduced spaces75F10F3C1577

Proof. We prove (1) and the first assertion of (2) together. Let $u$ be locally plurisubharmonic on $X$, let $F\colon Y\to X$ be holomorphic, and let $y\in Y$. Choose a local model of a neighborhood of $F(y)$, with ambient coordinates $w$ and a ball $V$ around $F(y)$ on which $u$ is the restriction of a function $v$ that is plurisubharmonic or identically $-\infty $, and an arbitrary local model of a neighborhood of $y$, with ambient coordinates $z$. By the definition of holomorphic maps and of $\mathcal O_ Y$, the components of $F$ in the $w$-model extend holomorphically to a neighborhood of $y$ in the $z$-ambient space, and their joint extension $G$, being continuous with $G(y)=F(y)$, maps a smaller ball $V'$ around $y$ into $V$. By proposition 5.9.3, $v\circ G$ is plurisubharmonic or identically $-\infty $ on $V'$, and its restriction to $Y\cap V'$ is $u\circ F$. Hence $u\circ F$ is locally plurisubharmonic, with respect to every local model of $Y$. For $Y=X$ and $F$ the identity, with two local models at the same point, this is (1).

If $Y$ is a one-dimensional complex manifold, its holomorphic charts are local models. By what we have just proved, $u\circ F$ is therefore, in local coordinates near each point of $Y$, plurisubharmonic or identically $-\infty $, that is, subharmonic in the sense of definition 3.4.2. Subharmonicity is a local property, so $u\circ F$ is subharmonic in every local coordinate, and on the whole disc if $Y$ is a disc.

(3) Holomorphic charts are local models. So by (1), every point of $X$ has a coordinate ball on which a locally plurisubharmonic function $u$ is plurisubharmonic or identically $-\infty $, and by lemma 5.9.4, $u$ is plurisubharmonic or identically $-\infty $ on each connected component of $X$. Conversely, a function with this property is locally plurisubharmonic, with the charts as local models, since its expressions in charts with connected domains are plurisubharmonic or identically $-\infty $. The irreducible components of the manifold $X$ are its connected components, by corollary 9.3.16. Hence $\mathrm{PSH}(X)$, in the sense of definition 9.9.1, consists of the functions that are plurisubharmonic on each connected component of $X$, which are the plurisubharmonic functions in the sense of section 5.9, since plurisubharmonicity is a local property there. $\square$

Proof. Let $u$ be locally plurisubharmonic on the compact connected space $X$. It is upper semicontinuous, so it attains its maximum $M\in [-\infty ,\infty )$. If $M=-\infty $, then $u\equiv -\infty $. Suppose that $M$ is finite, and let $x$ be a point with $u(x)=M$. We show that $u=M$ near $x$. If $x$ is an isolated point, there is nothing to prove. Otherwise, let $X_1$ be a branch of $(X,x)$ of positive dimension, and take a parametrization of $(X_1,x)$, with its connected covering locus $X_{1,\delta }$, dense in the representative $X_1$. By proposition 9.5.2, applied to the subgerm consisting of the singular locus of $X$ and the points of $X_1$ outside $X_{1,\delta }$, there is a nonconstant holomorphic curve $\gamma $ with $\gamma (0)=x$ and $\gamma (t)\in X_{1,\delta }\cap X_{\mathrm{reg}}$ for $t\neq 0$. The function $u\circ \gamma $ is subharmonic by lemma 9.9.2(2), and it attains its finite maximum $M$ at $0$; by the maximum principle (proposition 3.4.7) it equals $M$ near $0$. Thus $u$ attains the value $M$ at points of $X_{1,\delta }$.

By lemma 9.9.2(2) and (3), applied to the inclusion of the connected manifold $X_{1,\delta }$, the restriction of $u$ to $X_{1,\delta }$ is plurisubharmonic or identically $-\infty $, hence plurisubharmonic, since it takes the value $M$. On a coordinate ball around a point where $u=M$, it is subharmonic by theorem 5.9.1, and it attains its maximum $M$ at the center, so it equals $M$ on this ball by proposition 3.4.7. Thus the set where $u=M$ is open in $X_{1,\delta }$; it is closed by upper semicontinuity and nonempty, so $u=M$ on $X_{1,\delta }$. Since $X_{1,\delta }$ is dense in $X_1$, upper semicontinuity gives $u\geq M$, hence $u=M$, on all of $X_1$. Doing this for each branch, we find that $u=M$ on a neighborhood of $x$.

Hence the set $\{ u=M\} $ is open. It is also closed, because $\{ u<M\} $ is open for an upper semicontinuous function bounded above by $M$. Since $X$ is connected, $u\equiv M$. $\square$

Proof. Let $u$ be locally plurisubharmonic on $X\setminus A$, and let $x\in A$. We bound $u$ over a circle that avoids the image of $A$ under a finite projection, and carry the bound inward along one-dimensional slices by the maximum principle. It suffices to bound $u$ from above near $x$ on each branch of $(X,x)$, and the restriction of $u$ to a branch is locally plurisubharmonic by lemma 9.9.2(2); so we may work on the representative $X_1$ of a branch on a parametrization polydisc. Let $X_2,\ldots ,X_ p$ be representatives of the other branches.

The hypothesis is inherited by $X_1$ near $x$. Indeed, let $y\in X_1$ be close to $x$, and let $Y$ be a branch of $(X_1,y)$. It has dimension $\dim X_1$, by proposition 9.3.9, and it lies in a branch $Y'$ of $(X,y)$, which lies in some $(X_ j,y)$. If $j=1$, then $Y'=Y$, because $Y$ is a branch of $(X_1,y)$. If $j\neq 1$, then $Y\subset (X_1\cap X_ j,y)$, which is impossible for $y$ close to $x$, because $\dim (X_1\cap X_ j,y)\leq \dim (X_1\cap X_ j,x)<\dim X_1$, by upper semicontinuity and proposition 9.3.9. Hence every branch of $(X_1,y)$ is a branch of $(X,y)$. The dimension $d$ of $X_1$ is at least two, since $A\cap X_1$ contains $x$ and has codimension at least two in $X_1$. Now write $X$ for $X_1$, and let $\pi \colon X\to \Delta '\subset \mathbb C^ d$ be the projection of the local parametrization (theorem 9.3.4); it is proper with finite fibers, and $\pi ^{-1}(0)=\{ x\} $.

Step 1: a cylinder avoiding $\pi (A)$. The image $\pi (A)$ is analytic by theorem 9.7.2, and its dimension is at most $d-2$ by lemma 9.3.15(3). In particular there is a holomorphic function $g\not\equiv 0$ near $0$ vanishing on $\pi (A)$. Choose a complex line through $0$ on which the first nonzero homogeneous term of $g$ does not vanish, and write the base coordinates as $(\zeta ,z')$, with $\zeta $ in the direction of this line. Then $\pi (A)$ meets the line only at $0$ near $0$, and by compactness there are $r>0$ and $\varepsilon >0$ such that the cylinder

is compactly contained in $\Delta '$ and disjoint from $\pi (A)$. Its inverse image is compact and disjoint from $A$, so $u$ has an upper bound $M<\infty $ there.

Step 2: slicing by curves. The projection to the $z'$-ball of $\pi (A)\cap \{ |\zeta |<r,\ |z'|<\varepsilon \} $ is proper, because the closure of this set does not meet $\{ |\zeta |=r\} $. By theorem 9.7.2, its image $S$ is analytic in the ball $\{ |z'|<\varepsilon \} $, and it has dimension at most $d-2$ by lemma 9.3.15(3), so it is nowhere dense in this $(d-1)$-dimensional ball. Fix $z'$ with $|z'|<\varepsilon $ and $z'\notin S$. Then the compact set $\pi ^{-1}(\{ |\zeta |\leq r\} \times \{ z'\} )$ is disjoint from $A$, and so is $C=\pi ^{-1}(\{ |\zeta |<r'\} \times \{ z'\} )$ for some $r'>r$.

The set $C$ is a pure one-dimensional complex space, or empty. Indeed, it is cut out of $X$ by the $d-1$ coordinate equations $z'_ k=\mathrm{const}$. At each stage, every branch of the current set at every point is mapped finitely, hence openly by lemma 9.5.5, to the space of the remaining base coordinates; so the next coordinate equation is not identically satisfied on it, and by proposition 9.5.6 the dimension drops by exactly one. Let $\nu \colon \widetilde C\to C$ be the normalization (theorem 9.4.6). By proposition 9.5.1, $\widetilde C$ is a one-dimensional complex manifold, and $\zeta \circ \pi \circ \nu \colon \widetilde C\to \{ |\zeta |<r'\} $ is proper with finite fibers, hence open on each connected component by lemma 9.5.5.

By lemma 9.9.2(2), the function $v=u\circ \nu $ is subharmonic on the one-dimensional manifold $\widetilde C$. Let $Q$ be a connected component of the open set $\{ |\zeta \circ \pi \circ \nu |<r\} \subset \widetilde C$; its closure is compact, by properness. A point of $\overline Q$ over the open disc $\{ |\zeta |<r\} $ has a connected neighborhood in $\{ |\zeta \circ \pi \circ \nu |<r\} $ meeting $Q$, and therefore lies in $Q$; so the boundary of $Q$ lies over the circle $\{ |\zeta |=r\} $, where $v\leq M$ by Step 1. This boundary is nonempty: otherwise $Q$ would be a connected component of $\widetilde C$, and its image would be open and, by properness, closed in the disc $\{ |\zeta |<r'\} $, hence equal to this disc, although it lies in $\{ |\zeta |<r\} $.

The upper semicontinuous function $v$ attains its maximum on the compact set $\overline Q$. If the maximum is finite and attained at a point of $Q$, the maximum principle (proposition 3.4.7), applied on coordinate discs, shows that the set where $v$ attains it is open in $Q$; this set is also closed, so $v$ is constant on $Q$, and by upper semicontinuity at a boundary point the constant is at most $M$. Otherwise the maximum is $-\infty $ or is attained on the boundary of $Q$, so again it is at most $M$. Hence $v\leq M$ on $Q$, and since $\nu $ is surjective, $u\leq M$ on $\pi ^{-1}(\{ |\zeta |<r\} \times \{ z'\} )$.

Step 3: the remaining slices. We have shown that $u\leq M$ on the part of $X\cap \pi ^{-1}(\{ |\zeta |<r,\ |z'|<\varepsilon \} )$ lying over $z'\notin S$. Let $y\notin A$ be a point of this set lying over some $z'\in S$, and let $Y_1$ be a branch of $(X,y)$. The set $T=\pi ^{-1}(\{ z'\in S\} )$ does not contain the germ $Y_1$: by purity, $Y_1$ has dimension $d$, so $\pi |_{Y_1}$ is open by lemma 9.5.5, while the image of $T$ is nowhere dense. Since $A$ is closed and $y\notin A$, proposition 9.5.2 gives a holomorphic curve $\gamma $ in $Y_1$ with $\gamma (0)=y$ and $\gamma (t)\notin A\cup T$ for $t\neq 0$. Then $u\circ \gamma $ is subharmonic near $0$, by lemma 9.9.2(2), and at most $M$ on the punctured disc, so $u(y)\leq M$ by the submean inequality on small circles. Thus $u\leq M$ on $X\setminus A$ near $x$. $\square$