9.10 Pluripotential theory on unibranch spaces5131A8B4E30A

Proof. Let $x\in \mathcal{X}(\mathbb {C})$. Then we claim that $\mathcal{X}$ is unibranch at $x$ if and only if $\mathcal{X}^{\mathrm{An}}$ is unibranch at $x$. By GAGA, there is a natural morphism of ringed spaces $(\mathcal{X}^{\mathrm{An}},\mathcal{O}_{\mathcal{X}^{\mathrm{An}}})\rightarrow (\mathcal{X},\mathcal{O}_{\mathcal{X}})$. Then since $\mathcal{O}_{\mathcal{X},x}$ is excellent, $\mathcal{X}$ is unibranch at $x$ if and only if $\widehat{\mathcal{O}_{\mathcal{X},x}}=\widehat{\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}}$ is integral ([Théorème 18.9.1, EGAIV4]). The latter implies that $\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}$ is unibranch by [Chapitre 5, Exercise 2.8, AC]. Running the same argument the other way round, we conclude that $\mathcal{O}_{\mathcal{X},x}$ is unibranch if and only if $\mathcal{O}_{\mathcal{X}^{\mathrm{An}},x}$ is. $\square$

Proof. Let $Z\subseteq X$ be the smallest nowhere dense analytic subset such that $\pi $ is an isomorphism over $X\setminus Z$. Assume that the fiber $F=\pi ^{-1}(x)$ over some $x\in X$ is disconnected. Since $\pi $ is proper, $F$ is compact. Choose a decomposition $F=F_1\sqcup F_2$ into two non-empty disjoint compact open subsets of $F$, and choose disjoint open neighborhoods $W_1,W_2$ of $F_1,F_2$ in $Y$. Since proper maps are closed, after shrinking an open neighborhood $U$ of $x$, we may assume that

Thus $\pi ^{-1}(U)$ is disconnected.

Since $X$ is unibranch, we may shrink $U$ further so that $U$ is irreducible. Then $U\setminus Z$ is connected (cf. [Chapter 9, §1.2, CAS]). Since $\pi $ is bimeromorphic, every irreducible component of $Y$ maps onto an irreducible component of $X$. Hence $\pi ^{-1}(Z)$ contains no irreducible component of $Y$ and is nowhere dense. On the other hand, write

as a union of two non-empty disjoint open subsets. Each $V_ i$ meets $\pi ^{-1}(U\setminus Z)$: otherwise $V_ i$ would be a non-empty open subset of $Y$ contained in the nowhere dense analytic subset $\pi ^{-1}(Z)$, which is impossible. Hence

is disconnected. This contradicts the fact that $\pi \colon \pi ^{-1}(U\setminus Z)\rightarrow U\setminus Z$ is an isomorphism and $U\setminus Z$ is connected. $\square$

Proof. Choose a smooth model $X'\rightarrow X$ on which $E$ is realized. By theorem 9.10.8, this proper bimeromorphic morphism is dominated by a modification $Y\rightarrow X$. After further resolving the pair if necessary, we may assume that $Y$ is smooth and that the strict transform of $E$ on $Y$ is smooth. This strict transform realizes the same prime divisor over $X$. $\square$

Proof. We first prove the uniqueness, which is a local problem on $X$. Let $\psi $ denote the function defined by the right-hand side of (9.10.15). If $\eta \in \mathrm{PSH}(X)$ is an extension of $\varphi $, then upper semicontinuity gives $\eta \geq \psi $. Conversely, take $z\in Z$ and a holomorphic map $f\colon \Delta \rightarrow X$ such that $f(0)=z$ and $f(\Delta )\not\subset Z$. From the subharmonicity of $f^*\eta $ and (9.10.15), we find that

where the second equality follows from theorem 5.14.10. Thus $\eta =\psi $, and uniqueness follows.

Having established the uniqueness of the extension, the existence also becomes a local problem. So we can freely shrink $X$ around any point in $Z$ in the proof.

2 Let $\pi \colon Y\rightarrow X$ be a resolution of singularities. By the smooth version of the extension theorem theorem 5.14.10, we know that $\pi ^*\varphi $ admits a unique extension to a psh function on $Y$, which we denote by $\eta $. Note that all fibers of $\pi $ are connected since $X$ is unibranch, thanks to theorem 9.10.5. Furthermore, the fibers are compact, since $\pi $ is proper. The restriction of $\eta $ to each fiber is psh or identically $-\infty $, hence is constant by the maximum principle applied to a resolution. It therefore descends to an upper semicontinuous function $\psi $ on $X$, and $\psi |_{X\setminus Z}=\varphi $.

We verify that $\psi $ is plurisubharmonic using theorem 9.10.13. Let $f\colon \Delta \rightarrow X$ be a holomorphic map. We assume that $f^*\psi \not\equiv -\infty $. It suffices to show that $f^*\psi $ is subharmonic at $0\in \Delta $. After replacing $\Delta $ by a finite ramified cover if necessary, the germ of $f$ lifts to $Y$. Indeed, consider the proper projection

Choose an irreducible analytic curve in $\Delta \times _ X Y$ passing through a point over $0$ and dominating $\Delta $; after shrinking $\Delta $ and normalizing this curve, we obtain a finite ramified cover of the disk. After replacing $\Delta $ by a smaller disk, there are an integer $k>0$ and a holomorphic map $f'\colon \Delta \to Y$ such that

Therefore, $\psi (f(t^ k))=\eta (f'(t))$ near $0$. Since subharmonicity descends under the finite map $t\mapsto t^ k$, it follows that $f^*\psi $ is subharmonic near $0$.

1 By the local description of complex spaces [Section 3.4, CAS], we may assume that there is a domain $\Omega \subseteq \mathbb {C}^ n$, a finite $s$-sheet branched covering $\Phi \colon X\rightarrow \Omega $ with branched locus contained in a proper analytic subset $V\subseteq \Omega $. We may assume that $X$ is connected, $n\geq 1$.

It suffices to show that $\varphi $ is locally bounded from above near $Z$, since then 2 applies. Suppose that this fails. Then we can find $z\in Z$ and $x_ i\in X\setminus Z$ ($i\geq 1$) converging to $z$ such that

Using (9.10.15) around each $x_ i$, we may further assume that $x_ i\in X\setminus (\Phi ^{-1}(\Phi (Z)\cup V))$.

Let $L$ be a complex line passing through $\Phi (z)$ intersecting $(\Phi (Z)\cup V)\cap \overline{B}$ only at $\Phi (z)$, where $B\Subset B'\Subset \Omega $ are two small open balls centered at $\Phi (z)$. Since $\Phi $ is finite, $\Phi (Z)$ still has codimension at least $2$ in $\Omega $. Hence, for each $i$, the set of directions of lines through $\Phi (x_ i)$ meeting $B'\cap \Phi (Z)$ is contained in a proper analytic subset of the projective space of directions. We can therefore choose lines $L_ i$ through $\Phi (x_ i)$ converging to $L$ such that $L_ i\cap (B'\cap \Phi (Z))=\varnothing $. Moreover, since $\Phi (x_ i)\notin V$, no such $L_ i$ is contained in $V$, and hence $L_ i\cap (B'\cap V)$ is discrete. Then $\Phi $ restricts to a branched covering over $B'\cap L_ i$ for all $i\geq 1$. Adding a constant to $\varphi $, we may assume that $\varphi |_{\Phi ^{-1}(L\cap \partial B)}<0$. We can then find an open neighborhood $U$ of $\Phi ^{-1}(L\cap \partial B)$ so that $\varphi |_ U<0$. For large $i$, we have $\Phi ^{-1}(L_ i\cap \partial B)\subseteq U$ and $\Phi (x_ i)\in B$. Put

Thus $C_ i$ is a one-dimensional complex space contained in $X\setminus Z$. Since $\Phi |_{C_ i}$ is finite over $L_ i\cap B'$, the inverse image $C_ i\cap \Phi ^{-1}(\overline B)$ is compact, and its boundary is contained in $\Phi ^{-1}(L_ i\cap \partial B)$.

Passing to the normalization of each irreducible component of $C_ i$, the pull-back of $\varphi $ is subharmonic. The one-dimensional maximum principle therefore gives

which is a contradiction. $\square$

Proof. Injectivity follows from the surjectivity of $\pi $. For surjectivity, let $u\in \mathrm{PSH}(Y,\pi ^*\theta )$. The assertion is local on $X$, so after subtracting a local potential of $\theta $ we may assume that $\theta =0$.

By theorem 9.10.5, the fibers of $\pi $ are connected. The restriction of $u$ to any compact fiber is psh, hence is constant by the maximum principle. Thus $u$ descends to a function $v$ on $X$. Since $\pi $ is proper, $v$ is upper semicontinuous.

It remains to check that $v$ is psh. By theorem 9.10.13, it suffices to test on holomorphic disks. Let $f\colon \Delta \rightarrow X$ be a holomorphic disk such that $f^*v\not\equiv -\infty $. As in the proof of theorem 9.10.14, after a finite ramified cover $\rho \colon \Delta \rightarrow \Delta $, the germ of $f\circ \rho $ lifts to $Y$. The pull-back $(f^*v)\circ \rho $ is therefore subharmonic, so $f^*v$ is subharmonic. Hence $v\in \mathrm{PSH}(X)$, and the general $\theta $-psh case follows by adding back the local potentials. $\square$