6.9 Analytic Lelong level sets and generic valuesDB8FD0974B86

Proof. We work on a bounded polydisc $\Omega $ around $x$ in a chart on which $\varphi $ is bounded above. Subtracting a constant changes neither side of (6.9.2), so we may assume $\varphi \leq 0$ on $\Omega $, and we write $\nu =\nu (\varphi ,x)$. The space $\mathcal H_{m\varphi }(\Omega )$ is not zero: choose $a\in \Omega $ with $\varphi (a)>-\infty $ and apply corollary 6.5.8. By theorem 6.8.9(3), $\mathcal I(m\varphi )_ x$ is generated by germs of elements of $\mathcal H_{m\varphi }(\Omega )$; since $\Omega $ is connected, a nonzero element of $\mathcal H_{m\varphi }(\Omega )$ has a nonzero germ at $x$, by the identity theorem. Hence the order $k$ of $\mathcal I(m\varphi )$ at $x$ is the smallest vanishing order at $x$ of the nonzero elements of $\mathcal H_{m\varphi }(\Omega )$.

For the lower bound, let $F\in \mathcal H_{m\varphi }(\Omega )$ vanish to order exactly $k$ at $x$, and let $P_ k\neq 0$ be its homogeneous term of degree $k$. By (6.7.3), computed in the coordinates of the chart, and since $\varphi \leq 0$, we have $\varphi (z)\leq \nu \log (|z-x|^2/r_0^2)$ for $0<|z-x|<r_0$ and a suitable $r_0>0$; hence $\mathrm e^{-m\varphi (z)}\geq c|z-x|^{-2m\nu }$ there, with $c=r_0^{2m\nu }$. By (6.8.8) and polar coordinates (3.1.2), for small $r>0$

with $c'>0$. The left side is finite, so $2k-2m\nu +2n>0$, that is, $k>m\nu -n$.

For the upper bound, we use that every element of $\mathcal H_{m\varphi }(\Omega )$ vanishes to order at least $k$ at $x$. Choose $R>0$ such that the polydisc $\Delta $ of polyradius $(2R,\ldots ,2R)$ centered at $x$ is relatively compact in $\Omega $. Let $\delta >0$ be half the distance from $\overline\Delta $ to the complement of $\Omega $. Put $A=|B(0,\delta )|^{-1/2}$. For $G\in \mathcal H_{m\varphi }(\Omega )$, (6.8.7) gives $\sup _{\overline\Delta }|G|\leq A\| G\| _{L^2(\Omega )}\leq A\| G\| _{m\varphi }$, and the coefficient bound in (5.2.1) gives $|c_\alpha |\leq (2R)^{-|\alpha |}\sup _{\overline\Delta }|G|$ for the Taylor coefficients $c_\alpha $ of $G$ at $x$. Since $c_\alpha =0$ for $|\alpha |<k$ and $|(a-x)^\alpha |\leq |a-x|^{|\alpha |}$, for $|a-x|<R$ we obtain

For the second inequality, note that there are $\binom {d+n-1}{n-1}$ multi-indices of length $d$; with $\rho =|a-x|/(2R)\leq 1/2$ we get $\sum _{d\geq k}\binom {d+n-1}{n-1}\rho ^ d\leq \rho ^ k\sum _{e\geq 0}\binom {e+k+n-1}{n-1}2^{-e}$, a convergent series. Now let $|a-x|<R$ with $\varphi (a)>-\infty $, and let $G$ be given by corollary 6.5.8, so that $G(a)=1$ and $\| G\| _{m\varphi }^2\leq C_\Omega \mathrm e^{-m\varphi (a)}$, where $C_\Omega $ depends only on $n$ and the diameter of $\Omega $. Then $1\leq C_ k^2C_\Omega |a-x|^{2k}\mathrm e^{-m\varphi (a)}$, that is,

here $C'=m^{-1}\log (C_ k^2C_\Omega )$, and the inequality holds trivially where $\varphi (a)=-\infty $. Since it holds for all $a$ with $|a-x|<R$, the characterization in proposition 6.7.1 gives $k/m\leq \nu $. $\square$

Proof. If $u=v+h=v'+h'$ are two such decompositions near $x$, then $v$ and $v'$ take the value $-\infty $ at the same points, and $v\leq v'+C$ and $v'\leq v+C$ near $x$, where $C$ bounds $|h|+|h'|$; so $\nu (v,x)=\nu (v',x)$ by proposition 6.7.1. If $X$ has dimension zero at $x$, then $v(x)$ is finite and there is nothing to prove. Otherwise let $U$ be a connected neighborhood of $x$ on which $v$ is plurisubharmonic and $|h|\leq M$ for a constant $M$. Applying lemma 6.9.1 to the weight $v$ on $U$ and $m=t$, we obtain $\operatorname {ord}_ x\mathcal I(tv)\leq t\nu (v,x)=0$. Hence $\mathcal I(tv)_ x$ contains a germ that does not vanish at $x$; this germ is a unit, so $\mathcal I(tv)_ x=\mathcal O_{X,x}$, that is, $\mathrm e^{-tv}$ is integrable on a neighborhood of $x$. Since $\mathrm e^{-tu}\leq \mathrm e^{tM}\mathrm e^{-tv}$ on $U$, the same holds for $\mathrm e^{-tu}$. For a weight $\varphi $, the integrability of $\mathrm e^{-t\varphi }$ near $x$ means that $1\in \mathcal I(t\varphi )_ x$. $\square$

Proof. Lelong numbers vanish on components of dimension zero, so we may assume that $X$ is connected of dimension $n\geq 1$. We describe $E_ c(\varphi )$ as an intersection of analytic sets defined by the orders of the ideals $\mathcal I(m\varphi )$ and then show that such an intersection is analytic. Let $\mathfrak a$ be a coherent ideal and $k$ an integer. The set of points $y$ where $\operatorname {ord}_ y\mathfrak a\geq k$ (with $\operatorname {ord}_ y\mathfrak a=\infty $ if $\mathfrak a_ y=0$) is analytic. Indeed, let $g_1,\ldots ,g_ r$ generate $\mathfrak a$ on a coordinate neighborhood $W$. A germ has order at least $k$ at $y$ if and only if its derivatives of order less than $k$ vanish at $y$, since these derivatives are multiples of its Taylor coefficients, by (5.2.1). Hence for $y\in W$ we have $\operatorname {ord}_ y\mathfrak a\geq k$ if and only if $\partial ^\alpha g_ i(y)=0$ for all $i$ and all $|\alpha |<k$: the condition is necessary because the $g_ i$ belong to $\mathfrak a_ y$, and sufficient because, by the Leibniz rule, every element $\sum _ ih_ ig_ i$ then vanishes to order at least $k$ at $y$. This is a finite system of holomorphic equations, and for $k\leq 0$ the set is all of $W$. By (6.9.2),

Indeed, if $\nu (\varphi ,x)\geq c$, the lower bound in (6.9.2) gives the integer inequality $\operatorname {ord}_ x\mathcal I(m\varphi )>mc-n$, that is, $\operatorname {ord}_ x\mathcal I(m\varphi )\geq \lfloor mc-n\rfloor +1$, for every $m$. If $\nu (\varphi ,x)<c$, the upper bound gives $\operatorname {ord}_ x\mathcal I(m\varphi )\leq m\nu (\varphi ,x)\leq mc-n<\lfloor mc-n\rfloor +1$ as soon as $m(c-\nu (\varphi ,x))\geq n$.

Thus $E_ c(\varphi )$ is an intersection of analytic subsets of $X$. An arbitrary intersection $A=\bigcap _\lambda A_\lambda $ of analytic subsets of $X$ is analytic. Since $A$ is closed, it suffices to show that each $p\in A$ has a neighborhood $W$ such that $A\cap W$ is the intersection of $W$ with finitely many of the $A_\lambda $. Let $I_\lambda \subset \mathcal O_{X,p}$ be the ideal generated by the germs at $p$ of local defining functions of $A_\lambda $. By proposition 9.2.4 the sum $\sum _\lambda I_\lambda $ is finitely generated, so it equals $I_{\lambda _1}+\cdots +I_{\lambda _ q}$ for finitely many indices; put $B=A_{\lambda _1}\cap \cdots \cap A_{\lambda _ q}$. Every local defining function of each $A_\lambda $ is then a combination of defining functions of $A_{\lambda _1},\ldots ,A_{\lambda _ q}$ near $p$, so $A_\lambda $ contains a neighborhood of $p$ in $B$. This neighborhood may depend on $\lambda $, and we now remove this dependence.

By corollary 9.3.16, the irreducible components of the analytic set $B$ form a locally finite family; so some neighborhood $N$ of $p$ meets only finitely many of them. Let $B_1,\ldots ,B_ l$ be those that contain $p$. The union of the other components meeting $N$ is closed and does not contain $p$, so there is an open neighborhood $W'\subset N$ of $p$ with $B\cap W'\subset B_1\cup \cdots \cup B_ l$. Fix $\lambda $ and $i$. The regular locus $(B_ i)_{\mathrm{reg}}$ is a connected manifold, dense in $B_ i$ (corollary 9.3.16), and its analytic subset $A_\lambda \cap (B_ i)_{\mathrm{reg}}$ has nonempty interior in it, because $A_\lambda $ contains a neighborhood of $p$ in $B_ i$ and $p$ lies in the closure of $(B_ i)_{\mathrm{reg}}$. By lemma 9.2.8, $A_\lambda \cap (B_ i)_{\mathrm{reg}}$ is therefore all of $(B_ i)_{\mathrm{reg}}$, and $B_ i\subset A_\lambda $ because $A_\lambda $ is closed. Therefore $B\cap W'\subset A_\lambda $ for every $\lambda $, and $A\cap W'=B\cap W'$. Applied to (6.9.6), this shows that $E_ c(\varphi )$ is analytic.

By proposition 6.7.1, $\nu (\varphi ,x)=0$ at every point where $\varphi (x)>-\infty $. Thus $E_ c(\varphi )$ is contained in the pole set of $\varphi $, which has measure zero because $\varphi $ is locally integrable (theorem 5.9.1), and therefore has no interior points; so $E_ c(\varphi )\neq X$.

Finally, let $T$ be a closed positive $(1,1)$-current on $X$. Every point of $X$ at which $X$ has positive dimension has a coordinate ball $U$ on which $T=\mathrm{dd}^{\mathrm{c}}u$ with $u$ plurisubharmonic, and $\nu (T,y)=\nu (u,y)$ for $y\in U$ (definition 6.7.4). By what we have proved, applied to the weight $u$ on $U$, the set $\{ y\in U:\nu (T,y)\geq c\} $ is an analytic subset of $U$ with empty interior. Since the components of $X$ of dimension zero do not meet $\{ y:\nu (T,y)\geq c\} $, and analyticity is a local property, this set is an analytic subset of $X$ whose intersection with every component of positive dimension is proper. $\square$

Proof. Put $\lambda =\nu (\varphi ,Z)$. For every integer $j\geq 1$, theorem 6.9.5 shows that $Z\cap E_{\lambda +1/j}(\varphi )$ is an analytic subset of $Z$, and it is a proper subset by the definition of $\lambda $. At every point of $Z$ outside the union of these sets, $\nu (\varphi ,x)<\lambda +1/j$ for all $j$, hence $\nu (\varphi ,x)=\lambda $.

We now show that the complement in $Z$ of an arbitrary countable union of proper analytic subsets of $Z$ is dense in $Z$; applied to the sets $Z\cap E_{\lambda +1/j}(\varphi )$, this shows that $\nu (\varphi ,Z)$ is attained on a dense subset of $Z$. The density is clear if $Z$ is a point, for then the proper analytic subsets of $Z$ are empty. Let $\dim Z>0$. By corollary 9.3.16, applied to the family consisting of $Z$ alone, $Z$ is an irreducible component of itself; so its regular locus $Z_{\mathrm{reg}}$ is a connected manifold, dense in $Z$.

Let $B$ be a proper analytic subset of $Z$. Then $B\cap Z_{\mathrm{reg}}$ is an analytic subset of $Z_{\mathrm{reg}}$, and it is proper, since otherwise the closed set $B$ would contain the closure $Z$ of $Z_{\mathrm{reg}}$. By lemma 9.2.8, every point of $B\cap Z_{\mathrm{reg}}$ has a connected open neighborhood $U$ in $Z_{\mathrm{reg}}$ with a holomorphic function $g\not\equiv 0$ on $U$ such that $B\cap U\subset \{ g=0\} $; shrinking $U$ to a coordinate ball, on which $g\not\equiv 0$ by the identity theorem, we see from lemma 5.5.1 that $B\cap U$ is a null set in the chart. Covering a chart $N$ of $Z_{\mathrm{reg}}$ by countably many such balls and by open sets disjoint from $B$, we see that $B\cap N$ is a null set. Consequently, a countable union of proper analytic subsets of $Z$ meets every chart of $Z_{\mathrm{reg}}$ in a null set, which has no interior points. Its complement is therefore dense in $Z_{\mathrm{reg}}$, and hence in $Z$. For a closed positive $(1,1)$-current the same argument applies, since theorem 6.9.5 holds for its Lelong numbers.

For the second assertion, note first that $\varphi \circ \pi $ is a weight on $Y$: by proposition 5.9.3 it is plurisubharmonic or identically $-\infty $ on each component of $Y$, and the second alternative is excluded. Indeed, since $\pi $ is bimeromorphic (section 9.4), there is a nowhere dense analytic subset $A\subset X$ such that $\pi ^{-1}(A)$ is nowhere dense in $Y$ and $\pi $ maps $Y\setminus \pi ^{-1}(A)$ biholomorphically onto $X\setminus A$. If $\varphi \circ \pi $ were identically $-\infty $ on a component $Y_0$ of $Y$, then $\varphi $ would be $-\infty $ on the nonempty open set $\pi (Y_0\setminus \pi ^{-1}(A))$, which is impossible because $\varphi $ is locally integrable.

Now let $E\subset Y$ and $E'\subset Y'$ be prime divisors on smooth models $\pi \colon Y\to X$ and $\pi '\colon Y'\to X$ that represent the same prime divisor over $X$, and let $\rho \colon Y''\to Y$ and $\rho '\colon Y''\to Y'$ be maps from a common smooth model on which the strict transforms of $E$ and $E'$ coincide, say with $E''$. By proposition 9.8.3, $\rho $ is biholomorphic over $Y\setminus S$ for an analytic subset $S\subset Y$ of codimension at least two, so it maps the nonempty open subset $E''\cap \rho ^{-1}(Y\setminus S)$ of $E''$ onto the nonempty open subset $E\setminus S$ of $E$, and $\nu (\varphi \circ \pi \circ \rho ,q)=\nu (\varphi \circ \pi ,\rho (q))$ for $q$ in this set, because Lelong numbers are invariant under biholomorphisms (proposition 6.7.1). By the first part, the generic value along $E$ is attained on a dense subset of $E$, so it equals the infimum of the Lelong numbers over any nonempty open subset of $E$; the same holds for $E''$. Hence $\nu (\varphi \circ \pi ,E)=\nu (\varphi \circ \pi \circ \rho ,E'')$. The same argument for $\rho '$ gives $\nu (\varphi \circ \pi ',E')=\nu (\varphi \circ \pi \circ \rho ,E'')$, since $\pi \circ \rho =\pi '\circ \rho '$. $\square$

Proof. Let $p\in Z$ be the given point and $\lambda =\nu (\varphi ,Z)$. Choose $r_0>0$, a constant $C$ and a neighborhood $W$ of $p$ in $Z$ such that for every $y\in W$ the closed ball $\overline{B(y,r_0)}$ lies in the coordinate domain and $\varphi \leq C$ on it. By (6.7.3) at $y$, we have $\varphi (z)\leq \nu (\varphi ,y)\log (|z-y|^2/r_0^2)+C$ for $0<|z-y|<r_0$. Take $z=(z',z'')$ with $y=(0,z'')\in W$ and $0<|z'|<r_0$. Then $|z-y|=|z'|$, the logarithm is negative and $\nu (\varphi ,y)\geq \lambda $, so $\varphi (z)\leq \lambda \log |z'|^2-\lambda \log r_0^2+C$. This is the asserted bound, with a constant independent of $z''$. Conversely, suppose that $\varphi (z)\leq c\log |z'|^2+C'$ for $z'\neq 0$ near a point $q$ of $Z$, where we may assume $c\geq 0$. For $y\in Z$ near $q$ we have $|z'|\leq |z-y|$, so $\varphi (z)\leq c\log |z-y|^2+C'$ for $z\notin Z$ near $y$. Since $Z=\{ z'=0\} $ is a Lebesgue null set in the chart, proposition 6.7.1 gives $\nu (\varphi ,y)\geq c$. By proposition 6.9.7 there are points $y$ arbitrarily close to $q$ with $\nu (\varphi ,y)=\lambda $, so $c\leq \lambda $.

Now let $\pi $ be the blowup of $x$, and let $n=\dim _ xX$. If $n=1$, then $\pi $ is the identity and $E=\{ x\} $, so there is nothing to prove. Let $n\geq 2$ and choose coordinates centered at $x$. The blowup is covered by $n$ charts; in the $k$th chart, $z_ k=w_ k$ and $z_ j=w_ kw_ j$ for $j\neq k$, so that

By (6.7.3), $\varphi \leq \nu (\varphi ,x)\log |z|^2+C$ near $x$. Pulling back, we get $\varphi \circ \pi \leq \nu (\varphi ,x)\log |w_ k|^2+C'$ on compact subsets of the $k$th chart near $E$; since $|w_ k|$ is at most the distance to any point of $E$ in that chart, $\nu (\varphi \circ \pi ,q)\geq \nu (\varphi ,x)$ for every $q\in E$, and $\nu (\varphi ,E)\geq \nu (\varphi ,x)$. Conversely, $E\cong \mathbb P^{n-1}$ is a compact connected submanifold of $Y$, so the first part of the proposition, applied to the weight $\varphi \circ \pi $ and $E$, gives $\varphi \circ \pi \leq \nu (\varphi ,E)\log |w_ k|^2+C$ near each point of $E$, in the coordinates of a chart containing it. Cover $E$ by finitely many such neighborhoods, on each of which $\log |w_ k|^2-\log |z|^2$ is bounded. Since $\pi $ is proper, the inverse image of a small neighborhood of $x$ lies in their union, so $\varphi (z)\leq \nu (\varphi ,E)\log |z|^2+C''$ near $x$, for $z\neq x$. Hence $\nu (\varphi ,E)\leq \nu (\varphi ,x)$. $\square$