Theorem 10.4.8 (Siu's divisorial decomposition)D905BB5BE91D

Proof. We first subtract one divisor, then show that only countably many divisors occur and that the series converges, and finally identify the residual current and prove uniqueness.

Step 1: one divisor. Let $D$ be a prime divisor and $c=\nu (T,D)$; we show that $T-c[D]\geq 0$. Near a regular point of $D$ choose coordinates with $D=\{ z_1=0\} $ on a coordinate polydisc $W$ with $|z_1|<1$, and a plurisubharmonic potential $u$ of $T$ on $W$. Since the Lelong numbers of $u$ at the points of $D\cap W$ are at least $c$, proposition 6.9.8 gives $u\leq c\log |z_1|^2+C$ near the point. Off $D$ the function $u-c\log |z_1|^2$ is plurisubharmonic, and it is bounded above near $D$; by theorem 5.11.1 it extends to a plurisubharmonic function $v$. Since $v=u-c\log |z_1|^2$ almost everywhere and $[D]=\mathrm{dd}^{\mathrm{c}}\log |z_1|^2$ near the point (theorem 5.5.6), we get $\mathrm{dd}^{\mathrm{c}}v=T-c[D]$ there. Hence $T-c[D]$ is positive on a neighborhood of $D_{\mathrm{reg}}$, and it equals $T$ on $X\setminus D$; so it is positive on the open set $X\setminus D_{\mathrm{sing}}$.

To extend the positivity across $D_{\mathrm{sing}}$, let $\phi $ be a strongly positive test form of bidegree $(n-1,n-1)$. Since $T$ and $[D]$ have measure coefficients (theorem 5.3.7), there is a real measure $m_\phi $ on $X$, a combination of the coefficient measures with the coefficients of $\phi $ as densities, such that $(T-c[D])(\chi \phi )=\int \chi \, \mathrm dm_\phi $ for test functions $\chi $. The restriction of $m_\phi $ to the open set $X\setminus D_{\mathrm{sing}}$ is a positive measure, because $(T-c[D])(\chi \phi )\geq 0$ for test functions $\chi \geq 0$ supported there; and its restriction to $D_{\mathrm{sing}}$ is that of the positive measure $\chi \mapsto T(\chi \phi )$, because $[D]$ gives no mass to $D_{\mathrm{sing}}$. Hence $(T-c[D])(\phi )=m_\phi (X)\geq 0$, that is, $T-c[D]\geq 0$.

Step 2: the series. By theorem 6.9.5, each set $E_ j=\{ x:\nu (T,x)\geq 1/j\} $ is an analytic subset of $X$ different from $X$. Being compact, $E_ j$ has finitely many irreducible components $Z$ by corollary 9.3.16. We first show:

An analytic subset of the connected manifold $X$ with an interior point is all of $X$, since the set of its interior points is open and, by the identity theorem applied to local defining functions, closed. Hence $E$ has no interior points, so its germs, and their branches, have dimension at most $n-1$, and $E_{\mathrm{sing}}$ has dimension at most $n-2$ at every point (proposition 9.3.9, corollary 9.3.13). At a point $y\in D_{\mathrm{reg}}$ the germ of $D$ has dimension $n-1$, so by monotonicity of dimension (see proposition 9.3.9 and the remarks following it) it is not contained in the germ of $E_{\mathrm{sing}}$; thus there are points $y'\in D_{\mathrm{reg}}\cap E_{\mathrm{reg}}$. Near such a point $E$ is a complex submanifold of dimension at most $n-1$ containing the $(n-1)$-dimensional submanifold $D$, so $D$ and $E$ coincide near $y'$.

Let $F$ be the connected component of $E_{\mathrm{reg}}$ containing $y'$. The analytic subset $D\cap F$ of the connected manifold $F$ has an interior point, so it is all of $F$, and $D$ contains the closure $Z$ of $F$, an irreducible component of $E$ (corollary 9.3.16). Since $D$ is compact and the components of $E$ form a locally finite family, $D$ meets only finitely many of them and is the union of its intersections with these. By induction on the number of sets, an irreducible set is not a finite union of analytic subsets different from itself; hence $D=D\cap Z'$ for a component $Z'$. Then $Z\subset D\subset Z'$, and since no component is contained in another, $D=Z$. This proves (10.4.10). In particular, two distinct prime divisors are not contained in one another, and every prime divisor $D$ with $\nu (T,D)\geq 1/j$, being contained in $E_ j$, is one of its finitely many components. Therefore at most countably many prime divisors have $\nu (T,D)>0$. Let $D_1,D_2,\ldots $ be an enumeration of them and put $S_ N=\sum _{k\leq N}\nu (T,D_ k)[D_ k]$.

We claim that $T-S_ N\geq 0$ for every $N$, and argue by induction; suppose that $T-S_{N-1}\geq 0$. Choose a point $x\in D_ N$ outside the proper analytic subsets $D_ N\cap D_ k$, $k<N$, and outside the exceptional sets of proposition 6.9.7 for $T$ and $T-S_{N-1}$ along $D_ N$. Such points exist, because by that proposition the complement in $D_ N$ of a countable union of proper analytic subsets of $D_ N$ is dense in $D_ N$. The currents $[D_ k]$, $k<N$, vanish near $x$, so $T-S_{N-1}=T$ near $x$, and therefore $\nu (T-S_{N-1},D_ N)=\nu (T-S_{N-1},x)=\nu (T,x)=\nu (T,D_ N)$. Step 1, applied to $T-S_{N-1}$, gives $T-S_ N\geq 0$. Consequently the trace measures of the $S_ N$ in every chart are bounded by that of $T$, and the masses $\int _ XS_ N\wedge \omega ^{n-1}$ are bounded by $\int _ XT\wedge \omega ^{n-1}$.

We now show that the sequence $(S_ N)$ converges weakly. Every test form is a finite linear combination of test forms $g\gamma $, where $g\geq 0$ is a test function supported in a chart and $\gamma $ is a strongly positive form with constant coefficients in that chart. To see this, use a partition of unity (proposition 5.2.6), the fact that strongly positive forms span the real forms (proposition 5.3.2), and write a real test function $g$ as $(g+M\psi )-M\psi $ with $\psi \geq 0$ a test function equal to one on $\operatorname{Supp}g$ and $M\geq \sup |g|$. For such $g\gamma $ the numbers $S_ N(g\gamma )$ increase with $N$, since $S_{N+1}-S_ N\geq 0$, and they are bounded by $T(g\gamma )$; hence they converge, and so does $S_ N(\phi )$ for every test form $\phi $. By the compactness assertion of theorem 5.3.7, a subsequence converges weakly to a closed positive current $S$, so $S_ N(\phi )\to S(\phi )$ for every $\phi $. Finally $R=T-S$ is closed and positive, as a weak limit of the closed positive currents $T-S_ N$.

Step 3: properties of $R$ and uniqueness. Fix a prime divisor $D$ and write $T=R+\nu (T,D)[D]+S_ D$, where $S_ D$ is the sum of the terms of the series for the other divisors. Lelong numbers are additive and nonnegative (proposition 6.7.9), and $\nu ([D],x)=\nu (\log |f|^2,x)=1$ at regular points $x$ of $D$, where $f$ is a local coordinate defining $D$. Hence, at a point of $D_{\mathrm{reg}}$ outside the countable union of proper analytic subsets where one of the currents $T$, $R$, $S_ D$ has a nongeneric Lelong number (such points exist by proposition 6.9.7, since $D_{\mathrm{sing}}$ is also a proper analytic subset of $D$),

Hence $\nu (R,D)=0$.

Now consider any decomposition $T=R'+\sum _ Db_ D[D]$ as in the statement, fix $D$, and let $S'_ D$ be the sum of the terms for the other divisors. We first show that the coefficient measures of $S'_ D$ give no mass to $D$. If $D'\neq D$ is another prime divisor, then $D\cap D'_{\mathrm{reg}}$ is an analytic subset of the connected manifold $D'_{\mathrm{reg}}$ (corollary 9.3.16) which is not all of it, since otherwise $D'\subset D$, and then $D'=D$ by (10.4.10). Hence $D\cap D'_{\mathrm{reg}}$ is a null set for the surface measure of $D'_{\mathrm{reg}}$ (lemma 5.5.1, in local coordinates of $D'_{\mathrm{reg}}$), and $[D']$ gives no mass to $D$. Now let $K\subset D$ be a compact subset of a chart, and for large $m$ let $\chi _ m$ be test functions with $0\leq \chi _ m\leq 1$, equal to one on $K$ and supported in the $1/m$-neighborhood of $K$ (lemma 2.5.1), all supported in a fixed compact subset of the chart, on which a test function $\psi $ equals one. For the trace measures in the chart, the weak convergence of the series gives $\mu _{S'_ D}(\chi _ m)=\sum _{D'}b_{D'}\mu _{[D']}(\chi _ m)$. As $m\to \infty $, $\chi _ m\to \mathbf1_ K$ pointwise, and dominated convergence, applied to the measure $\mu _{S'_ D}$ and to the series, whose terms are bounded by $b_{D'}\mu _{[D']}(\psi )$ with $\sum _{D'}b_{D'}\mu _{[D']}(\psi )=\mu _{S'_ D}(\psi )<\infty $, gives $\mu _{S'_ D}(K)=\sum _{D'}b_{D'}\mu _{[D']}(K)=0$. Since $D$ is covered by countably many such compact sets, the trace measures of $S'_ D$, and hence its coefficient measures (theorem 5.3.7), vanish on $D$.

By Step 1, $S'_ D-\nu (S'_ D,D)[D]\geq 0$; comparing the trace measures of both sides on $D$, where $[D]$ has positive trace mass, forces $\nu (S'_ D,D)=0$. Additivity of Lelong numbers at generic points of $D$, applied to $T=R'+b_ D[D]+S'_ D$, gives $\nu (T,D)=0+b_ D+0$. Thus $b_ D=\nu (T,D)$ for every $D$, and $R'=R$. $\square$