Theorem 10.2.15 (The ddc-lemma)D24B01620C94

Proof. Put $a=\int _ XT\wedge \omega ^{n-1}/\int _ X\omega ^ n$. Since $T\wedge \omega ^{n-1}=(\Lambda T)\omega ^ n/n$, the distribution $\Lambda T-na$ has mean zero, and lemma 10.2.5 gives a real distribution $f$ of mean zero with $Lf=-4\pi (\Lambda T-na)$. By lemma 10.2.12, $\Lambda \mathrm{dd}^{\mathrm{c}}f=\Lambda T-na$, so the closed current $U=T-a\omega -\mathrm{dd}^{\mathrm{c}}f$ has $\Lambda U=0$. By lemma 10.2.12 again, $U$ is harmonic, and by lemma 10.2.5 it is smooth. The Kähler form is harmonic as well, since $\mathrm d\omega =0$ and $*\omega =\omega ^{n-1}/(n-1)!$ is closed. Hence $H=a\omega +U$ and $f$ give the decomposition (10.2.16). If $T$ is smooth, then $\Lambda T$ is smooth, and so is $f$ by lemma 10.2.5.

An exact harmonic form $B$ vanishes: if $B=\mathrm dS$ for a current $S$, then

since $\mathrm d*B=0$. If $T=H'+\mathrm{dd}^{\mathrm{c}}f'$ is a second decomposition, then $H-H'=\mathrm d\, \mathrm d^{\mathrm c}(f'-f)$ is exact and harmonic, hence zero. Then $\mathrm{dd}^{\mathrm{c}}(f-f')=0$, so $L(f-f')=0$ by lemma 10.2.12; thus $f-f'$ is constant by lemma 10.2.5, and it vanishes because both have mean zero. The same argument proves the exactness criterion: if $T$ is exact, so is $H=T-\mathrm d\, \mathrm d^{\mathrm c}f$, whence $H=0$; conversely, if $H=0$, then $T=\mathrm d\, \mathrm d^{\mathrm c}f$.

Suppose now that $T\geq 0$. By theorem 5.4.3, theorem 5.9.1, lemma 6.14.2, every point of $X$ has a coordinate ball on which $T=\mathrm{dd}^{\mathrm{c}}u$ with $u$ plurisubharmonic and $H=\mathrm{dd}^{\mathrm{c}}h$ with $h$ smooth. Then $\mathrm{dd}^{\mathrm{c}}(f+h-u)=0$, so the scalar Laplacian of the distribution $f+h-u$ vanishes, and Weyl's lemma (theorem 3.1.11) shows that it is a smooth pluriharmonic function $p$. Thus $f=u-h+p$ is, on the ball, the class of a plurisubharmonic function minus a smooth function. On the overlap of two balls, the two local representatives $u_1-h_1+p_1$ and $u_2-h_2+p_2$ agree almost everywhere; after adding $h_1-p_1$ to both, both become plurisubharmonic, because $h_1-p_1-h_2+p_2$ is smooth with $\mathrm{dd}^{\mathrm{c}}(h_1-p_1-h_2+p_2)=H-H=0$, and plurisubharmonic functions that agree almost everywhere agree everywhere (theorem 5.9.1). Hence the local representatives define a global quasi-plurisubharmonic representative of $f$. It is the only one, since two quasi-plurisubharmonic functions that agree almost everywhere agree everywhere (lemma 6.14.4(1)). $\square$