PomeLemma 11.22.1.D9A37BB4C36B Let $Z$ be a complex manifold of dimension $m\geq 2$, and let $j:D\hookrightarrow Z$ be a smooth closed hypersurface. If a positive $(q,q)$-current $R$, $1\leq q\leq m$, is supported on $D$ and $\mathrm{dd}^{\mathrm{c}}R$ has order zero, then there is a unique positive $(q-1,q-1)$-current $S$ on $D$ such that $R=j_*S$. If $\mathrm{dd}^{\mathrm{c}}R=0$, then $\mathrm{dd}^{\mathrm{c}}S=0$. In particular, if $q=1$, $D$ is compact and connected, and $\mathrm{dd}^{\mathrm{c}}R=0$, then $R=c[D]$ for some $c\geq 0$.
Lemma 11.22.1D9A37BB4C36B
Proof. Use local coordinates $(z,w_1,\ldots ,w_{m-1})$ with $D=\{ z=0\} $. Both $R$ and $\mathrm{dd}^{\mathrm{c}}R$ have measure coefficients supported on $D$. Suppose first that $q<m$. For every compactly supported smooth form $\eta $ of bidegree $(m-q-1,m-q-1)$,
\[ 0=\langle \mathrm{dd}^{\mathrm{c}}R,|z|^2\eta \rangle =\langle R,\mathrm{dd}^{\mathrm{c}}(|z|^2\eta )\rangle =\langle R,\mathrm{dd}^{\mathrm{c}}|z|^2\wedge \eta \rangle , \]
where the first equality holds because $\mathrm{dd}^{\mathrm{c}}R$ is a measure supported on $D$, on which $|z|^2$ vanishes, and the last one because the terms containing $|z|^2$ or its first derivatives vanish on $D$. Thus $R\wedge \mathrm{i}\, \mathrm{d}z\wedge \mathrm{d}\bar z=0$, which removes the coefficients of $R$ containing neither normal differential.
Fix a constant positive Hermitian form $\tau $ in the $w$ variables. The current $M_\tau =R\wedge \tau ^{m-q-1}$ is positive of bidegree $(m-1,m-1)$, and $M_\tau \wedge \mathrm{i}\, \mathrm{d}z\wedge \mathrm{d}\bar z=0$. Its Hermitian matrix of measure densities therefore has zero diagonal entry in the normal direction, and positivity removes the corresponding row and column, so
\[ M_\tau \wedge \mathrm{d}z=0, \qquad M_\tau \wedge \mathrm{d}\bar z=0. \]
Put $s=m-q-1$. The powers $\tau ^ s$, as $\tau $ varies over the positive Hermitian forms, span the real horizontal $(s,s)$-forms. Indeed, a linear functional vanishing on these powers vanishes on a polynomial over an open cone; polarization makes it vanish on products of arbitrary real $(1,1)$-forms, and these products span the forms of that bidegree. Their products with a horizontal one-form span the complementary test forms for the mixed coefficients of $R$. The identities above therefore remove every coefficient containing exactly one normal differential. For $q=m$, every top-degree coefficient already contains both normal differentials, and this reduction is unnecessary.
The remaining coefficients are measures on $\{ z=0\} $ multiplying both normal differentials. They therefore define a unique current $S$ on $D$ with $R=j_*S$. Testing against extensions of strongly positive forms on $D$ proves positivity of $S$, and uniqueness makes these local currents agree on overlaps. Pushforward commutes with $\mathrm{dd}^{\mathrm{c}}$ and is injective for a closed embedding, so $\mathrm{dd}^{\mathrm{c}}R=0$ implies $\mathrm{dd}^{\mathrm{c}}S=0$.
When $q=1$, the current $S$ is a nonnegative distribution on $D$. The equation $\mathrm{dd}^{\mathrm{c}}S=0$ and Weyl's lemma make it a smooth nonnegative pluriharmonic function. If $D$ is compact and connected, the maximum principle makes it a constant $c\geq 0$, proving the last assertion. $\square$
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