PomeLemma 24.4.19.4865DD012C9C If a Hermitian form $\zeta $ satisfies $\mathrm d\zeta ^{n-1}=0$ and $\mathrm{dd}^{\mathrm{c}}\zeta =0$, then $\mathrm d\zeta =0$.
Lemma 24.4.194865DD012C9C
Proof. For $n=2$, the balanced condition is the assertion $\mathrm d\zeta =0$. Suppose $n\geq 3$ and set $\alpha =\partial \zeta $. Since $\zeta $ is balanced, $\alpha \wedge \zeta ^{n-2}=0$, so $\alpha $ is a primitive $(2,1)$-form. Applying $\partial \bar\partial $ to the balanced power gives
\[ 0=\partial \bar\partial \zeta ^{n-1} =(n-1)\partial \bar\partial \zeta \wedge \zeta ^{n-2} +(n-1)(n-2)\alpha \wedge \bar\alpha \wedge \zeta ^{n-3}. \]
The first term vanishes because $\zeta $ is pluriclosed. For a primitive $(2,1)$-form, pointwise Hermitian linear algebra gives
\[ -\mathrm i\, \alpha \wedge \bar\alpha \wedge \frac{\zeta ^{n-3}}{(n-3)!} =|\alpha |_\zeta ^2\frac{\zeta ^ n}{n!}. \]
Thus $\alpha =0$, and since $\zeta $ is real, also $\bar\partial \zeta =0$, so $\mathrm d\zeta =0$. $\square$
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