Let $\omega$ be the hermitian form for an hermitian metric on a compact complex manifold. Can $\omega^{n-1}$ be $\partial {\bar{\partial}}$-exact? (We know that our hermitian metric must necessarily be balanced and non-Kahler.)

## 1 Answer

Yes, this can happen. For a simple example, consider $M = \mathrm{SL}(2,\mathbb{C})/\Lambda$ where $\Lambda\subset \mathrm{SL}(2,\mathbb{C})$ is a discrete, co-compact lattice. Then $M$ is a compact complex $3$-manifold.

Let $\alpha_1,\alpha_2,\alpha_3$ be a basis for the right-invariant holomorphic $1$-forms on $\mathrm{SL}(2,\mathbb{C})$. Because they are right invariant, they are well-defined on the quotient $M = \mathrm{SL}(2,\mathbb{C})/\Lambda$, and one can choose these forms so that $$ \mathrm{d}\alpha_1 = \alpha_2\wedge\alpha_3\,,\quad \mathrm{d}\alpha_2 = \alpha_3\wedge\alpha_1\,,\quad \mathrm{d}\alpha_3 = \alpha_1\wedge\alpha_2\,.\quad $$ The positive $(1,1)$-form $$ \omega = \frac{i}{2}\left(\alpha_1\wedge\overline{\alpha_1}+\alpha_2\wedge\overline{\alpha_2}+\alpha_3\wedge\overline{\alpha_3}\right) $$ defines an Hermitian structure on $M$ but is not closed (of course). Meanwhile, we have $$ \tfrac12\omega^2 = \tfrac14\left( \alpha_2\wedge\alpha_3\wedge\overline{\alpha_2}\wedge\overline{\alpha_3} +\alpha_3\wedge\alpha_1\wedge\overline{\alpha_3}\wedge\overline{\alpha_1} +\alpha_1\wedge\alpha_2\wedge\overline{\alpha_1}\wedge\overline{\alpha_2}\right), $$ while the formulae for the exterior derivatives of the $\alpha_i$ yield $$ -i\,\partial\overline{\partial}\omega = \omega^2. $$