This question follows on from this one.

Let $(X, \omega)$ be a Hermitian manifold and define the Laplacians $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$ and $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$.

If $(X, \omega)$ is a Kähler manifold, that is $d\omega = 0$ (or equivalently $\partial\omega = 0$ or $\bar{\partial}\omega = 0$), we have $\Delta_{\bar{\partial}} = \Delta_{\partial}$.

More generally, on any Hermitian manifold we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]$ where:

- $[\bullet, \bullet]$ is the graded commutator;
- $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are the adjoints of wedging with the forms $\partial\omega$ and $\bar{\partial}\omega$ respectively; and
- $L$ is the Lefschetz operator, that is, wedging with $\omega$.

It is clear how the additional terms relating the Laplacians in the Hermitian case vanish if the metric is Kähler ($\partial\omega = 0$ and $\bar{\partial}\omega = 0$, so $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are both zero). What about the converse? That is:

If $\Delta_{\bar{\partial}} = \Delta_{\partial}$ on a Hermitian manifold $(X, \omega)$, is it necessarily Kähler?

The accepted answer in the linked question refers to *balanced* manifolds. These are manifolds with the property that $\Delta_{\bar{\partial}}f = \Delta_{\partial}f$ for any smooth *function* $f$. Not all such manifolds are Kähler. The above question is stronger as it requires equality for all smooth *forms*.