This question follows on from [this one](https://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible). 


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?
> 
> 
> 




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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*.



Does equality of Laplacians imply Kähler?
The answer is yes, as proved by Y. Ogawa in [this paper,](http://projecteuclid.org/euclid.jdg/1214429279) see Theorem 3.10. Apparently equality on functions and $1$-forms is enough to conclude that the metric is Kähler.


There is also a [related paper by C.C.Hsiung](http://dx.doi.org/10.1090/S0002-9947-1966-0189067-0), see Theorem 4.2.


