# Does equality of Laplacians imply Kähler?

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.

The answer is yes, as proved by Y. Ogawa in this paper, see Theorem 3.10. Apparently equality on functions and $1$-forms is enough to conclude that the metric is Kähler.
• Thanks YangMills. I have only had a brief look these papers, but it is not immediately clear to me that they answer my question. I'm sure they do, I just want to be certain before I accept your answer. My main point of unease is in Theorem 3.10 of Ogawa's paper which states that if the operator $\square$ (which I believe is $\Delta_{\bar{\partial}}$ in the above notation) is real for all 0 and 1 forms, then the (almost) Hermitian structure is Kähler. As $\Delta_{\partial}$ is real, we just need the remaining terms $\dots$ – Michael Albanese Jun 25 '12 at 12:04
• $\dots [\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]$ to be real. With this in mind, we have the following: $$\overline{[\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]} = [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]] - [\partial, [\Lambda_{\partial\omega}, L]] = -([\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]).$$ Does this show that $\Delta_{\bar{\partial}}$ is real iff $\Delta_{\bar{\partial}} = \Delta_{\partial}$? – Michael Albanese Jun 25 '12 at 12:21
• Yes, indeed in general you have that $\overline{\Delta_{\overline{\partial}}}=\Delta_{\partial}$, so the two Laplacians are equal iff any one of them is a real operator. Another way to see this is to note that your calculation shows that the error term is a purely imaginary operator. – YangMills Jun 25 '12 at 13:04
• As the papers are using almost Hermitian metrics, the relationship between $\Delta_{\partial}$ and $\Delta_{\bar{\partial}}$ would have additional terms. However, a similar argument must hold to show that the entire error term (as YangMills calls it) is a purely imaginary operator. – Michael Albanese Jun 28 '12 at 4:54
In addition to the papers mentioned by YangMills, there is also the earlier paper by A. W. Adler which shows that if $\Delta = 2\Delta_{\bar{\partial}}$ on a hermitian manifold, then it is Kähler.