# 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.

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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.

There is also a related paper by C.C.Hsiung, see Theorem 4.2.

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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
Thanks. That's what I was thinking, but I just wanted to make sure. –  Michael Albanese Jun 26 '12 at 2:18
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.

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