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On a Kahler manifold, the different Laplacians are compatible: $\Delta_d=2\Delta_{\bar{\partial}}=2\Delta_{\partial}$.

Are there non-Kahler Hermitian manifolds where the above identity holds?

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3 Answers

up vote 13 down vote accepted

Hermitian manifolds $M$ where $$\Delta_d f=2\Delta_{\bar{\partial}} f=2\Delta_{\partial} f$$ holds for every smooth function $f$ on $M$ are called balanced.

For more information, you can search for "balanced hermitian manifolds", Here, for instance, is a paper that reviews their basic properties and conditions to be Kahler.

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I thought that balanced manifolds were hermitian manifolds such that $d\omega^{dimX−1}=0$, where $\omega$ is the fundamental $(1,1)$-form associated to the hermitian metric. Is this equivalent to your definition? –  diverietti Jan 2 '11 at 2:27
    
Yes, these are equivalent definitions. –  Gjergji Zaimi Jan 2 '11 at 3:10
    
Could you sketch why, please ? It doesn't seem to me to be completely trivial... –  diverietti Jan 2 '11 at 16:21
    
There is a detailed proof of this result in an old paper of Gauduchon ams.org/mathscinet-getitem?mr=0486672 see Proposition 1, page 120 and the appendix. –  YangMills Mar 27 '12 at 14:52
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I just add some details to Zhao's answer.

Let $(X,\omega)$ be a hermitian manifold and $\tau$ be the operator of type $(1,0)$ and order $0$ defined by $$ \tau=[\Lambda_\omega,\partial\omega]. $$ Here $[\bullet,\bullet]$ is the (graded) commutator and $\Lambda_\omega$ the formal adjoint of the operator "wedge product with $\omega$". Often $\partial\omega$ is called the torsion of $\omega$ (which is $\equiv 0$ if and only if $\omega$ is Kähler) and $\tau$ the torsion operator. Then, we have the following identities: $$ \Delta_{\bar\partial}=\Delta_\partial+[\partial,\tau^\star]-[\bar\partial,\bar\tau^\star], $$ $$ [\partial,\bar\partial^\star]=-[\partial,\bar\tau^\star],\quad[\bar\partial,\partial^\star]=-[\bar\partial,\tau^\star], $$ and $$ \Delta_d=\Delta_\partial+\Delta_{\bar\partial}-[\partial,\bar\tau^\star]-[\bar\partial,\tau^\star]. $$ Therefore, $\Delta_\partial$, $\Delta_{\bar\partial}$ and $\frac 12\Delta_d$ no longer coincide, but they differ by linear differential operator of order $1$ only.

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Is it clear that $[\partial, \tau^{\star}] - [\bar{\partial}, \bar{\tau}^{\star}] = 0$ if and only if $\partial\omega = 0$? –  Michael Albanese May 20 '12 at 2:37
    
No, it is not clear. But I didn't state it! I just said that the torsion $\partial\omega$ is identically zero if and only if $\omega$ is Kähler. –  diverietti May 21 '12 at 7:06
    
Sorry, I was referring to Zhao's answer that you were expanding on. He claimed that the laplacians are compatible if and only if $\partial\omega = 0$. It is clear that if $\partial\omega = 0$ then $\tau = 0$ and hence $[\partial, \tau^{\star}] - [\bar{\partial}, \bar{\tau}^{\star}] = 0$. Do you know if the converse is true? If you only require the compatibility for functions, then the converse doesn't hold as there are balanced hermitian manifolds which are not K\"ahler. However, asking for this property to hold on functions instead of all possible forms is a much weaker restriction. –  Michael Albanese May 22 '12 at 6:13
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there is a theorem related to this:kahler is equivalent with the laplacian compatible equality above.Thus there no non-Kahler Hermitian manifolds where the above identity holds

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Do you have a reference for this theorem, or a sketch of the proof that $\Delta_{\partial} = \Delta_{\bar{\partial}}$ implies Kähler? –  Michael Albanese Jun 17 '12 at 9:13
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