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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?

flag	asked Apr 14 2010 at 9:22 Colin Tan 4,231●13●43

Gjergji Zaimi · Accepted Answer · 2010-04-14 10:14:46Z UTC

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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.

answered Apr 14 2010 at 10:14

Gjergji Zaimi
37.8k●2●75●151

	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 2011 at 2:27
	Yes, these are equivalent definitions. – Gjergji Zaimi Jan 2 2011 at 3:10
	Could you sketch why, please ? It doesn't seem to me to be completely trivial... – diverietti Jan 2 2011 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 2012 at 14:52

Quanting Zhao · Answer 2 · 2011-01-01 14:27:38Z UTC

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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

answered Jan 1 2011 at 14:27

Quanting Zhao
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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 2012 at 9:13

diverietti · Answer 3 · 2011-01-02 02:39:01Z UTC

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.

Non-Kahler manifolds where the different Laplacians are compatible

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