Non-Kahler manifolds where the different Laplacians are compatible - MathOverflow

Non-Kahler manifolds where the different Laplacians are compatible

2010-04-14T09:22:02Z

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?

Answer by Gjergji Zaimi for Non-Kahler manifolds where the different Laplacians are compatible

2010-04-14T10:14:46Z

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.

Answer by Quanting Zhao for Non-Kahler manifolds where the different Laplacians are compatible

2011-01-01T14:27:38Z

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

Answer by diverietti for Non-Kahler manifolds where the different Laplacians are compatible

2011-01-02T02:39:01Z

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.