Non-Kahler manifolds where the different Laplacians are compatible - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:55:25Z http://mathoverflow.net/feeds/question/21315 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible Non-Kahler manifolds where the different Laplacians are compatible Colin Tan 2010-04-14T09:22:02Z 2011-01-02T02:39:01Z <p>On a Kahler manifold, the different Laplacians are compatible: $\Delta_d=2\Delta_{\bar{\partial}}=2\Delta_{\partial}$.</p> <p>Are there non-Kahler Hermitian manifolds where the above identity holds?</p> http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible/21317#21317 Answer by Gjergji Zaimi for Non-Kahler manifolds where the different Laplacians are compatible Gjergji Zaimi 2010-04-14T10:14:46Z 2010-04-14T10:14:46Z <p>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 <em>balanced</em>.</p> <p>For more information, you can search for "balanced hermitian manifolds", <a href="http://cdsweb.cern.ch/record/422889/files/" rel="nofollow">Here</a>, for instance, is a paper that reviews their basic properties and conditions to be Kahler.</p> http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible/50861#50861 Answer by Quanting Zhao for Non-Kahler manifolds where the different Laplacians are compatible Quanting Zhao 2011-01-01T14:27:38Z 2011-01-01T14:27:38Z <p>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</p> http://mathoverflow.net/questions/21315/non-kahler-manifolds-where-the-different-laplacians-are-compatible/50897#50897 Answer by diverietti for Non-Kahler manifolds where the different Laplacians are compatible diverietti 2011-01-02T02:39:01Z 2011-01-02T02:39:01Z <p>I just add some details to Zhao's answer.</p> <p>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.</p>