Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:09:54Z http://mathoverflow.net/feeds/question/110140 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold Pavel 2012-10-20T10:24:19Z 2012-10-23T14:12:48Z <p>Hallo,</p> <p>I am reading the paper "Hyperkaehler structures on the total space of holomorphic cotangent bundles" by D.Kaledin and I am asking if it is possible to embedd a real-analytic Kähler manifold, isometrically, as a special Lagrangian in a Calabi-Yau manifold. Acctualy what I am looking for is the following: Start with a compact real-analytic Kähler manifold $(M, I, \omega)$ and in a neigbourhood of the zero-section in the cotangent bundle $T^{*}M$ there should exists a holomorphic $(n,0)-$form $\Omega$ (with respect to some complex structure on this neigbourhood) and a Kaehler form $\tilde{\omega}$ such that the forms $Im(\Omega)$ and $\tilde{\omega}$ vanishes when restricted to $M$ (the zero section) and $\tilde{\omega}^{2n} = C_{n} \Omega \wedge \bar{\Omega}$ for some constant $C_{n}$ that depends only on $n$. I know that one can do this. But I don't know some references where I can find a explanation of this. Is it sufficient just to read the paper of Kaledin or do I have also to switch to other references? By using Kaledin's paper what ingredients are necessary for a proof of this embedding problem? I am a beginner in Calabi-Yau manifolds and Hyperkaeler manifolds and I would be very thankfull if someone has the answers. I hope for a lot of replys and also hope that this question is not too trivial.</p> <p>Best Regards, Pavel</p> http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau/110246#110246 Answer by Peter Dalakov for Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold Peter Dalakov 2012-10-21T16:39:37Z 2012-10-21T16:50:45Z <p><strong>Disclaimer:</strong> I am not sure what kind of "explanation" you are looking for. I would guess that you are after the observation (due to Hitchin), that complex Lagrangian submanifolds become special Lagrangian after rotating the complex structure.</p> <p><strong>Observation:</strong> Let $X$ be a hyperkaehler manifold. Let $\{I,J,K\}$ be a triple of complex structures, satisfying the quaternionic identities, and let $\{\omega_I,\omega_J,\omega_K\}$ be the respective Kaehler forms. Let $M\subset (X,I,\omega_I)$ be a complex-lagrangian submanifold for the complex-symplectic form $\omega^c= \omega_J+i\omega_K$. Then $M$ is a special lagrangian submanifod of $(X,J, \omega_J,\Omega = (\omega_K+i\omega_I)^{\dim_{\mathbb{C}} M})$.</p> <p>(Actually, if $\dim_{\mathbb{C}} M$ is odd you must either take $i\Omega$ as your holomorphic volume form, or use the more relaxed definition of special Lagrangian. ) </p> <p>Here "complex-Lagrangian" means that $M\subset (X,I)$ is a complex submanifold and $\left. \omega^c\right|_M=0$.</p> <p>So given a real-analytic Kaehler manifold, you embed it as the zero-section of the cotangent bundle, take the Kaledin-Feix metric on a (formal) tubular neighbourhood, and rotate the complex structure.</p> http://mathoverflow.net/questions/110140/isometric-embedding-of-a-kaehler-manifold-as-a-special-lagrangian-in-a-calabi-yau/110432#110432 Answer by Mina for Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold Mina 2012-10-23T14:12:48Z 2012-10-23T14:12:48Z <p>How can you show then that, after a rotation, it satisfies the Calabi-Yau equation?</p>