# Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold

Hallo,

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.

Best Regards, Pavel

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try looking at papers of Stenzel: math.osu.edu/~stenzel.3/research/publications/ricci-flat.pdf – Spiro Karigiannis Oct 20 '12 at 12:48
Your question suggests that you are looking for an isometric embedding of the given Kähler manifold as a special Lagrangian in a Calabi-Yau manifold, but you don't mention this requirement in the text. I'll just point out that the induced metric on any special Lagrangian submanifold of a Calabi-Yau manifold is necessarily real-analytic, so it follows that it is not possible, in general to isometrically embed a given Kähler manifold as a special Lagrangian in some Calabi-Yau manifold. – Robert Bryant Oct 20 '12 at 22:19
ok, I see. lets assume that the given Kaehler manifold is also real analytic. Is it then possible? how can one explain this? what are the ingredients in showing this? – Pavel Oct 21 '12 at 6:01
I am not exactly sure what you mean by "explanation", but since you're asking for references, have a look at Birte Feix's thesis "Hyperkaehler metrics on cotangent bundles". There she constructs the HK metric in a different way. See also mathoverflow.net/questions/46752/… – Peter Dalakov Oct 21 '12 at 14:30

Disclaimer: 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.

Observation: 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 $(X,J, \omega_J,\Omega = (\omega_K+i\omega_I)^{\dim_{\mathbb{C}} M})$.

(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. )

Here "complex-Lagrangian" means that $M\subset (X,I)$ is a complex submanifold and $\left. \omega^c\right|_M=0$.

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.

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How can you show then that, after a rotation, it satisfies the Calabi-Yau equation?

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You should not ask a new question as an answer to an existing question. You can either create a new question or post a comment. But it's not at all clear what you mean. A hyperKahler manifold is Calabi-Yau in an $S^2$ worth of ways. This is clear, because the triple $\omega_I$, $\omega_J$, and $\omega_K$ are all parallel with respect to the Calabi-Yau metric, so the $\Omega$ that Peter defines is parallel, thus the pair $(\omega_J, \Omega = (\omega_K + i \omega_I)^{\dim_{\mathbb C} M}$ is a Calabi-Yau structure. – Spiro Karigiannis Oct 23 '12 at 18:19
Yes but does it follow then that $\omega_{J}^{n} = c_{n} \Omega \wedge \Omega$, where $c_{n}$ is a constant depending only on $n$, where $n = dim_{\mathbb{C}}M$? – Mina Oct 24 '12 at 12:26
ok I will post it as a question :). – Mina Oct 24 '12 at 16:46