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I have a question concerning the complexification of a real analytic Riemannian manifold. Let $(M,g)$ be a compact Riemannian manifold. It is a well known fact that in a neighbourhood $U$ of the zero section of the cotangent bundle $T^{*}M$, $M$ admits a complexification. By Stenzel and Guillemin this complexification is called the adapted complex structure. It is even Kähler. In this complexification, the phase function $\varphi$ solves the homogenous Monge-Ampere equation. My question is now: are there Kähler structures on a neighbourhood $U$ of the zero section in the cotangent bundle of $M$, which turn $M$ into a Lagrangian submanifold and such that $M$ is isometrically embedded in $U$, BUT differ from the adapted complex structure (up to biholomorphism)? Do all such structures come from an adapted complex structure? Is this already known or can it be derived easily?

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Try the paper "Symplectic geometry and the uniqueness of Grauert tubes" by D. Burns and R. Hind,… – Liviu Nicolaescu Dec 7 '12 at 13:45
I dont see how this answers my question? I am interested in: are there Kälhler structures in a neighbourhood of the zero section of some Riemannian manifold that are not adapted in the sense of Guillemin and Stenzel? – Dmitri Dec 7 '12 at 16:51

Suppose that $M \subset Z$ is a compact Lagrangian submanifold of a Kaehler manifold $Z$ with Kaehler form $\omega$. Take a function $f$ so that $\partial \bar\partial f=0$ at every point of $M$. Then for a small enough neighborhood of $M$ in $Z$, $\omega+i\partial\bar\partial f$ is a Kaehler form in which $M$ is a Lagrangian manifold with the same induced metric. In your case, take $Z=T^*M$ with the Stenzel metric.

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Is it clear that the new metric is not biholomorphic to the old one? – Sebastian Goette Mar 28 at 9:24
Biholomorphisms of a complex $n$-dimensional manifold depend on $2n$ functions of $n$ variables, while our function $f$ is one essentially arbitrary function of $2n$ variables, so it seems unlikely that a biholomorphism can identify such metrics. Of course, this isn't a rigorous argument. But $f$ can have compact support here, so it is easy to make a rigorous argument. – Ben McKay Mar 28 at 9:33

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