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Hallo,

I ave the following question: Due to Stenzel, Lempert, Szöke ect. we know that a Riemannian manifold $(M,g)$ admits a complex structure on an neighbourhood of the cotangent bundle. This complex structure $J$ is unique due to bruhat and whitney complexification method. Well, with this complex structure there comes a symplectic form $\omega$ that is Kähler and morover $M$ is a Lagrangian submanifold in this complexified neighbourhood of the zero section in the cotangent bundle. My question is: is this Kähler form $\omega$ unique? Does there exists a different (or in the same cohomology) Kähler form with the same properties? If its not unique to which extend does uniqueness fail? Or, how can one caracterize the space of Kähler forms on a neighbourhood of the zero section in the cotangent bundle that fixes $M$ as a Lagrangian submanifold? (actually in the last question I am very interested) I hope to get a lot of answers and I apologize if the question is too trivial/hard :).

Greetings Joan

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On a Kaehler manifold, two out of three structures --- $J$, $g$, $\omega$ --- determine the third. Since you are fixing $g$ and $J$, there doesn't seem to be any choice for $\omega$. –  Eugene Lerman Oct 6 '12 at 21:41
    
But I didn't fix $g$. $g$ comes with the symplectic form. –  Joan Oct 7 '12 at 4:24
    
If you start with a Riemannian manifold $(M, g_0)$ then $T^*M = TM$ acquires a natural Riemannian metric $g$ from $g_0$; this is what I thought you meant. Since this is not the case, are you asking: "I have a complex manifold $(N, J)$. What are all symplectic forms compatible with $J$." ? If this is the case, I believe there are lots of these forms. See arxiv.org/abs/math/0004122 which applies to your question when $M$ is a torus (so $T^*M = (C^\times)^n$). –  Eugene Lerman Oct 7 '12 at 11:47
    
But I am asking not only the forms compatible with $J$ but also that fixes $M$ as a Lagrangian manifold! –  Joan Oct 8 '12 at 7:06
    
Well, have you looked at the reference I gave you? There the real torus is Lagrangian. More concretely, take $M=S^1$. Then $T^*M$ is $\mathbb{C} -\{0\}$. There are many different Kaehler structures on $T^*M$. You can embed it in $\mathbb{C}$ with a flat metric. Or you can embed it in $\mathbb{C}P^1$ with the Fubini-Study metric. Basically toric manifolds provide plenty of examples. –  Eugene Lerman Oct 8 '12 at 11:33

1 Answer 1

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This is a rather tough question to answer.

Check out: D. Burns & R. Hind, Symplectic geometry and the uniqueness of Grauert tubes, Geom. Func. Anal. Vol. 11 (2001), 1-10.

In this paper, they prove that, if two Grauert tubes of $M$ in $TM$ with respect to two Riemannian metrics $g_1,\ g_2$ are biholomorphic, then the biholomorphism restricts to an isometry from $(M,g_1)$ to $(M,g_2)$. Thus, if no such isometry exists, then the resulting Grauert tubes are not biholomorphic. (If I am interpreting this theorem correctly).

In general, though, the Kaehler form is derived from the complex structure composed with the given Riemannian metric $g$. Thus, if we change the metric, we will be necessarily changing the Kaehler form. So, I would conclude that the Kaehler form is not unique.

It would be interesting to see how this symplectic structure compares with the standard symplectic structure of $T^*M.$

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