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Let $M$ be a symplectic manifold (not Kaehler). Does there exists in a neighbourhood of the zero section in the cotangent bundle $T^{*}M$ a Hyperkaehler structure? I know that by the paper by Feix on can construct such structures if $M$ is Kaehler. Does this also hold or is there a similar construction in the waker case the symplectic case?

Greetings mirta

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Are you going to require that there be any relation specified in advance between the usual symplectic structure on $T^\ast M$ and the hyperKähler structure? For example, whether $M^{2n}$ is symplectic or not, if it is parallelizable, then there is always a (flat) hyperKähler structure on a neighborhood of the zero section in $T^\ast M$, since some such neighborhood can be (openly) immersed in $\mathbb{R}^{4n}$, and one can pull back the flat hyperKähler structure on $\mathbb{R}^{4n}$ under such an immersion. – Robert Bryant Sep 23 '13 at 11:48
Also, do you require that the zero section be complex? If so, there would be an obstruction to your "not Kaehler" in this (to me) cryptic remark made by Feix in both his thesis (p. 2) and paper (§1): "If a hyperkähler metric on the cotangent bundle $T^*M$ of a complex manifold $M$ exists, $M$ must be real-analytic Kähler; this can be seen from the twistor interpretation." – Francois Ziegler Sep 23 '13 at 12:02
well I am not assuming anything. I just need $M$ to be a lagrangian manifold with respect to some structure. also I am not assuming that $M$ is parallelisable. – mirta Sep 23 '13 at 12:15
does there exist some construction? – mirta Sep 23 '13 at 12:18

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