# Is the cotangent bundle to a Kahler manifold hyperkahler?

Let me be more specific. Let $M$ be a Kahler manifold with Riemannian metric $g$ and complex structure $I$. Then $T^\ast M$ will also be Kahler with metric and complex structure induced from $M$ (I will give them the same name). It is also holomorphic symplectic, with canonical holomorphic symplectic form $\Omega _\mathbb C$.

If $M$ was an affine space with the standard metric I could define $\omega _J$ and $\omega _K$ on $T^\ast M$ by taking the real and imaginary parts of $\Omega _\mathbb C$ which would define a hyperkahler structure on $T^\ast M$ (everything is covariantly constant with constant coefficients).

Question 1: Does this work for a general kahler manifold $M$? It seems a bit unreasonable to me, as the construction of $\Omega _\mathbb C$ does not depend on the metric (but does depend on the complex structure, which is compatible with the metric...)

I also know that every hyperkahler manifold is holomorphic symplectic (with $\Omega _\mathbb C = \omega _J + I\omega _K$) and Yau's theorem implies that every compact holomorphic symplectic manifold is hyperkahler.

Question 2: Does $T^\ast M$ admit a hyperkahler metric, with the associated holomorphic symplectic form the canonical one (coming from the cotangent bundleness)?

Question 3: Is $g$ a hyperkahler metric for $T^\ast M$ at all? Or, does $T^\ast M$ admit a hyperkahler metric at all?

I don't know much about this sort of thing, but it seemed like a natural question to me, and I couldn't find an answer anywhere.

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I don't think that this is true in general. There are special cases where this is true, though. I think that if $M$ is a generalised flag manifold then yes, by results of Nakajima and also Biquard. Similarly if $M$ is a noncompact hermitian symmetric space, by results of Biquard and Gauduchon. There is also work of Kronheimer showing that there is a hyperkähler metric on the cotangent bundle of a complexified Lie group. –  José Figueroa-O'Farrill Nov 20 '10 at 17:29