Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer

up vote 9 down vote accepted

Such hyper Kaehler metrics do exist near the zero section, e.g. in a formal or an analytic tubular neighborhood of the zero section. After that one can use some homogeneity to spread them on the whole cotangent bundle but typically the resulting metrics are non-complete. One gets nice global metrics on the cotangent bundles of Hermitian symmetric spaces but this is pretty much it. This question was studied extensively. There are two different proofs of the existence: in this work of Birte Feix and this work of Dima Kaledin.

share|improve this answer
    
Thanks! I might have noticed Kaledin's paper if I searched for "Kaehler" instead of "Kahler"... –  Sam Gunningham Nov 20 '10 at 17:59
2  
This answer is pretty complete, but it is worth reading the paper of Calabi in Ann. Ec. Norm. Sup. 12 (1979) for an explicit construction of the HK metric on the cotangent bundle of complex projective space. The precise form of the metric is not obvious, and his approach (subsequently generalized to other HSS's) was to find the Kaehler potential. As in applications of Yau's theorem in the compact case, the HK metric is indeed compatible with the underlying holomorphic symplectic structure. –  Simon Salamon Dec 4 '10 at 19:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.