Is the cotangent bundle to a Kahler manifold hyperkahler? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:49:04Z http://mathoverflow.net/feeds/question/46752 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46752/is-the-cotangent-bundle-to-a-kahler-manifold-hyperkahler Is the cotangent bundle to a Kahler manifold hyperkahler? Sam Gunningham 2010-11-20T17:13:39Z 2010-11-20T17:42:29Z <p>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 $\Omega _\mathbb C$.</p> <p>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). </p> <p><strong>Question 1</strong>: 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...)</p> <p>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 <em>compact</em> holomorphic symplectic manifold is hyperkahler.</p> <p><strong>Question 2</strong>: Does $T^\ast M$ admit a hyperkahler metric, with the associated holomorphic symplectic form the canonical one (coming from the cotangent bundleness)?</p> <p><strong>Question 3</strong>: Is $g$ a hyperkahler metric for $T^\ast M$ at all? Or, does $T^\ast M$ admit a hyperkahler metric at all?</p> <p>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. </p> http://mathoverflow.net/questions/46752/is-the-cotangent-bundle-to-a-kahler-manifold-hyperkahler/46755#46755 Answer by Tony Pantev for Is the cotangent bundle to a Kahler manifold hyperkahler? Tony Pantev 2010-11-20T17:42:29Z 2010-11-20T17:42:29Z <p>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 <a href="http://www.google.com/url?sa=t&amp;source=web&amp;cd=4&amp;sqi=2&amp;ved=0CCMQFjAD&amp;url=http%3A%2F%2Fpeople.maths.ox.ac.uk%2Fhitchin%2Fhitchinstudents%2Ffeix.ps.gz&amp;rct=j&amp;q=Birte%20Feix&amp;ei=ogXoTKeTFsnMswahpLyyCw&amp;usg=AFQjCNEfkq4jK0h2fTxjUgWcH1_KS-uC8A&amp;cad=rja" rel="nofollow"> this work </a> of Birte Feix and <a href="http://arxiv.org/abs/math/0011256" rel="nofollow"> this work </a> of Dima Kaledin.</p>