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Yau's result for the Ricci curvature on Kahler manifold

  1. Oct 16, 2008 #1

    I've been reading through Yau's proof of the Calabi conjecture (1) and I was quite intrigued by the relation
    R_{i\bar{j}} = - \frac{\partial^2}{\partial z^i \partial \bar{z}^j } [\log \det (g_{s\bar{t}}) ]
    derived therein. [tex]g_{s \bar{t} } [/tex] is a Kahler metric on a Kahler manifold (I'm not entirely sure what that entails... I'm working on the assumption that it's like a complex version of a Riemannian metric.) [tex]R_{i\bar{j}}[/tex] is the Ricci curvature of said metric. I find it remarkable that the Ricci curvature is so readily computed for these Kahler manifolds. I've never seen a comparably simple relation in Riemannian geometry - I don't think there is one, right? Why would this be? What's different between Kahler manifolds and Riemannian manifolds that makes this result possible in the Kahler case?

    I'm also wondering to what extent Yau's result can cross over to Riemannian geometry. If I understand the paper correctly, Yau has basically solved the inverse problem of finding the Kahler metric that corresponds to a given Ricci form. In Riemannian geometry, we might pose a similar question: if R_ij is a symmetric matrix (that satisfies some criterion following from the Bianchi identities), is it the Ricci tensor for some Riemannian metric g'? Is g' unique?

    To me this is a tantalising question, because at least at some level, Einstein's equations for GR are posing a very similar problem, of finding the Riemannian metric that solves for a given stress-energy field.

    (1) Yau, Commun. Pure Appl. Math. 31 (1978) 339


  2. jcsd
  3. Oct 20, 2008 #2
    I likely know less than you on this subject but it turns out Peter Bergmann had some comments in THE RIDDLE OF GRAVITATION which I just read a few days ago:

    The author makes it sound like dealing with ten at a time is a lot simpler than dealing with twenty, especially making transformations from one coordinate system to another as the two tensors behave very differently...

    hope this helps a little.
  4. Oct 20, 2008 #3
    Cheers, thanks for the tip, you may well be right that the Weyl tensor and conformal transformations come into it somehow. Perhaps if I start writing things out it'll be clearer. Does anybody know offhand how the Ricci tensor changes under a conformal transformation?


  5. Oct 20, 2008 #4
    Kähler manifolds are hermitian manifolds with closed Kählerform. The Kählerform looks like the metrik tensor but is antisymmetric. if you impose that the exterior derivative of the kählerform vanishes you find that the metric can locally be expressed by the Kählerpotential. So if you got a Kählerpotential, the metric follows and with this the connection and the riemann tensor. i think this is basically the idea. you can take a look at nakahara...

    hope this gave you an idea
    Last edited: Oct 20, 2008
  6. Oct 20, 2008 #5
    More from Peter Bergmann:

    I have no idea what that means, but it doesn't sound encouraging!! good luck...
  7. Mar 25, 2009 #6
    Hi, I am a mathematician and I know a little bit about what you are talking about. An excellent book, is Einstein manifolds by Arthur Besse, this is a must. In this book, he explicitly computes how the full curvature tensor, the Ricci tensor, the Weyl tensor, change under a conformal change in the metric. He also talks about the Calabi-Yau problem.



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