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I've been reading through Yau's proof of the Calabi conjecture (1) and I was quite intrigued by the relation

[tex]

R_{i\bar{j}} = - \frac{\partial^2}{\partial z^i \partial \bar{z}^j } [\log \det (g_{s\bar{t}}) ]

[/tex]

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

Thanks,

Dave

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# Yau's result for the Ricci curvature on Kahler manifold

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