Yau's result for the Ricci curvature on Kahler manifold

[schieghoven]

Hi,

I've been reading through Yau's proof of the Calabi conjecture (1) and I was quite intrigued by the relation
<br /> R_{i\bar{j}} = - \frac{\partial^2}{\partial z^i \partial \bar{z}^j } [\log \det (g_{s\bar{t}}) ]<br />
derived therein. g_{s \bar{t} } 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.) R_{i\bar{j}} 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

 

[Naty1]

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 twenty components of the (Riemann-Christoff) curvature (tensor) in a four dimensional space can be grouped into two sets of ten components each, in a manner that is independent of any choice of coordinate system. One of these two sets involves the turning of vectors in the course of parallel transport...this set is usually referred to as the Ricci tensor...in a slight rearrangement the Ricci tensor may be called the Einstein vector. The..other ten components form..Weyl's tensor...the totality of of all components of curvature is called the Riemann-Christoff curvature tensor...

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.

 

[schieghoven]

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?

Regards

Dave

 

[ruwn]

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

 

[Naty1]

More from Peter Bergmann:

A Minkowski continuim may be conformally transformed to a curved continuium...one aspect remains unchanged by conformal transformations...that is the space like,time like or lightlike caharacter of each interval and direction...Weyl discovered that, under a conformal transformation, the Ricci tensor and the Weyl tensor behave very differently...The Ricci tensor changes in a very complicated way (which includes the possibility that it may vanish before the conformal transformation but be non zero after the transformation; the Weyl tensor remains unchanged.

I have no idea what that means, but it doesn't sound encouraging! good luck...

 

[hebertodelrio]

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.

Regards

Heberto

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?

Regards

Dave