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Is there a good notion of "Ricci curvature" in infinite dimensions?

My intuitive understanding of Ricci curvature is that it is some kind of an "average" of the curvature tensor over "different directions." In finite dimensions, taking the sum and taking the average makes little difference...

But in infinite dimensions, naively computing the curvature and taking the trace doesn't seem like it'll give anything convergent.

Is there some analogous notion of "average" curvature for Riemannian (Banach or Hilbert) Manifolds or some smart way of taking the "trace" of the "curvature tensor" so that it could still be called Ricci curvature and it'll be meaningful geometrically?

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Note that such a thing will be a section of a particular vector bundle over your manifold (hopefully a line bundle). It would be easiest to think about this in terms of the operations on vector bundles that take the metric and output the Riemann and Ricci tensors. – David Roberts Nov 3 2010 at 4:19

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Bowick and Rajeev computed the Ricci curvature of $Diff(S^1)/S^1$, which is an infinite dimensional Kahler manifold, in appendix A of their article: Nucl.Phys.B293:348,1987 (which can be found here). The Ricci curvature appears in the following setting. They consider the lift of the algebra of holomorphic vector fields on $Diff(S^1)/S^1$ to a line bundle which represents the bosonic string. This lift is not an isomorphism and the lifted action closes to a central extention (which is the celebrated Virasoro algebra). The Ricci curvature is the contribution to the central extention of the determinant line bundle (which is one component of the bosonic string's line bundle). It is interesting that for almost all Kahler metrics, the trace operation is finite and there is no need in a regularization of the sum.

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Ah, yes, in MR925071, Kirillov and Yurev computes the curvature tensor for the same metric. – BrainDead Nov 3 2010 at 17:01
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The PhD thesis of Dan Freed contains the computation of the Ricci curvature of an infinite-dimensional Kähler manifold. I don't have a scanned copy of the thesis with me, but it's probably condensed in his 1988 paper in the Journal of Differential Geometry: The Geometry of Loop Groups.

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I currently don't have access to the paper, but I'll get it soon. Do you know if the Ricci curvature happened to converge or he had to do some kind of regularization? – BrainDead Nov 3 2010 at 23:22
I think that the Ricci curvature is naively not convergent, but he regularised it. The details are murky -- it was a long time ago that I read this paper. – José Figueroa-O'Farrill Nov 3 2010 at 23:56
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There is a paper of Lott and Villani that defines the concept of nonnegative Ricci curvature in an arbitrary measured length space:

http://math.berkeley.edu/~lott/LottVillani.pdf

There is a dimension parameter which can be taken to infinity in that definition.

In his proof of the Poincare and geometrisation conjectures, Perelman used a heuristic argument in which he formally applied the Bishop-Gromov inequality to an infinite-dimensional, formally Ricci flat manifold to obtain the monotonicity of what is now known as the Perelman reduced volume; see Section 6 of

http://arxiv.org/abs/math.DG/0211159

or my notes on this at

http://terrytao.wordpress.com/2008/04/27/285g-lecture-9-comparison-geometry-the-high-dimensional-limit-and-perelman-reduced-volume/

EDIT: Some formal computations of Ricci and Riemann curvature on the (infinite dimensional) space of volume-preserving diffeomorphisms also appear in a famous paper of Arnold:

http://www.ams.org/mathscinet-getitem?mr=202082

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Thank you so much for providing me with so much information! I'll definitely check them out. – BrainDead Nov 3 2010 at 23:21

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