# Ricci Curvature in Infinite Dimensions?

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 '10 at 4:19

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 '10 at 23:21

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 '10 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 '10 at 23:56

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 '10 at 17:01

This is a comment on the Ricci flow aspect of Terence Tao's answer. Let $\frac{\partial g}{\partial\tau}=2\operatorname{Ric}$ be a solution to the backward Ricci flow on $M\times(0,T)$. For $N\gg1$ define the potentially infinite space-time metric $\tilde{g}\doteqdot g+\tilde{R}d\tau^{2}$, where $\tilde{R}=R+\frac{N}{2\tau}$. This metric is dual (i.e., it is modeled on shrinkers instead of expanders) to the one defined by Sun-Chin Chu (see Section 4 of arXiv:0211349) and is Perelman's metric without the potentially infinite dimensional $S^{N}$ factor. Perelman wrote in 6.4 of arXiv:0211159 that this is "a potentially degenerate Riemannian metric [on $M\times\mathbb{R}$], which potentially satisfies the Ricci soliton equation."

Let $\nabla$ and $\tilde{\nabla}$ be the Levi-Civita connections of $g$ and $\tilde{g}$, respectively. Let $T=\frac{\partial}{\partial\tau}$ and let other capital letters denote space vectors. Then $\tilde{R}^{-1}=O(N^{-1})$ and \begin{gather*} \tilde{\nabla}_{X}Y=\nabla_{X}Y-\frac{\operatorname{Rc}(X,Y)T}{\tilde{R} },\quad\;\tilde{\nabla}_{X}T=\operatorname{Rc}(X)+\frac{\left\langle \nabla R,X\right\rangle T}{2\tilde{R}},\\ \tilde{\nabla}_{T}T=-\frac{\nabla R+\frac{T}{\tau}}{2}+\frac{(\frac{\partial R}{\partial\tau}+\frac{R}{\tau})T}{2\tilde{R}}, \end{gather*} where $\frac{\partial R}{\partial\tau}=-\Delta R-2\left\vert \operatorname{Rc} \right\vert ^{2}$. The terms with $\frac{1}{\tilde{R}}$ factors comprise the dual of Hamilton's trace Harnack. The space-time Riemann curvature tensor is (Gauss equations): \begin{align*} \widetilde{\operatorname{Rm}}(X,Y,Z,W) & =\operatorname{Rm}(X,Y,Z,W)+\frac {\operatorname{Rc}(X,W)\operatorname{Rc}(Y,Z)-\operatorname{Rc}% (X,Z)\operatorname{Rc}(Y,W)}{\tilde{R}},\\ \widetilde{\operatorname{Rm}}(\cdot,T,Z,W) & =(\nabla_{W}\operatorname{Rc} )(Z)-(\nabla_{Z}\operatorname{Rc})(W)+\frac{Z(R)\operatorname{Rc} (W)-W(R)\operatorname{Rc}(Z)}{2\tilde{R}},\\ \widetilde{\operatorname{Rm}}(\cdot,T,T,\cdot) & =\Delta_{L}\operatorname{Rc} -\frac{\nabla^{2}R}{2}+\operatorname{Rc}^{2}-\frac{\operatorname{Rc}}{2\tau }+\frac{1}{\tilde{R}}(\frac{\nabla R\otimes\nabla R}{4}+(\frac{\partial R}{\partial\tau}+\frac{R}{\tau})\frac{\operatorname{Rc}}{2}), \end{align*} where $\cdot$ denotes a slot for a space vector. The space-time geometry has some sort of divergence structure in the sense that taking the space divergence of $\widetilde{\operatorname{Rm}}$ is essentially the same as taking one of the components to be $T$. One already sees this in Hamilton's matrix Harnack quadratic. The space-time second Bianchi identity can be used to efficiently compute the evolution of $\widetilde{\operatorname{Rm}}$, leading to an alternate proof of Hamilton's matrix Harnack estimate.

Let $\bar{\nabla}=\lim_{N\rightarrow\infty}\tilde{\nabla}$, $\overline {\operatorname{Rm}}=\lim_{N\rightarrow\infty}\widetilde{\operatorname{Rm}}$, and $\overline{\operatorname{Ric}}=\lim_{N\rightarrow\infty} \widetilde{\operatorname{Ric}}$. Define $\bar{V}=\frac{\partial}{\partial\tau }$. Then we obtain the potentially Ricci soliton equation $\overline {\operatorname{Ric}}-\bar{\nabla}\bar{V}=\frac{1}{2\tau}d\tau\otimes \frac{\partial}{\partial\tau}$. Furthermore, we have \begin{align*} (\bar{\nabla}_{X}\overline{\operatorname{Ric}})(Y,T)-(\bar{\nabla} _{Y}\overline{\operatorname{Ric}})(X,T) & =0,\\ (\bar{\nabla}_{X}\overline{\operatorname{Ric}})(T,T)-(\bar{\nabla} _{T}\overline{\operatorname{Ric}})(X,T) & =-\frac{1}{4\tau}\left\langle \nabla R,X\right\rangle ,\\ (\bar{\nabla}_{X}\overline{\operatorname{Ric}})(Y,T)-(\bar{\nabla} _{T}\overline{\operatorname{Ric}})(X,Y) & =\overline{\operatorname{Rm} }(X,T,T,Y)+\frac{1}{2\tau}\overline{\operatorname{Ric}}(X,Y). \end{align*} This is analogous to the gradient Ricci soliton equation $(\nabla _{X}\operatorname{Ric})(Y,Z)-(\nabla_{Y}\operatorname{Ric} )(X,Z)=\operatorname{Rm}(X,Y,Z,\nabla f)$.

We see some nice properties of the space-time metric; but for applications to Ricci flow, following Perelman one should expand in $N$ the path length functional (here, one may do without the $S^{N}$ factor, i.e., potentially infinite metric versus dimension). Let $\tilde{\gamma}(\tau)\doteqdot (\gamma(\tau),\tau)$, $\tau\in\lbrack0,\bar{\tau}]$, be a path. Then Perelman's $\mathcal{L}$-geometry arises from $$\operatorname{L}_{\tilde{g}}\left( \tilde{\gamma}\right) =\int_{0} ^{\bar{\tau}}\sqrt{\tilde{g}(\tilde{\gamma}^{\prime}(\tau),\tilde{\gamma }^{\prime}(\tau))}\,d\tau=\sqrt{2N\bar{\tau}}(1+\frac{\mathcal{L}(\gamma )}{2\sqrt{\bar{\tau}}}\,N^{-1}+O\left( N^{-2}\right) ),$$ where $\mathcal{L}(\gamma)=\int_{0}^{\bar{\tau}}\sqrt{\tau}(R\left( \gamma(\tau),\tau\right) +|\gamma^{\prime}(\tau)|_{g\left( \tau\right) }^{2})d\tau$. Another strong motivation for not only this, but also Hamilton's matrix Harnack estimate, is the work of Peter Li and Shing-Tung Yau (see Section 3 of Acta Math. 1986).

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Forgive me for the self advertising, but the following paper of mine is somewhat related to the question (sort of giving an answer in the negative direction).

The main gist of the paper is as follows:

As Terry Tao says in his answer, there has recently been much interest in the definition of lower Ricci curvature bounds via optimal transportation. It makes sense in an arbitrary metric space (so in particular, there is no need for finite dimensionality) with one caveat: there must be a fixed reference measure on the space as well. Changing this measure can entirely change the "Ricci lower bounds," so it is an essential part of the theory (*).

Thus, the following is a reasonable restatement of your question in the setting Lott--Villani--Sturm:

Find "reasonable" infinite dimensional metric measure spaces where the optimal transport definition of "lower Ricci curvature bounds" holds.

In particular, the problem is to find good measures to put on these spaces.

One thing that seemed like a reasonable possibility was a rather natural measure constructed by von Renesse and Sturm here which is a measure on 2-Wasserstein space over the unit interval (see also Sturm's higher dimensional generalization here).

Indeed, 2-Wasserstein space over the unit interval is "flat" in the sense of metric triangles (see Proposition 5.3 here). So, it seems likely that given a nice measure on it, it should also have nonnegative Ricci curvature.

However, as I show in the paper above, this does not hold! In fact, 2-Wasserstein space equipped with this measure does not have ANY generalized Ricci lower bounds (in spite of it being compact).

(*) This the measure needs to play a crucial role in Ricci lower bounds was anticipated before the work of Lott--Villani--Sturm and others by the following observation: if a sequence of (smooth) Riemannian manifolds converged to another smooth manifold in the Gromov--Hausdorff sense, lower Ricci bounds are not necessarily preserved. However, they are preserved if the measures weakly converge as well.

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