What is the best way to interpret, explain or somehow visualize the basic idea behind formal definition of Ricci Flow?

I am familiar with the hackneyed expressions like

"Ricci Flow is a non-linear analogue for the heat equation which smoothens metric",


"Ricci Flow describes the deformation of the Riemannian metric tensor on manifold".

However, I was looking for something similar to the balloon-under-pressure interpretation of the mean curvature flow, in particular of the surface tension flow:

enter image description here

Ultimately, I am looking for something that would make Ricci Flow concept clear for undergraduate students without diving too deep into technicalities of Ricci tensor and volume forms.

  • 5
    $\begingroup$ In general I don't think you can really get a better answer than "it's a co-ordinate independent heat equation for the metric" without getting technical. You can visualize Ricci flow of surfaces of revolution (see projecteuclid.org/download/pdf_1/euclid.em/1128371754) but the intrinsic nature of the flow makes it very hard to visualize in general. $\endgroup$ Apr 26, 2015 at 9:49
  • $\begingroup$ Hi, where does the nice image of the surface tension flow come from? Did you make it? $\endgroup$
    – unity
    Jul 6, 2018 at 18:56
  • $\begingroup$ @unity: picture source $\endgroup$
    – C.F.G
    Mar 17, 2020 at 5:12

3 Answers 3


The paper in the comment above gives some cool visualizations, but I thought I would mention some easy-to-understand computational interpretations, which can help give intuition (at least to me), even if they are not exactly physical in nature.

Shape Analysis in 3D

See Zeng et al, Ricci Flow for 3D Shape Analysis. Basically, the Ricci flow deforms an arbitrary 3d shape into one of $\mathbb{S}^2$, $\mathbb{H}^2$, or $\mathbb{R}^2$, depending on the shape's topology (and of course the resulting scalar Ricci curvature). The author's note that (in the 3D case) for a closed surface, if the total surface area is preserved during the flow, the Ricci flow converges to a metric with constant Gaussian curvature everywhere. Classic differential geometry is probably more intuitive.

Image Processing

See Appleboim et al, Ricci Curvature and Flow for Image Denoising and Super-Resolution. In it, the authors treat greyscale images as Riemannian manifolds and note that (with their particular construction of the image manifold) running the Ricci flow is essentially analogous to evolving (diffusing) the gradient of the image, rather than the pixel values.

Ito Diffusion Behaviour The behaviour of Brownian motion on manifolds is such that negative scalar curvature is known to accelerate diffusion outward from the starting point (while positive slows it down). See e.g. Debbasch et al, Diffusion Processes on Manifolds. Maybe with this in mind, one can imagine Ricci flow to be "smoothing out" the behaviour of a diffusing particle on the manifold, in a way.


We do not know asymptotic at infinity of $Ric +\nabla^2 f=Cg,\ C<0$ where $M$ is noncompact manifold. Then $g(t)=c(t)\phi(t)^\ast g(0)$ where $\phi$ is flow of $\nabla f$. Here $g(t)$ converges to flat when ${\rm dim}\ M=2$.


I'm late to the question, but this writeup by Terry Tao is the best intuitive, high level source on the subject that I know. I'll summarize it in my own words.

First we need to describe the notion of a Riemannian metric, since this is what is evolving under the Ricci flow. The Riemannian metric $g$ is a way to describe the shape of a manifold, by defining the lengths, areas and angles of that manifold locally. For example, standing on the surface of the earth, you only have a perspective of distances, angles and area with respect to that surface rather than the larger space $\mathbb{R}^3$.

The Ricci curvature of a point $x \in M$ tells us how "non-Euclidean" a Riemannian manifold is at $x$. Specifically, we can measure this by considering the area of a sector of a circle having angle $\theta$ and radius $r$. In Euclidean space, this area is $\frac{1}{2}\theta r^2$ and in a Riemannian manifold $M$, we can denote it by $|A(x,r,\theta,v)|$. The Ricci curvature of a surface is the (scaled) difference between these: $$ \text{Ric}(x)(v,v) = \lim_{r\rightarrow 0} \lim_{\theta \rightarrow 0} \frac{\frac{1}{2} \theta r^2 - |A(x,r,\theta,v)| }{\theta r^4/24}.$$

This is the formula for a 2-dimensional manifold (surface). More generally, the Ricci curvature measures the difference between the Euclidean volume and the manifold volume.

(Here is an image of how the area of a triangle differs in spherical, Euclidean, and hyperbolic space.)

The Ricci flow an intrinsic flow, meaning that it describes the evolution of a manifold using only its Riemannian metric (recalling that this is a local description of the shape at every point). It decreases the Ricci curvature in time. This means that a surface becomes "more Euclidean" at every point as the Ricci flow evolves it. Specifically, the Ricci flow is $$\frac{dg}{dt} = -2\text{Ric}$$ meaning that the Riemannian metric itself evolves, at a rate proportional to its negative Ricci curvature. This resembles mean curvature flow, in which a manifold evolves so that its mean curvature at every point "smooths out". The difference is that the Ricci is an intrinsic flow, measured by the local metric, as opposed to MCF which is extrinsic, where curvature is measured from the ambient Euclidean space.

As for your balloon-under-pressure interpretation I'm not sure. This probably has to do with the volume measured under the Riemannian metric. You might want to check out the surface tension viewpoint of minimal surfaces (surfaces that are stationary under MCF).


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