I'm beginning to study Ricci flow, but I'm stumbled by a basic question: Let's restrict ourselves to a smooth 2-manifold $M$ in $\mathbb{R}^3$. Let $g$ be the Riemannian metric of $M$ induced from the Euclidean metric. Given $g$, how can we tell the shape of $M$ (i.e. an explicit equation of the surface)?

For example, in spherical coordinates, the induced metric on $S^2$ is $d\phi^2 +$ sin$^2 \phi$ $d\theta^2$. I know the length of a vector $v$ is $|v| = g(v, v)^{\frac{1}{2}}$, but I don't see how this enables me to deduce the shape. Any help is greatly appreciated.

  • 1
    $\begingroup$ Following, because I never quite got that either! $\endgroup$ – RobertTheTutor Apr 14 at 19:13
  • 2
    $\begingroup$ Along these lines, there is a deep theorem, due to Pogorelov, that the intrinsic geometry of a smooth compact positively curved surface in $R^3$ (a convex surface) uniquely determines its extrinsic shape. However, this theorem is not at all explicit. This is how math frequently works: You know that something is possible but the proof will give you no clue on how to do this. (In this case, "this" means reconstructing the extrinsic shape from the intrinsic geometry.) $\endgroup$ – Moishe Kohan Apr 14 at 19:44

From the Riemannian metric, you can calculate the curvature, which reflects, roughly speaking, how non-flat the space is. For a Riemannian 2-manifold, this is just a scalar function, namely the Gauss curvature. This tells you how curved the space is itself intrinsically.

The Riemannian metric tells you how to measure the length of a tangent vector. This in turn allows you to compute the length of a curve. By minimizing the lengths of curves joining two given points, you can define the distance between any two points. This gives the Riemannian manifold a metric space structure. If it is complete, then there will be a curve whose length is equal to the distance between the two points. That's called a length-minimizing geodesic.

The metric also allows you to define the angle between two curves that meet at a common point. You can now define the concept of a right geodesic triangle and ask whether it satisfies the Pythagorean theorem. The fundamental fact is that all geodesic right triangles satisfy the Pythagorean theorem if and only if the Gauss curvature is zero everywhere.

This is all done intrinsically, which means it's all done using only the Riemannian metric and otherwise not on how the surface looks in $\mathbb{R}^3$.

Knowing Gauss curvature does give information about what the surface looks like in $\mathbb{R}^3$ but the shape is not uniquely determined. You can see this already when $K$ is zero everywhere. Any cylindrical surface is flat because you can change the shape of an inflexible piece of paper by bending it. Right triangles drawn on the paper still satisfy the Pythagorean theorem even if you bend the paper.

If you're learning differential geometry, you have or will learn how to calculate the Gauss curvature of a sphere and the spherical version of the Pythagorean theorem. More generally, there is a Pythagorean theorem for any curved surface, where the "error term" is the integral of the Gauss curvature over the interior of the triangle. This is a version of the Gauss-Bonnet theorem for a geodesic right triangle.

  • $\begingroup$ Thank you for your detailed answer. My question comes from understanding the animations of Ricci flows (such as a sphere collapsing to the origin in finite time). Should I interpret such animation as just one of the infinitely many possible processes? Like, at any time t, the manifold may not even be a sphere, and even if it is a sphere, its center may not be at the origin? Sorry that my question shifts a bit to Ricci flow. $\endgroup$ – Joseph Apr 14 at 21:22
  • 3
    $\begingroup$ @Joseph, it seems to me that trying to understand the Ricci flow before you know the details about how to get the shape of a sphere from its Riemannian metric is premature. There are a lot of things to learn along the way before you can really understand the Ricci flow. I suggest being patient and focusing on working your way carefully through the basics of differential and Riemannian geometry $\endgroup$ – Deane Apr 14 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.