Deduce the shape of a manifold from the Riemannian metric

Question

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.

Answer 1

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.