Does the curvature determine the metric?

Question

Here I asked the question whether the curvature deterined the metric. Since I am unfortunately completely new to Riemannian geometry, I wanted to ask, if somebody could give and explain a concrete example to me, as far as the following is concerned:

At the MO page (as cited above) I got the following answer to the question

Given a compact Riemannian manifold M, are there two metrics g1 and g2, which are not everywhere flat, such that they are not isometric to one another, but that there is a diffeomorphism which preserves the curvature? If the answer is yes: Can we chose M to be a compact 2-manifold?

On the positive side, if $M$ is compact of dimension $\ge 3$ and has nowhere constant sectional curvature, then combination of results of Kulkarni and Yau show that a diffeomorphism preserving sectional curvature is necessarily an isometry.

Concerning 2-dimensional counter-examples: First of all, every surface which admits an open subset where curvature is (nonzero) constant would obviously yield a counter-example. Thus, I will assume now that curvature is nowhere constant. Kulkarni refers to Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164, for a counter-example attributed to Stackel and Wangerin. You probably can get the book through interlibrary loan if you are in the US.

I looked up the example in Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164:

If we rotate the curve $x_3=\log x_1$ about the $x_3$-axis in space, we obtain the surface of revolution $X(u_1,u_2)=(u_2\cos(u_1), u_2\sin(u_1),\log(u_2))$, $u_2>0$. This is diffeomorphic to the helicoid $X(u_1,u_2) =(u_2\cos(u_1),u_2\sin(u_1),u_1)$.

I think, these manifolds are not compact (but I assumed compactness of the manifold in my question on MO). I don't understand, how to manipulate this example in order to get a compact manifold.

Thank you for your help.

Answer 1

Here's another example.

First, imagine a short cylinder $S^1\times [0,1]$ and a long cylinder $S^1\times [0,10^{10}]$, but with the same radii. Smoothly cap off both ends of the cylinders in the same way using spaces homeomorphic to discs.

The resulting manifolds are both homeomorphic to $S^2$, are not isometric (since one has a much larger diameter than the other), but there is a curvature preserving diffeomorphism between them.

To see this, just convince yourself there is a diffeomorphism $f:S^1\times[0,1]\rightarrow S^1\times [0,10^{10}]$ with the property that $f$ is an isometry when restricted to $[0,\frac{1}{4}]$ and $[\frac{3}{4},1]$. This condition allows you to extend $f$ to a curvature preserving diffeo of both compact manifolds.

Answer 2

Imagine a sphere with several identical bumps. More precisely, the surface given in spherical coordinates by $\rho=F(\theta,\phi)$ where $F$ is equal to one except in some places that are far one from another. To be concrete, suppose there are four bumps uniformly distributed along the equator.

You can grab one of these bumps and slide it to the North Pole, or, if you wish, arrange them at the vertices of a tetrahedron. There is a diffeomorphism that maps the initial surface onto the new one, and preserves curvature. Namely, the diffeomorphism is the isometry of corresponding bumps, extended in an arbitrary way to the rest of the sphere (where the curvature is constant).

The surfaces will not be isometric in general. For example, the diameter of the surface is greater when you have two bumps exactly opposite each other on the sphere.