Does the curvature determine the metric for all surfaces

In order to not make things even more confusing than they are, I split my two-in-one question into two parts. Here's the second part (the first part is here):

Here I asked the question whether the curvature deterined the metric.

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?

For concrete 2-dimensional counter-example see page 328 of Kulkarni's paper "Curvature and metric", Annals of Math, 1970. Weinstein's argument there shows that every Riemannian surface provides a counter-example (using flow orthogonal to the gradient of the curvature). Here is the Weinstein's argument: Consider a Riemannian surface $(S,g)$ and let $K$ denote the curvature function. Let $X$ be a nonzero vector field on $S$ orthogonal to the gradient field of $K$. (Such $X$ always exists.) Now, consider the flow $F_t$ along $X$. Then $F_t$ will preserve the curvature but will not be (for most metrics $g$ and vector fields $X$) isometric. For instance, $F_t$ cannot be isometric if genus of $S$ is at least $2$ or if $X$ has more than 2 zeros.

Since I am unfortunately completely new to Riemannian geometry, I wanted to ask, if somebody could explain that to me in detail, maybe with the help of a concrete example. Also I don't understand, why this actually answers my question, since it seems that we have only one metric (and not two) in Weinstein's counter-example for calculating the gradient.

Thank you very much.

• "it seems that we have only one metric". But you have two: the original one, and its pullback under $F_t$ (for any fixed $t>0$). – user31373 Jun 24 '12 at 16:57
• @Leonid: Thank you for your comment. – Bernhard Boehmler Jun 24 '12 at 18:48