Let $g$ and $g'$ be two $C^2$-smooth Riemannian metrics defined on neighborhoods $U$ and $U'$ of $0$ in $\mathbb R^2$, respectively. Suppose furthermore that the scalar curvature at the origin is $K$ under both metrics.
My question: Is there a coordinate transformation taking one metric to the other, such that they agree up to second derivatives at the origin? i.e., if $x : U \to U'$ is the transformation, we have
$g_{ij}' = g_{ab} ~x_i^a x_j^b$,
evaluating everything at $0$; there are similar equations for the first and second derivatives. Clearly this is false if the scalar curvatures aren't equal. I don't care what happens away from the origin.
In the excellent thread When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?, Greg Kuperberg says:
If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same".
If "local isometry" means that the metrics are equal on a neighborhood of the origin, then the metrics I have in mind are not locally isometric, since the only information I have is that their curvatures match at one point.
Edit: I'm pretty sure that Deane answered my question, but let me clarify. Let $g_{ij}$ be some "reasonable" metric, e.g. a bump surface metric, and consider a point $p$ where the scalar curvature is $K$. Let $g_{ij}'$ be an arbitrary metric on a neighborhood $U$ of the origin in $\mathbb R^2,$ with scalar curvature $K$ at $0$.
Then the question becomes: does there exist a coordinate change on the bump surface such that the equation $g_{ij}'(0) = g_{ab}(p) ~x_i^a x_j^b$ is satisfied, as well as the corresponding equations for the first and second derivatives? That is, there are $18$ pieces of pertinent information
$(\*)~~~g_{11}', g_{12}', g_{22}'; g_{11,1}', g_{12,1}', g_{22,1}', g_{11,2}', g_{12,2}', g_{22,2}'; g_{11,11}', g_{12,11}', g_{22,11}', g_{11,12}', g_{12,12}', g_{22,12}', g_{11,22}', g_{12,22}', g_{22,22}'$.
I want to change coordinates on my nice surface such that the metric and its derivatives line up with $(\*)$.
Best Answer
The answer is yes. Just use geodesic normal (also known as exponential) co-ordinates. If you have a book or two on Riemannian geometry, just look for that or a discussion of the exponential map.
[ADDITIONAL COMMENT] For a 2-dimensional metric, it's a nice exercise to figure all of this out using Jacobi fields. In fact, in my opinion, the best way to work with and understand the exponential map (which is a natural parameterization of all radial geodesics emanating from a point) is via Jacobi fields and the Jacobi equation. For example, it leads to an easy proof of a standard result, namely that the coefficients of the Taylor series of the exponential map at the origin contain only the Riemann curvature tensor and its covariant derivatives. The answer to your question follows from this theorem.