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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 $(\*)$.

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When you say "taking one metric to the other" you do not mean that to happen on any open set, do you? –  Mariano Suárez-Alvarez Jan 20 '10 at 23:57
    
Nope! I edited my question to clarify. –  Tom LaGatta Jan 21 '10 at 0:16
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3 Answers 3

up vote 10 down vote accepted

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.

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I'm going to try to explain the following claim, which I imagine is a theorem of Riemann: if $g$ is a Riemannian metric on $V:=\mathbb{R}^n$, the diffeomorphism-invariant information contained in the 2-jet of $g$ at $0$ is precisely what the curvature tensor $Riem$ (at $0$) sees. The case $n=2$ is, as I understand it, just what you wanted to verify.

First, let's look at the $0$th order term $g(0)$, modulo 1-jets of diffeomorphisms fixing $0$. Linear algebra tells us that we can make a linear transformation so that $g(0)=I$ (moreover, we have an $O(V)$'s worth of choices). Now let's try to arrange that $g=I+O(|x|^2)$. We'll do this by acting by 2-jets of diffeomorphisms $\phi(x)_i=x_i+a_i^{jk}x_jx_k$ to kill the $O(|x|)$ term in $g$. Well, the first order term of $g$ is in $Sym^2(V^\ast)\otimes V$ (because you look at $\partial g_{ij}/\partial x_k$) while the second order term of $\phi$ is in $V\otimes Sym^2(V^\ast)$. The diffeos act changing the 1st order term (according to my calculation) by $2(a_j^{ik}+a_k^{ij})$. The formula $a_i^{jk}\mapsto 2(a_j^{ik}+a_k^{ij})$ defines a surjective linear map $V^\ast\otimes Sym^2(V)\to Sym^2(V)\otimes V^\ast$, and that does the job. The linear map is then also injective, so we had no choices in this step.

The quadratic term of $g$ lies in $Sym^2(V^\ast)\otimes Sym^2(V)$, and we act by 3-jets of diffeos whose cubic term is in $V\otimes Sym^3(V^\ast)$ (because of the symmetry among the third-order partials). So the diffeomorphism-invariant part of the 2-jet of $g$ is the cokernel of a certain linear (symmetrization) map $$ f\colon V \otimes Sym^3(V^\ast) \to Sym^2(V)\otimes Sym^2(V^\ast). $$ The class of $g$ in $coker(f)$ is essentially the curvature at $0$. Indeed, $f$ turns out to be injective, so $coker(f)$ has dimension $$ \frac{1}{4}n^2(n+1)^2 - \frac{1}{6}n^2(n+1)(n+2) = \frac{1}{12}n^2(n^2-1): $$ precisely the dimension of the space of Riemann curvature tensors, as computed by looking at the standard symmetries. I've described this space as a representation of $O(V)$.

I think this approach might be in Spivak's multi-volume book. I heard it from Donaldson, and I hope I haven't mangled it too badly.

Edit: as Deane notes, the coordinates I've described are geodesic normal coordinates. I like the method I've sketched because it requires no technology; one doesn't even need to define a connection. The coordinate change (up to third order) is found by linear algebra, not solving ODE.

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Spivak probably comments in passing something along the lines of "in the years of yore, when students where familiar with classical invariant theory, everybody found this obvious" or something like that :P –  Mariano Suárez-Alvarez Jan 21 '10 at 2:00
    
Your argument is essentially correct (it might be totally correct but I haven't read every word). The act of changing from the original co-ordinates to geodesic normal co-ordinates performs the exact normalization that you are describing. Geometric arguments tell you exactly how to normalize. –  Deane Yang Jan 21 '10 at 2:01
    
@Tim: Since the goal is just a normal form up to second order at one point, you don't need to solve the ODE completely. You can simply write down the equations and solve them formally up to second order. The geometry (i.e., the Jacobi equation) provides a nice guide on how to do what you describe in an explicit way. Still, the approach and perspective you describe above is also very important and useful in certain contexts (which presumably is why Donaldson mentioned it). –  Deane Yang Jan 21 '10 at 15:21
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As far as I understand, your question and its natural generalizations belongs to the very developed theory (now it is a part of singularity theory, I think). This theory was started by Tresse:

Tresse, A., Sur les Invariants Differentiels des Groupes Continus des Transformations, Acta Mathematica, 1894, vol.18, pp.1-88.

See also an introduction and references to S. Dubrovskiy's paper "Moduli space of symmetric connections" http://arxiv.org/abs/math/0112291

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Thank you for the reference, Petya! –  Tom LaGatta Mar 7 '10 at 21:40
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