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I have a few questions, for which I have been trying to find a reference to no avail. They are somewhat related. Please provide references if possible.

In what follows $(U,g)$ and $(V,h)$ are two n-dim Riemannian manifolds and assume that $f:U\to V$ is a diffeomorphism. My questions are of local or semi-global nature, so please assume that all sets/manifolds are homeomorphic to a disk.

The first question seems to me to be be the real content of famous Theorema Egregium. It believe it is true, but I could not find a reference for it.

1) Assume that $n=2$ and that $K_g(x) = K_h(f(x))$ for all $x\in U$. Then is it true that $f$ is an isometry? Here, $K_g(x)$ stands for the Gaussian curvature of the metric $g$ at point $x$.

The second question relates to a generalization of 1) to higher dimensions

2) Assume that the pull back of the Riemann curvature tensor of $h$ under $f$ coincides with the Riemann curvature tensor of $g$. Is $f$ an isometry?

Now I will try to express question 2) in local coordinates, for a very simple case:

3) Assume $U=V \subset {\mathbb R}^n$ and $f=id$. Assume that the Riemann curvature tensor of the two metrics $g$ and $h$ coincide pointwise. Is then $g=h$?

Please note that a similar question can be formulated for the Riemann curvature tensor of (0,4) or (1,3) types. So two questions in one question:

3.1) If $R^i_{.jkl} (g)\equiv R^i_{.jkl}(h)$ for all indices, does it follow that $g=h$? Please note that there is no constancy assumption on curvature, only point-wise identities between the curvatures of the two metrics.

3.2) The same question as in 3.1), this time assuming $R_{ijkl} (g)\equiv R_{ijkl}(h)$.

4) Finally, coming back to the simplest case I am interested in: Assume $U=V \subset {\mathbb R}^2$ and that the Christoffel symbols $\Gamma^i_{jk}(g) \equiv \Gamma^i_{jk}(h)$, as a pointwise identity. Does it follow that $f=g$?

Of course 3.1) $\implies$ 4).

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    Most of the answers to your questions are here: http://mathoverflow.net/questions/100281/does-the-curvature-determine-the-metric – Renato G. Bettiol Nov 02 '13 at 00:02
  • "homemorphic to a disk" would force $:n=2;$. $;;$ Do you mean homeomorphic to a ball? $\hspace{1.19 in}$ –  Nov 02 '13 at 00:21
  • @RickyDemer: "disk" is sometimes also used as synonim with "ball" (as in "$n$-dimensional disk") – Qfwfq Nov 02 '13 at 09:59

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The answer to all questions is "no". Example for 1), 2): both spaces are discs in the plane with the standard metric. The curvature is 0. $f$ is any diffeomorphism of the disc, which is not an isometry. Example for 3): same unit disc. One metric is standard, another is $p(x,y)(dx^2+dy^2)$, where $p$ is any harmonic function. The curvature is $0$ for both metrics.