A curve in the plane is determined, up to orientation-preserving
Euclidean
motions, by its curvature function, $\kappa(s)$.
Here is one of my favorite examples, from
Alfred Gray's book,
*Modern Differential Geometry of Curves and Surfaces with Mathematica*,
p.116:

Q1. Is there an analogous theorem stating that a surface in $\mathbb{R}^3$ is determined (in some sense) by its Gaussian curvature?

I know such a reconstruction path (curvature $\rightarrow$ surface) is needed in computer vision, and so there are approximation algorithms, but I don't know what is the precise theorem underlying this work.

Q2. Are there higher-dimensional generalizations, determining a Riemannian manifold by its curvature tensor?

I have no doubt this is all well known to the cognoscenti, in which case a reference would suffice. Thanks!

**Addendum** (*4Oct11*). Permit me to augment this question with a relevant reference
which loosens the notion of "determines" and answers my **Q1** with that notion replaced
by "find some."
The paper by Gluck, Krigelman, and Singer, entitled "The converse to the Gauss-Bonnet Theorem in PL,"
*J. Diff. Geom*, 9(4): 601-616, 1974, poses this question:

Suppose that a closed smooth two-manifold $M$ and a smooth real-valued function $K \;:\; M \rightarrow \mathbb{R}$ are given, and that one is asked to find a Riemannian metric for $M$ having $K$ as its Gaussian curvature. [...] With these restrictions on $K$ [just elided], the problem has been completely solved for all closed smooth two-manifolds by: Melvyn Berger [...], Gluck [...], Moser [...], Kazdan and Warner [...]. Recently Kazdan and Warner have obtained a uniform solution. The problem for compact two-manifolds with boundary, however, seems not to have been addressed in the smooth category.

The MathSciNet review of this paper was written by Gromov.

notthe analogue of the curvature function of a curve. The latter is an extrinsic geometric invariant (i.e., it depends on how the curve is embedded in the plane), whereas the former is an intrinsic one (it depends only on the metric on the surface itself). So unless you make rather strong global assumptions (as in Jean-Marc's answer below), the Gauss curvature will in general not determine the surface uniquely. The simplest analogue of the 1-d results is that the first and second fundamental forms uniquely determine the embedding. $\endgroup$ – Deane Yang Oct 2 '11 at 13:401more comment