TITLE: Surface with prescribed first fundamental form
QUESTION [5 upvotes]: Consider a Riemannian manifold $(S,g)$ of dimension 2. 
What can we say about the possibility of an isometric immersion of this surface into $\mathbb{R}^3$?
Of course this is not unique, even up to rigid motions (like in the case of zero Gaussian curvature). How can we find at least one of the possible isometries? Do we have a condition for uniqueness? 
If you want to make it simpler, think about a $U \subset \mathbb{R}^2$ with assigned $g_{ij}\neq \delta_{ij}$ (with respect to the standard Cartesian coordinates).

REPLY [4 votes]: By the Cartan-Janet theorem it is possible to locally embed a 2-dimensional Riemannian manifold isometrically into $\mathbb{E}^3$, assuming that the metric is analytic. Note that this is a special case of the Nash embedding theorem. Two-dimensional Riemannian manifolds cannot always be globally immersed or embedded. A counter-example is given by Hilbert's theorem that states that the hyperbolic plane $\mathbb{H}^2$ cannot be immersed in Euclidean three-space.
However, I believe it is not easy to find in general an immersion for a given surface and metric, since this involves solving a system of three non-linear partial differential equations 
$$\sum_{i=1}^3 \frac{\partial u_i}{\partial x_j}\frac{\partial u_i}{\partial x_k} = g_{jk}(x),$$
where $u_1,u_2,u_3$ are the components of $\mathbb{E}^3$ and $x=(x_1,x_2)$ the coordinates on $U\subset \mathbb{R}^2$.
Also, the isometric embedding is not unique in general. A simple example is to think of a sphere with a dent in it. If you invert the dent (that is, make a bulge of it), you get an isometric surface that is not congruent (i.e. not the same up to isometries of the ambient three-space $\mathbb{E}^3$).
There are however some rigidity results. A well-known one is the theorem of Cohn-Vossen: two isometric ovaloids (compact surfaces with Gauss curvature $K>0$) are congruent, and hence the embedding is essentially unique.