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I've been scratching my head for quite some time on how one could find the surface embedded in 3D flat space that induces a given 2x2 metric. Usually one asks the opposite question: given a surface, find the corresponding metric (e.g. Finding the metric of a surface embedded in $\mathbb{R}^3$). But I'm interested in the inverse problem. To my surprise, almost nothing can be found online about this problem.

I can write down the partial differential equations that the parametrization of the surface should satisfy, but I am missing the boundary conditions. Can anyone help me with this?

Here is my reasoning: first suppose you have a 2x2 metric, so that $ds^2 = m_{\alpha\alpha}(\alpha,\beta) d\alpha^2 + m_{\alpha\beta}(\alpha,\beta) d\alpha d\beta + m_{\beta\beta}(\alpha,\beta) d\beta^2$. Now we claim that there exists a surface in flat 3D space that would induce this metric. Such a surface is parametrized by \begin{align} x &= f_x(\alpha,\beta)~,\\ y &= f_y(\alpha,\beta)~, \\ z &= f_z(\alpha,\beta)~. \end{align} From the constraints above it is quite direct to find \begin{align} dx &= \frac{\partial f_x(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_x(\alpha,\beta)}{\partial \beta}\partial \beta~,\\ dy &= \frac{\partial f_y(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_y(\alpha,\beta)}{\partial \beta}d\beta~, \\ dz &= \frac{\partial f_z(\alpha,\beta)}{\partial \alpha}d\alpha + \frac{\partial f_z(\alpha,\beta)}{\partial \beta}d\beta~. \end{align} Since we are in euclidean $\mathbb{R}^3$ we have $ds^2 = dx^2 + dy^2 + dz^2$. By inserting these constraints in the length element we find a set of 3 differential equations for 3 unknown functions so that the parametrization above induces the metric we had at the start \begin{align} \left(\frac{\partial f_x(\alpha,\beta)}{\partial \alpha}\right)^2 + \left(\frac{\partial f_y(\alpha,\beta)}{\partial \alpha}\right)^2 + \left(\frac{\partial f_z(\alpha,\beta)}{\partial \alpha}\right)^2 &= m_{\alpha\alpha}(\alpha,\beta)~,\\ \frac{\partial f_x(\alpha,\beta)}{\partial \alpha}\frac{\partial f_x(\alpha,\beta)}{\partial \beta} + \frac{\partial f_y(\alpha,\beta)}{\partial \alpha}\frac{\partial f_y(\alpha,\beta)}{\partial \beta} + \frac{\partial f_z(\alpha,\beta)}{\partial \alpha}\frac{\partial f_z(\alpha,\beta)}{\partial \beta} &= m_{\alpha\beta}(\alpha,\beta)~,\\ \left(\frac{\partial f_x(\alpha,\beta)}{\partial \beta}\right)^2 + \left(\frac{\partial f_y(\alpha,\beta)}{\partial \beta}\right)^2 + \left(\frac{\partial f_z(\alpha,\beta)}{\partial \beta}\right)^2 &= m_{\beta\beta}(\alpha,\beta)~. \end{align}

To solve these PDEs one would need boundary conditions of the type $f_x(\alpha,0)=g_x(\alpha)$, but I don't know how I should find this $g_x$ that defines the boundary condition for $f_x$ (of course, the same is true for the other boundary conditions).

Any help in how to continue this problem is greatly appreciated, same for any correction or different approach to solve this.

Alberto Rolandi
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    Not an answer, but this paper with a former student X. Wang, Clairaut Surfaces in Euclidean Three-Space, Tôhoku Math. J., 74 no. 2 (2022) 215-227, addresses the problem for metrics of the form $E(u),du^2+G(u),dv^2$. Loosely, we show that if an isometric embedding (in Euclidean 3-space) exists for this type of metric, there's an infinite-dimensional space of boundary conditions also giving isometric embeddings, which amount to specifying two components of the second fundamental form along a $u$-parameter curve. – Andrew D. Hwang Nov 02 '23 at 14:40
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    An isometric embedding/immersion in 3d Euclidean space is not always possible if you require $C^2$-smoothness. – Moishe Kohan Nov 02 '23 at 14:47
  • @MoisheKohan this is interesting, and i read that at some point. But from the above equations it seems that since we have 3 unknown functions and 3 differential equations there should always be a solution (except in some particularly pathological cases). Could you explain a little more? I don’t see what i am missing. – Alberto Rolandi Nov 05 '23 at 15:35
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    This is a very nasty and poorly understood system of PDEs. It can be rewritten as a scalar Monge-Ampère equation. If the Gauss curvature is strictly positive, then this equation is elliptic, and there is hope that there is a well posed boundary value problem. But, as far as I can recall, nothing this good is known. If the Gauss curvature is negative, then it is a nonlinear hyperbolic PDE, so there is potentially a well-posed initial value problem. If $K$ is neither always positive nor always negative, then even less is known. – Deane Nov 05 '23 at 16:21
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    An excellent survey is this: https://bookstore.ams.org/surv-130 – Deane Nov 05 '23 at 16:24
  • Counting the numbers of unknown and of equations is very naive except for the real analytic case when you get a local solution using the Cauchy- Kovalevsky theorem. For non-existence of a global solution, think of the Hilbert theorem or the fact that the flat torus admits no isometric immersions. – Moishe Kohan Nov 05 '23 at 16:34
  • @MoisheKohan, even the real analytic case is quite nontrivial. The system as written above cannot be solved using Cauchy-Kovalevsky. Some nontrivial work is needed to transform the system into an equivalent PDE that can be solved using Cauchy-Kovalevsky. – Deane Nov 05 '23 at 19:54

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