Integration of surface from compatible first and second fundamental forms

Question

Let $\Omega\subset\mathbf{R}^2$ be some simply connected domain. Given the functions $(E,F,G)$ and $(L,M,N)$ that satisfy the Gauss–Codazzi–Mainardi equations, we have, according to Bonnet theorem, a unique surface up to rotations and translations in $\mathbf{R}^3$. After an extensive search I have not found a single example where the integration of a surface, that is the obtaining the function $\mathbf{r}(x,y)\in\mathbf{R}^3$ from the fundamental form coefficients, is done $\textit{from scratch}$. There are examples where one a priori wants a surface of revolution and assumes a certain special form of $\mathbf{r}(x,y)$ and then proceeds, but in general one has no clue about how the surface will look like. So

How to integrate $\mathbf{r}(x,y)$ from the (GMC satisfying) fundamental forms coefficients $(L,M,N)$ and $(E,F,G)$, without assuming any special form of $\mathbf{r}(x,y)$ ? Are there analytical, non-trivial examples (beyond surfaces of revolution or any other symmetric shapes).

Answer 1

The Bonnet theorem indeed guarantees the existence of a surface given the first and second fundamental forms subject to the Gauss-Codazzi-Mainardi equations. However, the explicit integration to obtain the surface in terms of the position vector $\mathbf{r}(x, y)$ can be quite challenging and in general does not have a simple analytical form, especially for non-trivial cases beyond symmetric shapes like surfaces of revolution.

As you probably know, the coefficients of the first and second fundamental forms contain information about the metric (lengths, angles) and the curvature (shape) of the surface, respectively. The first fundamental form coefficients $E$, $F$, and $G$ are given by dot products of the tangent vectors to the surface, while the second fundamental form coefficients $L$, $M$, and $N$ come from dot products of the normal vectors.

In order to recover the surface, one essentially needs to integrate these forms. The integrability conditions (Gauss-Codazzi-Mainardi equations) ensure that this process is locally possible, but it doesn't tell us how to do it globally or provide an easy method for finding the integrated form.

The general procedure would be to find the tangent vectors to the surface, which can be obtained from the first fundamental form, and then integrate them to obtain the surface. But this process is often non-trivial and can be challenging due to the non-linear nature of the integrals involved.

Here is a rough sketch of how the process could go:

1. Start with the first fundamental form, which gives you the metric on the surface in terms of $E$, $F$, and $G$. This gives you the dot products of the tangent vectors $E = \mathbf{r}_x \cdot \mathbf{r}_x$, $F = \mathbf{r}_x \cdot \mathbf{r}_y$, and $G = \mathbf{r}_y \cdot \mathbf{r}_y$.

2. Solve these equations for the tangent vectors $\mathbf{r}_x$ and $\mathbf{r}_y$ as much as possible. This step can involve some ambiguity in the choice of solution due to the square roots that typically appear.

3. Once you have the tangent vectors, integrate them to obtain the position vector of the surface $\mathbf{r}(x, y)$. This is the step that is typically challenging due to the non-linear nature of the integrals, and in general an explicit solution may not be possible.

Unfortunately, the general case of this problem is notoriously difficult and typically requires the use of computer algebra systems for any chance of a solution. However, it's a beautiful illustration of the interplay between local and global properties in differential geometry. In many cases, the best we can do is to study the surface locally (i.e., in a small neighborhood of a point) rather than trying to find an explicit global parameterization.

Answer 2

Buried in the proof of this is the sequence of steps needed to construct the embedding. Here is one way to do it. Let $r(u^1,u^2)$ be the embedding we want to solve for, given a first fundamental form $$ g_{ij}\,du^i\,du^j $$ and second fundamental form $$ h_{ij}\,du^i\,du^j. $$ You can check that the equations satisfied by $r$ can be written as follows: \begin{align*} \partial_ir\cdot\partial_jr &= g_{ij}\\ \partial^2_{ij}r &= \Gamma_{ij}^k\partial_kr + h_{ij}n, \end{align*} where $\Gamma^k_{ij}$ are the Christoffel symbols of $g$ and $$ n = \frac{\partial_1r\times\partial_2r}{|\partial_1r\times\partial_2r|}. $$ We rewrite this as a system of first order PDEs for unknown vector-valued functions $r, p_1, p_2$: For all $1 \le i, j \le 2$, \begin{align*} \partial_ir &= p_i\\ \partial_ip_j &= \Gamma_{ij}^kp_k + h_{ij}\frac{p_1\times p_2}{|p_1\times p_2|} \end{align*}

Let's assume that we want to solve this in a neighborhood of $(u^1,u^2) = (0,0)$. Choose any values you want of $r(0,0), p_1(0,0), p_2(0,0)$ such that $$ p_i(0,0)\cdot p_j(0,0) = g_{ij}(0,0). $$ Along the line $u^2=0$, solve the following system of ODEs: \begin{align*} \partial_1r &= p_1\\ \partial_1p_1 &= \Gamma_{11}^1p_1 + \Gamma_{11}^2p_2 + h_{11}\frac{p_1\times p_2}{|p_1\times p_2|}\\ \partial_1p_2 &= \Gamma_{12}^1p_1 + \Gamma_{12}^2p_2 + h_{12}\frac{p_1\times p_2}{|p_1\times p_2|}. \end{align*} Then, for each value of $u^1$, solve the following ODEs: \begin{align*} \partial_2r &= p_2\\ \partial_2p_1 &= \Gamma_{21}^1p_1 + \Gamma_{21}^2p_2 + h_{21}\frac{p_1\times p_2}{|p_1\times p_2|}\\ \partial_2p_2 &= \Gamma_{22}^1p_1 + \Gamma_{22}^2p_2 + h_{22}\frac{p_1\times p_2}{|p_1\times p_2|} \end{align*} This gives you vector-valued functions $r(u^1,u^2), p_1(u^1,u^2), p_2(u^1,u^2)$. However, only a subset of the full system of equations was used to find these functions. The proof of the theorem requires checking that this process really does give a solution to the full set of equations. That's where the integrability conditions are needed.

So you can reconstruct the embedding by solving the two systems of ODEs given above. I'm not sure how well this will work numerically.