I would like to know the reasons why the existance of Chebyshev net in 3D-case is problematic.
This question boils down to the PDE described below.
(I do not know much about PDEs, so feel free to say something trivial.)


Set
$$\mathbb W^2 =\{\,(x,y,z)\in \mathbb R^3\mid x+y+z=0\,\}.$$
Given a smooth map $f: \mathbb W^2\to \mathbb R^3$ which is $C^\infty$-close
to the identity map,
I need to extend $f$ to a neighborhood of $\mathbb W^2$ so that it satisfies the following three equations:
$$\left|\frac{\partial f}{\partial x}\right|^2=\left|\frac{\partial f}{\partial y}\right|^2=\left|\frac{\partial f}{\partial z}\right|^2=1.$$


**Comments.**


* The 2D-case has a solution, this is due to Chebyshev and it is more than 100 years old.
* In general, 3D Chebyshev net is a solution of 3 equasions as above where $f$ maps a domain in $\mathbb R^3$ in a 3-dimensional Riemannian manifold and $|{\*}|$ is defined by its metric. I guess this case is just as hard as the special case described above.
* Assuming that $x+y+z>0$ and $f(x,y,z)$ is defined,
it is easy to see that $f(x,y,z)$ depends only on the values
$f(x',y',z')$ for $(x',y',z')$ in the triangle $\Delta\subset\mathbb W^2$ defined by the inequalities $$x'\le x,\ y'\le y,\ z'\le z.$$



Chebyshev net in 3D
The question asked is a special case of the following more general one: Let
$$
\mathbb{W} = \{x^1 + \cdots + x^n = 0\} \subset \mathbb{R}^n.
$$
Given an $n$-dimensional Riemannian manifold $M$ and a smooth map $f: \mathbb{W} \rightarrow M$, can $f$ be extended to a map $f: \mathbb{R}^n \rightarrow M$ such that
$$
|\partial\_1f|^2= \cdots = |\partial\_nf|^2 = 1\ ?
$$
Here, note that $\partial\_if(x) \in T\_{f(x)}M$, and the norm is taken with respect to the Riemannian metric on $M$.


The first remark, which I'm sure Anton and Robert already know, is that if $f$ happens to be an embedding (which is the case in the original question), then we can view it as a choice of all $n$ co-ordinate functions along a hypersurface in $M$ and the question is whether these functions can be extended to co-ordinate functions where the diagonal elements of the metric tensor written with respect to these co-ordinates are identically equal to $1$.


To determine what kind of PDE this system is, we can linearize it. Let $\dot{f}$ denote an infinitesimal variation of $f$. The linearized system is given by
$$
2\partial\_if\cdot\partial\_i\dot{f} = \dot{h}\_i.
$$
If $f$ is an immersion, then $\partial\_1f(x), \ldots, \partial\_nf(x)$ are always a basis of $\mathbb{R}^n$. Therefore, solving the linearized system for $\dot{f}$ is equivalent to solving for the functions
$$
u\_1 = \partial\_1 f\cdot \dot{f}, \ldots, u\_n = \partial\_nf\cdot \dot{f}.
$$
``Differentiating by parts'', the linearized system can be rewritten as
$$
2\partial\_i u\_i - g^{jk}u\_j\cdot\nabla\_i\partial\_kf = \dot{h}\_i.
$$
Note that the top order terms fully uncouple into ODE's in each of the co-ordinate directions. The coupling that occurs in the zero-th order terms prevents this system from being purely a system of ODE's. However, it is easy to check that this system is indeed a symmetric hyperbolic system, where initial smooth data posed on $\mathbb{W}$ can be extended to a unique smooth solution on all of $\mathbb{R}^n$. You can find more about this in a paper I wrote with DeTurck (Duke Math. J. Vol.51, No. 2
(1984), 243-260).


Using appropriate regularity estimates and the inverse function theorem with the appropriate Banach norm, this implies existence and uniqueness of a solution to the original nonlinear system on a tubular neighbhorhood of $\mathbb{W}$. I do not know whether there is a global solution; this requires a more careful analysis of this particular system.


ADDED: Actually, since the system is fully nonlinear, you cannot use the Banach space implicit function theorem to solve the system as given. You can, however, do one of two thing: Either "prolong" (which means roughly differentiate the system and add the partial derivatives of $f$ as unknown functions) the system into a quasilinear system or just use the Nash-Moser iteration argument. Either way, you still get what I describe above.


ADDED (in response to Robert's comment): Robert is right that if you specify $f$ along $\mathbb{W}$, the extension of $f$ is *not* uniquely determined, due to the full nonlinearity. Robert indicates that there are exactly 2 choices when $n = 3$. You have to do the linear algebra carefully along $\mathbb{W}$ to see this and count the number of possibilities in arbitrary dimension. I haven't done this yet.


ADDED: If along $\mathbb{W}$, you set $v = \frac{1}{n}(\partial\_1 f+\cdots \partial\_n f)$ and $u\_i = \partial\_if - v$, then the $u\_1, \dots, u\_n$ are tangential derivatives of $f$ along $\mathbb{W}$ and therefore given by the initial data. These vectors along $\mathbb{W}$ satisfy the equations
$$
u\_i\cdot v = \frac{1}{2}[1 - |u\_i|^2 - \frac{1}{n}(1 - \sum\_i |u\_i|^2)]
$$
$$
v\cdot v = \frac{1}{n}(1 - \sum\_i |u\_i|^2)
$$
Clearly, a necessary condition for a solution is that
$$
|u\_1|^2 + \cdots + |u\_n|^2 \le 1.
$$
If strict inequality holds on $W$, I believe that there are always exactly two solutions on $W$. These translate into a corresponding inequality that the tangential derivatives of $f$ along $\mathbb{W}$ must satisfy. And if strict inequality holds, then there are exactly two ways to extend $f$ onto a tubular neighborhood of $\mathbb{W}$ such that $f$ defines a Chebyshev net.


