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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.

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.

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. In particular, there are two possible solutions to each $|\partial_i f|^2 = 1$ along $\mathbb{W}$. So in fact, given $f$ along $\mathbb{W}$, there are exactly $2^n$ possible ways to extend $f$.

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.

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.

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. In particular, there are two possible solutions to each $|\partial_i f|^2 = 1$ along $\mathbb{W}$. So in fact, given $f$ along $\mathbb{W}$, there are exactly $2^n$ possible ways to extend $f$.