Determine the function from a simple condition on its Jacobian matrix.

[jostpuur]

Here's a summary of what I know and don't know. I assume that $\eta$ is some $N\times N$ symmetric matrix, which does not have zero as its eigenvalue. The mapping $\phi$ always means some mapping $\mathbb{R}^N\to\mathbb{R}^N$.

Definition: The mapping $\phi$ has a property AFF if it can be written in form $\phi(x)=Ax+a$ with some matrix $A$ and a vector $a$, and also
$$A^T\eta A = \eta.$$

Definition: The mapping $\phi$ has a property ISO if it satisfies
$$\big(\phi(x)-\phi(y)\big)^T\eta \big(\phi(x)-\phi(y)\big) = (x-y)^T\eta (x-y)$$
for all $x,y$.

Definition: The mapping $\phi$ has a property INF if it satisfies
$$\big(\phi(x+h)-\phi(x)\big)^T \eta\big(\phi(x+h)-\phi(x)\big) = h^T\eta h+o(\|h\|^2)$$
in the limit $h\to 0$ for all $x$.

Definition: The mapping $\phi$ has a property PDE if it satisfies
$$\big(\nabla\phi(x)\big)^T\eta\big(\nabla\phi(x)\big) = \eta$$
for all $x$.

I know that AFF$\Longleftrightarrow$ISO.

I know that INF$\Longleftrightarrow$PDE.

Also ISO$\Longrightarrow$INF is obvious.

So to summarise, we have $\Big($AFF$\Longleftrightarrow$ISO$\Big)\underset{!}{\Longrightarrow}\Big($INF$\Longleftrightarrow$PDE$\Big)$.

The final mystery to me is that how to replace the arrow marked with exclamation marks into an equivalent symbol, pointing in both directions.

 

[jostpuur]

So there was some confusion about the precise meaning of isometry. But I guess we are still wondering the same problem.

Let $\Omega\subset\mathbb{R}^2$ be some bounded open set containing the origin.

Suppose $\mathcal{L}_0$ is a collection of lines such that none of the lines in this collection intersect each other in $\Omega$, and also for every point $x\in\Omega$, there exists some line going through it.

In other words $L_1,L_2\in\mathcal{L}_0$, $L_1\neq L_2$ implies $L_1\cap L_2\cap \Omega = \emptyset$, and $\forall x\in\Omega\;\exists L\in\mathcal{L}_0$ such that $x\in L$.

First question. Are all lines in $\mathcal{L}_0$ parallel to each other neccessarily? The answer is no. For example, the lines could be of the form $\{(R-t\cos(\theta),t\sin(\theta))\;|\;t\in\mathbb{R}\}$ with some constants R (large) and $\theta$. The lines intersect in point $(R,0)$, but it is outside $\Omega$.

Suppose $\mathcal{L}_1$ is collection of curves such that for every point in $\Omega$ there exists a curve in $\mathcal{L}_1$ such that the curve goes through the point, and also that the curves do not intersect each other inside $\Omega$. Also let's assume that when a curve of $\mathcal{L}_1$ intersects a line of $\mathcal{L}_0$ the intersection is always perpendicular.

Continuing the previous example, the curves of $\mathcal{L}_1$ would need to be of the form $\{(R-t\cos(\theta),t\sin(\theta))\;|\;\theta\in\mathbb{R}\}$ with some constants $t$. So inside $\Omega$ these curves are pieces of circles, with center in $(R,0)$.

Final piece: Let's insist, that the curves of $\mathcal{L}_1$ must be straight lines too. Is it now neccessary, that lines in $\mathcal{L}_0$ are parallel to each other, and also lines in $\mathcal{L}_1$ are parallel to each other? I think the answer is yes. But how do you prove this?