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

[jostpuur]

With some assumptions the inverse \phi^{-1} can be assumed differentiable too. Then

<br /> (\nabla\phi)^T(\nabla\phi)=\textrm{id}<br />

will imply

<br /> \textrm{id} = (\nabla\phi^{-1})^T(\nabla\phi^{-1})<br />

I think the necessary pieces of puzzle are starting to be around now...

 

[jostpuur]

This condition seems to imply

<br /> \|\phi(x)-\phi(y)\|\leq \|x-y\|<br />

for all x,y.

In proof of this I used a triangle inequality that will not hold in Minkowski space. Damned thing...

Anyway, I would like to point out that the proof that micromass wrote in Hilbert spaces will also work in Minkowski type spaces. Assume that \eta is some N\times N symmetric matrix such that zero is not its eigenvalue (but the eigenvalues can be both positive and negative). If \phi:\mathbb{R}^N\to\mathbb{R}^N is a bijection that satisfies

<br /> \big(\phi(x)-\phi(y)\big)^T\eta \big(\phi(x)-\phi(y)\big) = (x-y)^T\eta (x-y),<br />

then the \phi will be an affine mapping \phi(x)=Ax+a. At least I didn't see anything that wouldn't work in these steps when they are modified: https://www.physicsforums.com/showpost.php?p=4417877&postcount=4

 

[jostpuur]

Continuity of the derivative should be enough to imply the approximation

<br /> \|\phi(x)-\phi(y)\|^2 = \|x-y\|^2 + o(\|x-y\|^2)<br />

in the limit \|x-y\|\to 0. Perhaps this infinitesimal isometry property can be extended some how.

I was doing things complicatedly here. The continuity of derivative is not needed. The result also generalizes.

Assume that \eta is some N\times N symmetric matrix. If \phi:\mathbb{R}^N\to\mathbb{R}^N is a differentiable mapping such that

<br /> (\nabla\phi(x))^T\eta (\nabla\phi(x)) = \eta<br />

for all x\in\mathbb{R}^N, then

<br /> \big(\phi(x)-\phi(y)\big)^T\eta\big(\phi(x)-\phi(y)\big) = (x-y)^T\eta (x-y) + o(\|x-y\|^2)<br />

in the limit \|x-y\|\to 0 (can be the Euclidian norm here, but it shouldn't matter). (To be rigorous, I think the limit must be interpreted so that other one of the parameters is constant in the limit.)

So the final critical gap is that how do you extend the infinitesimal isometry property to the whole space? That's something I didn't get yet.

 

[jostpuur]

Maybe I should have mentioned what I'm really trying to do. I want to prove rigorously that the isometry group of Minkowski spacetime (defined as $\mathbb R^4$ with the Minkowski metric) is the Poincaré group, defined as the set of all affine maps $x\mapsto \Lambda x+a$ such that $\Lambda^T\eta\Lambda=\eta$.

So this problem became solved in the other Mazur-Ulamn thread.

But this discussion also introduced a new problem which is that if the mapping has an infinitesimal isometry property, then the isometry property extends to finite regions and possibly the whole space. Did this become solved too? I didn't understand how it happened.

The problem I asked about came up after I proved that $\phi$ is an isometry if and only if its Jacobian $J_\phi(x)$ satisfies $J_\phi(x)^T\eta J_\phi(x)=\eta$ for all x.

This sounds like the precise problem I got stuck with. Are you sure you accomplished this?

 

[Fredrik]

Are you sure you accomplished this?

Yes, but keep in mind that I was talking about isometries in the sense of differential geometry, not in the sense of normed vector spaces. I included the proof in post #21.

So this problem became solved in the other Mazur-Ulamn thread.

Unfortunately no. Micromass' suggestion to use spacetime's exponential map is still the most promising approach. I haven't verified the details yet. I still don't even know how to define the exponential map of a Lorentzian manifold like Minkowski spacetime.

The technique that micromass suggested for Hilbert spaces works for Minkowski spacetime as well. It can be used to prove things like this: If $\Lambda:\mathbb R^4\to\mathbb R^4$ is a surjective map such that $g(\Lambda(x),\Lambda(y))=g(x,y)$ for all $x,y\in\mathbb R^4$, then $\Lambda$ is linear and its matrix representation with respect to the standard basis satisfies $\Lambda^T\eta\Lambda=\eta$. But that's not the problem I was having difficulties with.

If I could prove that "$\phi$ is an isometry" implies something like the assumption in the theorem I just stated, then I'm pretty much done. Unfortunately all I was able to prove by using the definitions in a straightforward way was that the Jacobian satisfies a condition like the one I want the function itself to satisfy.

But this discussion also introduced a new problem which is that if the mapping has an infinitesimal isometry property, then the isometry property extends to finite regions and possibly the whole space. Did this become solved too? I didn't understand how it happened.

I don't understand what you're referring to here, or what an infinitesimal isometry property is. Hm, "infinitesimal" often refers to things related to the tangent space. So maybe you meant something like an isometry in the sense of normed vector spaces, i.e. a condition like $g(\Lambda(x)-\Lambda(y),\Lambda(x)-\Lambda(y))=g(x-y,x-y)$? The problem I asked about in #1 doesn't start with a condition like that. It starts with the condition that $\phi:\mathbb R^4\to\mathbb R^4$ is such that $\phi^*g=g$, where $\phi^*$ denotes the pullback map. It's defined by $\phi^*g_{\phi(p)}(u,v)=g_p(\phi_*u,\phi_*v)$, where $\phi_*$ is the pushforward map, defined by $\phi_*v(f)=v(f\circ\phi)$. I'm using all these definitions in post #21.

 

[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
<br /> A^T\eta A = \eta.<br />

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

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

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

I know that AFF\LongleftrightarrowISO.

I know that INF\LongleftrightarrowPDE.

Also ISO\LongrightarrowINF is obvious.

So to summarise, we have \Big(AFF\LongleftrightarrowISO\Big)\underset{!}{\Longrightarrow}\Big(INF\LongleftrightarrowPDE\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?