Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension - MathOverflow

Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension

2010-10-14T18:19:45Z

In search for a Machian formulation of mechanics I find the following problem. In Machian mechanics absolute space does not exists, and the only real entities are the relative distances between the particles. As a consequence, the configuration space of a N-particle system is the set of the distances on a set of N elements. Actually these distances are usually required to be isometrically embeddable in $\mathbb{R}^3$. But if absolute space does not exists, this requirement appears to be not appropriate. The natural generalization it therefore to admit any possible distance as physically acceptable, and to find a preferred way to derive a 3-geometry, possibly non-flat, form a generic distance.

To be more specific, consider the following simple example. Let A be a metric space with 3 elements. There are infinitely many bi-dimensional riemaniann manifolds (surfaces) is which A can be isometrically embedded. There is however a preferred embedding, namely the embedding into a plane. The existence of a preferred embedding defines a preferred value for the angles between the geodetics joining the points, which in this case are simply the angles of the triangle defined by the distance between the points.

Suppose now that A has four point. In general this metric space cannot be isometrically embedded in a 2-plane. The problem therefore is the following: is there a preferred isometric embedding of this metric space in a 2-surface, or equivalently, there is a preferred way for defining the values of the angles between the geodetics?

In more forma way, the problem is the following: is there a preferred isometric embedding of a finite metric space in a riemaniann manifold of given dimension?

Answer by Suresh Venkat for Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension

2010-10-14T20:14:20Z

This is almost certainly not what you want, but it illustrates why you need a tighter specification of 'preferred'.

Any $n$-point metric space can be embedded isometrically in $\ell_\infty^n$.

Proof: (this is a well known result): Let the $r^{th}$ coordinate of $x_j$ be the distance from $x_i$ to $x_r$. BY triangle inequality, we know that for any triple $i,j,k$, $$ d(x_i,x_j) - d(x_k, x_j) \le d(x_i, x_k) $$ which establishes the correctness of the embedding.

One interesting notion of preferred therefore might be that the target space dimension is either independent of $n$, or at the very least depends sublinearly on $n$.

Answer by Anton Petrunin for Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension

2010-10-14T23:58:34Z

Your question is not well stated. In particular I did not understand why embedding into a plane is "preferred embedding".

Here are some associations...

4-point case.

A generic 4-point metric space can be isometrically embedded into two different model planes (i.e. simply connected surfaces of constant curvature $K$, which is eiter sphere, Euclidean plane or Lobachevsky plane depending on sign of $K$).

Thus you have two values of curvature $K_1\le K_2$ associated to (almost) any 4-point metric space. In this case the metric space can be isometrically embeded into a model 3-space of curvature $K_1\leq K\leq K_2$.

In fact for any 4-point metric space $M$ there is an subinterval $\mathbb I_M$ of $[-\infty,\infty)$ such that $M$ can be isometrically embedded into a model 3-space of any curvature $K\in \mathbb I_M$. (We assume that model space of curvature $-\infty$ is an $\mathbb R$-tree.)

Nearly all this was discovered by A. Wald in 1936 or so.