Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:53:47Z http://mathoverflow.net/feeds/question/42192 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42192/preferred-embedding-of-finite-metric-spaces-in-riemaniann-manifolds-of-given-dime Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension Bruno Galvan 2010-10-14T18:19:45Z 2010-10-14T23:58:34Z <p>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. </p> <p>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. </p> <p>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?</p> <p>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?</p> http://mathoverflow.net/questions/42192/preferred-embedding-of-finite-metric-spaces-in-riemaniann-manifolds-of-given-dime/42203#42203 Answer by Suresh Venkat for Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension Suresh Venkat 2010-10-14T20:14:20Z 2010-10-14T20:14:20Z <p>This is almost certainly not what you want, but it illustrates why you need a tighter specification of 'preferred'. </p> <blockquote> <p>Any $n$-point metric space can be embedded isometrically in $\ell_\infty^n$.</p> </blockquote> <p><strong>Proof</strong>: (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. </p> <p>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 $n$. </p> http://mathoverflow.net/questions/42192/preferred-embedding-of-finite-metric-spaces-in-riemaniann-manifolds-of-given-dime/42233#42233 Answer by Anton Petrunin for Preferred embedding of finite metric spaces in riemaniann manifolds of given dimension Anton Petrunin 2010-10-14T23:58:34Z 2010-10-14T23:58:34Z <p>Your question is not well stated. In particular I did not understand why embedding into a plane is "preferred embedding".</p> <p>Here are some associations...</p> <p><strong>4-point case.</strong></p> <p>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 $K$).</p> <p>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 $K_1\leq K\leq K_2$. </p> <p>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.)</p> <p>Nearly all this was discovered by A. Wald in 1936 or so.</p>