Re: embeddability of a finite metric space in a Riemannian manifold

From: Daniel Asimov (asimov.nospam_at_msri.org)
Date: 07/23/04 

Date: 23 Jul 04 11:14:40 -0400 (EDT)

Edwin Clark wrote

<<
... has anyone attempted to find out which finite metric spaces
might be embedded (isometrically) in which Riemannian manifolds (with
the metric given by the length of a shortest curve joining two
points)? For example, which finite metric spaces can be embedded in a
genus 0 Riemannian 2-dimensional manifold? All of them?

I am posting this question for Michael Rieck who recently raised the
question.
>>

Here's one metric space (F,d) that can't be isometrically embedded in
any complete Riemannian manifold (and probably not in any incomplete
ones either):

Let F = {x,y,z,w} with
d(x,y) = d(y,z) = d(z,x) = 2, and
d(x,w) = d(y,w) = d(z,w) = 1.
---------------------------------

Proof:
Assume that (F,d) embeds isometrically in some complete Riemannian
manifold M. Denote by bc some shortest geodesic between any points b
and c of M.

Consider wx, wy, wz, each of length 2. At least two of these -- WLOG
say wx and wy -- must meet at w with an angle < pi. Let S denote a
sphere in M with center w and radius eps > 0.

Then for sufficiently small eps, S intersects wx and wy in one point
each, say p and q, respectively. After possibly making eps smaller so
that an eps-neighborhood of w is sufficiently close to Euclidean, we
see that by following wx = xw from x only up to p, then going along
pq, and finally along wy only from q to y ... we trace a path from x
to y of length < 1+1 = 2 -- contradiction.
---------------------------------------------------------------

By contrast, MathWorld states that there is a metric u on R, inducing
the usual topology, such that any finite metric space embeds
isometrically in (R,u). It refers to this paper for a proof:

Ovchinnikov, S. "Universal Metric Spaces According to W. Holsztynski."
13 Apr 2000. http://arXiv.org/abs/math.GN/0004091/.