6
$\begingroup$

Given a finite metric space $(X,d)$, when does it hold that for all $y\in X$ and $r>0$, $\#B(y,r)$ does not depend on $y$? Here $ B(y,r):=\{x\in X: d(x,y)\le r\} $ denotes a ball of radius $r$ centered at $y$, and $\#A$ denotes the cardinality of $A$.

An example that does not satisfy the above property is $Z$-distance on $\{0,1\}^n$. It is defined as $ d_Z(x,y) := \max\{\#(\mathrm{supp}(x)\setminus\mathrm{supp}(y)),\#(\mathrm{supp}(y)\setminus\mathrm{supp}(x))\} $ for $x,y\in\{0,1\}^n$. Here $ \mathrm{supp}(x):=\{i\in[n]:x_i\ne0\} $ denotes the support (set of nonzero locations) of a vector $x\in\{0,1\}^n$. (The notation $Z$-distance is perhaps nonstandard. My motivation comes from coding theory, in particular error-correcting codes for $Z$-channels which only zero out bits but do not flip zeros to ones. One can check that $Z$-distance is indeed a metric.)

All normed spaces satisfy the above property.

It is easy to see that the above property is equivalent to the condition: for all $y,z\in X$ and $r>0$, $ \#(B(y,r)\setminus B(z,r)) = \#(B(z,r)\setminus B(y,r)) $. However, I couldn't further simply it. My question is: is there, or is it possible to have, a characterization of metrics satisfying the above property?

$\endgroup$
1
  • 1
    $\begingroup$ What kind of characterization are you looking for? All homogeneous space satisfy the condition, for example any group endowed with the length metric of a system of generators. $\endgroup$ Commented Jul 27, 2020 at 17:20

1 Answer 1

9
$\begingroup$

I do not know the answer in general, but in the case in which $X$ is a subset of $\mathbb{R}^n$ there are some known remarkable results:

If $X\subset\mathbb{R}^2$ is bounded, then $X$ consists of vertices of a regular $n$-gon or $X$ is a union of the vertices of two regular $n$-gons having having the same center and radius.

If $X\subset\mathbb{R}^n$ is bounded, then $X$ is contained in a sphere.

You can find proofs of these and other results in

B Kirchheim, D. Preiss, Uniformly distributed measures in Euclidean spaces. Math. Scand. 90 (2002), no. 1, 152–160.

The proofs are surprisingly difficult and deep. In view of these results I would expect that it is difficult to find a general and satisfactory answer.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .