A metric space $M$ is metrically homogeneous if for every pair of points $x, y \in M$ there is an isometry $f$ of $M$ onto $M$ such that $f(x)=y$. What is known about metrically homogeneous spaces? Any references? In particular, is there an explicit description of all metrically homogeneous subspaces (under the Euclidean metric) of the plane $\mathbb{R}^2$?

Remark. The same question in one dimension has an easy answer: A subset $M$ of $\mathbb{R}^1$ is metrically homogeneous if and only if $M$ is the union of two cosets $G+a\ $ and $G+b\ $ of an additive subgroup $G$ of $\mathbb{R}^1$.

  • 2
    $\begingroup$ I do not know about subsets of $R^2$, but many results on metrically homogeneous spaces follow from the solution of Hilbert's 5th problem, see e.g. arxiv.org/pdf/0908.4205v2.pdf . However, these results typically assume much more regularity about the metric space than you would have for arbitrary subsets of the plane. $\endgroup$ – Misha Jul 11 '13 at 22:40
  • 1
    $\begingroup$ No, sorry, older! $\endgroup$ – Włodzimierz Holsztyński Jul 12 '13 at 1:18
  • 1
    $\begingroup$ @WK, Leonard M. Blumenthal (or someone younger?) consider homogeneous spaces in one of a chapter. The title had something like "G-spaces", like geodesic spaces. (Also Borsuk considered this problem; S.Spież considered a related paper, where he proved that S×S cannot be homogeneous). $\endgroup$ – Włodzimierz Holsztyński Aug 24 '13 at 17:38

Here is an description of homogeneous subsets of ${\mathbb R}^2$ using group theory, similar to the one you have in dimension 1. Let $M\subset {\mathbb R}^2$ be homogeneous and let $G< I=Isom({\mathbb R}^2)$ be its group of isometries. Then $G$ fits into short exact sequence $$ 1\to T\to G\to S \to 1 $$ where $T$ is a group acting by translations on the plane and $S$ is a subgroup of $O(2)$. Unlike in 1-dimensional case, this sequence need not split, but you can still find a subset $C\subset G$ of coset representatives of $S$ so that $G=TC$ (I am using action on the left). Then $M=G\cdot m$ for some $m\in M$ and, hence, $$ M= T\cdot (Cm), $$ i.e., $M$ is a disjoint union of $T$-orbits, where $T$ is a subgroup of ${\mathbb R}^2$. This is an analogue of the description that you liked in the 1-dimensional case, except now $S$ is typically infinite. Thus, you have a (typically) infinite disjoint union, indexed by the set $S/G_m$, where $G_m$ is the projection of the stabilizer of $m$ in $G$ to the group $S$. I think, this is the best one can do without introducing further restrictions on the set $M$. Even in the case when $S$ is a finite dihedral group, there will be infinitely many possibilities for choosing a subgroup $T\subset {\mathbb R}^2$ normalized by $S$.

In order to get something more geometrically appealing, let's assume that $M$ is a closed subset of ${\mathbb R}^2$. This immediately implies that the group $G$ is a closed subgroup of the Lie group $I=Isom({\mathbb R}^2)$. Basic Lie theory tells you that $G$ is a Lie subgroup of $I$. Now, the problem essentially reduces to classification of Lie subgroups of $I$. Here it is:

  1. $dim(T)=2$, then $T$ acts transitively on ${\mathbb R}^2$ and, hence, $M={\mathbb R}^2$.

  2. $dim(T)=1$. Then:

a. Either $T$ is either isomorphic to ${\mathbb R}$, acting via translations along a line $L\subset {\mathbb R}^2$, or

b. $T\cong {\mathbb R}\times {\mathbb Z}$, where the first factor acts by translations along a line $L$ and the second acts by translations in the orthogonal direction.

In either case, clearly, $M$ is a product ${\mathbb R}\times M_1$, where $M_1$ is a discrete homogeneous subset of the real line (there are four possibilities for $M_1$ which you already know: single point, two points, orbit of the discrete group of translations or union of two such orbits).

3. $dim(T)=0$. Thus, $G$ is a discrete group of Euclidean isometries.

There are again 3 subcases here, depending on the rank of the free abelian group $T$ ($0, 1$, or $2$). If $T=0$ then $G< O(2)$ and, hence, $G$ is either 1-dimensional (in which case $M$ is a round circle) or finite cyclic or dihedral group; in the finite case everything reduces to vertex sets of regular or semiregular planar convex polygons. If $rank =1$ then $S$ is a subgroup of $Z_2\times Z_2$ and $M$ is either contained in one line or in two parallel lines.

The most interesting case is when $T$ has rank 2 and, hence, $G$ is a Euclidean crystallographic group (Russians call them "Fedorov groups"). If you so desire, you can now go through the well-known list of such groups and identify $G$-orbits in ${\mathbb R}^2$. Some will give you esthetically pleasing vertex sets of regular and semiregular "floor-tiling" patterns on ${\mathbb R}^2$. If I were Joseph O'Rourke, I would add some nice pictures here, but I am not.

| cite | improve this answer | |
  • $\begingroup$ Nice, Micha. But something seems missing: the group of rotations (that would be case 2c), producing circles. $\endgroup$ – Wlodek Kuperberg Jul 12 '13 at 17:55
  • $\begingroup$ @WlodzimierzKuperberg: You are right, I forgot one more case when $T=0$. $\endgroup$ – Misha Jul 12 '13 at 18:01
  • 1
    $\begingroup$ "But I am not". $\endgroup$ – Andrés E. Caicedo Jul 12 '13 at 18:11
  • 3
    $\begingroup$ @AndresCaicedo: Yes, to the best of my knowledge, I am not Joseph O'Rourke! $\endgroup$ – Misha Jul 12 '13 at 18:15

There's a complete classification that has been known since 1967: look past the Springer paywall here in the paper

Branko Grünbaum, L. M. Kelly. Metrically homogeneous sets, Israel Journal of Mathematics April 1968, Volume 6, Issue 2, pp 183-197.

If I recall correctly, there's a flaw in one of the lemmas as written, but a correction was published. I'll edit this answer once I find that correction.

Thanks to Francois Ziegler, we now have a link to the correction (see comment below). As pointed out in the other comments, the classification holds only for compact subsets of the plane.

Also, as pointed out in the comments, the definition of metric homogeneity used in this paper differs from the OP's as follows: in their definition, a metric space $(M,d)$ is homogeneous if for each pair $x,y$ of points, the set of distances $d(x,z)$ equals the set of distances $d(y,z)$ when $z$ ranges over points in $M$. This is weaker than the requirement that there be an isometry taking $x$ to $y$, so the objects of interest to the OP are certainly a subset of the classification and I hope that the paper is at least a starting point: the literature seems rather thin aside from this.

On the other hand, let's take the paper's definition of metric homogeneity. Fix $x,y$ in a space $(M,d)$ satisfying this definition and note that there exists a set-valued map $F:M \to 2^M$ defined as follows: for each $z \in M$, define $F(z) \subset M$ by

$$F(z) = \{w \in M \mid d(w,y) = d(x,z)\}$$

I wonder what conditions one must impose on $(M,d)$ -- aside from the requirement that $M$ is a subspace of the Euclidean plane -- in order to guarantee that there exists an isometry $f:M \to M$ which is a selection of $F$, i.e., $f(z) \in F(z)$ for all $z \in M$. All the selection theorems which I can think of crucially require convexity of $F(z)$, but maybe there are experts who know other ways.

| cite | improve this answer | |
  • 5
    $\begingroup$ The correction is here: ams.org/mathscinet-getitem?mr=268782 . Is it obvious that the OP is using the same notion of "homogeneous" as this paper? $\endgroup$ – Francois Ziegler Jul 12 '13 at 0:55
  • $\begingroup$ @Vidit Nanda: Their definition of metric homogeneity is not equivalent to the one in my question. They say "A subset of a metric space is called metrically homogeneous if the set of distances from a chosen point of the subset to all the other points of the subset is independent of the chosen point." $\endgroup$ – Wlodek Kuperberg Jul 12 '13 at 1:14
  • $\begingroup$ I would not call it a "complete classification" since it is restricted to compact sets. (In the context of the OP, their answer is: Circles or vertex sets regular polygons.) $\endgroup$ – Misha Jul 12 '13 at 1:30
  • 1
    $\begingroup$ @WlodzimierzKuperberg: True, but one can still extract a partial answer to your question from their result, after a bit of extra work. Once you know that the set is a smooth convex curve, the problem reduces to the classification of compact Lie subgroups of Isom(R^2). $\endgroup$ – Misha Jul 12 '13 at 1:32
  • $\begingroup$ If you assume compactness and convexity, the problem becomes easy. Assuming connectedness helps a lot, but does not make it trivial. I know that the only closed connected metrically homogeneous (MH) subsets of the plane are: single points, lines, and circles (oh, and the whole plane and also the empty set, if you insist). But this has not been published yet. Classifying all MH subsets seems to be much more difficult. $\endgroup$ – Wlodek Kuperberg Jul 12 '13 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.