MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A \subset X$ and $B \subset X$ be two isometric subsets of a metric space $X$. So there is an isometry $f: A \to B$. Say that a metric space $X$ has the superposition property (my terminology) if, for every pair of isometric subsets $A$, $B$, there is an isometry of $X$, $F: X \to X$, that superimposes $A$ onto $B$: $F(A) = B$, i.e. $F$ places $A$ on top of $B$.

Which metric spaces have this superposition property?

Euclidean space $\mathbb{E}^d$ has this property. But it seems the punctured plane does not: e.g. if $A$ is the point $(1,0)$ and $B$ is the point $(-2,0)$, then (I believe) there is not an isometry of the whole punctured plane that maps $A$ onto $B$.

Has this property been studied before? If so, under what name? I am (clearly) unschooled in this area. Thanks for pointers and/or examples!

share|cite|improve this question
In other words you're asking which spaces have the property that an isometry of subsets can be extended to an isometry of the whole space – Anthony Quas Jan 4 '13 at 1:18
It seems that if you start with a Riemannian manifold with this property it must be highly symmetric. For example, it must be a symmetric space: take a unit speed geodesic through a point $p$ and consider the space $A = \gamma([-\epsilon,\epsilon))$ and $B = \gamma((-\epsilon,\epsilon])$. Then the associated isometry should be a inversion around $p$. I have no idea if this is sufficient, although it would be pretty cool if it was. Also, I'm not quite sure what to do if you modified your definition to demand that $A,B$ are closed. – Otis Chodosh Jan 4 '13 at 1:47
@Anthony: I wonder if it is possible that $F$ maps $A$ onto $B$ but not identically to how $f$ maps $A$ to $B$ point by point...? For example, suppose $A$ and $B$ are congruent disks, and $f$ spins and translates $A$, but $F$ just translates $A$. Then $F$ is not an extension of $f$. (I am unsure of myself here...) – Joseph O'Rourke Jan 4 '13 at 2:18
up vote 10 down vote accepted

In other words, you are interested in the spaces where isometric subsets are congruent.

If the metric space is locally compact and intrinsic and simply connected then you get only spheres, Euclidean spaces and hyperbolic spaces. If not simply-conected, then in addition you get real projective spaces.

If the metric is not intrinsic you get discrete spaces and yet Cantor-like spaces build on them (who knows what else).

Yet in Urysohn universal space the property holds for compact subsets.

share|cite|improve this answer
@Anton: Thanks, especially for the connection to the Urysohn universal space! – Joseph O'Rourke Jan 4 '13 at 16:37
@Anton: isn't there a condition that the subsets be compact? – alvarezpaiva Sep 26 '14 at 6:27
@alvarezpaiva, yes, separable should be enough. – Anton Petrunin Sep 26 '14 at 22:50

If you restrict sets A, B to be simply points, then you are asking for spaces which have the property that for all points A,B\in X there exists an isometry T from X onto X so that T(A)=B (am I understanding you correctly?). Such spaces are called transitive. It is known that if a finite dimensional space is transitive then it is isometric to a Euclidean space, I think that this is a result of Mazur. It is an old, still open problem, whether every separable transitive Banach space is isometric to a Euclidean space. This problem is called Banach-Mazur problem and it goes back to 1930's. There has been a lot of work on this problem and it is connected to other interesting problems.

If you require that every 2 points can be mapped by a surjective isometry onto any other 2 points with the same distance, then I think the space is called 2-transitive. Similarly one defines n-transitive. I believe, but I am not 100% certain that 2-transitive Banach spaces have to be isometric to a Euclidean space. You might check work of V. Mascioni.

share|cite|improve this answer

A very natural concept arises if you should insist that $A$ and $B$ are small in some way, such as insisting that they are finite.

For example, the countable random graph under the shortest-path metric satisfies this version of the superposition property. The reason is that any partial isomorphisms of two finite induced subgraphs of the random graph extends to an automorphism of the random graph. (Meanwhile, the random graph does not have the full superposition property, since it is isomorphic to a proper subset of itself.)

We might define that a metric space has the $\omega$-superposition property, if any isometry of finite subspaces entends to an isometry of the whole space with itself. More generally, a space has the $\kappa$-superposition property, for a cardinal $\kappa$, when isometries of subspaces of size less than $\kappa$ extend to isometries of the whole space with itself.

This is a natural instance of what is known in model theory as a homogeneous structure, a structure for which any partial isomorphism of finitely generated substructures extends to an automorphism of the entire structure. This is essentially what is going on in your case, if you should restrict to finite $A$ and $B$ (and if also you should insist that the larger isometry $F$ agree with $f$ on $A$). More generally, we have the concept of a $\kappa$-homogeneous structure, for a cardinal $\kappa$, which means that any partial isomorphism of substructures of size less than $\kappa$ extends to an automorphism.

For example, structures arising as a Fraisse limit are always homogeneous. Also, every saturated structure is homogenous.

share|cite|improve this answer
@Joel: homogenous is a keyword here that I was missing (just out of ignorance). Thanks! – Joseph O'Rourke Jan 4 '13 at 1:50
A few references on homogeneity: Homogeneous structures in Prague…, A Survey of homogeneous structures, Homogeneous Structures by Lachlan These include numerous examples of homogeneous graphs, which can be regarded as metric spaces. – Joel David Hamkins Jan 4 '13 at 3:27

Like Euclidean geometry, also hyperbolic geometry has this extension property: an isometry defined on any subset extends to an isometry of the whole space. As I recall from long ago, in the book
Busemann & Kelly Projective Geometry and Projective Metrics
it is shown (among that class of geometries) there are very few of these spaces.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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