I want to obtain a metric characterization of the classical finite dimensional spaces of Euclidean geometry.

Motivation: Suppose $A$ and $B$ live in an $n$-dimensional Euclidean space. They are each assigned the task of constructing an equilateral triangle of side length 5. Subject $A$ finds first one point $p_1$. If $n>0$, he can then find another point $p_2$ at distance 5 from $p_1$. If $n>1$, he can then find $p_3$ at distance 5 from $p_1$ and $p_2$. Then $B$ proceeds similarly. He selects a point $q_1$ (probably different from $p_1$), then finds another point $q_2$ and $q_3$. We wouldn't expect him to get stuck before constructing $q_3$, since $n$ is the same for both subjects. I will say a space is all-equal if the possibility of completing a figure is independent of the starting points chosen. Are Euclidean spaces all-equal? Are there non Euclidean all-equal spaces?

Precise statement

Definition: A metric space $X$ is all-equal if every time $S$ and $T$ are isometric subspaces of $X$ (with a selected isometry $S\to T$), the isometry extends to an automorphism of $X$.

Question 0: Is every Euclidean space all-equal?

To state the second question we must first observe some limitations. Being connected our life intervals, we cannot visit more than one connected component of our space, and being imprecise our measurements, we cannot distinguish directly between a space and its completion. Finally, existing no natural unit of measurement, the correct category to state these questions is that of spaces in which the distance is defined up to scale. That is, it takes values in a 1-dimensional module $M$ over $[0,+\infty)$, with no multiplication structure (although lengths can be tensorised to obtain areas).

More definitions: A congruence space is a pair $(X,M,d)$ where $X$ is a set, $M$ is a 1-dimensional $[0,+\infty)$ module and d is a distance in $X$ taking values in $M$. The morphisms from $(X,M,d)$ to $(Y,N,e)$ will be the subsimilarities, that is, the pairs $(f,\alpha)$ where $f$ is a function from $X$ to $Y$ and $\alpha$ is a morphism from $M$ to $N$ such that for each $x$, $x'$ in $X$ we have $e(fx,fx')\leq\alpha(d(x,x'))$. Similarities are similarly defined using an $=$ sign instead of $\leq$. Subspaces are defined by restriction, and the identity of $(X,M,d)$ is the obvious $(id_X,id_M)$. It's easy to prove that every isomorphism is a similarity. A congruence space $X$ is all-equal if every time $S$ and $T$ are similar subspaces, the similarity extends to an automorphism of $X$.

Question 1: Is every connected complete all-equal space a finite dimensional real inner-product space?

Remarks: if the connectedness hypothesis is dropped, some discrete spaces would be all-equal. If we work in the usual category of metric spaces, projective and hyperbolic spaces could be all-equal, and also Euclidean spheres, both with their inner metric and with the subspace metric.

Partial results and lines of thought:

EDIT: As noted by Sergei, there are counterexamples. Generally, if $(X,d)$ is an all-equal metric space, and $p>1$, then $(X,d^{\frac 1p})$ could be an all-equal space. This is similar to the example in which However, I was thinking of spaces in which the distance from $x$ to $y$ is measured using a signal that travels form $x$ to $y$ at constant speed. Hence I require that the space have the following property:

If $x$ is a point and $a$ and $b$ are lengths, then $N_b(N_a(\{x\}))=N_{a+b}({x})$ (where the $N$ stands for open neighborhood of a set).

Related MO questions: Characterizations of Euclidean space Easy proof of the fact that isotropic spaces are Euclidean Characterization of Riemannian metrics

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    $\begingroup$ There are infinite-dimensional examples, e.g. Hilbert space and Urysohn universal space. $\endgroup$ – Sergei Ivanov May 8 '11 at 9:56
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    $\begingroup$ Also, $\mathbb R^n$ with distance $d(x,y)=\sqrt{|x-y|}$ is an example. $\endgroup$ – Sergei Ivanov May 8 '11 at 10:03
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    $\begingroup$ Question 0 the answer is yes. But (as noted) there are many non-Euclidean examples, too. A classic book with all this material: Projective Geometry and Projective Metrics by Herbert Busemann and Paul J. Kelly $\endgroup$ – Gerald Edgar May 8 '11 at 12:26
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    $\begingroup$ @Sergei: Not Hilbert space ... there is an isometry of the whole space with a proper subspace, and that of course does not extend to an automorphism. $\endgroup$ – Gerald Edgar May 8 '11 at 12:27
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    $\begingroup$ @Sergei: not Urysohn space either (there is also an isometry of the whole space with a proper subspace in that case) $\endgroup$ – Julien Melleray May 8 '11 at 15:42

I can complete Anton's plan with an additional assumption that geodesics do not branch. I also assume local compactness (otherwise there are too many technical details to deal with). More precisely, I prove the following:

Let $X$ be a geodesic space. Suppose that

  • For every finite subset $Q\subset X$, every similarity $Q\to X$ can be extended to a bijective similarity $X\to X$. A similarity is a map that multiplies all distances by a constant. (It is easy to see that, in the case of a length metric, the more complicated definition from the question reduces to this.)

  • $X$ is locally compact.

  • For every two distinct points of $X$ there is a unique line containing them. (A line is a subset isometric to $\mathbb R$. The existence of lines follows from the two other assumptions.)

Then $X$ is isometric to $\mathbb R^n$ for some $n$.

Proof. First observe that the group of similarities acts transitively on pairs (oriented line, point not on this line). Indeed, given two such pairs $(\ell_1,p_1)$ and $(\ell_2,p_2)$, it suffices to find similar triples $p_1,x_1,y_1$ and $p_2,x_2,y_2$ where $x_i,y_i$ is a positively oriented pair of points on $\ell_i$. And it is easy to see that such triples $p_i,x_i,y_i$ realise all similarity types of, e.g. isosceles triangles.

Next we construct perpendiculars. Lines $\alpha$ and $\beta$ intersecting at a point $q$ are said to be perpendicular if there exists an isometry that fixes $\alpha$ and maps $\beta$ to itself by reflection in $q$. It is easy to see that this relation is symmetric. Further, for every line $\ell$ and every point $p\notin\ell$ there exists a unique perpendicular to $\ell$ containing $p$. To prove existence, pick two points $x,y\in\ell$ such that $|px|=|py|$. There is an isometry that fixes $p$ and exchanges $x$ and $y$. It acts on $\ell$ by reflection in the midpoint $q$ of $[xy]$, hence the line $(pq)$ is perpendicular to $\ell$. Uniqueness follows from the fact that two distinct reflections of $\ell$ cannot fix $p$, otherwise their composition shifts $\ell$ along itself and fixes $p$, and some iteration of this shift would break the triangle inequality.

This argument also shows that the base $q$ of the perpendicular is the nearest point to $p$ on $\ell$, and the distance from $p$ to $x\in\ell$ grows monotonically with $|qx|$.

Clearly all right-angled triangles with given leg lengths $a$ and $b$ are isometric; denote their hypotenuse by $f(a,b)$. Then $f$ is strictly monotone in each argument and positively homogeneous: $f(ta,tb)=tf(a,b)$.

Next, we show that the sum of Busemann function of two opposite rays is zero. Or, equivalently, if $\gamma$ is an arc-length parametrized line, $q=\gamma(0)$ and a line $(pq)$ is perpendicular to $\gamma$, then $B_\gamma(p)=0$ where $B_\gamma$ denotes the Busemann function of $\gamma$. Suppose the contrary. We may assume that $|pq|=1$ and $B_\gamma(p)=c>0$. This means that for every point $x\in\gamma$ we have $|px|-|qx|\ge c$. Let $s\in\ell$ be very far away and let $p_1$ be the perpendicular from $q$ to $(ps)$. The above inequality implies that $|pp_1|\ge c$, hence we have an upper bound $|qp_1|\le\lambda<1$ where $\lambda$ is determined by $f$ and $c$ and does not depend on $s$. Let $q_1$ be the base of the perpendicular from $p_1$ to $(qs)$, then rescaling the above inequality yields that $|p_1q_1|\le\lambda|qp_1|\le\lambda^2$. The next perpendicular (from $q_1$ to $p_2$ on $(ps)$) has length at most $\lambda^3$ and so on. Summing up these perpendiculars, we see that $|ps|\le 1/(1-\lambda)$ which is not that far away, a contradiction.

The fact that opposite rays yield opposite Busemann functions implies that the Busemann function of a line $\gamma$ is the only 1-Lipschitz function $f$ such that $f(\gamma(t))=-t$ for all $t$. And, given local compactness and non-branching of geodesics, this implies that $X$ is split into lines parallel to $\ell$, where a line $\gamma_1$ is said to be parallel to $\gamma$ if the Busemann function of $\ell$ decays with unit rate along $\gamma_1$ (to construct a parallel line, just glue together two opposite asymptotic rays). Further, Busemann functions of parallel lines coincide, due to their above mentioned uniqueness.

Now it is easy to see that the level set of a Busemann function (which is also a union of perpendiculars to a given line at a given point) has the same isometry extension property and is a geodesic space. Then we can carry induction in the parameter $d$ defined as the maximum number of pairwise perpendicular lines that can go through one point. The cases $d=1$ and $d=2$ can be done by hand, then use isometries exchanging perpendicular lines to do the induction step.

  • $\begingroup$ Some remarks in my first attempt of reading: From the motivation paragraph, it is perhaps more natural to require that the automorphism group is transitive on each class of isometric finite subspaces, since the experimenter would not know more than this. Also, the Hilbert spaces would be counterexamples. However, in this answer, this possibility is eliminated by the additional hypothesis of local compactness. $\endgroup$ – Marcos Cossarini May 14 '11 at 0:09
  • $\begingroup$ I don't see that the triple $(p,x,y)$ realizes all similarity types, but I see that it realizes all types in which $d(p,x)=d(,p,y)$. This is enough to prove that all pairs (oriented line, point outside) are similar. In particular, $(\mathcal l,p)~(\mathcal l^{op},p)$. $\endgroup$ – Marcos Cossarini May 14 '11 at 0:29
  • $\begingroup$ @Marcos: 1. If you allow only isometries, how do you get rid of spheres and hyperbolic spaces? 2. Yes it is easier to consider only isosceles triangles, I'll edit the text accordingly. 3. Removing local compactness could be possible if one also removes finiteness in the transitivity assumption. The most essential use of local compactness is the existence of asymptotic rays. $\endgroup$ – Sergei Ivanov May 14 '11 at 5:54
  • $\begingroup$ Note that the Busemann functions here are defined as minus the Busemann functions of Springer's EoM. $\endgroup$ – Marcos Cossarini May 14 '11 at 16:49
  • $\begingroup$ 1. Yes, I meant similarities, not isometries. The emphasis was in the finite vs. infinite subspace. $\endgroup$ – Marcos Cossarini May 16 '11 at 4:13

First let me reformulate the problem.

Let $X$ be a complete length space such that $X$ is isometric to any of its rescalings and for any subset $Q\subset X$, any distance preserving map $Q\to X$ can be exteded to an isometry of $X$. Prove that $X$ is isometric to a Euclidean space.

I am sure that this can be done, but it does require some work. Below I wrote a sketch which might work; feel free to edit.

Consider a maximal set in $X$ such that distance between any pair of points in is equal to 1. Note that if this set (say $Q$) is finite, otherwise there are some distance preserving map from $Q$ to its subset and it can not be exteded to a (bijective!) isometry of $X$. Denote this number by $q(X)$; it will be used instead of dimension as the induction parameter.

It is easy to prove existance of lines (two-side infinite minimizing geodesics) in $X$.

Note that all level sets of all Busemann functions for any lines in $X$ are isometric and have the same property as $X$, one only has to find a way to show that it is length space. It should be true that for such level set say $X'$, we have $q(X') < q(X)$. Then from induction you get that you can sweep out $X$ by Euclidean spaces, this should not be far from the end of proof.

  • $\begingroup$ Why do the level sets of Busemann functions have the same property? $\endgroup$ – Sergei Ivanov May 13 '11 at 7:53
  • $\begingroup$ Well, maybe only for finite sets, but it might be enough. $\endgroup$ – Anton Petrunin May 13 '11 at 16:39
  • $\begingroup$ I also ask the same. It is easy to prove that spheres of finite radius have the property. Thanks for the concept of Busemann function. Also, the definition of $q(X)$ and its finiteness is nice. Now I'll read Sergei's answer. $\endgroup$ – Marcos Cossarini May 13 '11 at 22:22
  • $\begingroup$ Anton, I saw you wrote in mathoverflow.net/questions/42192/… about a real tree, that is a space of infinite negative curvature. I thought that it could be relevant, since the only thing that prevents the hyperbolic plane from being a counterexample is that it isn't similar to its rescalings (in which the curvature is scaled). $\endgroup$ – Marcos Cossarini May 13 '11 at 22:49
  • $\begingroup$ Since the infinite sign would absorb the scaling parameter, I thought it could be a counterexample, so I was expecting you to show up. But I guess the tree is not like a plane. Is there a standard real tree? (That is very symmetric?) $\endgroup$ – Marcos Cossarini May 13 '11 at 22:49

Perhaps Wikipedia's article titled distance geometry is relevant.


Since the post is getting too long, I will place here my partial results.

If a real Banach space is all-equal, I think that it is finite dimensional and inner-product. This can be proved (at least in the finite dimensional case) using the John ellipsoid (see Easy proof of the fact that isotropic spaces are Euclidean), even though I don't like to introduce such a cumbersome structure.

If a Riemannian manifold is all-equal, then it is homogeneous and with constant curvature (how do we deal with the fundamental group?). If a Finsler manifold is all-equal, then I would expect it to be Riemannian (again using the John ellipsoid). I have no idea how to proceed without differential structure.

The hypothesis added after edition is equivalent to saying that if $x$, $z$ are points and $a$, $b$, are lengths such that $a+b>d(x,z)$, then there exists $y$ such that $a>d(x,y)$ and $b>d(y,z)$.

Now it is easy to construct approximate midpoints, then midpoints (using completeness), then segments, then complete geodesics (a segment can be extended by moving the starting point to the midpoint and the midpoint to the end (EDIT:the starting point must go to the starting point, and the midpoint to the end)). Then every pair of points can be joined by a geodesic of the expected length.

Are geodesics unique, or is it possible that there is an ugly graph embedded in the space? (See Representability of finite metric spaces, an example of ugly graph is a 4-cycle with the path distance.)

Is it possible to define angles? That is, suppose $S$ and $T$ are rays with common origin $x$. If I consider $s$ and $t$ their respective intersections with a ball $B_r(x)$ and $s'$, $t'$ their respective intersections with the ball of radius $B_{r'}(x)$, is it true that $\frac{d(s,t)}{r}=\frac{d(s',t')}{r'}$?

Actually, since every sphere in an all-equal congruence space is all-equal (as a metric space; it doesn't have dilations), it would be nice to define an intrinsic metric in each sphere, so that the angle between $S$ and $T$ could be defined as $\frac{i(s,t)}{r}$, where $i$ is the intrisic metric of $B_r(x)$.


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