Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Normally, the Euclidean space is introduced as $\mathbb R^n$. However, I've now been thinking about how one might define the $n$-dimensional Euklidean space only from the properties of the metric. I've come up with the following conjecture:

A metric space $(M,d)$ is an $n$-dimensional Euclidean space iff it has the following properties:

Line segment (L): For any two points $A,B\in M$ and any number $\lambda\in [0,1]$, there exists exactly one point $C\in M$ so that $d(A,C)=\lambda\,d(A,B)$ and $d(C,B)=(1-\lambda)\,d(A,B)$.

Uniqueness of extension (U): If for any points $A,B,C,D\in M$ with $A\ne B$ we have $d(A,C)=d(A,B)+d(B,C)=d(A,D)=d(A,B)+d(B,D)$ then $C=D$.

Homogeneity (H): For any four points $A,B,C,D\in M$ with $d(A,B)=d(C,D)$ there exists an isometry $\phi$ of $M$ so that $\phi(A)=C$ and $\phi(B)=D$.

Scale invariance (S): For any $\lambda>0$ there exists a function $s\colon M\to M$ so that for any two points $A,B\in M$ we have $d(s(A),s(B)) = \lambda\,d(A,B)$.

Dimension (D): The maximal number of different points $P_1,\ldots,P_k$ so that each pair of them has the same distance is $n+1$.

Now my question: Is this correct? That is, do those conditions already guarantee that the metric space is an $n$-dimensional Euclidean space? If not, what would be an example of a metric space which is not Euclidean, but fulfils all the conditions above?

What I already found (unless I've done an error, in that case, please correct):

It is easy to see that it contains a full line for each pair of points: Given the points $A$ and $B$, the condition (L) already gives the points in between $A$ and $B$. Now for any $r>0$, (S) tells us that there exist two points $C,D$ so that $d(C,D) = (r+1)\,d(A,B)$. Then (L) guarantees the existence of a point $E$ with $d(C,E)=1$ and $d(E,D)=r$. And (H) guarantees us an isometry $\phi(C)=A$ and $\phi(E)=B$. Then the line segment from $A$ to $\phi(D)$ extends the line segment in the direction of $B$. (U) guarantees us that this extension is unique.

If we define a straight line $l$ as a set of points so that for any three points $A, B, C\in l$ the largest of their distances is the sum of the other two distances, then from we also get immediately that two lines can intersect at most in one point (because if they have two points in common, then (L) guarantees that all points in between are also common, and I just showed that the extension is also unique).

I can also use the law of cosines to define the angle $\phi = \angle ABC$ as $\cos\phi = \frac{d(A,C)^2-d(A,B)^2-d(B,C)^2}{2\,d(A,B)\,d(B,C)}$ (of course the law of the cosine assumes Euclidean geometry, but since I'm defining the angle, this just means that if the space is not Euclidean, the angle I just defined is not the usual angle). It is obvious that this angle is independent of scaling (because a common factor just cancels out).

I also think that with the definition of the angle above, I should get that the sum of angles in the triangle is always $\pi$ (because I can just map the three points individually on three points with the same distance onto a known Euclidean plane, and there I know that the angles add up to $\pi$).

However is that already sufficient to show that it is an Euclidean space? Or could there be some strange metric space where all this is true without it being an Euclidean space?

share|cite|improve this question
$U$ is not true in Euclidean space. – Chris Eagle Jul 6 '13 at 21:45
You might want to look at what an Euclidean vector space is. – Clement C. Jul 6 '13 at 21:47
@ClementC.: According top your link, it is "page not found" ;-) But anyway, my point was that I wanted to use only the metric. – celtschk Jul 6 '13 at 21:48
@ChrisEagle: Could you please give me a concrete counter example? – celtschk Jul 6 '13 at 21:49
(Strange; the link works for me; it also points out to (this one)[], which defines a Euclidean space and seems to do what you want) – Clement C. Jul 6 '13 at 21:50

Maybe this has some relevance: Cayley–Menger determinants.

(Most of this Wikipdia article was destroyed on November 11th by a user called "Toninowiki". I've restored much of what was destroyed. The original poster in this present thread has commented below that the article does not deal with higher dimensions. That is wrong. If you look at it and don't see anything on higher dimensions, then look at the version of the article that was there before November 11th. Or at the one I left there a few minutes ago.)

share|cite|improve this answer
Thank you for the link. It allows to show that my straight lines indeed are straight. Unfortunately the article doesn't give an extension to higher dimension. – celtschk Jan 25 '14 at 10:41
@celtschk : Sorry, you're wrong. It does go into higher dimensions. Just look at the article before the edits done by "Toninowiki" on November 11th. – Michael Hardy Jan 25 '14 at 21:09
@celtschk : I've restored the parts of the article that were destroyed by "Toninowiki" on November 11. – Michael Hardy Jan 25 '14 at 21:19
Ah, now it's much more useful. Thank you. – celtschk Jan 26 '14 at 0:02

I think the issue here is that there is some confusion between your use of the terms metric space, Euclidean and $\mathbb{R}^{n}$.

The most general object in the list above is that of a metric space.

The axioms given in the question are satisfied by any complete homogenous metric space -- for example the hyperbolic metric on the unit $n$-ball.

So if you start with the underlying space $\mathbb{R}^n$, and give $\mathbb{R}^n$ the standard metric, then the axioms are satisfied.

If you take a different underlying set, say the unit ball, and put a hyperbolic metric on it, again the axioms are satisfied!

So your axioms do not really distinguish between different metrics. They just give properties satisfied by many metrics on different spaces.

If you are asking if $\mathbb{R}^n$ can be given a metric that satisfies the axioms but where the cosine law say is different, then the answer is no.

Hope this helps!

share|cite|improve this answer
Does the hyperbolic metric on the unit ball really satisfy the scale invariance axiom (S)? – celtschk Jan 25 '14 at 10:11

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.