Normally, the Euclidean space is introduced as $\mathbb R^n$. However, I've now been thinking about how one might define the $n$-dimensional Euclidean 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?
Best Answer
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.)