31

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?

celtschk
  • 43,384
  • $U$ is not true in Euclidean space. – Chris Eagle Jul 06 '13 at 21:45
  • You might want to look at what an Euclidean vector space is. – Clement C. Jul 06 '13 at 21:47
  • 2
    @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 06 '13 at 21:48
  • 1
    @ChrisEagle: Could you please give me a concrete counter example? – celtschk Jul 06 '13 at 21:49
  • (Strange; the link works for me; it also points out to (this one)[http://planetmath.org/node/35743], which defines a Euclidean space and seems to do what you want) – Clement C. Jul 06 '13 at 21:50
  • Any points $A, B, C, D$ lying in that order in a straight line. – Chris Eagle Jul 06 '13 at 21:51
  • @ChrisEagle: Oops, you're right; I've forgotten a crucial condition ... I'll correct. – celtschk Jul 06 '13 at 21:52
  • @ChrisEagle: OK, now corrected. Thank you. – celtschk Jul 06 '13 at 21:55
  • 1
    @ClementC.: Strange, now your first link works. Must have been a temporary server problem. Anyway, even your second link is not what I want because it also refers to an additional structure, namely the Euclidean vector space defined in your first link. – celtschk Jul 06 '13 at 21:58
  • What about the following: all norms are equivalent on $\mathbb{R}^n$. This is not true in general. You could use this as a starting point for a definition for $\mathbb{R}^n$. – Cameron Williams Jul 06 '13 at 22:01
  • @CameronWilliams: So how do you define "all norms" if all you have is a set $M$ and a single(!) function $d\colon M\times M\to M$? – celtschk Jul 06 '13 at 22:03
  • $d:M\times M\rightarrow \mathbb{R} $ ;) but I'm saying that suppose you had a family of norms (eliminating those that are equivalent up to isomorphism) and that they are all equivalent then the space is Euclidean. You may need more detail to have a proper definition but it's a good alternative. Now this uses norms instead of metrics so it's a bit more general but along the lines you're interested in. Also there is more than one norm on $\mathbb{R}$. – Cameron Williams Jul 06 '13 at 22:08
  • @CameronWilliams: There are surely many different ways to define Euclidean spaces. However I'm specifically interested to do it from the metric. – celtschk Jul 06 '13 at 22:19
  • You actually need even less than that: http://www.google.com/search?q=characterization+of+euclidean+space+as+metric+space&ie=UTF-8&oe=UTF-8&hl=en&client=safari. It characterizes $\mathbb{R}$ and (if you use that to construct $\mathbb{R}^3$, you can characterize all Euclidean spaces by Theorem 2. – Brian Rushton Jul 06 '13 at 22:32
  • @BrianRushton: Your link is to a Google search. Which of the many links (which almost certainly show up in different order for me than for you) is the one you're talking about? – celtschk Jul 06 '13 at 22:35
  • Haha, sorry about all the link trouble today. http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1102868444&page=record – Brian Rushton Jul 06 '13 at 23:03
  • @BrianRushton: Thanks, that's a very interesting read. I wonder if it would be possible to replace the euclidean four point property by some "tetrahedron inequality" (or inequalities). OTOH the need for this four-point property makes me suspect that my conditions may not be sufficient, because it shows that for dimension $3$ there obviously are additional freedoms which I've not explicitly taken into account. I wonder what a convex metric space without the euclidean four point property would look like. – celtschk Jul 06 '13 at 23:29
  • Ah, I've now seen the $n=2$ counterexample for corollary 2.3 in the article; that one would violate my (U) condition, e.g. for $A=(-1,0), B=(0,0), C=(1,0), D=(0,1)$. I wonder if there exist counterexamples which don't violate (U). – celtschk Jul 06 '13 at 23:37
  • How would you show that $|cos \phi |\leq 1$? – DanielWainfleet Sep 01 '15 at 05:27
  • 1
    @user254665: According to the triangle inequality, $d(A,C)\le d(A,B)+d(B,C)$, therefore $$d(A,C)^2-d(A,B)^2-d(B,C)^2\le((d(A,B)+d(B,C))^2-d(A,B)^2-d(B,C)^2 = 2d(A,B)d(B,C)$$ and thus $\cos\phi \le 1$. Further be wlog $d(A,B)\ge d(B,C)$. Again due to the triangle inequality $d(A,B) \le d(A,C)+d(C,B)$, thus $d(A,C) \ge d(A,B) - d(B,C)$, thus $d(A,C)^2 - d(A,B)^2 - d(B,C)^2 \ge -2 d(A,B) d(B,C)$, thus $\cos\phi\ge-1$. – celtschk Sep 01 '15 at 12:16
  • 1
    Stepping back to $\mathbb R^2$, found a metric space approach - https://math.stackexchange.com/a/1991372/432081 – CopyPasteIt Jan 20 '19 at 02:43
  • @CopyPasteIt: Thank you, that's an great find. Indeed, while that axiomatization first defines betweenness separately, axiom 9 actually could serve as its definition, making it indeed a purely metric space approach. And indeed, the only things that depend on the dimension seem to be axiom 1, (which basically specified the dimension to be larger than 1), axiom 8 (which is not hard to generalize), and the definition of parallel lines, which enters the parallel axiom (that definition clearly doesn't work in more than two dimensions). – celtschk Jan 20 '19 at 08:11

4 Answers4

5

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.)

  • 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 00:02
3

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!

1

In order to show that your axioms are correct, it suffices to show that they are equivalent to Tarski's axioms of Euclidean geometry, which have been proven to be correct. Below I have written them out in the context of a metric space, to make the equivalence easier to verify.

Tarski's axioms only have one primitive notion, that of points, and only two relations, congruence and betweenness.

Restricting to metric spaces, it is very straightforward to define an appropriate notion of congruence to substitute into Tarski's axioms: points $p_1, p_2, p_3, p_4$ are congruent (written $p_1 p_2 \equiv p_3 p_4$) if and only if $d(p_1, p_2) = d(p_3, p_4)$.

We can also define an appropriate notion of betweenness in entirely metric terms. (The inspiration for this definition comes from convex metric spaces, of which Euclidean space is an example. However, see also 1 or 2.) We say that the point $p_2$ is between the points $p_1$ and $p_3$ if and only if equality is achieved in the triangle inequality: $d(p_1, p_2) + d(p_2, p_3) = d(p_1,p_3)$.

Note that, with these definitions of congruence and betweenness, axioms 1,2,3, and 6 in this numbering become redundant, so we don't need to mention them. Thus, in the case of metric space models for Euclidean geometry, the number of axioms reduce automatically from 11 to 7. This leads to the following metric space axiom list for Euclidean geometry (with new numbering):

  1. Segment Extension: Given points $p_1,p_2,p_3,p_4$: $$\begin{array}{l}\text{There exists a point }p_5\text{ such that: }\\ d(p_1,p_2) + d(p_2,p_5) = d(p_1,p_5)\text{ and }d(p_2,p_5) = d(p_3,p_4) \,. \end{array} $$
  2. Five Segment Axiom: Given points $p_1,p_2,p_3,p_4$ and $q_1,q_2,q_3,q_4$, with $d(p_1,p_2) > 0$: $$ \begin{array}{l} \text{If:} \\ \begin{array}{ll}d(p_1,p_2) = d(q_1,q_2) \,, & d(p_2,p_3) = d(q_2,q_3) \,, \\ d(p_1,p_4) = d(q_1,q_4) \,, & d(p_2,p_4) = d(q_2,q_4) \,, \\ d(p_1,p_2)+d(p_2,p_3) = d(p_1,p_3) \,, & d(q_1,q_2)+d(q_2,q_3) = d(q_1,q_3) \end{array} \\ \text{Then } d(p_3,p_4) = d(q_3,q_4) \,. \end{array} $$
  3. (Inner) Pasch Axiom: Given points $a,b,c$ and $p,q$: $$ \begin{array}{l} \text{If } d(a, p)+d(p, c) = d(a,c)\text{ and }d(b, q)+d(q, c) = d(b,c) \\ \text{Then there exists a point }x\text{ such that: }\\ d(p,x)+d(x,b) = d(p,b)\text{ and }d(q,x)+d(x,a) = d(q,a) \,. \end{array} $$
  4. Lower Dimension Axiom ($n \ge 2$): There exist points $a,b,c$, and $n-1$ distinct points $p_1, \dots, p_{n-1}$, such that: $$ \begin{array}{l} d(a,b)+d(b,c) < d(a,c) \,, d(b,c) + d(c,a) < d(b,a) \,, d(c, a)+ d(c, b) < d(c,b) \,, \text{ and} \\ d(a, p_1) = d( a, p_2) = \dots = d( a, p_{n-1})\,, d(b, p_1) = d( b, p_2) = \dots = d( b, p_{n-1})\,, \\ \text{and }d(c, p_1) = d( c, p_2) = \dots = d( c, p_{n-1}) \,. \end{array} $$
  5. Upper Dimension Axiom ($n \ge 2$): There exist points $a,b,c$, and $n$ distinct points $p_1, \dots, p_{n-1}, p_n$, such that: $$ \begin{array}{l} \text{If } d(a, p_1) = d(a,p_2) = \dots = d(a, p_n)\,, d(b, p_1) = d(b, p_2) = \dots = d(b, p_n) \,, \\ \text{and }d(c, p_1) = d(c, p_2) = \dots = d(c, p_n) \\ \text{Then at least one of }d(a,b)+d(b,c) = d(a,c)\,, d(b,c)+d(c,a) = d(b,a) \,, \\ \text{ or }d(c,a)+d(a,b) = d(c,b) \text{ is true.} \end{array} $$
  6. Euclid Axiom: Given points $p_1, p_2, p_3, p_4$ and $t$: $$ \begin{array}{l} \text{There exist points }x,y\text{ such that: } \\ \text{If }d(p_1,p_4) >0\text{, }d(p_1,p_4)+d(p_4,t) = d(p_1,t)\text{, and }d(p_2,p_4)+d(p_4,p_3) = d(p_2,p_3) \\ \text{Then } d(p_1,p_2)+d(p_2,x) = d(p_1,x)\text{, }d(p_1,p_3)+d(p_3,y) = d(p_1,y)\,,\\ \text{and }d(x,t)+d(t,y) = d(x,y) \,. \end{array} $$
  7. Continuity Schema

Note: The Five Segment Axiom is a "rewording" of the Side-Angle-Side postulate. Taxicab geometry is an example of a geometry satisfying most (if not all) of the axioms of Euclidean geometry except for the Side-Angle-Side postulate. The Five Segment Axiom therefore as a result is necessary to make the notion of angle in Euclidean geometry a useful (isotropic) one.

The Euclid axiom is a "rewording" of the Parallel Postulate. Thus it has the same geometric significance as the Parallel Postulate, which I assume is well-known to you.

The Inner Pasch Axiom is equivalent to the Pasch Axiom (also called sometimes the Outer Pasch Axiom) which is also equivalent to the Plane Separation Axiom. It is necessary to prevent really degenerate spaces from also being models of Euclidean geometry. The interpretation of the axiom's geometric significance is easiest to understand when we assume additionally that the metric space is complete (see below, this case also makes Axiom 7 redundant): see here or here.

If we assume additionally that the metric space is (Cauchy) complete, we get the full second-order logic continuity axiom (a la Hilbert) and a clean relationship with the real numbers. As a result, Axiom 7 becomes redundant when we assume a complete metric space. See (1)(2)(3).

Note also that the lower dimension axiom, in the case that $n=2$, simplifies to the following:

Lower Dimension ($n = 2$): There exist points $a,b,c$ such that: $$ \begin{array}{l} d(a,b)+d(b,c) < d(a,c) \,, \\ d(b,c)+d(c,a) < d(b,a) \, \\ d(c,a)+d(a,b)<d(c,b) \,. \end{array} $$

Since three points are collinear if and only if one of the points is between the other two, this is equivalent to the statement "there exist three non-collinear points".

In the case of $n = 1$ (which amounts to an essentially coordinate-less characterization of a space homeomorphic to a real closed subfield of $\mathbb{R}$, or of a space homeomorphic to $\mathbb{R}$ if we demand a complete metric space) one has the following simpler versions of both the upper and lower dimension axioms:

Lower Dimension Axiom ($n = 1$): There exist two distinct points $a,b$.

Upper Dimension Axiom ($n = 1$): Given three points $a,b,c$, at least one of the following is true: $$ \begin{array}{c} d(a,b)+d(b,c) = d(a,c)\\ d(b,c)+d(c,a) = d(b,a) \\ d(c, a)+d(a,b) = d(c,b) \end{array} $$ In other words, any three points $a,b,c$ are collinear.

For $n= 0$, all of the axioms can be replaced by a single axiom:

Upper Dimension Axiom ($n = 0$): For any points $a,b$, one has that $a = b$.

Chill2Macht
  • 20,920
  • 1
    Thank you, that is definitely useful. If I'm not mistaken, Segment Extension is effectively what I've proved in my question when showing that there's a complete line for each two points. I initially thought Cauchy-completeness should follow from my axioms, but I now notice that there's nothing preventing extra infinitesimally close points (I demand the existence of points based on real numbers, thus preventing e.g. $\mathbb Q^n$, but I don't demand the non-existence of points that cannot be reached that way. – celtschk Nov 12 '17 at 12:16
1

Metric Space Axioms: Between every two points $A$ and $B$ is a distance $AB$ such that:
Axiom 1: For all points $A$: $AA = 0$,
Axiom 2: For all points $A$ and $B ≠ A$, $AB > 0$,
Axiom 3: (Symmetry) For all points $A$ and $B$: $AB = BA$,
Axiom 4: (The Triangle Inequality) For all points $A$, $B$ and $C$: $AB + BC ≥ AC$.

Euclidean Axioms:
Axiom 5: (Infinite Divisibility) Any distance can be divided by any number. Given any two points $A$ and $B$, and any number $N = 2, 3, 4, ⋯$, there is a point $C$ such that $$AC + BC = AB = N·AC.$$ Axiom 6: (Infinite Extensibility) Any distance can be multiplied by any number. Given any two points $A$ and $B$, and any number $N = 2, 3, 4, ⋯$, there is a point $C$ such that $$AB + BC = AC = N·AB.$$ Axiom 7: (Generalized Pythagorean Theorem) If $A$, $B$ and $C$ are three points such that $AB + BC = AC$, then for any point $D$: $$AB·CD^2 - AC·BD^2 + AC·AD^2 = AB·AC·BC.$$ Axiom 8: (The Counter-Zeno Axiom) Let $A_1, A_2, A_3, ⋯$ be a sequence of points such that for all integers $n > 1$: $$A_n A_{n+1} ≤ ½ A_n A_{n-1}$$ then there exists a point $A$, such that for all integers $n ≥ 1$: $$A_n A ≤ 2 A_n A_{n+1}.$$

For Axiom 8, such sequences may be termed Zeno Sequences. The axiom asserts that a Zeno Sequence always converges a point that lies within twice the distance of any of its segments. Every Cauchy sequence has a Zeno subsequence, and every Zeno sequence is Cauchy. So, Zeno sequences serve just as well as (and better than) Cauchy sequences.

Dimensionality Axioms
Axiom 9: ($N^-$) No $N + 2$ points are equally distant from one another.
Axiom 10: ($N^+$) There exists $N + 1$ points that are equally distant from one another.

Axioms 1-10 provide an equivalent description of the Euclidean space contained in $ℝ^N$, for any $N = 1, 2, ⋯$. If you stretch the meaning of $0^+$ to say that at least one point exists, and the meaning of $0^-$ to say that no two points exist at all, then this also applies to the case $N = 0$ and to the single-point zero-dimensional Euclidean space.

Theorem 1 Existence of the affine operator $[A, r, B]$.
Given points $A$ and $B$ and a number $r ∈ ℝ$, one must show that there exists a unique point $C ≡ [A, r, B]$ such that $AC = |r| AB$ and $BC = |1 - r| AB$. Axioms 5 and 6 cover the case where $r$ is rational, Axiom 8 fills in the gaps for the case where $r$ is irrational and Axiom 7 ensures that the point is unique.

Affine Space Structure
Just as well, you can also show that: $$[A, 0, B] = A, [A, 1, B] = B$$ and $$[A, rt(1-t), [B, s, C]] = [[A, rt(1-s), B], t, [A, rs(1-t), C]].$$ This one operator, and these three properties, alone, happen to provide an equivalent axiomatization for affine spaces over fields of size 4 or more, not just $ℝ$. (Over fields of size 3, these 3 properties equivalently define a commutative quandle).

Inner Product Space Structure
Pick any point $O$. Define $rA = [O, r, A]$, $A + B = 2 [A, ½, B]$, and $<A, B> = ½(OA² - AB² + OB²)$, and use Axiom 7 to show that this results in a vector space with an inner product. One can then also prove that $$[A, r, B] = (1 - r) A + r B.$$

Axioms 1-4, Axiom 7 and Theorem 1 equivalently define real inner product spaces.

Metric Space Zeno-Completion
For any two Zeno sequences $a = (a_n: n = 1, 2, ⋯)$ and $b = (b_n: n = 1, 2, ⋯)$, a unique number $ab$ exists such that $|a_n b_n - ab| ≤ 2\left(a_n a_{n+1} - b_n b_{n+1}\right)$ for all $n = 1, 2, ⋯$. This satisfies all the properties of metric spaces, except Axiom 2: $ab = 0$ is possible if $a ≠ b$; and also satisfies Axiom 8. It is upwardly compatible with the distance function of the underlying metric space by the conversion of points $A$ to constant sequences $(A, A, A, ⋯)$. So, for metric spaces satisfying only Axioms 1-4, but not Axiom 8, reducing the Zeno sequences by equivalence modulo the zero-distance relation $ab = 0$, results in an upward extension of the original metric space to one that also satisfies Axiom 8: the Zeno Completion of the metric space, for lack of a better name.

None of the Axioms 5-7 are required for this. Therefore, Axioms 1-4 and 8 therefore equivalently define complete metric spaces.

Axioms 1-8 provide an equivalent description of real Hilbert spaces, since they are inner product spaces, which are complete as metric spaces.

Axioms 1-4, 7 along with the 3 axioms, listed above, for Affine spaces, along with the property $$[A,r,B][A,r,C] = |r| BC$$ provide an equivalent description of a real Banach spaces, provided one chooses a point $O$ as its $0$ and defines $|A| = AO$, for all points $A$.

NinjaDarth
  • 540
  • 2
  • 4