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In trying to understand what actually constitutes a "geometry" I came across many definitions of Euclidean spaces and geometries. Euclidean space is defined as an affine space with an inner product space acting on it. I was wondering if it could be defined in an equivalent and more natural manner, by relying only on the fundamental notion of distance, without the need for an inner product.

A set $E$ is an Euclidean space iff there is a function $d:E\times E\to \mathbb R$ that satisfies the following axioms:

(1) $d(a,b)+d(c,b)\ge d(a,c)$, for every $a,b,c\in E$.

(2) $d(a,b)=d(b,a)$, for every $a,b\in E$.

(2') $d(a,b)=0$ iff $a=b$.

(3) For every $p_1,p_2$ in $E$ there always exists a unique set $P$ of points that contains $p_1,p_2$ such that for any points $a,b,c\in P$ if $d(b,c) <= d(a, c)= >d(a,b) $ then $d(a,c)=d(a,b)+d(b,c)$.

(4) For any such set $P$ and for any point $p\notin P$ there is always a unique set $P_2$ (for which (3) holds) and that contains $p$, such that for every pair $(p_1,p_2)$ where $p_1\in P$ and $p_2\in P_2$, $D \le d(p_1, p_2)$, and for every $p_1 \in P$ there exists $p_2 \in P_2$ such that $d(p_1, p_2) =D$, $D \in \mathbb{R}$.

With the variation of this last property geometry should become non-Euclidean.

First two axioms define a usual metric, third defines geodesics, and last defines parallel geodesics.

For continuity there could be a requirement that for every geodesic $P$, for any real number $r$, there always exists a pair of points $p_1, p_2$ on $P$ such that $d(p_1, p_2) =r$.

Edit : I appreciate all the comments and suggestions, I would just like to say that I wouldn't refer to this as "my axioms" foe geometry as it was not my intention to just randomly come up with some generic new axioms.

It seemed reasonable to me to ask whether various geometries could be interpreted as sets on which there is a quantitative distance relation between points that can define geodesics through triangle equality. In this case Euclidean geometry.

  1. for any geodesic P and p in P and any r in R, there are p1, p2 such that d(p1, p) =d(p2, p) =r

After i posted the question i wanted to add this right away but then I didn't want to change the OG question too much.

Kugutsu-o
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    This question isn't quite clear to me. What exactly is the goal - that is, what would it mean for this proposed definition to be (un)satisfactory? Do you want to pin down $\mathbb{R}^2$ as a metric space up to isometry? Or something else? (Separately: I'm on the fence, but I think this might be more appropriate for math.stackexchange.) – Noah Schweber May 29 '21 at 16:24
  • Is this equivalent to the starboard axiomatisation? – Kugutsu-o May 29 '21 at 16:31
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    No, it is not nearly enough. I suggest you read Birkhoff's axioms here https://en.m.wikipedia.org/wiki/Birkhoff's_axioms – Moishe Kohan May 29 '21 at 16:39
  • Why isn't this equivalent to brikhoffs axioms that's the question basically. – Kugutsu-o May 29 '21 at 16:45
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    For continuity, don't you want to fix $p_1$ and $r$, and then ask for (possibly two) points $p_2$ at distance $r$? Otherwise, I fear there could simply be holes in the plane. But deleting just one point would cause a problem with the uniqueness property in (4). – Joel David Hamkins May 29 '21 at 17:11
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    but (3) does not hold for Euclidean space: such $P$ is not unique – Fedor Petrov May 29 '21 at 17:12
  • Yea I messed that up, I edited it how I meant to put it, I won't edit agaain.. That was my bad – Kugutsu-o May 29 '21 at 17:44
  • @Kugutsu-o I think that Fedor intends to refer to the possibility that a subset of the line might also satisfy your condition---for example, discrete multiples of the difference between the two points (or indeed just the two points themselves). This violates uniqueness. Perhaps you want to say that there is a unique maximal set like that. – Joel David Hamkins May 29 '21 at 17:46
  • @GeraldEdgar Doesn't his continuity condition ensure this? – Joel David Hamkins May 29 '21 at 18:03
  • Oh yea but in that case there that appendix axiom for continuity that answers both these two comments – Kugutsu-o May 29 '21 at 18:04
  • @Kugutsu-o Your formulation of continuity does not address Fedor's point, since any subset of a line will satisfy your condition in (3), and so even the Euclidean plane does not satisfy your (3). – Joel David Hamkins May 29 '21 at 18:06
  • By appendix axiom I mean the last sentence of the description. – Kugutsu-o May 29 '21 at 18:13
  • Yes, I get that. Fedor's point is that uniqueness in (3) will fail in the Euclidean plane, since any subset of the line containing $p_1$ and $p_2$ will also satisfy your condition. So there is no unique $P$ but many of them. This is why I had suggested that you might want (3) assert that there is a unique maximal such $P$. – Joel David Hamkins May 29 '21 at 18:20
  • Oh so you mean that requiring that a subest has pairs of points of any arbitrary real distance between them still doesn't imply it's maximal. In that case sure OK, it can be further required for it to be maximal. – Kugutsu-o May 29 '21 at 18:39
  • @jorl David hampings I think you're right, I didn't think much on that last "appendix" axiom, I wanted to edit it even before Iv seen you comment but I didn't want to change up the question too much. – Kugutsu-o May 29 '21 at 20:23
  • I know details are important, but some are really slight adjistments and don't automatically mean the whole question crushes, I presume. – Kugutsu-o May 29 '21 at 20:36
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    @Kugutsu-o Your overall attitude in this comment chain and the one on the answer you received is quite strange, and frankly quite rude. As "Jorl David Hampings" suggests, there's quite a few of these slight adjustments to make before the question becomes sensible or interesting to others -- that does not mean that the adjustments are not worth making, but you should probably let your question mature some more before you demand others to adjust their answers to every new edit you make to this question. – Carl-Fredrik Nyberg Brodda May 30 '21 at 11:49

1 Answers1

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Even with the most charitable interpretation of the posed question (which keeps evolving), the answer is negative. Examples are given by $\ell_p$-planes, $p\in (2,\infty)$. (I borrowed the example from this answer.)

The only thing which is not immediate is that geodesics in $\ell_p$-spaces are affine lines. The proof is not difficult, see Proposition I.1.6 in

Bridson, Martin R.; Haefliger, André, Metric spaces of non-positive curvature, Grundlehren der Mathematischen Wissenschaften. 319. Berlin: Springer. xxi, 643 p. (1999). ZBL0988.53001.

where it is proven that if $B$ is a strictly convex Banach space equipped with the metric $d(x,y)=||x-y||$ then affine lines in $B$ are the only geodesics in $(B,d)$. It is also a pleasant exercise to show that an $\ell_p$-plane is not isometric to the Euclidean plane unless $p=2$.

An axiomatic system for planar Euclidean geometry based on the notion of a metric space was given by Birkhoff, see here for axioms and references.

My favorite reference is

Moise, Edwin E., Elementary geometry from an advanced standpoint., Reading, MA: Addison-Wesley. 450 p. (1990). ZBL0797.51002.

A nice and freely available treatment of Euclidean geometry from the metric viewpoint is given in

A. Petrunin, "Euclidean plane and its relatives. A minimalist introduction." Arxiv, 1302.1630.

Postulates of angle measure and similarity are missing in the set of axioms proposed by OP.

Incidentally, the following is a cute open problem due to Keith Burns:

Suppose that $X$ is a Riemannian surface (complete, simply connected, without conjugate points) which satisfies the Playfair's axiom. Does it follow that $X$ is flat?

In this setting, all Birkhoff's postulates hold except for, possibly, the similarity postulate.


Edit. Here is a clean interpretation of OP's question.

Let $(X,d)$ be a metric space. (No need to define this, since such a definition is a standard undergraduate material.)

Definition. 1. A map $\gamma: {\mathbb R}\to (X,d)$ is called an isometric embedding if $d(\gamma(s), \gamma(t))=|s-t|$ for all $s, t\in {\mathbb R}$.

  1. A line in $(X,d)$ is the image of an isometric embedding $\gamma: {\mathbb R}\to (X,d)$.

Now, one can state OP's axioms:

A1. $(X,d)$ is a metric space containing at least two distinct lines. (This axiom was presumably simply forgotten by OP.)

A2. Every two distinct points in $(X,d)$ belong to one and only one line.

A3. For every line $L$ in $(X,d)$ and a point $p\in X\setminus L$ there is one and only one line $M$ in $(X,d)$, containing $p$ and disjoint from $L$.

Moishe Kohan
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  • I wouldn't say that this question should require some charity or that it is evolving. Only change I made is some standard and trivial parts of the metric definition. – Kugutsu-o May 29 '21 at 20:20
  • Your problem is interesting – Kugutsu-o May 29 '21 at 20:24
  • I think that the angle measure, could be derived from the axioms. – Kugutsu-o May 29 '21 at 20:25
  • Also I don't think your countexamole us too clear, il appreciate if the answer was more self contained. – Kugutsu-o May 29 '21 at 20:27
  • @Kugutsu-o The angle measure axiom by itself is quite useless. In my example you can take it to be the Euclidean angle (by putting an auxiliary Euclidean metric on the underlying affine plane). – Moishe Kohan May 29 '21 at 23:13
  • I think your answer is quite interesting, and quite involved for math stack exchange. Presuming even that it is right, I don't know why the question should be closed. – Kugutsu-o May 30 '21 at 08:35
  • But u don't see how banach spaces matter here, as there is not inner product here. – Kugutsu-o May 30 '21 at 08:38
  • @Kugutsu-o: Could you proofread/edit your comments for typos? I do not know about "u" but I do not see how inner products should be relevant, since you only asked for a metric space and I explained how to define one given a Banach space. Actually, I agree, your question is not research-level. – Moishe Kohan May 30 '21 at 10:12
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    @Kugutsu-o: As for the claim that my answer is "quite involved for math stack exchange," it is simply false. I posted over 1700 answers there and most are much more complex than this one. – Moishe Kohan May 30 '21 at 10:27
  • It's not just about complexity. This is not a very complex question. But it doesn't mean it can't still be quite subtle. – Kugutsu-o May 30 '21 at 10:38
  • So why I said you answer is to involved for math SE is because you require i know what an Lp space is and then refer to a graduate to research level paper. So that by definition then is an answer on that level so how then is the question not provided the answer is suitable for it? – Kugutsu-o May 30 '21 at 10:41
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    @Kugutsu-o Part of the charity that is required is that you never fixed axiom (3), which is not true for Euclidean space, because in Euclidean space there are many $P$ with your property, including $P={p_1,p_2}$. The way your axiom is stated, it implies that every line has only two points. – Joel David Hamkins May 30 '21 at 10:45
  • But that's OK, it's a good answer attempt I'm not sure yet if it's correct. – Kugutsu-o May 30 '21 at 10:48
  • But look at the edit. – Kugutsu-o May 30 '21 at 10:49
  • That counter example didn't work even with the original formulation. – Kugutsu-o May 30 '21 at 10:50
  • Even originally I had one additional axiom for continuity, as I posted I realised it could have been stated in a better way but then there were already comments and I didn't want to change the question too much. – Kugutsu-o May 30 '21 at 10:54
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    The problem was with (3), not continuity (which had its own issues), since (3) is false for the Euclidean plane. You can't fix that problem by changing another axiom. But I've said the same thing three times with no effect, so I guess I'll give up. – Joel David Hamkins May 30 '21 at 10:59
  • You said geodesics here have only 2 points, that's what the continuity axiom negates and that's what you said is wrong with 3 by itself, that that it allows non unique geodesics.. – Kugutsu-o May 30 '21 at 11:04
  • I don't get the toxicity.. You give up cause you're wrong, I hate to say it but it's true.. Why are you being so dismissive... – Kugutsu-o May 30 '21 at 11:12
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    @Kugutsu-o: Please can I encourage you to reflect on your attitude here? You are posting huge numbers of comments, most quite argumentative with other users, dismissing their concerns (e.g. disputing that your question had been “evolving” significantly, even though multiple people noted this), and holding their arguments to a high degree of detail (e.g. asking this answer to give more details of its counterexample, even though it already gives a fair bit of detail and references) while giving your own claims with little or no detail (e.g. saying you think angle measure is derivable). [cont’d] – Peter LeFanu Lumsdaine May 30 '21 at 12:51
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    I think you have, at base, a good question and a good mathematical attitude to approaching it. Your argumentative comments and repeated edits have undermined that. It’s probably too late to rescue this question, but for next time you ask, I’d urge you to make your comments and edits fewer but more carefully thought-out. Each time you would make an edit/comment, stop and think it over a bit longer. [cont’d] – Peter LeFanu Lumsdaine May 30 '21 at 12:51
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    That would help you avoid making some of the sillier comments you’ve made here; condense some of the good-but-incremental ones into a few more substantial ones (which is much easier for everyone to follow); and flesh out the good-but-unsubstantiated ones. Overall it will help you live up to MathOverflow’s usual good standards of clarity and collegiality. If you can do this then you will have a much better experience on future questions! – Peter LeFanu Lumsdaine May 30 '21 at 12:51
  • I appreciate each comment, no matter how parts of some even yours seem more than silly. – Kugutsu-o May 30 '21 at 15:40
  • I could have expanded on The angle measure, but it would be In vain since as you said the question is dismissed. – Kugutsu-o May 30 '21 at 15:41
  • I mean how would you approach a this, if I see a detail that needs to be polished what do I do? – Kugutsu-o May 30 '21 at 15:42
  • I only said to moishe kohan that saying this question is not research level and than answering it by a research level text reference seems contradictory. – Kugutsu-o May 30 '21 at 15:44
  • I did point out every time I patched or updated something and asked if it could be pardoned in the interest of the question. I still think it wasn't changed much.. – Kugutsu-o May 30 '21 at 15:46
  • I will take note on the substantially of the comments, but it wasn't and isn't my intention really to dismiss anyone or go into arguments. – Kugutsu-o May 30 '21 at 15:48
  • @moishe kohan I don't see anything missing from your reinterpretation except from the A3. Isn't this supposed to be a version of the playfairs axiom? I mean, and maybe someone said this, my question could (almost) be stated as "is every geodesic complete metric space with playfairs axiom an euclidian space"? – Kugutsu-o May 31 '21 at 08:54
  • There are a few more subtle things that are effectively different. – Kugutsu-o May 31 '21 at 09:07
  • One more on the more obvious side, shouldn't a line still be, albeit a degenerate, case of a Euclidean space on which one can do Euclidean geometry? That's why I didn't have A1 – Kugutsu-o May 31 '21 at 11:05
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    @Kugutsu-o: I am done with this question and have no desire to discuss it any further. – Moishe Kohan May 31 '21 at 12:49
  • You were done before you answered it.. – Kugutsu-o May 31 '21 at 16:49
  • In the open problem, it should be complete, simply connected, but NOT of curvature ≤0; otherwise, the statement is evident. – Anton Petrunin Sep 12 '21 at 02:30
  • @AntonPetrunin What is evident? – Moishe Kohan Sep 12 '21 at 02:38
  • The open problem should be about complete Riemannian metric on $\mathbb{R}^2$ with Playfair's axiom; if you add that curvature is ≤0, then it is triviallity. – Anton Petrunin Sep 12 '21 at 02:42
  • @AntonPetrunin How do you prove it? – Moishe Kohan Sep 12 '21 at 02:51
  • Assume there is a point $p$ with negative curvature, choose a line $\ell$ thru $p$; choose a point $q\notin\ell$ and observe that left and right asymptotically parallel lines to $\ell$ thru $q$ do not coiside. – Anton Petrunin Sep 12 '21 at 20:29
  • @AntonPetrunin Oh, you are right of course. I probably misremembered and it was the NCP assumption instead of NPC! – Moishe Kohan Sep 12 '21 at 21:59
  • I assume that NCP = no conjugate points --- yes. – Anton Petrunin Sep 12 '21 at 22:25
  • @AntonPetrunin Yes, no conjugate points. – Moishe Kohan Sep 12 '21 at 23:16