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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.

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    $\begingroup$ 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.) $\endgroup$ Commented May 29, 2021 at 16:24
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    $\begingroup$ No, it is not nearly enough. I suggest you read Birkhoff's axioms here en.m.wikipedia.org/wiki/Birkhoff's_axioms $\endgroup$ Commented May 29, 2021 at 16:39
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    $\begingroup$ 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). $\endgroup$ Commented May 29, 2021 at 17:11
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    $\begingroup$ but (3) does not hold for Euclidean space: such $P$ is not unique $\endgroup$ Commented May 29, 2021 at 17:12
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    $\begingroup$ @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. $\endgroup$ Commented May 30, 2021 at 11:49

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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 $L$.

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    $\begingroup$ @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. $\endgroup$ Commented May 30, 2021 at 10:45
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    $\begingroup$ 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. $\endgroup$ Commented May 30, 2021 at 10:59
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    $\begingroup$ @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] $\endgroup$ Commented May 30, 2021 at 12:51
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    $\begingroup$ 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] $\endgroup$ Commented May 30, 2021 at 12:51
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    $\begingroup$ 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! $\endgroup$ Commented May 30, 2021 at 12:51

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