-2

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 on which an inner product space acting on it. I was wondering if I could define in an equivalent way, without relaying on the inner product.

A set $E$ with a function $d:E \times E \to \Bbb{R}$ is an Euclidean space iff it satisfies the following axioms :

(1) $d(a,b)+d(c,b)\geq(a,c)$, for ever $a, b, c \in E$

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

(3) For every $p_1, p_2$ in E there always exists a set P of points that contains $p_1,p_2$ such that for any points $a, b, c \in P$ if $d(b,c)\lt d(a,c) \gt d(a,b)$ than $d(a,c)=d(a,b)+d(b,c)$

(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 $(p_1,p_2)$ where $p_1$ is from $P$ and $p_2$ is from $P_2$, $D$ <= d(p1, p2), and for every p1 in P there exists p2 in P2 such that d(p1, p2) =D

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.

Edit:

For continuity there could be a requirement that for every geodesic P, for any real number r, there always exists a pair of points p2, p2 on P such that d(p1, p2) =r

Kugutsu-o
  • 226
  • You don't define orthogonality in this way, which is a fundamental aspect of Euclidean spaces. Furthermore, an Euclidean space is a finite dimensional vector space, which means that a simple metric space (i.e. with a distance, as yo consider it) cannot fit into the frame. – Jean Marie May 19 '21 at 14:54
  • 1
    @JeanMarie We could define orthogonality just based on the metric: $P$ is orthogonal to $Q$ if for every $p\in P$, the closest point $q \in Q$ to $P$ is the intersection point of $P$ and $Q$. – Misha Lavrov May 19 '21 at 14:57
  • 2
    Consider a set of $n\geq 4$ points with distance $1$ between any two different points and sets $P$ consisting of two points each. If I understand your axioms correctly, it satisfies them. – Litho May 19 '21 at 14:59
  • See this – Jean Marie May 19 '21 at 15:09
  • Yea, orthogonality can be defined from this in various ways, only I don't know with how much computational practicality – Kugutsu-o May 19 '21 at 15:20
  • I edited the last axiom for more clarity – Kugutsu-o May 19 '21 at 15:21
  • @litho so it's a bit loose in continuity but i can tweak it a little so it does become strict on that – Kugutsu-o May 19 '21 at 15:26
  • 2
    @KingOfgames So if your question is "if I keep tweaking the axioms, will I eventually get Euclidean geometry?" the answer is probably yes. But that makes it very hard to answer. I have already adjusted my answer once in response to changes in your axioms, and I don't want to keep doing that. – Misha Lavrov May 19 '21 at 15:28
  • OK no problem.. – Kugutsu-o May 19 '21 at 15:33
  • @misha I guess I'm on the right track then, and it's just a few clarifications and trivial additions I made to make it more precise and concise, not a major rewrite.. – Kugutsu-o May 19 '21 at 15:36

1 Answers1

2

No, this set of axioms is still missing some things. It has many models that are very different from Euclidean space.

The affine-space-whose-vector-space-is-a-real-inner-product-space definition bakes in coordinates over $\mathbb R$, which is very limiting. One simple non-Euclidean definition you could give that satisfies your axioms is to take the Euclidean metric on $\mathbb Z^2$, instead.

The parallel postulate you've given doesn't quite match the usual parallel postulate. The two differences are:

  1. Usually we have a less restrictive definition of "parallel": we only ask that two parallel lines don't intersect, not that there be a constant distance between them. (In Euclidean geometry, the first implies the second.)
  2. Usually we have a more restrictive requirement on parallel lines: we ask that given a line $P$ and a point $p \notin P$, there should be only one line through $p$ parallel to $P$.

As a result, there are models with "weird parallel lines", as pointed out in the comments: take any number of points, and define the distance between two distinct points to be $1$. Then any set of two points forms a line, and any two non-intersecting lines are parallel.

Some other things to worry about:

  • Take Euclidean space, and throw out all irrational points. This satisfies all your axioms so far, but doesn't respect our usual notions of angle: Even though $(5,0)$ and $(3,4)$ are equidistant from $(0,0)$, there's no rotation about $(0,0)$ that takes $(5,0)$ to $(3,4)$.
  • Take Euclidean space, and keep only the algebraic points. This satisfies all your axioms so far, and anything you may add to patch the former problem, but doesn't have continuity - a line doesn't look like $\mathbb R$.
  • Take Euclidean space, and keep only the points at distance $<1$ from the origin. Again, this satisfies all your axioms, but doesn't have the Archimedean property.
Misha Lavrov
  • 142,276
  • Thanks for the answer, I'm not sure what you mean by the first part though. I edited the last axiom to make it more clear. – Kugutsu-o May 19 '21 at 15:32
  • For the first part, I mean take $E = \mathbb Z^2$, and $d(x,y) = \sqrt{(x_1 -y_1)^2 + (x_2 - y_2)^2}$. – Misha Lavrov May 19 '21 at 15:38
  • But there are no coordinates here – Kugutsu-o May 19 '21 at 15:39
  • I don't think the first part is relevant for this question – Kugutsu-o May 19 '21 at 15:42
  • Oh I didn't see you edited, now it's more clear – Kugutsu-o May 19 '21 at 15:43
  • OK so I am missing the requirement for uniqueness, so I'll ad. !, before existance quantified. – Kugutsu-o May 19 '21 at 15:45
  • 2
    A big thing you're still missing is continuity. Nothing you've done forces the math to be done over the reals, rather than say the rationals. You might want to compare your approach to Hilbert's axiomatic approach. These are different axioms of course, but you might ask yourself how many of Hilbert's axioms can be proven from yours. – Misha Lavrov May 19 '21 at 16:07
  • OK iv seen you re edit, you make some great points. I think there is any easy fix I'll tweak it again lol, you don't need to respond again, because this might go indefinitely perhaps. So basically I'd just ad a metrical definition of non compactness and simple connectedness. I think that should do. – Kugutsu-o May 19 '21 at 16:34
  • I was going for the least amount of axioms possible. – Kugutsu-o May 19 '21 at 16:34