# Arrow Space Construction

Is there a paper or book that has rigorously constructed the space of "arrow vectors" and shown that it is a vector space?

By "arrow vectors" I mean oriented line segments in Euclidean n-space. This space will be over the field of real numbers and the operations of vector addition and scalar multiplication are defined as usual:

• Vector Addition is defined via the parallelogram method
• Scalar multiplication is defined by scaling a line segment by the amount of the scalar. Where multiplication by positive numbers preserves direction and negative numbers reverses it.

I'm just wondering how far anyone has followed the heuristic.

• What do you mean by an "arrow vector" - actually good definitions are the first part of a rigorous treatment, because a good definition makes sure we know what we are talking about. And also are you thinking about the plane, or about $3$-dimensional space, or more generally? Dec 21, 2014 at 17:02
• @MarkBennet Arrow vectors = oriented line segments in $n$-dimensional Euclidean space. Dec 21, 2014 at 17:41
• What do you propose to be the sum of two oriented line segments in $n$-dimensional Euclidean space? The data required to specify a vector space is more than just a set of elements. One must also specify an addition operation and a scalar multiplication operation (and verify the vector space axioms hold for these operations). So your question makes no sense unless you can specify what these operations might be. Dec 22, 2014 at 14:09

All what follows is purely speculative, since I don't have enough time to check all the details, but I'd be glad to help if someone is stuck in a proof.

Let's first define what the arrow space is. Given an Euclidean space $$S$$ (a model of Tarski's axioms ), consider the following equivalence relation $$\operatorname{E}$$ on $$S^2$$ :

$$(P,Q) \operatorname{E} (P',Q') :\Longleftrightarrow$$ the segments $$[PQ']$$ and $$[P'Q]$$ have the same middle point (ie if $$PQQ'P'$$ is a parallelogramm).

A vector is an element of $$S^2/\operatorname{E}$$ (an equivalence class). For convenience, one writes $$\vec{PQ}$$ for the equivalence class of $$(P,Q)$$.

You'll need the following lemma :

lemma For any points $$P,Q,R$$, there is a (unique) point $$S$$ such that $$\vec{PQ} = \vec{RS}$$.

eg, here's a way to define the addition of 2 vectors $$\vec{PQ}$$ and $$\vec{P'Q'}$$ :

Let $$S$$ be the point such that $$\vec{QS} = \vec{P'Q'}$$, one defines $$\vec{PQ} + \vec{P'Q'} := \vec{PS}$$.

As for product of a vector by a scalar, you'll need a suited field. If you work with a model $$(S,\operatorname{B},\equiv)$$ of Tarski's axiom, then take any line $$\mathcal{D} = (PQ)$$. We'll work with the following (ordered) field $$(F,+,\cdot,\leqslant)$$ :

• domain : $$F := \{\vec{AB} \,|\, A,B \in \mathcal{D}\}$$ . Hence, $$F = \{\vec{PA} \,|\, A \in \mathcal{D}\}$$ .
• addition : vector addition, as above. $$(F,+)$$ is a commutative group with neutral element $$\vec{PP}$$.
• order : the order is induced by the betweenness relation $$\operatorname{B}$$ and the addition $$\vec{PA} \leqslant \vec{PB} :\Longleftrightarrow \operatorname{B}(P,C,Q) \vee \operatorname{B}(P,Q,C) \textrm{ where }C \textrm{ is such that} \vec{AB} = \vec{PC}$$
• multiplication : that's the trickiest part, it mimics Thales's theorem. Let $$x,y \in F$$, say $$x = \vec{PA}$$ and $$y = \vec{PB}$$. Here's how one computes $$x \cdot y$$ :

Up to swaping $$x$$ and $$y$$, we can wlog assume that there is $$C \notin \mathcal{D}$$ such that $$CQ \equiv PB$$. Let $$D$$ be the intersection of $$(PC)$$ and the parallel to $$(CQ)$$ passing trough $$A$$. The product $$x \cdot y$$ will be equal to $$\vec{PE}$$, where $$E \in \mathcal{D}$$ satisfies $$PE \equiv AD$$. There are two possibilities for $$E$$, $$\vec{PE} \leqslant \vec{PP}$$ or $$\vec{PE} \geqslant \vec{PP}$$:

1. If both $$\vec{PA}$$ and $$\vec{PB}$$ are $$\leqslant \vec{PP}$$ (resp. $$\geqslant \vec{PP}$$ ) one takes $$E$$ such that $$\vec{PE} \geqslant \vec{PP}$$.
2. Otherwise, one takes $$E$$ such that $$\vec{PE} \leqslant \vec{PP}$$.

proposition $$(F,+,\cdot,\leqslant)$$ is a real closed field with zero $$\vec{PP}$$ and unit $$\vec{PQ}$$.

remark $$F=F_{P,Q}$$ seems to depend on the choice of $$P$$ and $$Q$$, but for any $$P' \neq Q' \in S$$ such that $$PQ \equiv P'Q'$$, one gets $$F_{P,Q} \cong F_{P',Q'}$$ canonically.