8
$\begingroup$

Here is a simple question: Why does it not make sense to define addition on points in geometry?

To be a bit more specific: assume we are talking only about standard Euclidean geometry.

It is clear that a barycentric combination of points is well-defined.

If I understand correctly, it also does not make sense to define the addition of points like this: $P+Q := O + \vec{OP} + \vec{OQ} $ (i.e., just use the position vectors), because the result would not be invariant wrt. different choices of $O$.

But why would it be impossible to define the addition of points in any meaningful way? Or is it?

$\endgroup$
  • 2
    $\begingroup$ To see the difficulties, just try coming up with a sensible translation-invariant addition rule that satisfies (for instance) $(P+Q)+R=P+(Q+R)$ for all points $P,Q,R$. $\endgroup$ – TonyK Nov 17 '19 at 15:34
  • $\begingroup$ Well, if you don't mind an origin to be distinguished, then your definition (the addition of position vectors relative to $O$) makes perfect sense. Otherwise you're left with affine combinations only, including the ternary Malcev operation $m(A,B,C) := \overrightarrow{OA} - \overrightarrow{OB} + \overrightarrow{OC}$. $\endgroup$ – Berci Nov 17 '19 at 17:32
  • $\begingroup$ Well, I wouldn't mind an origin to be distinguished so lang as the addition operation is invariant wrt. the choice of the origin, i.e., the result should always be the same point in space (albeit with different coordinates). $\endgroup$ – Gabriel Nov 17 '19 at 23:11
  • $\begingroup$ 10 pages of handwritten notes from July 1986, unread since, purport to show: (1) an Abelian group $V$ (of "vectors") acting transitively and faithfully on a set $L$ (of "points") can be identified with a ternary operation $P$ on $L$ such that $Paab=b$, $PabPbac=c$, and $PaPbcdd=Pabc$; (2) if $F$ is the free Abelian group generated by $L$, and $F_0$ the subgroup of formal sums $\sum_{p\in L}n_p[p]$ with $\sum_pn_p = 0$, there is a homomorphism $F_0\to V$ whose kernel $Z$ is generated by all sums $[a]-[b]+[c]-[d]$ with $d-a=c-b$; (3) all identities in $L$ and $V$ reduce to trivialities in $F/Z$. $\endgroup$ – Calum Gilhooley Nov 19 '19 at 1:51
  • $\begingroup$ Not that that necessarily makes any sense! I haven't checked it in 33 years. Also, in desperately trying to condense 10 yellowing pages of my scrawl into 600 characters, I think I erred in writing "faithfully", when I should have written "freely". Finally, the sum of two points in this algebra isn't a point, so it may not be relevant to your particular question. On the other hand, if you would find it satisfying to be able to write the relation between the vertices of a parallelogram $OPRQ$ as the perfectly ordinary equation $R = P - O + Q$, in an Abelian group, then perhaps it is of interest. $\endgroup$ – Calum Gilhooley Nov 19 '19 at 2:29
6
$\begingroup$

Any definition of addition will imply the definition of a special origin point $O$, because $O$ can be defined as the point so that $O + A = A$ for any $A$. In Euclidean geometry, we traditionally do not consider an origin point - e.g., it is not defined anywhere in Euclid's Elements. I believe it was not until Cartesian geometry that this idea was considered.

Euclidean geometry is typically considered to be invariant under isometries - transformations of the plane that do not change the distances between two points, such as reflections, rotations and translations. That is, if you take any theorem and proof in Euclidean geometry and apply an isometry to every line and point, the theorem and proof will still be valid. Nothing is assumed about any "absolute" location on the Euclidean plane. Since the result of an addition of two points depends on their location relative to the origin, which may change when you apply such a transformation, it is typically not considered in traditional Euclidean geometry.

tl;dr: Addition is undefined in Euclidean geometry because Euclid never thought of using coordinates.

$\endgroup$
3
$\begingroup$

With the barycentric concept, we are able to give a meaning to $A+B$, but it's no longer a "pure" point but a weighted point $2C$ where $C$ is the midpoint of $[AB]$.

What is this concept of "weighted points". We are going to see that one can build a universal space (see figure below ) mixing points and vectors, giving in particular the ability to define new points out of already existing ones by means of barycentric combinations.

Let us consider the simplest case.

The following linear combination

$$C:=aA+bB \tag{1}$$

where $A$ and $B$ are known points and $a,b \in \mathbb{R}$, makes sense because it defines unambiguously a geometrical entity with two cases :

  • (in general) a new point $C$ if $a+b=1$

  • (exceptionally) a new vector $C$ if $a+b=0$ (indeed, if $b=-a$, we have $C=a(B-A)$ which is nothing else than our familiar $\vec{AB}$...

In fact, the case $a+b=1$ defining points can be given a different setting : we will say that point $C$ is defined unambiguously by $A$ and $B$ through the (more general) following definition :

$$(a+b)C=aA+bB, \ (1) \ \ \text{whatever} \ a,b \ \text{such that} \ a+b \neq 0$$

(the equivalence with the initial definition is achieved by dividing LHS and RHS of (1) by $a+b$.).

let us give three examples :

1) $a=b=\frac12$ : $C=\frac12A+\frac12B$ is the midpoint of $[AB]$. If one prefers, this relationship can be written $A+B=2C$.

2) $a=\frac13$, $b=\frac23$ : $C$ is inside line segment $[AB]$ twice more attracted by $B$ than by $A$ (see figure below).

3) $a=2$, $b=-1$, $C$ is situated on line $(AB)$ but outside line segment $[AB]$, "attracted" by $A$ but "repelled" by $B$ (as the signs of $a$ and $b$ indicate it) but where is it exactly ? Transforming $C=2A-B$ into $2A=C+B$, we see that $C$ is situated in a place such that $A$ is the midpoint between $U$ and $B$. Another way to transform $C=2A-B$ is to write it $C-A=A-B$ i.e., $\vec{AC}=-\vec{AB}$ confirming the position already found just before.

There is a model for the operations we have been achieving based on an "embedding" of plane $\mathbb{R}^2$ into space $\mathbb{R}^3$ (the "universal space") with reciprocal transformation :

$$(wx,wy,w) \mapsto (x,y) \ \tag{2}$$

(where the third axis accounts for weights $w$).

Understanding (2) is essential for the justification of the operations we have done).

enter image description here

In fact, (2) can be explained in the following way (see figure) : a (non horizontal) line issued from the origin is an equivalence class of weighted points with weight $w$. The representatives of these equivalence classes can be taken at $w=1$, the set of ordinary points (weight one) constituting a "workplane" placed at "one meter high", the ground plane being a kind of mirror of the latter plane, devoted to vectors. This is known as the "projective interpretation". The barycentric interpretation is as follows.

Let us define for example $3C=1A+2B$. How can be visualized this relationship ? Let us build the "parallelogram of forces" (as a physicist would consider it) with weighted points $1A$ and $2B$, resulting into the weighted point $3D$. Now it remains to consider line $OD$ : its intersection with the plane of points gives point $C$. Briefly said, "in a natural way" $C$ is the weighted sum of $A$ (weight $1$) and $B$ (weight $2$).

$\endgroup$
  • 2
    $\begingroup$ A weighted average makes perfect sense and has a role in classical Euclidean geometry. But we have to honor the restriction $a+b=1,$ where the original question is why we can't set $a=b=1.$ $\endgroup$ – David K Nov 17 '19 at 18:12
  • $\begingroup$ This looks as if it could be closely related to those scribblings of mine, from 33 years back, that I wrote about in some comments on the question a moment ago. (If not, sorry!) $\endgroup$ – Calum Gilhooley Nov 19 '19 at 2:49
  • $\begingroup$ @Calum Gilhooley I would be pleased to have pointers on these comments. $\endgroup$ – Jean Marie Nov 19 '19 at 7:53
  • $\begingroup$ @David K You are right. With the barycentric concept, we are able to give a meaning to $A+B$, but it's no longer a "pure" point but a weighted point $2C$ where $C$ is the midpoint of $[AB]$. I will include it as a preamble to my answer. $\endgroup$ – Jean Marie Nov 19 '19 at 7:58
  • 1
    $\begingroup$ I'll try, although I may be slow getting round to it! Mercifully, I can drop most of the stuff about ternary operations, because it is probably fairly easy to reconstruct from these references: Torsor, redirecting to Principal homogeneous space; heap; Heaps and semi-heaps. (But I see nothing about my combined associative-commutative axiom, or the nicer notation for ternary ops I later devised.) $\endgroup$ – Calum Gilhooley Nov 19 '19 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.