Thread by @ProfKinyon, Rethinking Vector AdditionorHow I Learned to Stop Worrying and Love NonassociativityA thread [...]

Rethinking Vector Addition
or
How I Learned to Stop Worrying and Love Nonassociativity

A thread in 29 tweets

(0/28)

Let's revisit a classical and elementary topic: the addition of vectors. Now when I say "vector", I don't mean an element of an abstract vector space. I instead mean a good old fashioned directed line segment. (1/28)

What I want to explore here is the role that Euclid's 5th Postulate, the Parallel Postulate, plays in our understanding of vector addition. If you think about the classical geometric construction, the parallelogram law, the 5th Postulate plays a role right from the start! (2/28)

Here is the classic diagram. If we want to add the vectors OA and OB, we draw a line parallel to OA through B and a line parallel to OB through A. Let C denote the intersection of the parallel lines. Then OA + OB = OC. All well and good. (3/28)

You may have learned a slightly different version: draw a parallel copy of OB based at A, and then connect O to the head C of the copy. This treats a vector not as an individual directed line segment, but rather as an equivalence class of parallel directed line segments. (4/28)

Historically, the Parallel Postulate was always controversial, so in that spirit, let's consider the following problem:

Can we remove the dependence of vector addition on the Parallel Postulate?

(5/28)

Because of this, we will ignore the vector-as-equivalence-class point of view, which has parallelism built into it, and instead focus on vectors as single directed line segments, all with a common base point O as origin. (6/28)

In the parallelogram law, we draw two parallel lines, so we seem to be invoking the Parallel Postulate twice. In fact, it's easy to reduce this to just once. Draw a line through A parallel to OB. Then mark off the line segment AC congruent to OB. (7/28)

We can do this with ruler and compass essentially by Book I, Prop. 2 of the Elements, which does not require the Parallel Postulate.

This form of the parallelogram law is asymmetric but still constructs the same resultant vector as before. (8/28)

But to fully eliminate the Parallel Postulate, we need to look at what it says. The best known form is Playfair's Axiom:

In a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the point.

(9/28)

The key words there are "at most one". The *existence* of a parallel line does not require the Postulate. One construction is in the Elements, Book I, Prop. 31. An equivalent one will be explained shortly. The Postulate guarantees *uniqueness* of the parallel line. (10/28)

Absolute geometry is geometry without the Parallel Postulate. In absolute geometry, parallels exist. In Euclidean geometry, they are unique. In hyperbolic geometry, they are not.

Our problem reformulated:

Can we make sense of vector addition in absolute geometry?

(11/28)

I wouldn't be this deep into a thread if the answer weren't yes. But if there are multiple lines through a point parallel to a given line, how do we decide which one to use to construct our vector sum?

Answer: we'll bypass this question! (12/28)

Here is how we will add the vectors OA and OB:

1) Reflect point B about the point O to get point B'.
2) Find the midpoint M of OA.
3) Reflect B' about M to get C.
4) OA + OB = OC

(13/28)

The vector AC is parallel to OB in absolute geometry. The line AC is actually the same line that Euclid's Book I, Prop. 31 constructs. I won't prove that here because it's not the point of this thread. We are interested in OC and in the addition operation we just defined. (14/28)

Let's see what this construction looks like in the hyperbolic plane! Here is a picture in the Poincare model. We choose O to be the origin, so our vectors are straight line segments to us outside Euclidean observers. Note the hyperbolic line segment through B', M and C. (15/28)

I need to pause here to mention that in the pictures that follow, the labeling of points is less than optimal. This is because GeoGebra is very limited in what characters it allows as labels. Hopefully it will be clear from context. (16/28)

You can probably already guess that our operation + is not commutative. This picture illustrates that. It is not a coincidence that x+y and y+x differ by a rotation. More on that later. (17/28)

If we use -x to denote the vector obtained from reflecting x about the origin, then the following identity holds:
-(x+y) = (-x)+(-y). Here is an illustration. (18/28)

As you have probably guessed from the title of this thread, + is not associative in the hyperbolic case. In the picture, (OA+OB)+OC is labeled ABpC and OA+(OB+OC) is labeled ApBC.

Here is a takeaway for you:

*** Associativity is Euclidean ***

(19/28)

What is true about +, then, besides -(x+y)=(-x)+(-y)?
Well, we do have (-x)+(x+y)=y.
These are actually illustrated by the same picture!

The second identity implies that equations of the form a+x=b can always be solved uniquely for x:
x=(-a)+b. (20/28)

Even more, we have the following generalization of associativity:
(x+(y+x))+z = x+(y+(x+z))
This is called the Bol identity (after Gerrit Bol). Unfortunately the picture is very cluttered, so I won't blame you if your eyes just glaze over it. (21/28)

Summary of properties:
0+x=x=x+0
-x+(x+y)=y, x+(-x)=0, -(x+y)=(-x)+(-y)
(x+(y+x))+z = x+(y+(x+z))
A magma (D,+) with these properties is called a Bruck loop (and also other names), after R.H. Bruck. Let me explain what loops are. (22/28)

A quasigroup is a magma (Q,*) such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y in Q. A quasigroup is a loop if it has an identity element. (23/28)

The structure we are discussing is a loop. The identity element is the fixed origin. We already discussed why a+x=b is uniquely solvable for x. The other equation y+a=b is also uniquely solvable for y, but it is messier to illustrate. (24/28)

Now the Poincare model is actually the unit disk D={z:|z|=1} in the complex plane. So you may wonder if there is a formula for our new vector addition. There is indeed! I have denoted our new operation by a dotted plus. (25/28)

Notice that the noncommutativity is measured by a rotation, that is, an element of the circle group S^1, exactly as the picture told us.

What isn't obvious is that the same element measures nonassociativity! (26/28)

This thread is already quite long, so here is a summary and some final remarks.

Vector addition makes sense in absolute geometry but it only gives an abelian group in the Euclidean case. In the hyperbolic case, it gives a nonassociative Bruck loop! (27/28)

Bruck loops are related to many other mathematical structures, including quandles and symmetric spaces. More generally, quasigroups and loops show up in many places in mathematics. I hope to touch upon these in future threads.

That's all for now. Thanks for reading! (28/28)

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