The infinity ab initio

 After a month of silence, I’ve returned to sharing my thoughts with two new posts: “June Circles” and “From Spheres and Circles to Spacetime — Evolving Coordinates.” But to be honest, it didn’t go well. That month away seems to have cost me my coherence. A few loyal readers were glad to see me back—but less enthusiastic about the content. What I wrote came out muddled and confusing. So, it’s time to begin again—from scratch.

Interestingly, in theoretical chemistry, the phrase ab initio ("from the beginning") often appears in paper titles. I rarely see it in physics or mathematics, but here, it feels just right. This will be my third attempt, and I’m starting ab initio, from chaos toward order. As the saying goes, “On the third knock, the door opens.” Let’s see if it does.

My plan is to discuss the infinity. The infinity point of space and time. It all started with projective geometry of an ordinary two-dimensional plane. Parallel lines should meet at some point "at infinity". Different bundles of parallel lines meet at infinity at different points. Thus projective geometry added "the line at infinity". If we replace 2D plane by 3D space, we need to add a point at infinity for each bundle of parallel lines in ℝ³. This is widely used in computer aided graphics. Here is an excerpt from the paper "Beyond the Celestial Sphere: Oriented Projective Geometry and Computer Graphic" by Kevin G. Kirby,  Mathematics Magazine, Vol. 75, No. 5 (Dec., 2002), pp. 351-366.

To infinity ...

You are driving a virtual car in a computer game. Look out through the windshield. The trees that line  the highway  are rushing past you.  They are nicely  displayed in perspective: they begin far away as dots, but they grow taller as you race toward them. Next, look ahead at the horizon. You see the sun. Unlike the trees, the sun never gets closer to you. Still, it is certainly subject to some transformations: you turn your car left, and the sun veers right.
The trees and the sun need to be represented inside the program somehow. Ultimately, this depends on attaching parts of them to points in a virtual space. One might think at first that a useful way to represent any point would be to use Cartesian coordinates, a triple (x, y, z) of real numbers. Taking your car to be located at the origin, a tree might be centered at, say, (-20.42,  10.63, -94.37).  Where is the sun? It is very far away, perhaps at (9.34⨉10⁹, 2.71⨉10⁹, -1.23⨉10¹¹). But there is  something strange about these large numbers. It seems pointless to waste time (and numerical precision) decrementing such big numbers by a few tens as our car drives on toward the sunset. We would like to simplify things by somehow locating the sun at infinity.

One uniform way to represent both ordinary points and points at infinity is to use four numbers instead of three. Here's how it works. Take the point (2, 3, 4).  Instead of representing it as a column vector in ℝ³, we tack a 1 on the end and represent it as a column  vector  in  ℝ⁴  : x  = [2 3 4  1]^T. This representation is  not  meant  to be unique: we can multiply this column vector by any positive number and we will say it represents the same point.(...)

You are driving a virtual car in a computer game

That may be good for traveling in space. But we want to travel in space and in time. For lines in space-time we sometimes use the term world line. There are three kinds of world lines: we can travel with a speed that is slower than the speed of light, faster than light, or with exactly the speed of light.   There will be, perhaps, three kinds of infinity points. There are two known mathematical ways to achieve that, using a trick similar to that used in projective geometry. The first way, the standard one,  is to start with space-time and add two extra dimensions, that is to add a hyperbolic plane with signature (+1,-1), and then to study the projective null cone there. This is called Möbius geometry. The second way, the way I am following, is the Lie sphere geometry. We start with 3D space alone, and time emerges automatically, related to the radius of a sphere. We add three extra dimensions to the three dimensions of space, we add extra dimensions with signature  (+1,-1,-1), and take the projective null cone there. The end result is mostly the same - we end up with six dimensions and signature (+1,+1,+1,-1,+1,-1). But, in the Lie sphere geometry approach, we obtain a reacher structure. I like this second approach better, as it fits the possibility of taking into account the possible  'ether', or 'quantum vacuum', with a preferred reference frame.

In the previous two posts I was oscillating between circles and spheres. Finally I decided to take spheres, and restrict to circles only when it will be more convenient for graphical illustrations. So, let us start, ab initio.

There are two ways of playing with geometrical constructs. The first way goes back to Descartes - we use coordinates. The second way goes back to Euclid - it is coordinate-free. Nowadays, in practice, we often first use coordinates, and only after the result is obtained, we work on expressing our result in a coordinate-free way. I will follow this path. We will be using coordinates at first, and only then search for a way to understand what  have been done in a more elegant and, perhaps, deeper way.

Notation.

Let V be a 6-dimensional real vector space with the quadratic form Q of signature (4,2).

Note: In previous post I have used Q to denote the Lie quadric. But since now we start ab initio, therefore the symbol Q is being cleared from its previous meaning. We will use X,Y,... to denote vectors of V. The scalar product in V will be written as X·Y.  Thus

Q(X)=X·X.                (1)

We denote by N the null cone in V:

N = {X∈V: Q(X) = 0}.                (2)

In V⟍{0} (V with the removed origin) we introduce the equivalence relation

X∼Y if and only if Y = λX, λ>0.                (3)

We denote by PV and PN the sets of equivalence classes of vectors in V⟍{0} and in N⟍{0}:

PV = V⟍{0} / ℝ^×_>0,                (4)

PN = N⟍{0} / ℝ^×_>0,                (5)

where ℝ^×_>0 is the multiplicative group of all strictly positive real numbers.

We denote by π the natural projection from V to PV. It maps every non-zero X in V to its equivalence class [X].

V is 6-dimensional, N⟍{0} is 5-dimensional, and so PN is 4-dimensional. That is the central object of our studies. As it will be seen in the future, it is the (doubly) compactified Minkowski space, diffeomorphic to the product S³⨉S¹. It was denoted Q⁺ in previous posts.

We can  use π to equip PN with a natural topology and with a differentiable structure. We will do it later.

The flat Minkowski space-time M will be identified as a particular open, dense set in PN. First we will do it using coordinates. For this we will use orthonormal bases for M and for V. Using an orthonormal basis in M we identify M with ℝ^3,1. Thus the scalar product in M can be written as

x·y = x^Tηy,                (6)

where η is the diagonal matrix η=diag(1,1,1,-1). The quadratic form q of M is then

q(x) =  x·x = (x¹)² + (x²)² + (x³)² - (x⁴)².                (7)

Note: sometimes we may use the letters x¹=x, x²=y, x³=z, x⁴=t. We will also use the notation x = (x,t), where x is a vector in ℝ³,  and t is a real number (time). In this case x·x = x² - t².

A basis E_A in V, (A=1,2,...,6), is called orthonormal if E_A·E_B =G_AB, where G is the diagonal matrix

G = diag(1, 1, 1, -1, 1, -1).                (8)

We denote by X^A the coordinate of X with respect to such a basis: X = X^A E_A.

The embedding

We will now define the embedding of M into PN. It is defined by the following map from M to V

X(x) = ( x, ½(1 - q(x, t)), -½(1 + q(x, t)) ).                (9)

Then the embedding, which will be denoted by  τ is defined by

τ(x) = [X(x)].                (10)

Exercise 1. Prove that if τ(x) = τ(x'), then x=x'. 

We notice that if X = X(x), q = q(x,t), then X⁵ - X⁶ = 1, X⁵ + X⁶ = -q. We define 

X⁰ ≐ X⁵ - X⁶.                 (11)

X^∞ ≐ X⁵ + X⁶,                 (2)

Thus the embedding formula looks simpler if we use, instead of the two basis vectors E₅ and E₆,  vectors E₀ and E_∞ defined as

E₀ = ½(E₅ - E₆),                (13)

E_∞ = ½(E₅ + E₆).                (14)

Using these coordinates the embedding formula takes the form:

X(x) = x^μE_μ - q(x)E_∞ + 1 E₀.                (15)

The vectors E₀ and E_∞ are now in N (Why?), therefore they define two points p₀ = [E₀], and p_∞ = [E_∞] in PN. We also notice that 

E₀·E_∞ = 1/2.                 (16)

The point p₀ is the image τ(0) of the origin x = 0 of M. The point p_∞ does not correspond to any point in M, it is not in the set τ(M). It is one of the points of the infinity set, which we define as

M_∞ = {[X]: X⁰ = 0} = {[X]: X⁵ = X⁶}.                (17)

In the next post we see that PN is a disjoint union of three sets

PN =  M₊ ∪ M_- ∪ M_∞,                     (18)

where

M₊ = {[τ(x)]: x ∈ M},                (19)

M_- = {[-τ(x)]: x ∈ M}.                (20)

Thus PN, the doubly compactified Minkowski space, consists of two copies of M, and of the infinity set M_∞. We will discuss in details the structure of M_∞, and also the intuitive meaning of these 'points at infinity'.

In all this I am borrowing ideas from what is called 'oriented projective geometry', as described in the computer graphics review paper by Kirby, mentioned at the beginning. In oriented projective geometry one cares about the direction of the line. For space-time that means that we want to distinguish, in particular, between the future and the past. This is not usually done in the standard conformal compactification of the Minkowski space. I am not so sure about the necessity of distinguishing between left and right, but if it gives better algorithms for the computer graphics, perhaps, for efficiency reason, it is also exploited in the Nature.

P.S. 26-06-25 17:09 Anna was kind to send me the following information:

"I know you're busy and excited about your new post, but I can't resist showing you these dual turrets of Varlamov, because it seems to me that they express exactly what you said about two-sided Universe. These are «two three-dimensional projections (SO(4, 2)-towers) of the weight diagram of the Lie algebra so(4, 4) of the rotation group SO(4, 4) of the eight-dimensional pseudo-Euclidean space R(4,4)."

It is a puzzle how the conformal group SO(4,2) enters Varlamov's picture!

P.S. 03-07-25 13:55 Corrected signs in (9) and (15).