Tuesday, July 1, 2025

Tuesday Special - Tetractys and Lattice Infinity

 

9 + 16 = 25

32 + 42 = 52.


a2 +b2 = c2.

Real magic. The Pythagoras Theorem. Except that it does not belong to Pythagoras. Pythagoras was a mystic, not a mathematician.

“There is not a single mathematical sentence that can be attributed with certainty to Pythagoras as an individual. The image of Pythagoras as a mathematician and scientist is a construction of later times. In contrast, the belief in the transmigration of souls and the religious character of the Pythagorean community is consistently and early attested. The historical Pythagoras appears not as a scientist, but as the founder of a way of life.”
Walter Burkert, Lore and Science in Ancient Pythagoreanism, trans. Edwin L. Minar Jr., Harvard University Press, 1972, p. 113

"Pythagoras appears as a figure of the shaman type, especially in his claim to possess a soul of a special kind and in his ability to recall its previous incarnations. He was thought to have a miraculous, semi-divine origin; he performed healings and miracles; he was able to travel great distances in a moment, to be in two places at once, and to communicate with the dead. These are traits which correspond to shamanistic ideas and practices found in Central Asia and the Near East."
— Walter Burkert, Lore and Science in Ancient Pythagoreanism, trans. Edwin L. Minar, Jr. (Harvard University Press, 1972), p. 298

Tetractys

But, it is said that the Pythagorean Brotherhood worshipped the Tetractys: 1- the Monad, 2 - Duality, 3 - the Triad (beginning of form), 4 - the Tetrad (the World).

Back to (3,4,5). Pythagorean triples ware know to Babylonians ~1800 BCE. They knew (3, 4, 5), (5, 12, 13), (7, 24, 25), and wrote it in cuneiform. They knew even larger triples, such as (119,120,169). But in the Pythagorean Brotherhood there were initiation rites and levels of access. I think that the members of its inner circle knew that beyond the popular Triad a2+b2=c2, at a deeper level, there is the well balanced Tetrad:

  a2 + b2 =c2 + d2, (a,b,c,d integers).            (1)

Of course tetrads contain triads, when one of the numbers is zero. So, those who know Tetrads, know also Triads, but not the other way. Of course we are interested in primitive solutions, that is in cases where a,b,c,d have no common divisor.  I could not find anything about tetrads in Babylonia, but I found them on math.stackexchange: Diophantine equation a2 + b2 =c2 + d2. The complete solution can be found in the textbook L.J. Mordell, "Diophantine Equations", Academic Press 1969, on p. 15.

Well, it is not explicitly complete there, it is somewhat sketchy, but here it is (I skip the proof).

Proposition 1. Every primitive solution of  (1) is of the form

a = (mp+nq)/2,
b = (np-mq)/2,
c = (mp-nq)/2,
d = (mq+np)/2,

where m,n,p,q are integers. Conversely, for any integers m,n,p,q such that a,b,c,d are integers, the formula above provides a solution of  a2 + b2 =c2 + d2.

What it has to do with the compactified Minkowski space?

Recall the quadratic form Q from The infinity ab initio:

Q(X) = (X1)2 + (X2)2 + (X3)2 -(X4)2+ (X5)2 - (X6)2,

and the equation of the null cone N:

(X1)2 + (X2)2 + (X3)2 - (X4)2 + (X5)2 - (X6)2 = 0.

Skip two dimensions X2 and X3:

(X1)2  - (X4)2+ (X5)2 - (X6)2 = 0.

Rewrite it as

(X1)2 + (X5)2  = (X4)2 + (X6)2 .            (2)

Then set X1 = a, X5 = b, X4 = c, X6 = d, and you get (1). In (2) we are interested in the projective cone PN. Each tetrad a,b,c,d satisfying (1) determines a unique point (a',b',c',d') = λ(a,b,c,d)  on PN for which a'2+b'2 = c'2+d'2 = 1. We get a point on the torus S1S1. That is the image of our compactified Minkowski space with suppressed two space-like dimensions. By assuming the X components are integers we assume that ℝ4,2 is a regular discrete lattice.

And so I did. Using Proposition 1 I generated about 1 mln of tetrads (a,b,c,d) and plotted them on a torus. Here is the result:

627169 + 627169 + 4446 points

There is some pattern there that can be seen. I do not understand where this pattern comes from, except, perhaps, of two circles around the torus. We will meet these circles soon. They represent  infinity.


Wednesday, June 25, 2025

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 3. 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. KirbyMathematics 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 (9.34109, 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

XY if and only if = 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 S3S1. 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 ℝ3,1. Thus the scalar product in M can be written as

x·y = xTηy,                (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 EA in V, (A=1,2,...,6), is called orthonormal if EA·EB =GAB, where G is the diagonal matrix

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

We denote by XA the coordinate of X with respect to such a basis: X = XA EA.

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 ≐ X+ 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

E0½(E5 - E6),                (13)

E½(E5 + E6).                (14)

Using these coordinates the embedding formula takes the form:

X(x) = xμEμ - q(x)E + 1 E0.                (15)

The vectors E0 and Eare 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.

Monday, June 23, 2025

From Spheres and Circles to Spacetime — Evolving Coordinates

Here we continue from the June circles.  There will be changes in the notation. Changes always create temporary chaos. We have to learn how to survive when our environment undergoes change. Survival of the fittest — that rule seems to govern all evolution in the universe. Survival of the fittest and adaptation. A cactus that retains water well survives in arid deserts. A person who learns from failure may “survive” mentally and socially better than better than someone who rigidly resists change. Will my readers survive after this post?



But why the changes? Here is the reason. The Lie sphere geometry has been discovered while studying the geometrical properties of spheres. It was pure mathematics, but, surprisingly, it has found application in relativistic physics. The idea goes as follows. Consider the simple equation of a circle on the plane:

x2 + y2 = r2.

Now rewrite it as

x2 + y2 - r2=0.

Now substitute r=ct:

x2 + y2 - c2t2=0.

This is the equation of the light cone of special relativity. The radius of the circle expands with the velocity of light. The value of this radius can serve as a time coordinate of an event. If a circular wave was created at a point x and time t, when it reaches the origin x=0, its radius has the value r=ct. Therefore it should not be surprising that studying circles we study spacetime of special relativity. This is will how it will go.

We will denote the coordinates of the 2+1 dimensional Minkowski space by x,t, and coordinates in the 5-dimensional space R3,2 of Lie spheres by ξ = (ξ01245) with the quadratic form

ξ·ξ = 0)2 + (ξ1)2  + (ξ2)2  - (ξ4)2 - (ξ5)2.                 (0)

With

q(x,t) = x2-t2,             (1)

the formula (12) of the previous post embeds the Minkowski space in the projective quadric Q+ would read:

ξ(x,t) =  ½(1 − q(x, t)) e0 + x  + t e4 + ½(1 + q(x, t)) e5,                 (2)

Note: In what follows I  will use the term 'projective' to mean 'semi-projective', where instead of lines we consider half-lines in R3,2 (λ>0, instead of λ≠0 in the equivalence relation).

Environmental change — the fittest survive.

To obtain (2) we used a particular convention for representing oriented spheres (circles in our case) as points of a space with 1+2 more dimensions. Let us compare our formula with the formula from the paper "On organizing principles of discrete
differential geometry. Geometry of spheres
" by A.I. Bobenko and Yu. B. Souris, Russian Math. Surveys 62:1 1–43. Here is the corresponding extract from this paper:




To relate it to our Eq. (2) we replace, in the formula (4) of the paper, c by x, r by t, and substitute (3) to obtain

ξ= - ½(1 − q(x, t))eN+1 + x + teN+3 + ½(1 + q(x, t))eN+2.

We see eN+3 corresponds to our e4, eN+2 corresponds to our e5, and eN+2 corresponds to our -e0.  It will be more convenient for us in the future to follow the convention used in this paper, particularly regarding the sign of e0. On the other hand, it will also be convenient to number the indices as 1,2,3,4,5. Thus, in what follows, we will use the basis E1=e1, E2=e2, E3=e4, E4=e5, E5=-e0. We will write X=(XA), A=1,...,5):

X = XAEA

with

X·X = (X1)2 + (X2)2 -(X3)2+(X4)2- (X5)2.

For this reason in the future we will replace our embedding formula (2) with the following one:

X(x,t) =  x  + t E3 + ½(1 + q(x, t)) E4 - ½(1 − q(x, t)) E5 ,                 (2a)

In our coordinates it reads:
X1
(x, t)  = x1
X2
(x, t)  = x2,                 (3)
X3
(x, t) = t,
X4
(x, t) = ½(1 + q(x, t)),
X5(x, t) = -½(1 − q(x, t)) ,         

We notice that

X4(x, t) - X5(x, t) = 1.                (4)

However, this last formula is not invariant under selecting an element from the equivalence class defining the projective space. The invariant formula is, instead:

X4(x, t) - X5(x, t) > 0.                (5)

Notice that we have

X(x,t)·X(x,t) = 0,                (6)

and this equation defines the projective null cone in the projective space — our Q+ universe. It then follows from the definitions that Q+ splits into a disjoint union of three sets:

Q+ = M+ ∪ M- ∪ M,                (7)

where

M+ = {[X]: X0+X5 >0},            (8)

M- = {[X]: X0+X5 <0},            (9)

M  = {[X]: X0+X5 = 0}.            (10)

The sets M+ and M- are  open in Q+, the set M  is closed (Why?). The map (x,t) ⟼ [X(x,t)] is bijective from R4 to M+. Similarly the map  (x,t) ⟼ [-X(x,t)] is bijective from R4 to M-. The map [X] ⟼ [-X] is a bijection from M+ onto M-.

Exercise 1. Prove the above statements.

Exercise 2. Prove that M+M- = ∅.


Conformal structure

 New post on my other blog