An SO(2,2) Iterated Function System Part 5 - Back to July 1

 Researchers believe that cats have an abstract understanding of numbers, often up to about seven. Some claim that mother cats can count as high as six or seven, though three or four is more likely. In contrast, humans seem able to count much further. George Cantor, for example, explored numbers reaching to infinity—and even beyond. Personally, I find numbers challenging, both in practice and theory. Number theory has always been a painful experience for me. 

Still, when there’s a need, there’s always a way. So, in this blog post, we draw a final dot to mark the end of our journey through the ring of integer.

Let us start with a quote from July 1 post "Tuesday Special - Tetractys and Lattice Infinity":

 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.

There I quoted a Proposition from Mordell's book, without a proof. But now, with Proposition 1 from the previous post,  we have a complete proof of Mordell's statement. Let us discuss this in details.

Suppose a,b,c,d are integers satisfying a² + b²  = c² + d². Let x be a vector in R^2,2 with components (a,b,c,d). Then x is a null vector: Q(x) =  a² + b²  - c² - d² = 0. The matrix x^, defined as in  Part 2, is

x^ = {{c+a,b+d},{b-d,c-a}}                (1)

is, automatically, of determinant zero, with integer  components. Thus we can apply Proposition 1 from Part 4 to deduce that there are vectors v,w with integer components such that

x^ = vw^T.                 (2)

Let (p,-q) be the components of v, and let (m,n) be the components of w. Then (2) takes the form

{{c+a,b+d},{b-d,c-a} = {{pm,pn},{-qm,-qn}}            (3).

From (3) we immediately get

a = (pm+qn)/2,

b= (pn-qm)/2,

c = (pm-qn)/2,

d = (pn+qm)/2,

which are exactly the formulas from Proposition 1 in Mordell's text. In order to get the exact correspondence we have set (p,-q) to be the components of v, but that does not really matter, since if q runs through all integers, so does -q.

An SO(2,2) Iterated Function System Part 4 - Integer spinors

 The great merchant, Hypatia of Alexandria, was dead. In her will, she left her vast fortune not to a person, but to a puzzle. Her wealth was locked in a magnificent chest with not one, but three complex locks.

Her will stipulated: "My fortune shall go to the first of my former students who can provide the number of coins in this chest. To aid you, I can say this: the number is more than 100 but less than 200. If you attempt to count the coins by twos, threes, fours, up to tens, there will always be a specific number of coins left over."

A young scholar, Euclid, was the first to arrive. He listened to the executor read the conditions and immediately asked for the remainders.

"The remainders are as follows," the executor said, unrolling a scroll.

Divisible by 2: Remainder 1

Divisible by 3: Remainder 2

Divisible by 4: Remainder 3

Divisible by 5: Remainder 4

Divisible by 6: Remainder 5

Divisible by 7: Remainder 0

Divisible by 8: Remainder 1

Divisible by 9: Remainder 2

Divisible by 10: Remainder 3

Euclid's face fell. "This is chaos! The conditions are not coprime. Four, six, eight, nine, and ten are all composite! This will take me a week of calculations!"

Just then, his rival, the quick-witted Sun Tzu, arrived. He glanced at the list and chuckled. "You are thinking like a laborer, Euclid, not a mathematician. You are looking at the locks; I am looking at the key."

"And what key is that?" Euclid scoffed.

"The key," said Sun Tzu, "is to see what the number almost is, not what it is. Look at the list. What do you see?"

I will skip the rest of the story. Stories from the past are fine, but we have a story of the present,  and it needs your full attention. Here it is.

In this note we continue the discussion started in An SO(2,2) Iterated Function System Part 3 - Weyl spinors and null vectors, but now replacing the field of real numbers R by the ring of integers Z. We will show that the following Proposition holds:

Proposition 1. Let A =  A = {{a,b},{c,d}} be a non-zero matrix with integer entries, A ∈ Z^2⨉2, with det(A) = ad - bc = 0. Then there exist vectors v,w ∈ Z² such that

A = vw^T.                (1)

The factorization is unique up to a sign: if A = vw^T, then A = (-v)(-w^T) is the only other (trivial) variation.

The proof of this Proposition uses the concept of the Greatest Common Divisor (gcd). The standard definition of gcd is:

Definition. For any two integers a,b, at least one of which is nonzero,  we denote by gcd(a,b) their greatest common divisor, that is the greatest integer d such that there exist integers e,f such that a = de, b = df. In particular, if (a) is nonzero, we have  gcd(0,a) = gcd(a,0) = |a|.

But that definition is tailored for integers, while the concept of gcd is more general, and it has to do with the Unique Factorization Property. So here is a deeper and more formal definition (for non-zero integers)

In other words gcd(a,b) is the product  it's the product of all common primes, each raised to the power of the smallest exponent found in either number. The signs of the original integers do not affect the result.

Before proving the Proposition, we first prove the following Lemma

Lemma 1. Let a,b,c,d be integers such that gcd(a,b)=1 and ad = bc. Then there is an integer k such that

c = ka,

d = kb.                (1)

Proof. First note that if a = 0 and b = 0, then gcd(a,b) is undefined, therefore this case is implicitly excluded by the assumption that gcd(a,b) = 1. Suppose a = 0 and b ≠ 0. Then gcd(0,b) = |b|, and, since gcd(a,b) = 1, it follows that b = ± 1. From ad = bc it follows that c = 0. Take k =  ±d. Then d = kb and c = ka. Similarly if b = 0 and a ≠ 0. Let us now assume that both a ≠ 0 and b ≠ 0. By assumption  gcd(a,b) = 1 and ad = bc. Think of the unique prime decomposition of (b) on the right hand side of ad = bc. None of these primes entering the composition of b is in (a), since gcd(ab)=1. Therefore they must be all contained in d. Therefore there is an integer k such that d = kb. But then, from ad = bc, it follows kab = bc, and, since b ≠ 0, we have c = ka. QED                                                                                      

We can now return to proving Proposition 1.

Proof (of Proposition 1).  Since A is a nonzero matrix, at least one of it rows is nonzero. Suppose it is the first row: (a,b). For the second row the proof goes in a complete analogy.

Let g = gcd(a,b). Since at least one of a,b is nonzero, g = gcd(a,b) is well defined. Thus g is a positive integer, and there exist integers a',b' such that

a = ga',
b= gb',                (2)

and gcd(a',b') = 1. Since ad = bc, and g is positive, from ad = bc it follows that a'd = b'c. We can now apply Lemma 1 to deduce that there exist an integer k such that

c = ka',
d = k b'.            (3)

From (2) and (3) it follows that A is of the form:

A = {{ga',gb'},{ka',kb'}}. Setting v = (g,k)^T, w = (a',b') we have A = vw^T. QED

It follows also from the proof that the decomposition is unique if we require the two components of (w) to be coprime (i.e. if their gcd is 1).

Exercise 1. Find the decomposition of the matrix  x^, where x = (3,14,6,13) as mentioned  in the Example Part 2.

To be continued...

An SO(2,2) Iterated Function System Part 3 - Weyl spinors and null vectors

 Introduction.

In the early 1960s, Penrose was deeply preoccupied with a fundamental problem in physics: how to describe the geometry of spacetime in a way that naturally incorporated quantum mechanics and the behavior of light. He was frustrated with the standard mathematical tools and felt there must be a more profound, elemental description of reality.

The pivotal moment came in 1963. Penrose was a visiting professor at the University of Texas at Austin. He was not in his office, but was driving with a colleague (some accounts say it was the physicist Ivor Robinson) outside of the city.

As he was gazing out the car window at the flat, featureless Texas landscape, his mind began to wander.

The long, straight highway and the vast, open horizon triggered a shift in his perspective. He started thinking about the paths of light rays—how they could be seen as fundamental, and how points in spacetime might be a derived concept from the way these light rays intersect.

The key insight was this: Instead of thinking of space as the primary concept and light rays moving through it, what if he reversed the roles? What if the light rays (the "null lines" or paths of photons) were the primary objects, and a "point" in spacetime was defined as the set of all light rays passing through it?

This was the genesis of twistor theory. In that moment, he realized he needed a new mathematical space—what he would later call twistor space—where each point represents a light ray in our physical spacetime. The geometry of our universe could then be encoded in the complex geometry of this twistor space.

The Humorous Aftermath
The anecdote often includes a charmingly human detail. The flash of inspiration was so intense and all-consuming that Penrose, excitedly trying to explain his radical new idea to his colleague, began scribbling equations and diagrams on the car's dashboard.

One can imagine the driver's mixed feelings about having their car used as a blackboard for groundbreaking theoretical physics!

We continue from An SO(2,2) Iterated Function System Part 2. The two-dimensional real vector space R² of Weyl spinors for Spin(2,2) carries an SL(2,R)-invariant bilinear form ε. In order to distinguish spinors from vectors, from now on, we will use Greek letters ψ etc. to denote the elements of R² endowed with this form. Thus

ε(φ,ψ) = φ^Tωψ.                (1)

It is then natural to introduce what physicists call the Dirac conjugated spinor

φ^bar = φ^Tω,                 (2)

so that the invariant bilinear form ε can be written as

ε(φ,ψ) = φ^bar ψ.                (3)

Exercise 1. Verify that φ^bar φ = 0 for all φ.

Null Vectors from Weyl spinors

Now we we are ready redefine the construction of determinant zero matrices from spinors. We define now

X(φ,ψ) = φ ψ^bar = φ ψ^Tω.                (4)

Now, for (S,T) in SL(2,R)⨉SL(2,R), we have

X(Sφ,Tψ) = S X(φ,ψ) T^-1,                 (5)

the standard transformation law of vectors x represented by 2⨉2 matrices x^.

Exercise 2. Verify Eq. (5).

Note. Our "spinors" are spinors of the group SO(2,2), which is the conformal group of R^1,1. Therefore we should, in fact, call them "twistors". They are of course our toy baby  twistors. The "adult" twistors of Roger Penrose are spinors of SO(4,2).

Spinors from null vectors.

In the construction below we will first take a purely pragmatic approach, without discussing its geometrical meaning. So, let

A={{a,b},                    (6)
      {c,d}}

be any nonzero matrix of determinant zero. We will show that A is necessarily of the form

A =  φ ψ^{bar                (7)}

for some φ, ψ.

Since it is at non-zero matrix, at least one of its elements must be non-zero. Suppose it is the first row, first column, element A₁₁ = a. Let us define  φ' to be the column vector equal to the first column of A:

         a
φ' = (    ),                (8)
        c
and let ψ'^bar be the first row

ψ'^bar = (a,b).                (9)

Construct A' =φ' ψ'^bar . This the matrix

A' ={{aa,ab},{ca,cb}}.                (10

The zero determinant condition implies ad = bc. We can thus substitute cb in A' by ad. Then A' becomes

A' = {{aa,ab},{ca,ad}} = a {{a,b},{c,d}} = a A.                (11)

Therefore setting

         1
φ = (    ),                (12)
       c/a

ψ'^bar = (a,b),                (13)

solves our problem.

Exercise 2. Can a similar reasoning be used assuming, for instance, that it is b ≠ 0, instead of a ≠ 0 as above?

Exercise 3. Show that the decomposition (7) is essentially unique, that is that if  φ', ψ' is another solution of (7), the there is a constant λ≠0, such that

φ' = λφ, ψ'= (1/λ)ψ.                (14)

Hint: the exercise may need the concepts of a kernel and a range of a matrix considered as a linear operator. Using AI for help is allowed, and even encouraged.

In the next post we will solve the same problem but replacing real numbers with integers. Since division is not allowed within integers, we have have to use a more sophisticated approach in that case.

To be continued...