Conformal transformations in 1+1 dimension and SO(2,2) - Part 1

 In a recent paper by Jack Sarfatti: "Dark Energy, Dark Matter, Low Power UAP Warp Drive Time Travel Poincare Gauge Theory" he suggests to use the Poincaré group gauge theory to explain Dark Matter, Unidentified Aerial Phenomena and Time Travel.

Dark Matter, Unidentified Aerial Phenomena and Time Travel

While Poincaré  Group Gauge Theory includes torsion naturally, in my opinion it is not the right choice. A better choice would be Conformal Group Gauge Theory. Conformal group contains the Poincaré group as a subgroup. While the discrete mass spectrum of elementary particles  seems to break conformal invariance, electrodynamics is naturally conformally invariant, and it seems to me  that it is light that is at the foundations of all being, a bridge between material and non-material. So this post will take us closer to the very heart of the conformal group, even if it will be only for 1+1-dimensional spacetime that we are playing with in our sandbox here.

The pdf below is an extract from Notes, just the last chapter. The whole document is here, while this is a link to the part shown below. It is just the beginning, In the next part we will go from SO(2,2) to SL(2,R)xSL(2,4), and then we have a look at graphical representation of transformations both in the Minkowski space and in its conformal double compactification 

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...

An SO(2,2) Iterated Function System Part 2

 Everybody knows Pythagorean triples a² + b² = c². (3,4,5) is a beautiful example. But triples are a particular case of balanced quadruples a² + b² = c² + d². When d=0 (or a=0, or b=0, or c=0) we have a triple. But what if neither of them is zero? What would be the simplest example? Of course a=c and b=d is always a solution, but that would cheating. Perhaps (3,14,6,13) - check it! - and its variations, is a good try? Relatively easy to remember, though not as easy as the famous triple (3,4,5) of Pythagoras.

L.J. Mordell, in his textbook  "Diophantine Equations", Academic Press 1969, provides, on p. 15, a general formula for generating all balanced quadruples using a rather straightforward number-theoretic reasoning. I have quoted the formula in the last post "An SO(2,2) Iterated Function System Part 1". Here we will derive essentially the same formula using algebra and geometry of Weyl spinors of the group SO(2,2).

Let us start with recalling some of the formulas that we have already discussed. We realize the pseudo-Euclidean space R^2,2 as Cl(2) ≈ Mat(2,R). In Mat(2,R), which is 4-dimensional, we have selected  a basis (cf. Eqs. (1)-(4) in the Notes):

e₁ = {{1,0},{0,-1}};
e₂ = {{0,1},{1,0}};
e₃ = {{1,0},{0,1}};
e₄ = {{0,1},{-1,0}};

Each vector x = (x¹,x²,x³,x⁴) is then represented by the matrix x^:

x^ = x¹ e₁ + x² e₂ + x³ e₃ + x⁴ e₄ =
    = {{x¹ + x³, x² + x⁴},
         {x² - x⁴, -x¹ + x³}}.

We have then:
 - det(x^) = (x¹)² + (x²)² - (x³)² - (x⁴)²,

so that the quadratic form of R^2,2 determining its geometry is encoded in the determinant of the matrix.

Example:  x = (3,14,6,13),
x^ =
{{9, 27},
  {1, 3}},
det(x^) = 0. Columns of x^ are linearly dependent: the second column is 3 times the first column. Rows are linearly dependent. The first row is 9 times the second row.

There is also another way in which we can arrive at the scalar product of R^2,2. For this we use the volume element ω = e₁e₂ of the Clifford algebra Cl(2). In our case it coincides with e₄ matrix. Its square is minus the identity, matrix, so that ω^-1 = - ω. If, for any 2⨉2 matrix A, we define

ν(A) = -ω A ω,

then

-½ Tr( x^ ν(y^) ) = x¹y¹ + x²y² - x³y³ - x⁴y⁴,

which is precisely the scalar product (x,y) in R^2,2.

We will use the matrix ω later on, for a different purpose though.

We are particularly interested in the null cone of R^2,2, that is in the set of all vectors x in R^2,2 for which (x,x) = 0, that is for which det(x^) = 0. In fact we are interested in the "projective null cone" PN consisting of all equivalence classes of non-zero null vectors, where we identify any two null vectors if one is proportional to the other with a strictly positive proportionality constant. But this will come later. For now we are interested in the null cone, excluding the trivial zero vector. Every non-zero vector in N defines then a quadruple of real numbers such that

(x¹)² + (x²)² - (x³)² - (x⁴)² = 0.

If the components xⁱ happen be all integers, we have a "balanced quadruple". So the question is: how can we generate all determinant zero matrices with integer coefficients? And, even before that, how can we generate all non-zero 2⨉2 matrices of determinant zero?

We first notice that if v is a column vector (a,b) and w is a column vector (c,d), then A(v,w) = vw^T is a matrix {{ac,ad},{bc,bd}} of determinant zero. Can every 2⨉2 matrix of determinant zero can be written in this way? The answer is "yes", and, for a non-zero matrix, v and w are unique up to a scaling. While this would work for generating the elements of N, there is one important little detail that needs to be taken into account. And it is better to take care of this little detail now, rather than later on.

Action of SL(2,R)

We know that the group Spin(2,2) is isomorphic to SL(2,R)⨉SL(2,R). If (S,T) is in SL(2,R)⨉SL(2,R), then it acts on Mat(2,R) by

(S,T): A ⟼ SAT^-1.

On the other hand SL(2,R) acts naturally on R²:

S: v⟼ Sv.

If A = vw^T, if v⟼ Sv and w⟼ Sw, then A⟼SAT^T, instead of A⟼SAT^-1, as we would like to have. What should we change to achieve the desired transformation law? To answer this question we need to understand the geometrical meaning of the action of SL(2,R) on R². The situation is somewhat (but only "somewhat") similar to that we have in Minkowski signature and the group SO(3,1). In Minkowski signature we have four-component Dirac spinors and two-component Weyl spinors. Here we also have Weyl spinors, and this is our R² with SL(2,R) action. What we need is an invariant scalar product in the space of Weyl spinors. Thus we need a non-degenerate bilinear form, let us call it  ε(v,w) that is invariant under the action of SL(2,R). It is a simple exercise to see that, up to a proportionality constant, there is only one such form, and it is

ε(v,w) = v^Tωw.

Exercise: Show that for every S in SL(2,R) we have ε(Sv,Sw) = ε(v,w) . In other words S^TωS = ω.

To be continued....

An SO(2,2) Iterated Function System Part 1

 This is the first, introductory part of a short series of posts explaining the meaning of the picture below:

The last part of "Notes on Clifford algebra Cl(2,0)" ended with two one-parameter subgroups of SL(2,R): LT(a) and RT(a). They are left- and right-triangular matrices with1 on the diagonal and parameter a in the lower left (resp. upper right) corner. Cf. Eqs. (47) and (48) in the notes. LT and RT can act on Cl(2) matrices from the left or from the right, with left and right actions commuting.

Now let us collect together several facts.

1. Our work area is R^2,2 - the Möbius extension of R^1,1, where  R^1,1 is the Minkowski space-time with only one space dimension. The signature of  R^2,2 is (++--).
2. The we restrict our attention to the null cone N in R^2,2. It consists of points with coordinates  (x¹,x²,x³,x⁴) which satisfy (x¹)² + (x²)² - (x³)² - (x⁴)² = 0, or (x¹)² + (x²)² = (x³)² + (x⁴)².
3. The note Tuesday Special - Tetractys and Lattice Infinity we have discussed "balanced tetrads: 
   a² + b² = c² + d², (a,b,c,d integers),           (1)
   and the algorithm of generating all of them:
   

   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.

4. Points on N with all four integer coordinates  are "balanced tetrads" as in 3.
5.  Each SL(2,R) transformation induces an SO(2,2) transformation of R2,2, that maps the null cone N into itself.
6. SO(2,2) matrices with integer coefficients map points of N with integer coefficients to other points of N with integer coefficients.  
   
7. Sl(2,R) matrices RT(2a), LT(2a), for a = 1 and a = -1, induce SO(2,2) transformations with integer coefficients.
   

The last observation follows from the explicit formulas (45) in Notes. Denoting by I₂ the 2x2 identity matrix, we have:

For left actions:

Λ(I₂, LT(2a)) = {{-1, -a, -a, 0}, {a, -1, 0, -a}, {-a, 0, -1, a}, {0, -a, -a, -1}};
Λ(I₂, RT(2a)) = {{-1, a, -a, 0}, {-a, -1, 0, -a}, {-a, 0, -1, -a}, {0, -a, a, -1}};

For right actions:

Λ(LT(2a), I₂) = {{-1, -a, a, 0}, {a, -1, 0, -a}, {a, 0, -1, -a}, {0, -a, a, -1}};
Λ(RT(2a), I₂) = {{-1, a, a, 0}, {-a, -1, 0, -a}, {a, 0, -1, a}, {0, -a, -a, -1}};

We choose a=1 and a = -1 and obtain altogether 8 SO(2,2) matrices with integer coefficients. We can use these eight matrices to construct an iterated function system on the torus as follows.

Iterated function system (IFS) from eight SO(2,2) matrices.

We have obtained eight SO(2,2) matrices, let us call them M₁,...,M₈:

M₁ = {{1, 1, -1, 0}, {-1, 1, 0, 1}, {-1, 0, 1, 1}, {0, 1, -1, 1}},

M₂ = {{1, -1, 1, 0}, {1, 1, 0, -1}, {1, 0, 1, -1}, {0, -1, 1, 1}},

M₃ = {{1, -1, -1, 0}, {1, 1, 0, 1}, {-1, 0, 1, -1}, {0, 1, 1, 1}},

M₄ = {{1, 1, 1, 0}, {-1, 1, 0, -1}, {1, 0, 1, 1}, {0, -1, -1, 1}},

M₅ = {{1, 1, 1, 0}, {-1, 1, 0, 1}, {1, 0, 1, -1}, {0, 1, 1, 1}},

M₆ = {{1, -1, -1, 0}, {1, 1, 0, -1}, {-1, 0, 1, 1}, {0, -1, -1, 1}},

M₇ ={{1, -1, 1, 0}, {1, 1, 0, 1}, {1, 0, 1, 1}, {0, 1, -1, 1}},

M₈ = {{1, 1, -1, 0}, {-1, 1, 0, -1}, {-1, 0, 1, -1}, {0, -1, 1, 1}}.

We can start now the IFS-game. We select an initial point x₀ on N with integer coordinates. For instance x₀ = (0,1,0,1) is a good candidate. We apply to x0 each of the matrices Mi to obtain 8 new points x_i = M_i x₀. To each of the new point we apply each of Mi. We obtain 64 points x_ji = M_jx_i. And so on. At step n we obtain 8ⁿ points. Each of them is a point on N with integer coordinates, thus defining a "balanced tetrad" of the type a² + b² = c² + d².

To be continued...

Knowledge protects, ignorance endangers

 

As they say "Knowledge protects, ignorance endangers." 

And so recently, thanks to AI, I have learned that what we are doing here is being done by other people, though with a somewhat different goal. The domain of research that concerns us is called "Conformal Field Theory". Here are two pages from the PhD Thesis "Relations between 2D and 4D Conformal Quantum Field Theory", by Daniel Meise, University Göttingen, 2011.

We will derive here formulas similar to (2.58), but not by the method described in there. It is in order to have simple formulas like (2.58) that I have changed my formulas for Clifford algebra generators e_μ. Then I doubted if I have it all correct, but today I realized that all seems to be ok.

So, I will be adding new stuff to my "Notes".

Afternotes

17-08-25 19:21 I know. I promised a new post. But I was busy. Finally, using our triangular matrices, I managed to create this "fractal".

Scottish torus

In the next post, hopefully on Monday, I will describe how I got it. The torus is, of course, our conformally compactified 1+1 dimensional universe.

18-08-25 19:06 I will not finish writing my new post today. Will finish tomorrow.