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....

Afternotes:

01-09-25 13;23 Again delay. After starting writing a new note I realized that I do not understand sufficiently the subject that I was writing about. So, still fighting with a very slow progress. In the meantime my new substack post appeared today: "The Art of Criticism: How Easy It Is to Find Faults in Science, Even Among Experts."

01-09-25 13:35 I am experimenting with AI. Yesterday DeepSeek AI constructed a beautifully written (but false!) proof of a mathematical theorem in algebra and  number theory. It took me a while to pinpoint the error. One should NEVER rely on AI! NEVER! It can lie to you shamelessly! The reasoning may look very logical, but the evil devil may be hiding behind inside.AI can also be a cause of a degenerating process in our brain functioning. It CAN helpful, for sure, but it is also extremely dangerous.
Here is the link to the Theorem and its false proof. All written by DeepSeek
Exercise: Find the error in the proof.