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 R2,2 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 R2,2 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 R2,2. 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 R2,2.
We will use the matrix ω later on, for a different purpose though.
We are particularly interested in the null cone of R2,2, that is in the set of all vectors x in R2,2 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) = vwT 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 = vwT, if v⟼ Sv and w⟼ Sw, then A⟼SATT, 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) = vTωw.
Exercise: Show that for every S in SL(2,R) we have ε(Sv,Sw) = ε(v,w) . In other words STωS = ω.
To be continued....
