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) = φTωψ. (1)
It is then natural to introduce what physicists call the Dirac conjugated spinor
φbar = φTω, (2)
so that the invariant bilinear form ε can be written as
ε(φ,ψ) = φbar ψ. (3)
Exercise 1. Verify that φbar 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 ψbar = φ ψTω. (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 R1,1. 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 ψbar (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...
