Spin+(2,2)

 While preparing this note over the past few days, I had extensive conversations with Grok 4. Most of them were helpful. I was just testing. For instance Grok solved Exercise 2 below. Even if it was easy, it required logical thinking, and that Grok does pretty well. On the other hand, Grok sometimes behaves like a hound dog. You ask it to bring you a duck, and it happily brings you a hen. Then, when you are upset, it apologizes: 'Yes, indeed, that is not exactly what you asked for, and I know it.'

This post is a continuation of The three involutions of Cl(2,2). Our aim is to examine the group Spin(2,2) - the double covering group of SO₀(2,2). SO₀(2,2) is the connected component of the identity of the group O(2,2) - the conformal group of our toy spacetime with only one space dimension. We need to start with some definitions. The conventions may differ from one author to another, so I am choosing just one, suiting my personal taste at the moment, but mostly I am trying to follow the notation used in D.J.H. Garling, “Clifford Algebras: An Introduction”,  London Mathematical Society, CUP 2011. A good introduction to Clifford algebra can be found in the lecture notes “Math 210C: Clifford algebras and spin groups” by Brian Conrad (Stanford University, available as a free PDF at http://math.stanford.edu/~conrad/210CPage/handouts/clifford.pdf).

General Definitions:

The Clifford norm Δ(a)

For any Clifford algebra Cl(V,q) , V-vector space, q - quadratic form on V, we define

1) The quadratic norm Δ(a)

Δ(a) = ν(a)a,            a∈Cl(V,q)            (1)

2) The Clifford group Γ(V.q)

The group Γ(V,q) is the group of invertible elements g∈Cl(V,q) having the property

α(g) v g^-1 ∈ V if v ∈ V.                (2)

Here α is the main automorphism: α(x) = x for even elements, α(x) = -x for odd elements of the Clifford algebra. In the previous post we have found the explicit form of α for Cl(2,2) represented as Mat(4,R), but that is not important at the moment.

Exercise 1. Carefully verify that Γ(V,q) is a group.

Exercise 2. Using the properties of α show that if g∈Γ(V.q) , then α(g)∈Γ(V.q).

We will be interested in the group Spin+(V,q) defined as:

Spin⁺(V,q) = {g ∈ Cl(V,q)⁺: Δ(g) = 1, g ∈ Γ(V.q) }.            (3)

Note: In (3) , the condition Δ(g) = 1, 1  should be understood as the Clifford algebra element 1. In a matrix representation it will be the identity matrix, not a number.

Spin+(2,2) Explicitly

With these foundations in place, let's apply them to our case of Cl(2,2).

Let us now restrict ourselves to the case at hand, and find an explicit form of elements of  Spin⁺(2,2). An element g of Cl(2,2) should belong to the even Clifford algebra Cl(2,2). In our matrix representation Cl(2,2) = Mat(4,R), while Cl⁺(2,2) consists of block diagonal matrices of the form {{A,0},{0,B}}. Then we need the condition Δ(g) = 1. In The three involutions of Cl(2,2) we have found that

ν(x) = Ux^TU^-1,            (4)

^where

U = -e₁e₂.            (5)

In our representation generators are given by matrices

e₁ = {{0, 0, 0, 1},
       {0, 0, 1, 0},
       {0, 1, 0, 0},
       {1, 0, 0, 0}}

e₂ = {{0, 0, 1, 0},
       {0, 0, 0, -1},
      {1, 0, 0, 0},
      {0, -1, 0, 0}}

e₃ = {{0, 0, -1, 0},
       {0, 0, 0, -1},
       {1, 0, 0, 0},
       {0, 1, 0, 0}}

e₄ = {{0, 0, 0, -1},
       {0, 0, 1, 0},
       {0, -1, 0, 0},
      {1, 0, 0, 0}}

Calculating U we get, in the block matrix form

U = {{ε,0},{0,ε}},               (6)

where ε = {0,1},{-1,0}}.

We can now calculate Δ(g) for g = {{A,0},{0,B}}. The result is the block diagonal matrix

 Δ(g)  = {{det(A)I₂,0},{0,det(B)I₂}}.            (7)

Computational Verification

For similar symbolic calculation we can use free Maxima software. Here is the Maxima code that does our calculation:

e1: matrix([0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]);
e2: matrix([0, 0, 1, 0], [0, 0, 0, -1], [1, 0, 0, 0], [0, -1, 0, 0]);
e3: matrix([0, 0, -1, 0], [0, 0, 0, -1], [1, 0, 0, 0], [0, 1, 0, 0]);
e4: matrix([0, 0, 0, -1], [0, 0, 1, 0], [0, -1, 0, 0], [1, 0, 0, 0]);
ω: e1 . e2 . e3 . e4;
/* Define the 2x2 matrices A and B */
A: matrix([a11, a12], [a21, a22]);
B: matrix([b11, b12], [b21, b22]);
/* Define 2x2 zero matrix */
Z: zeromatrix(2, 2);
/* Create the block matrix u = [[A, 0], [0, B]] */
a: mat_unblocker(matrix([A, Z], [Z, B]));
u:  -e1 . e2;
u . transpose(a) .  invert(u) . a;

From (7) we see that in  the matrices g = {{A,0},{0,B}} in Spin+(2,2) we must have det(A)=det(B)=1.

Exercise 3. Write Maxima code  that verifies that for g = {{A,0},{0,B}} with det(A) = det(B) = 1, the third condition  g ∈ Γ(V.q) is automatically satisfied. (You can also use online AI to write the code for you!).

Hint: Represent the generators e1–e4 as matrices, compute α(g) v g^{-1} for each basis vector v in V, and check if the result remains in V."

So we have a truly beautiful result:

Spin+(2,2) = { g = {{A,0},{0,B}}: det(A)=det(B)=1}.

Thus we see that, as a group, 

Spin⁺(2,2) = SL(2,R)×SL(2,R).

In the next post, we'll explore the 2:1 homomorphism Spin+(2,2) → SO₀(2,2), followed by an examination of minimal left ideals in Cl(2,2) where spinors reside.

P.S. 15-07-25 11:58 Preparing the next post I was conversing with Grok 4. Here is the Report (written in .tex by Grok himself!) on this conversation: