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 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...

Notes on the Clifford algebra Cl_{2,0}. II

 

Once in a while we need to re-organize our environment and our thinking. Sometimes it is good, sometimes it is a disaster. It is much like with the Law of Three:

“There is good, there is evil,
and there is the specific situation
that determines which is which.”

So I reorganized my Notes, and added some new stuff and new exercises.

Here is the link to pdf.

 Today's post is a bit of a wild adventure—an experimental dive into the world of AI-assisted writing that blends education, recreation, and a touch of mathematical whimsy. Imagine stepping into a laboratory where human creativity meets artificial intelligence, not as rivals, but as playful collaborators. That's exactly what we're doing here today. I decided to hand over the reins to Grok 4, xAI's latest marvel, and asked it to craft a complete blog note on deriving the image of Spin(2,2) matrices in SO(2,2). 

What you'll see below is the raw, unedited output—straight from the AI's "mind" to your screen, like a digital genie granting a wish without any polishing.Why experiment like this? Well, it's educational on multiple levels. First, it showcases how AI tools like Grok can handle complex topics in Clifford algebras and spin groups, breaking down abstract concepts into clear, step-by-step explanations. For math enthusiasts, it's a reminder of how these algebraic structures underpin everything from quantum mechanics to computer graphics—think of Spin(2,2) as a "double cover" that elegantly navigates the symmetries of a 4-dimensional space with a mixed signature (two positive, two negative dimensions). Recreationally, it's just plain fun! It's like playing a game of "telephone" with an super-smart robot: you whisper a prompt, and out comes a fully formed article. But here's the twist—I won't make this a habit. Personal writing is where the real joy lies; it's the spark of individuality, the quirky turns of phrase, and the satisfaction of crafting something uniquely yours that AI can't replicate. Experiments like this keep things fresh, spark curiosity, and remind us that technology is a tool, not a replacement. Who knows? It might inspire you to try your own AI collaborations—perhaps generating poetry about Lie groups or doodling fractal art with code.That said, the AI's output isn't flawless (what creation ever is?). The block-diagonal form it presents for the Spin(2,2) element g has a slight quirk in its presentation—it's stylized in HTML tables, which works but isn't the sleek LaTeX rendering I might prefer for matrices. Also, the variable names for the elements of matrices A and B (like p, q, r, s for A and t, u, v, w for B) are a tad generic; I might have opted for something more mnemonic, like using Greek letters or coordinates tied to the algebra. I could have nudged Grok to refine it, but nah—this post is all about raw demonstration. It's a peek behind the curtain of AI possibilities, warts and all, to highlight how far we've come in natural language processing and symbolic math handling. Think of it as a recreational puzzle: spot the improvements you'd make, and maybe even tweak it yourself!And now, without further ado, here's the note Grok 4 produced. Enjoy the ride through the world of spinors and orthogonal groups—it's a surreal blend of rigor and revelation.

Deriving the Image of Spin(2,2) in SO(2,2)

As readers of this blog, you're familiar with the basics of Clifford algebras and spin groups. Today, we'll derive the explicit formula for the image of a Spin(2,2) matrix in SO(2,2) under the standard homomorphism. We'll recall key concepts for clarity and proceed step by step.

Recalling the Clifford Algebra Cl(2,2)

The Clifford algebra Cl(2,2) is generated by four elements e₁, e₂, e₃, e₄ satisfying:

• e₁² = e₂² = 1,
• e₃² = e₄² = -1,
• e_i e_j + e_j e_i = 0 for i ≠ j.

This corresponds to the quadratic form Q(x) = x₁² + x₂² - x₃² - x₄² on ℝ^2,2.

In the matrix representation we're using, the generators are 4×4 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

The Spin⁺(2,2) Group

Spin⁺(2,2) is the connected component of the spin group, consisting of even elements g in Cl(2,2) with norm g ˜g = 1, where ˜ denotes the reverse (or Clifford conjugate, which for even elements is the transpose in this matrix rep, but we use the standard inverse for the homomorphism).

In this representation, elements of Spin⁺(2,2) are block-diagonal matrices of the form:

g =

A		0
0		B

where A and B are 2×2 matrices with det(A) = det(B) = 1.

Let A =

p	q
r	s

and B =

t	u
v	w

, with ps - qr = 1 and tw - uv = 1.

The Homomorphism to SO(2,2)

The double cover homomorphism ρ: Spin⁺(2,2) → SO⁺(2,2) is given by ρ(g)(v) = g v g^-1, where v is a vector in the grade-1 part, i.e., v = ∑ x_i e_i.

This induces an orthogonal transformation L on ℝ^2,2 such that ρ(g)(∑ x_k e_k) = ∑ x_k g e_k g^-1 = ∑_j,k x_k L_{j k} e_j.

Thus, g e_k g^-1 = ∑_j L_{j k} e_j, so the columns of L are the coefficients of the conjugated basis vectors.

Deriving the Formula

To find L explicitly, compute g e_k g^-1 for each k=1,2,3,4, then express as a linear combination of the e_j.

Since the e_i are linearly independent in the 4-dimensional subspace of matrices, we can vectorize the matrices (flatten into 16-vectors) and solve the system: basis_matrix · column_k(L) = flatten(g e_k g^-1), where basis_matrix has columns flatten(e_j).

Performing this computation symbolically (e.g., using SymPy), assuming det(A)=det(B)=1 for the inverses, yields the following matrix L:

L = (1/2)

p t - q u - r v + s w	-p u - q t + r w + s v	p u - q t - r w + s v	-p t - q u + r v + s w
-p v + q w - r t + s u	p w + q v + r u + s t	-p w + q v - r u + s t	p v + q w + r t + s u
p v - q w - r t + s u	-p w - q v + r u + s t	p w - q v - r u + s t	-p v - q w + r t + s u
-p t + q u - r v + s w	p u + q t + r w + s v	-p u + q t - r w + s v	p t + q u + r v + s w

This is the explicit formula for the image of the Spin(2,2) element in SO(2,2). You can verify it preserves the metric and has determinant 1.

Conclusion

This derivation highlights the power of the Clifford algebra representation in computing group homomorphisms explicitly. In future posts, we might explore applications or extensions to other signatures.

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.

The three involutions of Cl(2,2)

  In The infinity ab initio and The infinity ab initio 2 we have introduced the six-dimensional space of oriented spheres in R³. R³ is endowed with the quadratic form q of signature (3,0). We use coordinates x¹,x²,x³ for R³. Going to the six-dimensional space of oriented spheres in R³ involves adding 3 more dimensions, with coordinates x⁴,x⁵,x⁶, and added signature (1,2).  x⁵ can be interpreted as time, x4 and x⁶ were extra two dimensions. We have used Q to denote the resulting quadratic form in R^4,2.

Creator realized that Cl(1) is not good

But after that, in Sunday special: Conversing with Grok about the Clifford algebra Cl(2,2), we have decided to play first with a toy model, suppressing two space dimensions x² and x³. Then space became 1-dimensional. Spheres in one dimension, oriented or not, are somewhat special. What is a sphere in R¹? It is just a pair of points (-r,+r). (Why?) It is good to have in mind this particularity, but it is not a problem in our construction.

While introducing the Clifford algebra Cl(2,2) we decided to rename the coordinates and to switch from the signature (+-+-) to equivalent (++--), as it is more convenient for working with the Clifford algebra. We have also introduced a particular matrix representation of the Clifford algebra, where the matrices representing an orthonormal basis in R^2,2 are given by:

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}}

These matrices will play the role of Dirac gamma matrices, but now adapted to the signature (2,2). In our case they are all real! While discussing the Clifford algebra of R3, we have mentioned the ideas developed by David Hestenes: the imaginary unit in quantum theory is related to the fact that the element ω = e¹e²e³ of Cl(3) is related to quantum-mechanical imaginary unit "i", since its square is -1. But now, in Cl(2,2),  ω = e¹e²e3e4 is given by

ω = diag(1,1,-1,-1).

and ω² = 1. Does that mean that for a  1-dimensional space quantum mechanics would be all real? I do not think so. One-dimensional harmonic oscillator, for instance, is studied in all quantum-mechanical textbooks, and the Hilbert space there is always complex. So, I am not sure anymore that Hestenes' idea, although elegant and attractive, really hits the target. But we will never know the truth until we understand what spin is and what electric charge is. On the other hand, perhaps Creator realized that Cl(1) is not good, Cl(2) is also not good (both do not lead to complex structures), tried the next option, Cl(3), and saw that it was good?

Taking all different products of ei we span the whole 16-dimensional algebra. In our realization of Cl(2,2) we have

Cl(2,2) = Mat(4,R).

So Cl(2,2) is isomorphic to Cl(3,1) - as an algebra. But Clifford algebras come with extra structure, and it is this extra structure that differentiates between Cl(2,2) and Cl(3,1). The vector space R^2,2 is realized in Mat(4,R) differently than the vector space R^3,1. With different embedding of underlying vector spaces come different form of the three Clifford involutions. We will discuss them now.

Clifford involutions for the matrix realization of Cl(2,2)

We have already met involutions for Cl(3) in The Spin Chronicles (Part 10) - Dressing up the three involutions. These were

1. The main automorphism π, changing the signs of the vectors.
2. The main anti-automorphism   τ, reversing the order of multiplication in Cl(V).
3. Their composition π∘τ = τ∘π, which we will call ν.

We will now do the same for Cl(2,2), but we will change the notation.

• The main involution will be denoted by α: x ⟼ α(x).
• The main anti-authomorfism will be denoted by x ⟼ x^~.
• The Clifford conjugation, the composition of the main automorphism and the main anti-automorphism, will be denoted by ν: x ⟼ ν(x).

Three involutions by Raphael

The point is that we want to save π and τ for other purposes.

The main involution is easy to guess. Since ω² =1, and ω anticommutes with all generators, the formula

α(x) = ωxω, x in Cl(2,2),

does the job in any representation.

The main anti-automorphism (Clifford reversion) is more subtle. The natural anti-automorphism in Mat(4,R) is the transposition. But the main Clifford anti-automorphism should leave the generators invariant. This works fine with e₁  and e₂ , but fails for e3 and e4 since they are represented by antisymmetric matrices. We have to compensate for this fact. The anti-automorphism will be of the form

x^~ = Sx^TS^-1,    x in Mat(4,R),

where S is a matrix commuting with e₁,e₂, and anticommuting with e₃,e_4.

There is an easy solution. Let

S = e₃e₄.

Then S² = -1, so S^-1 = -S, S commutes with e₁ and e₂,  and anticommutes with e₃ and e₄.

AI Note. I tried AI with this problem. Grok 3 was repeatedly proposing wrong solutions for reversion. When I was pointing out why his solution is wrong, he was acknowledging his error, and then proposing another solution, also wrong. Then I tested with the new Grok 4. Grok 4 understood the problem, was thinking correctly, used Python for calculations, and came out with the right solution. But he did not realize that his solution can be expressed as e₃₄. When asked explicitly if e34 is a good solution - he was reasoning correctly, like a good student. While Grok 3 was giving (often wrong) answers immediately, Grok 4 was a very slow thinker. I am also slow, so I think I like Grok 4.

Now we need Clifford conjugation. It will be of the form:

ν(x) = Ux^TU^-1.

Composing α with reversion (in any order), we get

ν(x) = ωx^~ω^-1 = ωSx^TS^-1ω^-1 = e₁e₂e₃e₄e₃e₄x^T(e₁e₂e₃e₄e₃e₄)^-1 =(-e₁e₂)x^T(-e₁e₂).

So

U = -e₁e₂

is a solution.

Thus we have found the explicit form of the three graces of the Clifford algebra Cl(2,2).

In the next post we will examine the group Spin(2,2), the double cover of SO₀(2,2)

Sunday special: Conversing with Grok about the Clifford algebra Cl(2,2)

Over the past few days, I’ve engaged in an extensive dialogue with Grok, the AI developed by xAI, about the Clifford algebra ( Cl(2,2) ). 

Note: For those new to the topic, the Clifford algebra Cl(2,2)  is a mathematical structure used to describe geometric and physical properties of a 4-dimensional space with a metric signature (+,+,−,−). It’s particularly relevant in physics for modeling spacetime symmetries and spinors, offering a rich framework for connecting geometry to quantum mechanics.

Grok was remarkably helpful, anticipating my questions

Grok was remarkably helpful, anticipating my questions, assisting with Mathematica calculations, and discussing relevant publications on Cl(2,2)  and related topics. However, Grok also made a series of mathematical errors, from proposing incorrect generators for Cl(2,2)  to suggesting flawed minimal left ideals. At one point, Grok even claimed that (−1)×(−1)=−1! Each time I caught these mistakes, Grok graciously apologized. This led me to hypothesize that Grok might be programmed to test the mathematical acumen of its interlocutors, subtly checking whether I truly understand the subject! Despite the errors, the experience was both fun and enlightening. So, here’s my new blog post on Cl(2,2). I take full responsibility for its content, and if there are any mistakes, I invite my readers to point them out so I can make corrections. After all, I might be making errors even more frequently than Grok!

There is a clash between what is natural or just convenient for mathematics and what is natural or convenient for physics. We have to deal with this clash somehow, and it may create a confusion. So, let us meet this clash without fears. For space-time we have chosen coordinates x¹,x²,x³,x⁴ and metric signature (+++-). Then we add extra coordinates x5,x6 with signature (+,-). So we end up with coordinates x¹,...,x⁶ and signature  (+++-+-) . We have total signature (4,2), but not in mathematically preferred natural order (++++--). The isometry group of our extended space-time is isomorphic to O(4,2), but not exactly O(4,2). It is not a big problem, it is just an annoyance that may lead to a confusion here and there. Since we will be dealing with mathematics a lot, and physics will come only later, I will choose signature (++++--). At first we will discuss the toy model with suppressed two space dimensions. So, we will have signature (++--), with isometry group O(2,2). To avoid unnecessary confusion I will use coordinates y¹,y²,y³,y⁴. The relation with previously discussed extended space-time coordinates is

y¹ = x¹, y³ = x⁴, y² = x⁵, y⁴ = x⁶.

In particular the conformal infinity condition  x5 = x6, which arises in physical contexts like twistor theory, that we have discussed in the previous post, becomes y² = y⁴ in this framework, preserving the geometric structure.

Clifford algebra Cl(2,2)

We have already had some experience with Clifford algebras in the Spin Chronicles series of posts, where we were discussing Minkowski space spinors. Minkowski space metric has signature (3,1). Now we have signature (2,2).  Our discussion will be in parts similar, but but we will discover new features. For Minkowski space we ended with complex structures and biquaternions. Now everything will be real.  No complex numbers needed. In the beginning I will follow the way Cl(2,2) Clifford algebra is discussed in D.J.H. Garling, "Clifford Algebras: An Introduction",  London Mathematical Society, CUP 2011. But Garling has a opposite sign convention in the definition of Clifford algebras, so I will adapt the discussion from his book to our needs.

A Clifford algebra is a mathematical structure that encodes the geometry of a space through generators that satisfy specific multiplication rules, reflecting the space’s metric. For Cl(2,2) we need four mutually anticommuting generators e₁,...,e₄, with e₁² = e₂² = 1, e₃² = e₄² = -1. For Cl(2,2), the generators e₁,e₂,e₃,e₄ correspond to an orthonormal basis of a 4-dimensional space with signature (+,+,−,−). While we could proceed in purely algebraic way and search for minimal left ideals, we will take an easy way and start with a particular matrix realization of the algebra. Of course a curious Reader would ask: how this particular realization can be derived?  Well, it can, like we have derived Pauli matrices in Spin Chronicles, but it would take us unnecessarily long time. It is much easier to work with a particular realization while remembering that it is not necessarily the best for all purposes, and that there are infinitely many other realizations. This matrix representation, where the generators are 4x4 real matrices, leverages the isomorphism Cl(2,2) ≈ Mat(4,R).  While other representations exist (e.g., via different matrix forms), this choice simplifies our exploration of the algebra’s structure and its spin group.

So here are the four matrices representing the orthonormal basis, listed row by row.

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}}

Matrices e₁ and e₂ are symmetric, e₃ and e₄ are anti-symmetric. Written in a block matrix form, with 2x2 blocks, e₁  and e₂ are of the form

0	X
X	0

while e₃ and e₄ are of the form

0	X
-X	0

Let us examine the volume element ω = e₁e₂e₃e₄. It is the diagonal matrix, with (1,1,-1,-1) on the diagonal:

ω = diag(1,1,-1,-1).

For the Clifford algebra of space Cl(3) it was commuting with all elements of the C(3), and its square was -1. This gave us the complex structure akin to the imaginary unit ( i ), which was crucial for spinor representations in 3D space. Now ω² = 1, and ω anti-commutes with all e_i, which implies a real structure, aligning with the split signature ((2,2)) and leading to real spinors, which we’ll explore in the context of Spin(2,2). We have an essentially  different situation now. But there is a hope! In the case of Cl(3) we were discussing the whole Clifford algebra. Now we are interested mainly in the automorphisms of our extended mini space-time, that is in the group Spin(2,2), which consists of elements of the even Clifford algebra CL⁺(2,2).

The whole algebra splits into

1) scalars, span of 1 - real numbers (1-dimensional subspace)
2) vectors - span of e_i (4-dimensional subspace)
3) bivectors - span of e_ie_j - e_je_i (i<j) (6-dimensional subspace)
4) 3-vectors - span of ωe_i (4-dimensional subspace)
5) pseudo-scalars - span of ω (1-dimensional subspace)

Together 1+4+6+4+1 = 16 = 2⁴ dimensions. It spans the whole algebra of real 4x4 matrices Mat(4,R).

The even Clifford algebra Cl⁺(2,2)

The even subalgebra Cl⁺(2,2) is crucial because it contains the Spin group Spin(2,2), the double cover of the isometry group SO(2,2). This group describes transformations that preserve the (2,2) metric, and its elements are key to understanding spinors and symmetries in our toy model.The even Clifford algebra Cl⁺(2,2) is the span of scalars, bivectors, and pseudo-scalars. It is 8-dimensional. While the generators ei anti-commute with ω, products of even number of generators all commute with ω. We can see it from the block form of generators: they will always be of the block-diagonal form

X	0
0	Y

This is, in fact, a general form of elements in Cl⁺(2,2) , where X and Y can be arbitrary real 2x2 matrices:

Cl⁺(2,2) = Mat(2,R) ⨁ Mat(2,R).

Since Mat(2,R) is 4-dimensional, we get 4+4 = 8 dimensions, as needed for Cl⁺(2,2).

The next step will be, as we have dome it before for Cl(3), to identify the three involutions of Cl(2,2). We will do it in the next post. In the next post, we’ll explore the three main involutions of Cl(2,2) —grade involution, reversion, and Clifford conjugation—and how they act on our matrix representation. These involutions are essential for defining Spin(2,2) and understanding spinor transformations, paving the way for applications in physics.

This is, in fact, a general form of elements in Cl⁺(2,2) , where X and Y can be arbitrary real 2x2 matrices:

Cl⁺(2,2) = Mat(2,R) ⨁ Mat(2,R).

Since Mat(2,R) is 4-dimensional, we get 4+4 = 8 dimensions, as needed for Cl⁺(2,2).

The next step will be, as we have dome it before for Cl(3), to identify the three involutions of Cl(2,2). We will do it in the next post. In the next post, we’ll explore the three main involutions of Cl(2,2) —grade involution, reversion, and Clifford conjugation—and how they act on our matrix representation. These involutions are essential for defining Spin(2,2)
 and understanding spinor transformations, paving the way for applications in physics.