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 R4,2.
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| 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 R2,2 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 R2,2 is realized in Mat(4,R) differently than the vector space R3,1. 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
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).
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| 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~ = SxTS-1, x in Mat(4,R),
where S is a matrix commuting with e1,e2, and anticommuting with e3,e4.
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) = UxTU-1.
Composing α with reversion (in any order), we get
ν(x) = ωx~ω-1 = ωSxTS-1ω-1 = e1e2e3e4e3e4xT(e1e2e3e4e3e4)-1 =(-e1e2)xT(-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)
