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.
 |
| 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).
 |
| 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)
P.S. 13-07-25 21:29 Screnshot from my conversation with Grok 4 today. Grok's reply:
If ν(x) = UxTU-1 = ωx~ω-1 = (-e1e2)xT(-e1e2),
ReplyDeletethen U does not look like being U = -e1e3.
FWIW.
Fixed. Thank you!
DeleteSince the file "Notes on Lie Sphere Geometry" on your google drive has not been updated since June 1st, and you've taken another, a bit different route of approach to the topic, would it be correct to assume that the file is completed for the time being regarding the topic of Lie Sphere geometry, in a sense that you won't be adding new things to it?
ReplyDeleteIn fact I have in mind adding a lot to these notes, perhaps also revising what is already there. Simply the priorities changed. Now, that I signed the contract for the second addition of quantum fractals, I have a lot of work to do, and with a (soft) deadline. What I am putting on the blog now is directly related to this new plan. I have in mind creating new fractals on the torus. Perhaps it will work, perhaps not. Hard to say in advance. And at the same time I am getting a better idea what to put in the "Notes". So, it is a multitasking experience for me. For the book I need to also add some mathematics of fractals. We have fractals on the computer, but mathematical theorems proving that they are really fractals are still iffy.
DeleteOh, didn't know that, then will wait for the new, enhanced edition to come up.
DeleteWishing you luck and lot of fun with all these things on your plate. And if you'd like some 'low level' help with all that, please don't be shy to let me know how I could be of service to you.
The requirement for a second edition means that the interest to the book has exceeded what was originally planned, great! Fractals are one of the most intriguing topics, but unfortunately the Book has not been read yet, perhaps it is too difficult for my level. Therefore, I would be very grateful if someting on this topic appear in the Blog in an adapted form.
ReplyDeleteI will keep it in mind. It was my intention to post a new note today, but I was having an extensive conversation with Grok. Hre is a small part of it. Discussing the Clifford norm (it will appear in the new post). This is to give you a atste of my conversation.
DeleteArk: But before you said that the map x to N(x) is always a homomorphism to the multiplicative group of the base field. It seems to contradict the example that you have just calculated? Please explain this apparent contradiction.
Grok: The apparent contradiction arises from a nuance in the general definition versus the specific structure of the even subalgebra Cl0(2,2)\mathrm{Cl}^0(2,2)\mathrm{Cl}^0(2,2)
. In many standard treatments of Clifford algebras Cl(V,q)\mathrm{Cl}(V,q)\mathrm{Cl}(V,q)
over a field like R\mathbb{R}\mathbb{R}
, especially when the algebra (or subalgebra) is central (i.e., its center is precisely the base field R\mathbb{R}\mathbb{R}
), the norm map N:x↦xx~N: x \mapsto x \tilde{x}N: x \mapsto x \tilde{x}
(or analogous forms using other involutions) is indeed a homomorphism from the group of units to the multiplicative group of the base field R×\mathbb{R}^\times\mathbb{R}^\times
. This is because (N(x)) lands in the center, which is identified with R\mathbb{R}\mathbb{R}
.However, the even subalgebra Cl0(2,2)≅M2(R)⊕M2(R)\mathrm{Cl}^0(2,2) \cong M_2(\mathbb{R}) \oplus M_2(\mathbb{R})\mathrm{Cl}^0(2,2) \cong M_2(\mathbb{R}) \oplus M_2(\mathbb{R})
is not central—its center is R⊕R\mathbb{R} \oplus \mathbb{R}\mathbb{R} \oplus \mathbb{R}
(scalars in each component). Thus, the norm (N(x)) for x=(A00B)x = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}x = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}
lands in this larger center, specifically as (det(A)I200det(B)I2)\begin{pmatrix} \det(A) I_2 & 0 \\ 0 & \det(B) I_2 \end{pmatrix}\begin{pmatrix} \det(A) I_2 & 0 \\ 0 & \det(B) I_2 \end{pmatrix}
, which can be identified with (det(A),det(B))∈R⊕R(\det(A), \det(B)) \in \mathbb{R} \oplus \mathbb{R}(\det(A), \det(B)) \in \mathbb{R} \oplus \mathbb{R}
. The map is a homomorphism to the units of the center, (R××R×)(\mathbb{R}^\times \times \mathbb{R}^\times)(\mathbb{R}^\times \times \mathbb{R}^\times)
, not just R×\mathbb{R}^\times\mathbb{R}^\times
. The general statement I made earlier applies primarily to central (simple) algebras, but requires adjustment for semisimple cases like this, where the center is extended.
Unfortunately the math is messed up in copy and paste, even if it is beautifully rendered my screen. But it give you a taste. Grok 4 is a good student, but it still relies too much on what it finds on the "trusted" sources of information. When it calculates - it does it quite well. I will mention it in my new post. It provided a nice counterexample to my hypothesis today. I knew the hypothesis is false, so I asked Grok to find the simplest counterexample. And he found it! In Clifford algebra Cl(3,2).
Added a screenshot in P.S.
Delete"This works fine with e2 and e2" -->
ReplyDeleteThis works fine with e1 and e2
Thanks!
Delete"and ω^2 = 1. Does that mean that for a 1-dimensional space quantum mechanics would be all real?"
ReplyDeleteAnd couldn't we interpret this case as an algebra of double numbers (or hyperbolic numbers, also perplex numbers) based on a hyperbolic unit j satisfying j^2 = 1?
https://en.wikipedia.org/wiki/Split-complex_number
Not quite, since in Cl(2,2) ω anticommutes with the generators instead of commuting.
DeleteBut indeed the algebra generated by 1 and ω alone is the same as hyperbolic numbers.
DeleteOh yes, indeed, the algebra of hyperbolic numbers is commutative. This is because it has only two generators: 1 and j, and everything commutes with 1.
DeleteThe hyperbolic numbers are isomorphic to R+R, and I assumed that they might be suitable for describing the vector space R2,2, but alas, this is not true.
If l am not mistaken, one more (geometric) interpretation of the highest basis element e1e2e3 is oriented volume. How can we relate volume to imaginary unit is still a puzzle to me...
ReplyDeleteYou mean in Cl(3)?
DeleteNot only in Cl(3). If we consider e_1, e_2, ..., e_n as differentials dx_1, dx_2, ..., dx_n of an n-dim vector space, than n-form dx_1 ^ dx_2 ^ ... ^ dx_n makes n-dim oriented volume in this vector space, doesn't it? But i don't know whether anybody relates this n-form to imaginary unit like in case n=3.
DeleteDoes anyone know about relation between imaginary unit and volume, which makes me so anxious...? There may be some deep physical meaning, with the air of duality: in order to define an algebra, we need basis elements and the highest element (co-unity), which gives the relation between them. In order to define a physical system, we need some space and a metrics, i.e., a scale on it. The highest element e1e2...en seems to be like elementary volume, giving the scale (or, probably, the curvature) of the space.
Delete"we will never know the truth until we understand what spin is and what electric charge is".
ReplyDeleteI would like to add also 'mass' into the list. Mass and spin of particles are definitely related.
For example, Carlos Castro relates them all in his paper "The Charge–Mass–Spin Relation of Clifford Polyparticles, Kerr–Newman Black Holes and the Fine Structure Constant" https://philpapers.org/rec/CASTCR-4. The Abstract looks complicated, but could be of interest.
Thanks. Will read.
DeleteThe intro passage of this post/note has been bothering me, as it seems that some things do not add up.
ReplyDeleteYou say that we started from 3-dim space R3 of signature (3,0), and then added 3 more dimensions of added signature (2,1) where one of them could be interpreted as time, to end up with 6-dim space (of presumably signature (4,2)). And then dropping out two space dimensions we arrived at our toy model with signature (2,2).
How did we get that second time dimension in our story, i.e. the other dimension of negative signature, if we only added one of them to the original R3, while the other two were with positive signature denoting two spacelike dimensions as noted in the intro passage?
Thanks. I have corrected this misleading passage.
DeleteThank you!
DeleteAs far as I understand, we are talking about algebraic structures associated with the conformal group of the space $R^{p,q}$, and this is the group $O(p+1,q+1)/Z_{2}$. On the other hand, a conformal group is generated by a conformal compactification, where the conformal compactification is a mapping $R^{p,q}$ to the
ReplyDeleteS^{p}\times S^{q} with 'antipodal points' identified.
Do I understand correctly?
That's correct. But I am playing with the toy model now, p=1, q=1.
DeleteThe model can be also interpreted as Lie sphere geometry for R^1. I may use this interpretation later on, at some point.
DeleteIn this simplest case, conformal compactification is a folding of the pseudo-Euclidean plane onto the projective plane, where the projective line is an element of a compact space, the prototype of which is the winding of the torus.
DeleteYes, this is the torus shown on the picture in https://ark-jadczyk.blogspot.com/2025/07/the-infity-ab-initio-2.html
DeleteIt would be interesting to know how the closed windings of a torus and a spinor are related.
Delete