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,
- ei ej + ej ei = 0 for i ≠ j.
This corresponds to the quadratic form Q(x) = x12 + x22 - x32 - x42 on ℝ2,2.
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 | ||
Let A =
| p | q |
| r | s |
| t | u |
| v | w |
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 = ∑ xi ei.
This induces an orthogonal transformation L on ℝ2,2 such that ρ(g)(∑ xk ek) = ∑ xk g ek g-1 = ∑j,k xk Lj k ej.
Thus, g ek g-1 = ∑j Lj k ej, so the columns of L are the coefficients of the conjugated basis vectors.
Deriving the Formula
To find L explicitly, compute g ek g-1 for each k=1,2,3,4, then express as a linear combination of the ej.
Since the ei 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 ek g-1), where basis_matrix has columns flatten(ej).
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.
P.S. 16-07-25 17:26 All above is written by Grok, including the introduction.


This comment has been removed by the author.
ReplyDeleteOK, compared my result with AI (first column), found one error in sign, it should be
ReplyDeleteg e1 g^{-1} = 1/2((N2+N3)e1 + (N1-N4)e2 - (N1+N4)e3 - (N2-N3)e4)
The correspondence between my N's and AI's letters is the following:
N1 = qw - pv
N2 = pt - qu
N3 = sw - rv
N4 = rt - su
But I didn't like the AI's writing at all. It is terrible - dry and dull, especially compared to what we were used to. After this experience, I appreciate Ark's unique style and rare ability to explain deep and complex things in easy and artistic way even more.
When comparing the first column of L in the post to your result in the comment, can't see any difference. Am I missing something?
Delete@Anna I think I can teach Grok to imitate my personal writing style, including occasional spelling errors and typos, so that it would be virtually impossible to say who wrote it. But, I believe, the texts written by AI would still feel different.
Delete@Anna But I do like illustrations made by AI. I am still learning, by practice, the art of creating them.
DeleteOnly now I noticed the three hands on the illustration at the top!
Delete@Saša,
Delete"comparing the first column of L in the post to your result in the comment, can't see any difference"
Here, I added minus sign before e4 component: -(N2-N3)e4
At the previous post the components were
((N2+N3)/2, (N1-N4)/2, -(N1+N4)/2, (N2-N3)/2)
Thanks to AI. Nevertheless, the third hand is not a pleasant image. I better like the cartoon-style picture of the previous post, with emotions of each character conveyed perfectly.
I remember Vladimirov admitting that duality is good, but triality is even better!
Delete@Anna " the third hand is not a pleasant image". This time I explicitly asked for a surrealistic painting. So we got three arms and melting clock.
Delete@Anna
DeleteThanks for explanation.
@Ark
Well, didn't noticed three arms until reading your today's comments just now. Probably the upset cat caught my attention.
To my taste, it looks more like Kunstkamera (cabinet of curiosities) than the triality which Vladimirov likes :)
DeleteI am sure that even if Ark will train AI for a long time, it will never become so creative and exquisite as his creator, and will not be able to "возбуждать улыбку дам огнем нежданных эпиграмм".
@Anna
Delete""возбуждать улыбку дам огнем нежданных эпиграмм".
Where is it from?
V Мы все учились понемногу Чему-нибудь и как-нибудь, Так воспитаньем, слава богу, У нас немудрено блеснуть. Онегин был, по мненью многих (Судей решительных и строгих), Ученый малый, но педант.[5] Имел он счастливый талант Без принужденья в разговоре Коснуться до всего слегка, С ученым видом знатока Хранить молчанье в важном споре И возбуждать улыбку дам Огнем нежданных эпиграмм. VI Латынь из моды вышла ныне: Так, если правду вам сказать, Он знал довольно по-латыни, Чтоб эпиграфы разбирать, Потолковать об Ювенале, В конце письма поставить vale,[6] Да помнил, хоть не без греха, Из Энеиды два стиха. Он рыться не имел охоты В хронологической пыли Бытописания земли; Но дней минувших анекдоты, От Ромула до наших дней, Хранил он в памяти своей.
Delete@Igor. Thanks.
Deletehttps://ilibrary.ru/text/436/p.2/index.html
Enlish translation. Not too bad!
Delete" with an epigram-surprise,
of kindling smiles in ladies' eyes."
https://lib.ru/LITRA/PUSHKIN/ENGLISH/onegin_j.txt
I knew you would find it!😊
DeleteA very good translation indeed!
Delete@Anna I am sure I read it in school. But the only piece I remember from my school years is this one:
DeleteБуря мглою небо кроет,
Вихри снежные крутя;
То, как зверь, она завоет,
То заплачет, как дитя,
Great! It is not surprising. The first line of the poem "буря мглою" is so recognizable among Russian speakers that it is used as an idiom to express real poetic style.
DeleteBy the way, that my mentioning Kunstkamera -- was it clear that I meant the museum at St.Petersburg started in XVIII from collection of anatomical preparations of a Holland scientist? https://www.kunstkamera.ru/en/museum/kunst_hist/5/5_2/
DeleteAt the previous post Saša asked about the origin and meaning of the dimensions of our toy 4-dim space. Now I am thinking about some alternative interpretation of these dimensions. Why necessarily space-time? Perhaps, we could consider 4d space with (2,2) signature as a PHASE space (x,p) of a 2d system. Lagrangian L=T-U seems to be a quadratic form of the right signature. Vector product [x,p] and Poisson bracket {x,p} are both anticommutative and probably can be tailored somehow to be product operation.
ReplyDeleteI like your idea very much!
Delete@Anna
ReplyDelete"Lagrangian L=T-U seems to be a quadratic form of the right signature."
If my memory serves me well, potential energy U in Lagrangian in principle is not a quadratic form neither in coordinates nor in their derivatives. FWIW.
You are right, in general. But specifying to the harmonic oscillator Anna's idea should work.
DeleteL=T-U is a quadratic form for the case ddx = -x, where 'dd' is the second time derivative. Then, T = 1/2(dx dx) and U = 1/2(x x). In this case, phase trajectories are circles. Ellipse-like trajectories are topologically the same as circles, so probably they can also be included.
DeleteYes, that kind of L works in case where Hooke's law is applicable. Although, it has this quadratic form in case of simple 1-dim h. o., not sure if there would not be some mixed terms (like products of two coordinates) if we'd go to 2-dim system with two springs for example, and check arbitrary displacements.
DeleteIn other words, even for Hooke's law systems, in 2-dim we might not get a clear signature (2,2) quadratic form in the lagrangian of that systems.
If we were to look for similar quadratic forms of signature (2,2), diagonal elements of Maxwell's stress tensor might be nice candidates as well.
P.S.
DeleteAs you repeatedly warned us, Ark, the devil often hides in the details.
I am citing from V. I. Arnold "Mathematical methods":
ReplyDelete"In case of small oscillations of spherical pendulum, U = 1/2(x1^2 + x2^2), the level surface is a 3d sphere in 4d space. Phase curves are big circles of this sphere. The set of phase curves on the level surface comprises 2d sphere. More presicely, Hopf mapping
w = (x1 + i y1)/(x2 + i x2) maps 3d sphere (level surface) to 2d sphere = complex plane + infinitely distant point".
Next goes the elipse-like cases and Lissajous curves.
These are all familiar concepts. But I'm not sure they have any relevance to the Ark's story. Similarity can be misleading.
Thanks Anna, this was lurking in the back of my mind while writing previous comment. In case of pendulum there is basically only gravity acting on the system, and it's always directed downwards, while in case of 2-dim h. o. with two perpendicularly placed springs we would have two orthogonal forces for which I'm not sure from the top of my head that there would not be some mixing of two coordinates in the lagrangian of that system for arbitrary displacements.
DeleteMy question about dimensions in the comment of previous Ark's post was incorrectly informed by that misleading intro passage. After Ark's correction, it seems that his toy model involves 1-dim space and 1-dim time, both of positive signature, while those two added dimensions of negative signature, for now at least in my case, do not have a clear physical meaning, but seem to be there for playing and perhaps educational purposes.
Please correct me if there are flaws in my understanding of this subject.
Another example of sort of a realistic 2-dim h.o. might be pendulum with a spring attached, or two coupled pendulums by a spring between them, although I'm not sure if that would be in the same ballpark as what you've been discussing here. The resulting space-time curve would be so called beats, not a nice "monochromatic" sinusoidal oscillations curve. FWIW.
DeleteThe group O(2,2)/Z_2 is generated by the conformal compactification of the pseudo-Euclidean plane onto the projective plane, so we need to consider the oscillations of the pendulum on the projective plane. To visualize such oscillations, we need to turn to the following physical model. We put a bead on the meridian of the sphere, sliding along the meridian without friction, and spin this meridian along the axis connecting the north and south poles. Then, in zero gravity, the trajectory of the bead describes a curve whose latitude obeys the dynamic equation of oscillations of a mathematical pendulum and oscillates between the polar circles. The parameters of this pendulum are the angular velocity of the meridian and the linear velocity of the bead along the meridian at the latitude of the equator. Details here
Deletehttps://www.researchgate.net/publication/325226826_Chaotic_dynamics_of_an_electron
Thank you very much Igor for the exemplatory explanation!
DeleteSaša, thank you for your attention to my search! It is nice when the lurking becomes obvious. You provide valuable details, though my idea is more vague and abstract - I am trying to understand whether Clifford algebras are applicable to describe the phase space or not. The key concept is SYMPLECTIC space: on the one hand, it is a universal physical space (underlying both classical and quantum systems), on the other hand, its structure is very similar to the structure of Clifford algebras, so the latter should be ubiquitous in physics. However, the algebraic approach is not generally accepted in physics, and Cl-algebras have not become widespread, and there is a reason for this. What is it?
DeleteI wouldn't know, really. Although I might guess, from experience of listening theoretically more inclined colleagues around me, that they cling more to the Lie (and Poincare) algebras, even though Clifford ones as presented by Ark here seem so natural and straightforward vis-a-vis their logical and math-phys implications. Only recently, in last ten years or less, have seen and heard that some of more prominent theoreticians have been involved in studying conformal symmetries, but again no mentioning of Clifford and basically only in context of some "squeeze it" in the quantum gravity framework of research.
DeleteFrankly, it's a very sad state of affairs the theoretical physics has been in for quite some time, at least it seems so, and Ark's work presented here feels like a much needed, or maybe better said, desperately yearned for, breeze of fresh air based on logical and quite "simple" natural reasoning.
Of course, not to forget, your, Igor's and Bjab's contributions are also highly valuable and help a lot in better understanding of the topics discussed. Thank you all!
Anna, you asked an interesting question about the symplectic structure and the Clifford algebra. As far as I understand, the Clifford algebra is generated by the scalar product, and the symplectic structure is the volume form of the phase space. Let's look at the simplest example to see how the volume form can be related to the scalar product. If we take the scalar square of a vector (x,y) in orthonormal coordinates of the pseudo-Euclidean plane, we get the expression x^2 - y^2. At the same time, if we take the square of a vector in isotropic coordinates (X,Y), it has the volume form XY. Isn't this where the dog is buried? At least doubling the Minkowski space by adding the signature space (-,-,-,+) to the signature space (+,+,+,-) could extend the coordinate space to the phase space.
DeleteSasha, thank you so much for the kind words about our role here. This blog is like an ark for me that took me on board after many years of lonely floundering in phys-math sea in desperate attempt to comprehend Reality.
ReplyDeleteIgor, i am very glad that you find my question interesting! And your proposal of doubling space-time to obtain phase space is exciting indeed. Especially since we will obtain the (4,4) space of Varlamov’s construction. It's a pity Ark is busy today by taming Linux and training Grok. His opinion would be invaluable.
ReplyDeleteYeah you actually get two volume forms XYZT and xyzt. The original before dimensional reduction pseudoscalar XYZTxyzt volume form would end up the annihilation version of the Higgs scalar. Clifford algebra is great in that the doubling gives a natural creation-annihilation structure for a Fock space but also can have a middle grade where you can do differential geometry volume form, metric, connections kind of things. It's very EEQT. I never thought I'd be able to appreciate EEQT beyond being a GRW-like interpretation. Seeing the symmetry makes things more easy to see kind of like Grok's mentioning of Feynman using symmetry to clarify things in Ark's new post.
ReplyDeleteIn the case of doubling the Minkowski space, it becomes possible to endow this space with a symplectic structure, namely, a 2-volume form X/\X'+Y/\Y'+Z/\Z'+T/\T' where isotropic coordinates of double Minkowski space are used
Delete