We’re on a journey. A journey to where, you ask? Honestly, I’m not entirely sure myself. Is the ultimate goal to solve all the mysteries of the universe? (No pressure, right?) Or maybe just crack open one of them, like an intellectual piñata? Or perhaps it's something subtler, something closer to René Magritte's enigmatic wisdom: “Each thing we see hides something else we want to see.”
But Magritte, ever the paradoxical philosopher, also declared: “I think we are responsible for the universe, but that doesn’t mean we decide anything.” Really, René? Nothing? Well, I’ll take my victories where I can, starting with today’s decision to write this blog. I have no clue what cosmic ripples this choice will send out (but hey, I hear butterflies start hurricanes), yet I’ll cling to the notion that I’m somehow fulfilling my obligations to the universe by clicking away at my keyboard.
So, where were we? Ah, yes—the journey. It turns out, we’re not exactly on the straight path. We’ve taken a detour. A scenic route, if you will. After Part 17, the plan was to dive straight into options (3) and (4) for two-sided actions of the Clifford group on Cl(V):
Instead, I decided to veer off-road. Why? Because sometimes you’ve got to set aside the GPS of mathematical rigor and take a little detour to explore something... different. Something fun. So, let’s play with the action of the Clifford group on the null cone in Minkowski space and the induced action on the -plane.
At first blush, this might sound like one of those “because I can” exercises in pure abstraction. But hold on—this isn’t some random doodle in the margins of physics. It has a gorgeous physical interpretation! The null cone in Minkowski space represents light emanating from a single spacetime point. Picture a laboratory at time , where light bursts forth in all directions. By , that light forms a sphere around the lab. Think of it as the universe’s way of saying, “Let there be geometric algebra!”
On this sphere, we can project the stars of the universe—the celestial sphere—much like Flammarion’s famous engraving, which tantalizingly invites us to peek beyond the cosmic veil:
Now, let’s spice things up a bit. What happens if we apply a Lorentz transformation to our lab? A rotation is straightforward—the sky spins, and the stars gracefully whirl like a cosmic waltz. But a Lorentz boost? That’s a whole different party. Enter relativistic aberration: the stars’ positions shift dramatically, bending and dancing under the influence of spacetime’s warping lens.
We’re going to calculate this starry choreography using the Clifford group. (And if you’re getting déjà vu, it’s because I hinted at this back in my September 4, 2024 post, “Monkeys, UFOs, and Quantum Fractals: A Mind-Bending Journey through Relativity and Imagination.”)
Here’s the twist: we’ll go beyond. We’ll project the celestial sphere onto the -plane of our lab. But don’t expect a straightforward mapping. The projection originates from the South Pole of the celestial sphere—a point conspicuously absent in Flammarion’s image. (That’s because it’s “beneath” the feet of our starry-eyed astronomer. Practical, but disappointing.)
This is the plan. And in the next post, we’ll dive into the algebra to unravel what happens to the projected images of stars. Will they remain stars? Will they smear into streaks of cosmic glory? Stay tuned. Who knows—this little exploration might even shake up your view of the heavens. Or at least provoke a disagreement or two.
P.S. 04-12-24 Updated the pdf file.
"... On the other hand, our earth turns very fast, and days pass away like dreams. I hope, nevertheless, to give myself the scientific pleasure of studying a portion of these mysteries, and perhaps what one man cannot do may be done by others. Every one may bring his little stone to assist in those construction of a future pyramid.Every author is in charge of souls. We ought only to tell what we know. Perhaps we ought not always to tell all we do know; but even in our every-day life we ought never to tell what we do not know. ·Then let us lay up knowledge, let us work and hope."
light rays emanating ->
ReplyDeletelight emanating
or
light energy emanating
or
electro-magnetic distortion traveling
I changed to "light emanating" as it is the shortest of possibilities. Thank you.
ReplyDelete"And what is the use of a book… without pictures or conversations?" Oh no, the 'Geometric algebra' paper is very useful indeed, but... looks a bit ascetic after the art fest at https://ark-jadczyk.blogspot.com/ we have used to already.
ReplyDeleteFortunately, our wandering round the Spinning Wonderland is going on, and what i like most of all is its unpredictability.
Lorenz attractor attracts because of its unpredictability. Normal feature in the fractal land.
DeleteHelp!
ReplyDeleteOn one hand:
e1e1e2 = e2
on the other hand:
e1e1e2 = e1(e1e2)
= e1(e1^e2 + e1·e2)
= e1(e1^e2 + 0)
= e1(e1^e2)
= e1^(e1^e2) + e1·(e1^e2)
= 0 + 0
= 0
The formula (5) in the file assumes that x,y are vectors. If one of them is a bi-vector, we proceed like in "On one hand".
DeleteI will add a comment about it in the updated file.
DeleteI think (but might be wrong) that e1^(e1^e2) is not 0, in 2nd row from the bottom in Bjab's comment, but equals e2, if "^" is like a cross product. Maybe the definition "^" might be useful, FWIW (haven't seen it explicitly stated in the file).
Delete@Saša OK. I see that I should include the Grassmann algebra "^" as well. Thanks.
Delete@Bjab I will have to add the Grassmann algebra part first to make it all clear. It will take a while. So, please, be patient.
Delete@Saša:
Delete"I think (but might be wrong) that e1^(e1^e2) is not 0, in 2nd row from the bottom in Bjab's comment, but equals e2"
Indeed, but if so then e1^(e1^e2) = MINUS e2
so something is still wrong.
In J. Sanchez's paper, it is mentioned that Cl(3,0) and Cl(0,3) would both give time as a natural dimension, not an ad hoc added one by hand. Here we work with Cl(3,4), based on things presented in Part 12, right? And we also got time naturally, as scalar x0 like you said, or its Hodge dual i. How would Cl(3,0) and Cl(0,3) look like and how would they relate to what's been presented here (presumably Cl(3,4))?
ReplyDeleteWe work with Cl(3,0) all the time. Where did you get (3,4) from?
DeleteApparently a wrong conclusion based on 4 quadratic (bilinear) forms at the end of Part 12, introduced with involutions in Part 8.
DeleteI obviously need to reread everything from the start with a fresh mind and 'new' pair of eyes.
DeleteBefore doing that, can you help me understand Cl(0,3)? Vector space would be 0-dim? How to 'visualize' that, like a point? And which 3 bilinear quadratic forms "acting" on it?
Cl(0,3) means that the square of x is negative instead of positive as in Cl(3,0). Cl(3,0) works with signature +++, Cl(0,3) with signature ---.
DeleteThank you.
DeleteYou are welcome. But one has to be careful and check with each author. Denoting by Q(x) the standard positive definite Eucldean norm squared, some authors define Clifford algebra requiring xx = Q(x), while there are also other well known authors requiring xx=-Q(x). For them the meaning of Cl(3,0) and Cl(0,3) would be reversed. It can be confusing, but that's life.
DeleteYeah, noticed that, for example in:
Deletehttps://mathworld.wolfram.com/CliffordAlgebra.html
v^2=-Q(v).
The following seemed like a nice intro for construction of Cl(3,4):
https://hal.science/hal-03015551v2/file/The_Clifford_algebra_book-24%20(1).pdf
but apparently way above my current level (and it basically didn't mention any Cl(p,q) with p+q=n > 4).
In this paper:
https://arxiv.org/pdf/2011.05171
Cl(3,4) is mentioned in context of symmetry breaking in strong force sector, but it also said about including QCD in Clifford algebra (in contrast to SM approach to keep the gauge group outside of it):
"It seems to require a rather radical reinterpretation of the strong force as being not so much a force as a quantisation
of space itself. Such an interpretation is certainly unconventional, but is perhaps
hinted at by the phenomenon of asymptotic freedom."
Any advice for a good literature for construction of Cl(3,4)? Or to rather drop it and stick to basics for starters?
On
https://en.wikipedia.org/wiki/Classification_of_Clifford_algebras#Classification
it says the 'generators' of Cl(3,4) would be M_8(C), which somehow seems a bit 'too much' just for having with. :/
Oh well.
DeleteIn a wavy way, from Cl(3,4) or Cl(4,3) depending who's watching, over isotropic quadratic forms in field theory, came to p-adic numbers which seem to touch upon an issue wrestled with about 3 years ago, so it seems there's sort of a starting point.
Any suggestions for a jump-start literature?
Under (4):
ReplyDeletewith Λ0(E) = R
or
with Λ0(E) = C ?
After thinking about it - though R
ReplyDeleteI will also have to make comments about this R and C issue. What I am doing here is not quite standard, so some warnings will be needed.
DeleteIt does seem that we are using the 3 basis vectors of Cl(3) to make 8 multivectors (1 scalar, 3 vectors, 3 bivectors, 1 trivector) that are then made into basis vectors themselves (for bilinear forms, E8, Cl(8)).
ReplyDeleteWhat would it mean for a trivector time to be reused as a basis vector for a bigger structure instead of a scalar time? Could the bivectors be used for space instead of vectors for the basis vector reuse? These questions make more sense to me for the cellular automata that Tony Smith used as a Clifford algebra analogy since 3 bits there resulted in 256 rules.
For the analogy a large number of bits (vectors) kept spacetime from messing up the identity of the small bit things like translations and special conformal transformations.
Interestingly, along those lines was the initial flash idea for Cl(3,4): 3 (unit) vectors, 3 bivectors (unit quaternions) and the trivector or pseudoscalar, i.e. imaginary unit i. At least the signature would be OK, (+++,----). And the leaving of the scalar part out might be philosophically argued that there's no absolute nothingness, only its reflection in the imaginary realm.
DeleteHow does that sound to you?
But your notation Cl(3,4) suggests that we are taking Cliffird algebra os a 7-dimensional space. Dimension of such a Clifford algebra would be 2^7 = 128. But this is not what you seem to have in mind. Though I do not know. There is nothing wrong about Clifford algebra of another Clifford algebra...
DeleteI'm aware of that, it would be 7d space spanned by the basis of our Cl(3,0) (minus the scalar), and the Clifford geometric algebra over it would have 128 basis elements, like you pointed out.
DeleteIf also the scalar would be included, like John G hinted to, would we be dealing with Cl(8) or Cl(4,4) in that case?
In a sense, it would be like 'observing the observation', if that makes any sense, where our Cl(3) would represent the Universe observing Itself and Cl(3,4) would represent the consciousness units (of the "universal origin") observing that "original observation", including themselves (but not actually the Origin Itself, just Its observation). Kind of like a circular action, where the next step/round in line/spiral might be the Clifford algebra over 2^7 - 1 = 127 dimensional space (if needed to go on with 'circularity'). Would that make any sense?
DeleteVery interesting idea.
DeleteA comment on the nested hierarchical structure of Clifford algebras. As known, the real Clifford algebras C(p,q) form eight classes that define the Brauer-Wall group (directly related to the Cartan-Bott periodicity). The action of the Brauer-Wall group generates a fractal-like chessboard structure on the representation system of the Lorentz group, similar to the Sierpinski carpet. The fractal dimension (Besicovitch-Hausdorff) of this structure is D = ln63/ln8 = 1.9924. For details, see http://msm.univer.omsk.su/jrns/jrn35/varl1503.pdf
DeleteThanks. Mathematicians seem to have infinite imagination. How it applies to the real world? This question is not bothering them too much. Just once in a while they have dreams about their free constructions to have SOME relation to reality.
Delete
DeleteThe main ideas of an 'infinite chessboard' have been known from ancient times. It was not occasionally that Indians placed the World-modeling game on a 8x8 checkered board.
Infinite perodicity was added by Buddha, who set two mirrors one opposite another. A sujet worth depicting somewhere in the corner of one of the beautiful illustrations that Ark draws for us.
An old and unsolved question for me is how the fractal dimension is defined and determined, still eager to learn this out.
I had to numerically estimate the fractal dimension for my book p.169. What I remember now is that that are several definitions and they do not always agree.
DeleteSo, is it true that:
ReplyDeletee1^(e1^e2) = -e2
and is it true that:
e1·(e1^e2) = 0
e1^(e1^e2) = 0 since exterior product is associative and e1^e1=0.
Deletee1·(e1^e2) = 0 is true because e1 and e1^e2 have different grades.
I am assuming that the dot in e1·(e1^e2) stands for the scalar product.
DeleteBut
Deletee1×(e1×e2) = -e2
"I am assuming that the dot in e1·(e1^e2) stands for the scalar product."
DeleteYes, it does.
So which of the below equations isn't correct?
DeleteOn one hand:
e1e1e2 = e2
on the other hand:
e1e1e2 = e1(e1e2)
= e1(e1^e2 + e1·e2)
= e1(e1^e2 + 0)
= e1(e1^e2)
= e1^(e1^e2) + e1·(e1^e2)
= 0 + 0
= 0
This is not true:
Deletee1(e1^e2)
= e1^(e1^e2) + e1·(e1^e2)
And it is not true because e1^e2 is a bi-vector, not a vector. I will give the formula for calculating such expressions later today.
In fact I can give it now:
Deletexu=x^u + i_x u
where i_x has been defined in Part 6. In particular
i_x y = x·y,
i_x(v∧w) = i_x(v)w - i_x(w)v,
for a vector x and arbitrary u,v
So, on one hand:
Deletee1e1e2 = e2
on the other hand:
e1e1e2 = e1(e1e2)
= e1(e1^e2 + e1·e2)
= e1(e1^e2 + 0)
= e1(e1^e2)
= (e1·e1)e2 - (e1·e2)^e1
= 1e2 - 0e1
= e2 - 0
= e2
(And everything is OK. So thank you Ark.)
You are welcome. There will be yet another method of getting the same result using another formula - in the new post that should be ready within an hour or so.
Delete