Circles and spheres. Spheres and circles. My mind is spinning. Circles seems to be more free than spheres. Two circles can entangle each other without touching. Spheres can't do such things. But sphere have inside and outside. Circles do not know about such concepts. There are also planes. They can be thought as spheres of infinite radius. Planes have two sides, but neither of them deserves to be called outside. Or both can be called such. Straight lines can be considered as circles of infinite radius. Points are spheres of zero radius - poor degenerate beings. So far talking about space I was dealing with points and linear concepts - like multivectors. It is time to get beyond linearity, and we will start our new adventure. It will be about the music of spheres. We will delve into the Lie sphere geometry. Wikipedia has an article on this subject. You can find some relevant references there. Wikipedia has also an article on Music of the Spheres. There we read:
The musica universalis (literally
universal music), also called music of the spheres or harmony of the
spheres, is a philosophical concept that regards proportions in the
movements of celestial bodies—the Sun, Moon, and planets—as a form of
music. The theory, originating in ancient Greece, was a tenet of
Pythagoreanism, and was later developed by 16th-century astronomer
Johannes Kepler. Kepler did not believe this "music" to be audible, but
felt that it could nevertheless be heard by the soul.
 |
| The music of the spheres |
My soul (if I have one) is mathematical. So that is how we will be proceeding - through the math, I am only learning this subject. I wanted to learn it long ago, having a deep feeling of its importance, but I was always postponing it - for later. And now it comes to the fruit. Whatever I learn, I will write about it. Little by little, step by step. At first it will be just geometry. With time algebra (geometric algebra) will join in. Spinors will reappear in new clothes. So, for a while, we leave aside Spin Chronicles. For those curious I will start with the approach described in the paper "Lie Sphere Geometry in Hilbert Spaces" by Walter Benz, Result. Math. 40(2001), 9-38. It is a real poetry.
In the article on Musica Universalis Wikipedia quotes William Shakespeare:
Sit, Jessica. Look how the floor of heaven
Is thick inlaid with patines of bright gold:
There's not the smallest orb which thou behold'st
But in his motion like an angel sings,
Still quiring to the young-eyed cherubins;
Such harmony is in immortal souls;
But whilst this muddy vesture of decay
Doth grossly close it in, we cannot hear it.
Sit, Jessica. Look how the floor of heaven
And indeed heaven outside, soul inside, and yet they are one. Like in Lie sphere geometry that will come.
P.S. 26-03-25 9:11 Yesterday I have attended a seminar talk by G. Shipov:
The talk was very well prepared and the presentation was very clear. Shipov was talking about the difficult path of inventors in the world ruled by money. He talked about his own path, about Tesla, patents, and the present day uncertain situation. A part of his talk was about his own ideas
The discussion after the talk was lively, altogether more than two and half hours. Igor Bulyzhenkov, who was in charge of the seminar did a very good job preventing the discussion to become a war between some strongly opinionated participants. Shipov himself was quite open-minded and appreciating alternative ideas. I have noticed that he was however somewhat hesitant in admitting a possible "experimenter effect" - a possibility mentioned during the discussion.
That was a really captivating overture! The audience waits with bated breath for the curtain to rise.
ReplyDelete"Two circles can entangle each other without touching. Spheres can't do such things". - Even in the n>3 dimensions?
"Even in the n>3 dimensions?"
DeleteGood question. Do you know the answer?
But here, in Lie sphere geometry, by a "sphere" we always mean "hypersphere", that is of dimension n-1.
Does n-1 hypersphere for n=3 mean a disk, infinitesimally flat, like a filled circle?
DeleteIf so, then by extrapolating from n=3 and n=4 cases to higher number of dimensions, it would seem our n-1 hyperspheres would not be able to entangle each other without touching.
"by a "sphere" we always mean "hypersphere", that is of dimension n-1"
DeleteAs i have suspected. So, 'circle' is always a 1d closed curve and 'sphere' has dimension n-1. Ok. Then, it looks like Saša is right and "n-1 hyperspheres would not be able to entangle each other without touching". While circles are able. Such a strange exclusion of 1d from all other dimensions.
Well, maybe it would not be an exception.
DeleteFor n=2, n-1 hypersphere would be circle, but staying in those 2 dimensions of origin it still would not be able to entangle each other without touching. It would become possible to entangle without touching only if a third dimension, perpendicular to those two of origin, would be used.
And for n=1, n-1 hypersphere would be a simple point.
At least that's how I understood it.
"staying in those 2 dimensions of origin it still would not be able to entangle each other without touching"
DeleteActually, that is what i meant from the beginning: if we take two 2d spheres not in 3d but in 4d space, wouldn't they behave as 1d circles in 3d and entangle without touching?
From morning i'm staring into the theory knots and links (encountered a proposition by Artin about enclosure of S2 into R4) but still cannot figure out an unambiguous answer to the question: can two n-spheres entangle each other without touching in any dimension?
DeleteРазве это не очевидно? Раз окружность и 2-сфера не запутываются в нашем 3-мерном пространстве, то как могут запутаться две 2-сферы.
DeleteIt is evident that two 2d-spheres cannot get entangled in 3d, but similarly cannot two 1d-circles in 2d, as Saša noted above. What about generalization - is it true that no entangling is possible for (n-1)spheres in Rn? And what happens in still higher dimensions?
DeleteВы не поняли мою логику. Если 2+1 не запутываются в 3, то 2+2 не запутаются тем более. Иначе говоря, если бы 2+1 запутывались, то можно было бы ожидать, что 2+2 тоже запутаются. А так, увы.
DeleteIgor, i understand your considerations, but i was asking about higher dimensions: 2-spheres in, say, 4 dimensions
DeleteHi, found a typo : " I wanted to larn it" and maybe a missing word : "At first it will just geometry." ??
ReplyDeleteAs always, thanks for sharing, although the maths are way above my actual knowledge, I really enjoy getting a glimpse of what you're all pursuing here. Hoping someway, somehow we'll get there.
Olivier
Thank you!
DeleteFinally found what have been looking for. Didn't turn out as expected, but even simpler and more elegant, and still preserved the complex nature of things. Pythagorean spiral was the key that unlocked the "scroll", as expressed in the style of John Carter.
ReplyDeleteIf you find it worthy of your time, I'm asking for a cross-check with what you see as our shared reality.
It's an 8-dim space composed of two "identical" 4-dim subspaces of opposite signs, (+/-). The subspace bases are not linear, but basically dimensional; a scalar and let's say a line, a plane and a volume. Simple basis representation would be powers of Sqrt(i), with "i" being imaginary unit. I'm currently using letter "h" as Hamilton's original notation for imaginary unit for basis vectors of this space, if you have a more suitable idea, I'm all ears.
So, the 8 basis vectors, orthonormal if we consider the norm of a complex number in a usual way, that is ||z||^2=zz*=z*z, with "*" being complex conjugation, are h_n=[Sqrt(i)]^n = Sqrt(i^n) = i^(n/2):
h0=1, h1=Sqrt(i), h2=i, h3=Sqrt(-i) and
h4=-1, h5=-Sqrt(i), h6=-i, h7=-Sqrt(-i).
Applying complex conjugation we get relations between subspaces:
h1=Sqrt(i)=-iSqrt(-i)=-h5=(h3)* and h3=Sqrt(-i)=iSqrt(i)=-h7=(-h5)* and h2=h6*, so the two subspaces seem to be entangled just fine even without the "mixing of the squares" of individual basis vectors;
(h1)^2 = (h5)^2 = h2 and (h3)^2 = (h7)^2= h6, while (h2)^2=(h6)^2=h4 and finally (h4)^2 = (h0)^2 = h0.
The multiplication rules follow straightforwardly from multiplication of powers as i^(n/2), that is addition of exponents; and the same with division, that is subtraction of exponents.
If there would be interest in this kind of representation of our reality, when the basis vectors multiplication / division table would be completed, could post it here for further discussion.
It seems, as complex numbers commute, that multiplication and division would be commutative. Since basis vectors form a set of linearly independent vectors, as different powers of imaginary unit, the addition and subtraction of general vectors represented as linear combination of basis vectors, would follow usual Pythagorean addition or we would add or subtract arbitrary vectors in quadrature.
What do you think guys, might this kind of representation be useful?
A question to those a bit better versed in algebraic math; would this 8-dim vector space described above allow for a construction of let's say proper algebra over it, in the sense that we have been doing and discussing here at Ark's for last several months?
DeleteApologies for errors in math expressions earlier today, was a bit too enthusiastic to share the idea and findings, so was not using precise language. The elements of the "basis" set are not vectors yet as they have been presented in previous comment nor they could then be orthogonal to each other. That will come with multiplication rules and table, as well as other sets of numbers, probably, apart from naturals or integers if we take both halves (+/-) of the "basis" set together. Like said in the comment before, this set presented a "basis" for dimensions building, and as can be seen they are in fact 1/2 dimensional in nature which allows also for fractal dimensions, or at least some of them, to be naturally acounted for in the picture. And obviously to call them scalar, line, plane and volume was also a bad misnomer or even an error. More to the point would be to just say that for now we have 4 dimensions sort of in outer world, the + sign half of the set, and 4 dimensions in inner world, the - sign other half, with both halves probably being finely interwoven between each other by relationships presented in previous comment.
DeleteTo your knowledge, would I be inventing hot water or the wheel with all this?
Sasha, the Lie algebra of vector killing fields of an 8-dimensional space with a neutral metric is the Lie algebra of biquaternions.
Delete@Saša, i didn't understand how these 8 elements form an orthonormal basis: norms squared z*z should be = 1, but what is conjugate to sqrt(i)? And why the elements are orthogonal, what is the inner product in your algebra?
Delete@Ark, this construction proposed by Saša seems to have something in common with the 8-spinors of Gustav Mie, how do you think?
I say basis vectors a lot when I probably need to say basis elements. The biquaternions have a 3-dim basis and then you could certainly use XYZT plus xyzt (internal) as the basis for Cl(8).
DeleteI actually call XYZT, GMAC because it was easy to remember and refers to plus, minus, translations, and conformal though G is greater than and A is anti-Desitter though Desitter is kind of more the real situation.
You could have elements of XYZT along the left and elements of GMAC along the top and make a Hodge Star map of sorts with this. It's an off the beaten path wheel perhaps because Cl(8) is rather large.
@John
DeleteI understand your point, but frankly Cl(8) has 2^8=256 basis elements which seems a bit too much, and in the corner of the mind there was what the C's suggested when they said that Clifford algebra is "close", but didn't say "Yes", and that octonions would be "better" than quaternions, and that currently there's still no such mathematical concept or object that would properly do the job. If also add to that the suggestion that there are 4 spatial and 2 temporal dimensions, this h_n=i^n/2 "octuplet" of basically rotating unit vectors in complex polar plane looks rather promising, especially when compared to Pythagorean spiral and checked out how our ordinary x-y-z 3-dim space can be created from three unit vectors rotating in 3 perpendicular planes, creating essentially a 3-dim sphere or a solid ball.
In Pythagorean spiral, each new right triangle is created by adding a unit length leg in perpendicular direction to the hypotenuse of previous right triangle, representing in principle new "orthogonal" unit element to the sum of all already existing elements in basis set. In that sense, yeah, we could say the basis set has infinite number of elements, each new one being perpendicular to the sum of all previous ones.
When checked our 3-dim space concluded by comparison that first or starting unit vector in Pythagorean spiral rotates in x-y plane, second that's perpendicular to the starting one rotates in y-z plane, and third one perpendicular to sum of previous two rotates in x-z plane. Then realized that first after rotation for +Pi/2 in standard mathematical counterclockwise fashion points in initial direction of the second, second after rotation for +Pi/2 in initial direction of the third, and third after rotation for +Pi/2 in opposite direction of the first. The it dawned on me that the opposite direction to any radius vector is in fact towards inside or towards inner dimension or world.
After some additional sketching and checking for possible mathematical expressions for the progression series in Pythagorean spiral, and seeing the Sqrt(7) line crossed just a bit in the complete opposite direction of the starting unit line, decided to have fun with various possibilities for an "octuplet" basis composed of two "quadruplets" with opposite signs. As a result arrived at this h_n=i^n/2 guys that make use the fact that imaginary unit i has the periodicity 4 in its powers or exponent. And they just might fit the bill, as would explain in next comment as a reply to Anna's questions.
The Cs did say "geometric algebra" for a good math and that's kind of Clifford algebra preferring to be over the reals. Even with Cl(8), you symmetry break down to SO(4,2). Usually you get a 2-dim time disk and 4-dim space hypersphere from the SO(4,2) and the 4 imaginary dimensions do point to the interior (from each of the disk and hypersphere boundaries to each of the two interiors) while the boundaries provide the 4 real dimensions. Infinity for Clifford algebras comes via the tensor product.
Delete"representing in principle new "orthogonal" unit element to the sum of all already existing elements in basis set. ...the basis set has infinite number of elements, each new one being perpendicular to the sum of all previous ones".
DeleteSasha, thanks a lot for attracting attention to Pythagorean spiral, i see that it is a really interesting object despite of its age:)
And it exactly fits my intuition that spinors (fermions) accounts for quality, while bosons - for quantity in our world.
The 8-spinor would be a triality-Spin(8)-SO(8)-Cl(8) thing too.
DeleteJohn, what do you mean by triality? i see magic of triality in the exact sequences of groups, and search for them everywhere...)
DeleteFrom http://deferentialgeometry.org/papers/loun112.pdf
DeleteOn the level of representation spaces, triality could be viewed as permuting the vector space R8 and the two even spinor spaces, that is, the minimal left ideals Cl8+..., which are sitting in the two-sided ideals Cl8+...=Mat(8,R) of Cl8+=2Mat(8,R). This means a 120 degree rotation of the Coxeter–Dynkin diagram of the Lie algebra D4 :
Rather than permuting the representation spaces, triality permutes elements of Spin(8), or their actions on the vector space and the two spinor spaces.
Because of its relation to octonions, it is convenient to view triality in terms of... the paravector space... having an octonion product... and the primitive idempotents...
The action of u on the left ideal Cl0,7... results in the matrix representation Spin(8) > u =
U1 0
0 U2
where U1,U2 < SO(8).
http://deferentialgeometry.org/papers/loun112.pdf
The formatting is of course better in the Lounesto book excerpt which I only understand the general idea of anyways. It was nice Lounesto was in to large Clifford algebras like Cl(7) and Cl(8). His primitive idempotents for Cl(8) provided a little check when I was trying to make a Hodge Star map of sorts; I wanted the primitive idempotents to be on the diagonals and they were with a particular order for the Cartan subalgebra of SO(8).
John, thank you very much for elaborating about triality and for reference to Lounesto, I will read and try to underestand. At first glance his triality has more relation to threefold Dyson's way rather than to exact sequences. Nevertheless, it is worth delving into the matter!
DeleteThere's a Wikipedia page for it too but it doesn't go into details for the spinor/Clifford algebra part.
Deletehttps://en.m.wikipedia.org/wiki/Triality
@Anna
ReplyDeleteIf we write h1=Sqrt(i) in polar notation, it gives exp[i(Pi/4+k*Pi)] or when write that as a complex number, z = +/- 1/Sqrt(2) (1+i), which is also h5=-Sqrt(i) of these h_n=i^n/2 guys, so the complex conjugate would be as usual z* = +/- 1/Sqrt(2) (1-i) or exp[i(3Pi/4+k*Pi)], which would correspond to h3=Sqrt(-i) and h7=-Sqrt(-i).
By appropriate choice between so called degenerate cases so that angles when expressed as exponentials reflect the indexes, we can write h_n = i^n/2 = exp[i(n*Pi/4 + 2k*Pi)], and then complex conjugate of h1=Sqrt(i) is simply h1*=h3=Sqrt(-i) and then also the other example in previous comment is correct and valid, h5=h7*, and complex conjugation with these two pairs of elements works as basically rotation for Pi/2.
From polar form of h_n we see that we essentially got 8 unit radius vectors in complex polar plane separated by the angle of Pi/4 between two elements with neighboring indexes, which means that neighboring elements with even indexes are separated for Pi/2, that is two closest neighbors are perpendicular, which is true also for odd indexes. Coupled with the fact that we have basically two subsets of identical, but opposite elements, sort of reflections of each other, maybe there's something to work with, like perhaps to have two planes spanned by two even and two odd elements for each subset. Need to think about that a bit (any ideas are welcome), because odd indexes or half integer exponents elements when expressed as complex numbers already contain scalar parts as (1+/-i) which maybe could be "converted" to two additional unit vectors like Hamilton's j and k for example, so we get our usual 3-dim space notation.
As for the multiplication rules, since we are dealing with the products of exponents, then we just basically add the indexes, that is h_i*h_j = h_(i+j) with (i+j) written in modulo_8 notation now due to periodicity of 8 in exponents.
Might return to Pythagorean spiral to see if some additional information could be obtained there that might shed some more light on how and if to proceed with this idea further.
Sasha, thank you very much for the explanation! i realized that i've forgotten elementary rules of dealing with imaginary numbers, surely, as soon as sqrt(i)=exp[i(Pi/4+k*Pi)], all the further, the complex conjugation, etc., is clear.
ReplyDeleteBut i'm still in doubt concerning orthogonality: every basis element is orthogonal to any other, right? How can we show this? The angle between two elements with neighboring indexes is pi/4, not pi/2.
How can we get zero inner product for any pair of elements?
Well, the short answer would be: we don't.
DeleteI mean, of course, we could postulate some things like people do with QM and ideals or with relativity and speed of light. But here we deal essentialy with dimensions, weaving space in a way, and not representing dimensions in some already existing space as a stage for our actors.
If we are really adamant to force this view into sort of a linear perspective and a singular polar plane, it might be helpful to note that odd index elements are in fact doublets, in polar notation h1=h5 and h3=h7, that is their period is k*Pi, as like they are oscillating or vibrating between the directions of even index elements. Something like oscillation of inside out and outside in, perhaps, while rotating. Simple relativity argument, nicely pictured in a particular Quantum Quirk image on SOTT, says that we can't really distinguish between rotation of the outsides and rotation of the insides, that is rotation of one side would be perceived by the other side as its own rotation in opposite direction. So even index elements with 2k*Pi periods would be representing rotation, while odd index elements like describing reflection on the surface between inner and outer, or maybe more to the point, reflection on the rotating "plane" spanned by even index elements.
But, that would still be in deficit compared to broader overall perspective, as we would still be sticking to linearity while in fact knitting the multidimensional fabric of perceived reality. Sort of "experimenter effect", squeezing the observations into preconcieved expectation and comprehension boxes.
FWIW.
Sasha, ok, thank you! Least of all i would like to force views into preconcieved boxes, just wanted to understand how do you introduce orthogonality of the basis elements. Now, when you elaborated this point, the idea appears to be even more unusual than i thought at first glance.
Delete"...we deal essentialy with dimensions, weaving space in a way"
This phrase is like balm on my soul since in my worldview dimensions play fundamental role in structuring the reality.
If understand correctly what we've been discussing here, we do not need orthogonality with basis elements. Our general basis multivectors were not orthogonal to each other, for example so called pseudoscalar i=e123 was not orthogonal to other basis multivectors, but was a "product" of e1 and e2 and e3, if remember right, just like e12 was e1e2. There are few more steps between having a basis elements like multivectors and an orthonormal basis "vectors" like our usual (e0,e1,e2,e3).
DeleteOur basis Ei of multivectors is, in fact orthogonal. See part 11 and Part 12.
DeleteYes, I see, after we introduced three involutions and basically scalar product, when turning things to Cl(V). Before that, with only Grassmann exterior algebra when we first met multivectors, we didn't yet have defined the product to be in position to talk about orthogonality. At least that's how I understood things.
DeleteIs it OK with you to post here how things advance with this idea and to ask questions in that respect or help when needed?
Of course you can ask questions. Usually I do not understand your questions, but it seems that other participants of the discussion do not have this problem, and they usually react and try to help.
DeleteI earnestly looked through Parts 11 and 12 but still do not see in what sense our basis multivectors are orthogonal. Probably, in pure geometrical sense, three vectors Ei and three planes Eij are perpendicular pairwise? But what about scalar 1 and pseudoscalar e_123 then?
Deleteunless we resort to metaphysical arguments, like that elements of different dimensions are always orthogonal because they cannot interact on equal footing? :)
DeleteIf our basis multivectors are biquaternions, they should be orthogonal; not that I can show that.
Delete"Of course you can ask questions. Usually I do not understand your questions, but it seems that other participants of the discussion do not have this problem, and they usually react and try to help."
DeleteThank you. I'm pretty confident you will understand this stuff as it concerns just math, usual math, not the pranalytical things. Only issues that I see could arise is due to HTML format and limitations of the comment box.
Starting from Pythagorean spiral, we have right triangles R_n - R_(n-1) - 1, where n is the number of steps taken from the start of the spiral, that is |R1|=1. So in vector form we have the relation, R_n - R_(n-1) = 1, where vector 1 is pointing from R_(n-1) to R_n and is perpendicular to vector R_(n-1). Multiplying this equation by vector R_(n-1), we get,
|R_n| |R_(n-1)| cos(φ_n) - |R_(n-1)|^2 = 0,
with φ_n being the angle between R_n and R_(n-1).
Using that |R_n|=Sqrt(n) yields relation for cos(φ_n),
Sqrt(n-1) [Sqrt(n) cos(φ_n) - Sqrt(n-1)] = 0,
which for n=1 gives also cos(φ_n)=0, suggesting that vector R1 is perpendicular to vector R0, which is zero vector and thus correctly gives that zero vector is perpendicular to any direction we choose to take in real 3-dim space or 2-dim plane in case of Pythagorean spiral.
For general n, the expression for cosine is
cos(φ_n) = Sqrt(n-1)/Sqrt(n) = |R_(n-1)|/|R_n|.
Multiplying the triangle relation by vector R_n, which might be more correct as R_n is not zero vector for any n greater or equal to 1, we get,
|R_n|^2 - |R_n| |R_(n-1)| cos(φ_n) = |R_n| cos(π/2 - φ_n) = |R_n| sin(φ_n),
because the angle between vectors R_n and 1, as they all with R_(n-1) form a right triangle, is (π - π/2 - φ_n)=(π/2 - φ_n).
Using again the magnitudes of R_n and R_(n-1), gives relation:
Sqrt(n) - Sqrt(n-1) cos(φ_n) = sin(φ_n) = 1/Sqrt(n) = 1/|R_n|,
which confirms that the angle between R1 and R0=0 is indeed φ1=π/2.
That's it about the general vectors from Pythagorean spiral, next comes moving to polar coordinate system and complex number representation.
Our vector R_n in polar coordinates looks like,
R_n = |R_n| [cos(ϑ_n) + isin(ϑ_n)],
where ϑ_n = Sum(φ_k) - π/2, is the total angle between R_n and R0 rotated for π/2 to have x-axis or real axis start at 0 as usually denoted in polar coordinates, that is Sum(φ_k) goes from k=1 to k=n.
Now our R_n can freely be denoted at complex number Z_n,
Z_n = |Z_n| {cos[Sum(φ_k) - π/2] + isin[Sum(φ_k) - π/2]},
and after using the relations sin(x-π/2)=sin(x) and cos(x-π/2)=sin(x), we get,
Z_n = |Z_n| sin[Sum(φ_k)] (1+i),
which suggests that the last two factors in fact make a unit vector Z_n/|Z_n|.
Now, writing last term (1+i) as a product like (A+iB)(A-iB) to reflect the nature of the usual complex number norm as a product zz*, with A=+/-1 and B=+/-Sqrt(i), and using that |Z_n|=|R_n|=Sqrt(n)=1/sin(φ_n), we get the form that looks like,
Z_n = {sin[Sum(φ_k)] / sin(φ_n)} [(+/-1 +/- iSqrt(i)) (+/-1 -/+ iSqrt(i))],
where +/- iSqrt(i) are h3 and h7 from previous comments, and which then gives for the first step Z1 several different h_n combinations like for example,
Z1 = (h0 + h3)(h0 - h3) or (h4 - h7)(h4 + h7).
Please if somebody can crosscheck the correctness of the math, before introducing complex logarithm and looking for the suitable definition of the scalar product.
By the way, maybe we can allow the angle φ_k to be solid angle and not just lying in the same plane all the time, but that would come later when we really get adequate geometric algebra in proper shape.
Thanks.
@ Anna March 27, 2025 at 7:24 PM
DeleteIn Part 11 we see the matrix of the bilinear form Re(B0). It is diagonal.But it is nothing else that the matrix of scalar product of basis vectors for this particular scalar product. The matrix is diagonal, so the basis is orthogonal.
In Part 12 we see the matrix of the bilinear form Re(Bν). Again diagonal matrix. Thus our basis is orthogonal also with respect to this scalar product.
The same with respect to the bilinear form Re( Bπ) there.
P.S. Found an error in calculations, used incorrect trig. identities, so the last part is faulty. Will go over it again to get the right expression for the complex logarithm application.
DeleteWould be grateful though if somebody can check at least the vector part and the polar representation. Thanks.
"we see the matrix of the bilinear form Re(Bν). Again diagonal matrix. Thus our basis is orthogonal also with respect to this scalar product".
DeleteSo, those bilinear forms are diagonal only because we initially treat scalar E_0 as an independent fourth component of the 8-multivector, the three two-vectors -- as independent duals to three vectors, and the three-vector E_123-- as one more independent component dual to scalar. Their orthogonality is nothing more than our convention.
Of course you can treat the construction of the Grassmann algebra and the Clifford algebra as "our convention". But mathematicians have for certain conventions the term "natural". The bilinear forms we consider are "natural", and the basis we consider is also "natural" according to this mathematical convention. I thought this was already clear, no one complained where we were discussing these issues. But it is never too late! Good that you came back to this subject.
DeleteI remember once I wanted to think of rotations and boosts as the basis/Cartan subalgebra for SO(3,3). It seemed wrong when I thought about it a lot but it fit with what I wanted to do overall so I did it. Don't know how natural that was?
Delete"But it is never too late! Good that you came back to this subject".
Delete@Ark, thank you for your patience. Our elephant is so big, and i'm walking so slowly around it... It takes time to come to the new viewpoint and see what i didn't notice beforehead.
True. But once we climb the elephant, we can shoot quantum tigers:
Deletehttps://djvu.online/download/MRjCPo9l2QbzR
p. [115]
Dear all, have you heard the news?
ReplyDeleteKashiwara, a professor at Kyoto University, Japan, received the award “for his fundamental contributions to algebraic analysis and representation theory, in particular the development of the theory of D-modules and the discovery of crystal bases”.
https://www.newscientist.com/article/2473330-mathematician-wins-2025-abel-prize-for-tools-to-solve-tricky-equations/
Peter Woit wrote about it more in his new post:
Deletehttps://www.math.columbia.edu/~woit/wordpress/?p=14448
His two previews posts were about politics, and made me wonder how an otherwise bright mind can be so clueless and brainwashed when it comes to politics.