Spin Chronicles Part 43: Feeding GNS doll

 Grothendieck, the great and strange mathematician, complained about the nature of mathematics:

"I think there's an inherent flaw in the mathematical way of thinking and I have the impression that it's also reflected in this Manichaean vision of human nature. On the one hand, there's the good, and on the other, the bad, and in the best case scenario, we see both living side by side."

Grothendieck, disappointed,  has left mathematics and went into seclude in Pyrenees. I do not understand it. Personally I do not see anything wrong with mathematics, and with yes-no Aristotelian logic:

There is good and there is evil and there is the specific situation that determines which is which.  What's wrong with this?

From The Guardian article about Alexander Grothedieck of Aug 31, 2024: "He was in mystic delirium’: was this hermit mathematician a forgotten genius whose ideas could transform AI – or a lonely madman?"

"I open a first page at random. The writing is spidery; there are occasional multicoloured topological diagrams, namechecks of past thinkers, often physicists – Maxwell, Planck, Einstein – and recurrent references to Satan and “this cursed world”

Alexander Grothendieck - forgotten genius
whose ideas could transform AI

So I stay with the devil of the algebra and the angel of geometry, and with their happy marriage - geometric algebra. Which geometric algebra? This is a good question. Certainly time is not the same as space. Einstein and Minkowski merged space with time, but their reasoning was highly questionable. Maxwell equations were taken as a starting point. 10-parameters Poincare group came out rather easily as an invariance group, and it was set on a pedestal. It was given the absolute power over all other possible interactions, for no good reason. Then it was discovered that Maxwell equations are invariant under the 15-parameters conformal group, containing the Poincare group as a subgroup leaving the "conformal infinity" (Dupin cyclide)  as the unmoved "absolute", but it didn't help us with understanding the nature of time. Is time "real" or "imaginary"? Or both? There is a lot that needs to be done in physics before jumping ahead into the future of fancy mathematics with no "yes-no"  dichotomy. Some physicists promote the view that the universe is a quantum computer - like it would explain anything. It does not explain anything at all, it just sweeps problems under the carpet, since we do not understand what quantum theory is about. Quantum theory does not explain consciousness. We need to understand what "consciousness", "information", and "measurements" are first. Then, perhaps, we will be able to get some idea about quantum theory, the theory of "measurements".

Let's go back to algebra, geometric algebra. The mystery of spin is hidden there. We do not need to mix space and time from the start (see Peter Woit blog "The Mystery of Spin"). If time comes out of space and spin - that would be a nice surprise. Geometric algebra A = Cl(V) of space happens to be isomorphic to complex quaternions that can be faithfully represented by 2×2 complex matrices Mat(2,C). Unexpected meeting of geometry with the formalism of quantum mechanics. Coincidence? Or there is some deep meaning hidden in this fact? Can we read/decode the message?  

Hamilton invented quaternions when trying to make space into an algebra. He was forced to add the fourth dimension. What is the meaning of this fourth dimension? Is it just pure mathematics that has nothing to do with reality? Certainly the fourth dimension of Hamilton is not the "real time" of special relativity theory. So what is it? For a while we have a toy that we are playing with, discovering new functionalities. I talked to Laura about all these mysteries, and she told me that as a little girl she owned a doll that could even pee. Of course you had first to fill the doll with water. What can our Cl(V) produce if properly filled with  proper stuff? Shall we get spinors?

Feed 1.

First we have fed A with an orthogonal  projector p = (1+n)/2,  p = p* = p². It produced the right eigensubspace

{u: up=u} = Ap,

which is a non-trivial left ideal of A. It carried an irreducible representation of A,  where "spinors" live. We also got its orthogonal complement A(1-p). Spinors can reside also there.

Feed 2.

Then we have chosen a different approach, through the Gelfand-Neumark-Segal (GNS) construction. This time we fed A with a "state" - a positive linear functional f on A. It produced a left ideal I_f:

I_f = {u: f(u*u)=0}

which we didn't like (as it consists of norm zero vectors), so we have taken the quotient Hilbert space H_f = A/I_f, build as the set of equivalence classes [a], and we got a representation  ρ_f of A on H_f with cyclic vector Ω_f = [1] such that

f(a) = (Ω_f,ρ_f(a)Ω_f). 

If f is a pure state, then the representation ρ_f is irreducible (we did not show it yet). So elements of H_f are also candidates for spinors. The relation to Feed 1 is still waiting for an explanation.

Feed 3.

Here again we start with a state f, but this time we add some salt - we use the fact that our A is not only a *-algebra, but also a Hilbert space on its own. This allows us to represent f through an algebra element, denoted by the same letter, so that

f(a) = <f*, a>_{.                             (1)}

Taking the positive square root f^½  of f  we get

f(a) = <f^½, L(a)f^½ >,

where L is the left regular representation of A on A.

Connecting Feed 2 to Feed 3.

We can now happily use Proposition 4.5.3 from the GNS Promenade to connect the two constructions. For this to take place, Feed 3 should result in a cyclic representation. As it stands, the representation is not necessarily cyclic in A. Let us understand this problem better. Let us look at Eq. (1), but from the right to the left. If f, on the right,  is a positive element of the algebra A, then (1) defines a positive functional. What are the simplest possible positive elements of A? The answer is: they are orthogonal projections p=p*=p². So, what happens if we take f = p = (1+n)/2? Since p = p* = p², it follows that f^½ = p. In that case L(A)f^½ = Ap is our left ideal of Feed 1. It is certainly not the whole A - for this choice of f. Thus, in this case
f^½  is not cyclic in A. But it is cyclic in the left ideal Af^½ ! Define H'_f = L(A)f^½,
Ω'_f = f^½, ρ'_f = L restricted to H'_f. Then, according to Proposition 4.5.3,  there exists an isometry U from H_f to H'_f, that maps  Ω_f  into
f^½, and U intertwines the two representations:

Uρ_f(a) = L(a)U   on H_f.

The GNS construction (Feed 2) is not really needed in our case. The point is that GNS construction is quite general, it works with any C*-algebra. Here we have a particular case, when our algebra is already a Hilbert space (perhaps too big), so we can apply Feed 3.

We have already made a connection to Feed 1 by taking f = p, but this needs to be made clear. We will do it in the next post.

As for spinors - they are lurking, always nearby, but still not as main characters.

P.S. 04-02-25 From my correspondence: "... Allais' effect only worked when his pendulum was triggered by a man. Allais' effect was officially recognized by the French academy of sciences."

Interesting. Quoting from Wikipedia

https://en.wikipedia.org/wiki/Allais_effect

Allais' paraconical pendulum

Maurice Allais states that the eclipse effect is related to a gravitational anomaly that is inexplicable in the framework of the currently admitted theory of gravitation, without giving any explanation of his own.^[30] Allais's explanation for another anomaly (the lunisolar periodicity in variations of the azimuth of a pendulum) is that space evinces certain anisotropic characteristics, which he ascribes to motion through an aether which is partially entrained by planetary bodies.

His hypothesis leads to a speed of light dependent on the moving direction with respect to a terrestrial observer, since the Earth moves within the aether but the rotation of the Moon induces a "wind" of about 8 km/s. Thus Allais rejects Einstein's interpretation of the Michelson–Morley experiment and the subsequent verification experiments of Dayton Miller.^[31][32]

Spin Chronicles Part 33: Consciousness and conscious choices

 Once upon a time, guided by a whisper of intuition—or perhaps a playful nudge from fate—we set out on a journey. At first, our quest seemed clear: to uncover the mysteries of the enigmatic spinors. We had a map (or so we thought) and a destination in mind. But as we wandered deeper into the unknown, the wide road faded into a meandering trail, and the trail became a wisp of a path. Before we knew it, we were in a forest—dense, shadowy, and alive with secrets.

The forest wasn’t on the map, but here we were. And while our grand quest felt like a distant memory, the forest itself had other lessons to teach. At first, we worried: how would we ever find our way? But then we noticed the sweetness in the air, the earthy scent of moss, and the rustling leaves whispering ancient songs. We saw plump berries, glistening with dew—some delicious, others mysterious. And then, as if out of a dream, a gentle roe deer emerged, its soft eyes urging us to follow. It led us to a crystal-clear lake, where the water was cool and refreshing, as though the forest itself offered us a blessing.

... the forest itself offered us a blessing.

Forests, after all, are not just places to get lost; they are places to be found. They nourish the soul, if only we stop to look. So we paused, took a deep breath, and began to notice both the towering trees and the soft carpet of the forest floor. In this moment of stillness, we remembered our original quest. Yes, we were here to understand spinors, but perhaps the forest—the journey—was as important as the destination.

This particular forest is called Geometric Algebra A. It is a simple land, yet rich with wonder. To truly know it, we must not just walk its trails but see its beauty, smell its air, touch its textures, and listen to its tales. Some stories are soft whispers; others roar like waterfalls. This is one of those stories, told by the forest itself.

So, dear traveler, let us begin.

We are in geometric algebra A. It is simple, but it has a rich structure. We need to feel this structure by sight, by smell and by touch. We need to be able to hear the stories it says to us, sometimes silently, sometimes in a really loud voice. So this is on of these stories.

A is simple. In algebra saying this has a precise meaning: an algebra is simple if it has no non-trivial two-sided ideals. A two-sided ideal is a subalgebra that is at the same time left and right ideal. We did not consider two-sided ideals yet (and we will not in the future), but it is not difficult to show that A is indeed simple. But we did consider left ideals, and those of a particular form. To construct such an ideal we select a direction (unit vector) n in V, from this we construct p, with p=p*=pp:

p=(1+n)/2

Then we define, let us call it  I_n:

I_n = {u: up=u}.

This is a left ideal.

Now, the defining equation up=u is equivalently written as un=u (convince yourself that this is indeed the case!). Since n² =1 (n is a unit vector), and n*=n, it follows that n, considered as an operator acting on A from the right, has two possible eigenvalues +1 and -1. So the equation  un=u means that I_n consists of eigenvectors of n belonging to the eigenvalue +1. This is our left ideal under consideration.

The first thing we notice is that p itself is an element of I_n. But that is not all I_n. I_n is a complex two-dimensional space. Thus there are two linearly independent (even mutually orthogonal) vectors in I_n.

Thus we proceed as follows: we choose an oriented orthonormal basis e¹,e²,e³ in V in such a way that e³ coincides with n. Then e² and e³ are perpendicular to n. Then we define a basis E₁,E₂ in I_n by choosing:

E₁ = p
E₂ = (e¹ - ie²)/2,

Then magic happens: in this basis the left action of e¹,e²,e³ on  I_n is given exactly by the three Pauli matrices!

We came to the lake in a forest and it is time to fore out the thinking machine in our brains. We have arrived naturally at Pauli matrices, which is very rewarding. Except for the fact that there is nothing "natural" in this process! First we had to select a direction n in otherwise completely isotropic space V. This cannot be deterministic. No deterministic process can lead to breaking a perfect symmetry. It can be done only by a conscious choice. So consciousness is entering here (or it can be a random choice, but then consciousness is needed to define what precisely a "random choice" is). In practice the choice of a reference direction, and of an orthonormal basis is being made by a conscious "observer" (or by a machine programmed by a conscious "observer"). You can, of course, replace "observer" by "experimental physicist" or "an engineer", but that will not change the idea.

Then we decided to define E₂ the way we did above. Another application of consciousness. Thus, temporarily, I am associating the right action of the algebra on itself, the action of p in up=u, with consciousness. It is not needed for further considerations, but it something that should be thought about: we have left and right actions of operators on our Hilbert space. In ordinary quantum theory only left actions are being considered, What can be the meaning of right actions, if any? But let us abandon philosophy and return to math.

We have the basis E_α (α=1,2) in I_n, we have the basis eⁱ (i=1,2,3) in V and they are related by Pauli matrices σ_i by the following relation

eⁱ E_α = E_β (σ_i)_βα.       (*)

Notice that I write the right hand side by putting the basis vectors first, and coefficients after. That has the advantage that matrices transforming the components are transposed to those transforming the basis vectors. This way I do not have to transpose anything.

But what happens if we replace our basis E_α by some other orthonormal basis in I_n? Then the whole beauty and simplicity of Pauli matrices will be spoiled. And we like Pauli matrices so much! And here is the place to demonstrate the power of the desire. We want Pauli matrices, whatever the cost would be! So, we start thinking. When there is a desire, there must be a way! So we start looking at our equation (*) from a different point of view. The elements E_α  and eⁱ are related (or "correlated") by the Pauli matrices. If change E_α, perhaps eⁱ also need to be changed so, that the correlation stays the same? We try our great idea of saving our love - the sigmas. The result is condensed in the following statement:

Proposition. There is one and only one way to have Eq. (*), with Pauli matrices in it, valid for all orthonormal bases in I_n. It goes as follows: if E_α is replaced by E'_α related to E_α by a 2 by 2 unitary matrix A of determinant 1 (element of SU(2)):

E'_α = E_β A_βα,

then e_i must be replaced by e'i, related to e_i by a real orthogonal 3 by 3 matrix R(A) (an element of SO(3)):

e'_i = e_j R(A)_ji,

where the relation between A and R is

Aσ_iA* = σ_j R(A)_ji.

Proof. Left as a straightforward, but needing use of indices,  exercise.

And this way we have accomplished something that was left unexplained in October 16 post  Part 3: Spin frames.

What we see now is that this correlation between spin  frames Eα and orthonormal frames e_i is not so "natural" at all. It requires certain arbitrary human-made choices. It has little to do with the "true state of affairs". Spin frames and orthonormal frames are two different realities. Yes, they can be "correlated", but this correlation is artificial. So, the question remains: what are spinors? Elements of a left ideal? But which one? And why this one, and not some other?

Exercise 1. Do the calculations needed to prove the Proposition.

Exercise 2. For any x in A denote by Ax the set

Ax = {ax: a in A}

Show that Ax is a left ideal. Show that it is the smallest left ideal containing x. With I_n and p =(1+n)/2  show that I_n = Ap. Why Ap is not the same as An? What is An?

Exercise 3a.  If u in A is invertible, it cannot be contained in any of the  I_n's.

Exercise 4. Show that the * operation transforms every left ideal into a right ideal, and conversely.

Exercise 5. If Il1 and Il2 are two left ideals, is their intersection also a left ideal? If Il is a left ideal and Ir a right ideal, is their intersection a two-sided ideal?

The Spin Chronicles Part 32: Ideal exercises

 The Spin Chronicles Part 30:

Solutions to exercises of Part 30

This post is all written by Saša. Saša agreed to present the results of our common work (Saša, Anna, Bjab). So here they are.

Saša with ideals