Spin Chronicles Part 26: Imagine Clifford Algebra

 The song is in C major. Even if I like C minor better, I like the song.

"Imagine all the people

Livin' life in peace

You"

Can you imagine it? I can, even though it is not in my memory banks. Imagination is a powerful mechanism. I am reading "The Story of the Tower of Babel: Human Aspirations and Dreams".

There:

"Human imagination is a unique and powerful ability that allows us to conceive of things that do not exist, create new concepts, explore possibilities, and even transcend the current physical limitations in spirit. This ability is indeed rare in the natural world, and, at least based on our current knowledge, humans are unique in their use of complex symbolic thinking and creative imagination."

Imagination and creativity are close friends. I am imagining what I will write today. Then I start writing. The result will probably be not quite as I have imagined it.

"You may say I'm a dreamer

But I'm not the only one

I hope someday you'll join us

And the world will be as one"

You may say I'm a dreamer....

Yes, you may say I'm a dreamer. But I'm not the only one. Last night I dreamed about Hilbert Algebra. The dream had its roots in my past memory. My PhD thesis was partly about equilibrium states in quantum thermodynamics, the algebraic approach, with KMS (Kubo-Martin_Schwinger)  states. Tomita and Takesaki, the two Japanese mathematicians, developed the theory of Hilbert algebras and modular automorphisms. Much later, in 1994,  Alain Connes and Carlo Rovelli  applied it  to the "flow of time":

"We consider the cluster of problems raised by the relation between the notion of time, gravitational theory, quantum theory and thermodynamics; in particular, we address the problem of relating the “timelessness” of the hypothetical fundamental general covariant quantum field theory with the “evidence” of the flow of time. "

But now is now, and we are dealing with the Ur atom of space - the geometric Clifford algebra of 3D space Cl(V). We have an algebra. Buit where is "Hilbert"? Searching for "Hilbert Algebra" leads us to Wolfram MathWorld. There we find conditions 1, 2, 3, 4, 5 that need to be satisfied.

In our case only condition 3 counts. All other are automatically satisfied. So, we should have an algebra with involution (we have one), which is a Hilbert space (and Cl(V) is a Hilbert space, as we will see soon),  with a scalar product <a,b> satisfying

<ba,c> = <b,ca*>.                    (1)

We need to define Hilbert space scalar product on Cl(V). We have already done it in Part 11,

where we have defined:

"... But then we have also the second natural "bilinear" form:

B_τ(u,v) =t_n(τ(u)v),

In fact, this form is, when considered as a form on a complex vectors space,  Hermitian: it is complex linear in v, but complex anti-linear in u."

In Part 12 we further wrote:

B_τ(u,u) = |p₀|² + |p|².

This is our Hilbert space scalar product - the scalar part of τ(u)v: we set

<u,v> = the scalar part of the product τ(u)v. If u = (p₀,p), v = (q₀,q), then

<u,v> = p_0^*q₀ + p^*· q.

Then we need an involution in our algebra, we need a *-algebra. We take τ for our involution. We just set u* = τ(u). There may arise some confusion now. Because in my blog I often used the star to denote the complex conjugation. And now I am using the same symbol to denote the conjugation in the algebra! Well, the meaning will be defined by the context, and when confusion can arise, it will have to be explicitly addressed.

Let us recall the action of τ:

If u = (p₀,p), then τ(u) = (p₀*,p*), thus u* =  (p₀*,p*). In other words:

(p₀,p)* = (p₀*,p*).

Not too bad! I suppose we can live with that.

Now comes the crutial point:

Exercise 1. Prove (1).

Hint: there are at least two ways of proving it. Hard way, using definitions, and not-so-hard, if you are creative.

P.S. 18-12-24 7:48 Under Part 24 Anna asked a number of questions. Here are these questions repeated:

AnnaDecember 17, 2024 at 8:51 PM

So late, but i have got that result x'=x+a and y'=y+b.

Still not happy, since it appeared as a rabbit from hat, i don't see the meaning of this result.

and have an entire list of questions:

1. Why for the transformations (1)-(4), which seem quite similar, the results differ so drastically?

2. Why Sasha spoke about boosts and rotations though we use their mixture when taking exp(e1+i e2), etc.

3. Conformal transform is inversion + translation + inversion. Hence, translation can be a special case of conformal transform if we eliminate the operation of inversion, i.e., take q=1.

These are vectors on a unit circle, and it seems to be the stable point of conformal transform. But what does it mean physically? Somewhat a mass shell if we work in the momentum space?

4. Finally, Ark, about your remark on a vanishing denominator. In our particular case, denominator is (1 + x a + y b)^2, and it vanishes when the scalar product of vectors (x,y) and (a,b) is minus 1, i.e. the stable point is (-a,-b), and somehow we must account for normalization, right?

I will be addressing these questions one by one, though I am not sure that I have good answers to all of them. It will take me several sessions to reply. Certainly they are all valid questions and it is very good that they have been asked. Let me start with the "rabbit from hat".

Indeed this is how it looks like, because I did not reveal the trick. And the trick is described in the conformal group literature. It is known among the specialists that the conformal group for a space of signature (r,s) is the group SO(r+1,s+1), We add one space-like dimension and one time-like dimension. The Euclidean plane (x,y) has r=2, s=0. Thus its conformal group is SO(2+1,1)=SO(3,1) - the Lorentz group. Normally one would need to take the Clifford algebra Cl(3,1) for that. But the group itself is realized, by standard procedures through the even subalgebra of Cl(3,1). And the even subalgebra of CL(3,1) is isomorphic to the whole algebra Cl(3,0), our Cl(V). This general knowledge was in my mind all the time, so I was interested how it is all implemented explicitly. That was my reason for the calculations and exercises.

Now let me address the point "2. Why Sasha spoke about boosts and rotations though we use their mixture when taking exp(e1+i e2), etc." Certainly e1 generates boosts, and i e2 generates rotations. But they do not commute, so exp(e1+i e2) is not the same is exp(i e2)exp(e1). But what it is then? To answer this question it is convenient to think of Cl(V) as the algebra of 2x2 complex matrices. We already know that these two algebras are isomorphic, and it is more convenient now to make use of this isomorphism. Playing with matrices is, in this case,  easier than using abstract Clifford algebra tools. In matrix representation e1,e2,e3 are represented by the three Pauli matrices. So exp(t(e1+i e2)), is a matrix. It is a matrix of determinant 1, thus invertible. If we have access to a computer algebra program, we can find this matrix explicitly. Every inversible matrix A has a unique polar decomposition

A = UT

where U is unitary and T is positive. I was doing such manipulations in my Quantum Fractals book. Unitary matrices represent rotations, positive matrices represent boosts. Computer algebra programs can easily find this decomposition explicitly. One can calculate explicitly the boost direction and rapidity from T, and the rotation axis and angle of U. U and T happen to be very particularly related, so that there is only one fixed point for the whole transformation. I am not saying that all is evident here. But that is how it is, in essence.

I will continue in the next P.S.

P.S. 18-12-24 11:38 Replying to "3. Conformal transform is inversion + translation + inversion. Hence, translation can be a special case of conformal transform if we eliminate the operation of inversion, i.e., take q=1."

First recall the embedding formula from Part 20:

"ζ₀ = (1+x·x)/2, ζ₁ = x₁, ζ₂ = x₂, ζ₃ = (1-x·x)/2.

Now ζ₀ + ζ₃ = 1. If we have any point ζ' on the cone with ζ'₀ + ζ'₃ > 0, we can always rescale it by a unique positive λ to get ζ₀ + ζ₃ = 1, and then read the coordinates from ζ (the rescaled ζ') the two coordinates on the plane. "

Now, conformal inversion is a discrete operation. It is not in our group G. In our language it corresponds to the change of sign of   ζ₃ . The determinant of such a transformation is -1, while our group elements are all of determinant +1. Nevertheless it is a well defined operation, it is, in fact, and algebra automorphism, since it preserves the fundamental relations: anti-commutativity and squares. Let us see what it does with our embedding. We embed x to get ζ as above. Now we change the sign of   ζ₃ . We get

ζ'₀ = (1+x·x)/2, ζ'₁ = x₁, ζ'₂ = x₂, ζ'₃ = -(1-x·x)/2.

But now ζ'₀ + ζ'₃ = q, instead of 1. So we have to rescale,  to divide by ζ'₀ + ζ'₃ = q. The result is

x' =x/q, y'-y/q.

The conformal inversion. Now we know howe we can get translations from special conformal transformations and special conformal transformations from translation within our framework

P.S. 18-12-24 15:03 I am still thinking and can't find a satisfactory reply to 4.

P.S. 18-12-24 17:10 The denominator becomes zero for x = -a/(a.a). The only stable point for

x' =(x + qa)/(1+ q (a²+b²) + 2(xa+yb)),
y' =(y + qb)/(1+ q (a²+b²) + 2(xa+yb)).

is the origin x=y=0.

P.S. 18-12-24 17:30 1. Why for the transformations (1)-(4), which seem quite similar, the results differ so drastically?

They do not differ drastically at all if we are on the sphere. It is only when we select one point on the sphere to project the sphere on the plane, only then they start look differently - with respect to this particular selected point