Spin Chronicles Part 29: Don't be cruel

 Every good story deserves a happy ending. After all, nobody wants to be left with frustration—especially during the holidays! So, on this cheerful Christmas Day, I bring you the happy conclusion to the journey we embarked on in Part 28. 

Happy conclusion

If you recall, I ended that post with a bit of a cliffhanger:

It would be cruel of me to ask the Reader, on Sunday, two days before Christmas Eve,  to prove that, in fact, we have

R(A) = L(A)',

L(A) = R(A)'.

So, I leave the proof for the next post. But, perhaps it is not so cruel to ask the following

Exercise 5. Show that L(A)∩R(A) = C, where C denotes here the algebra of cI, where c is a complex number and I is the identity matrix.

Now, I must confess—despite my best intentions, I may have accidentally channeled a little too much academic spirit right before the holidays. As Elvis Presley, a favorite in our home, would croon, “Don’t be cruel.” But cruel I was, unintentionally!

Thankfully, Saša rose to the challenge with some impressive attempts to crack the commutator identities. In mathematics, as in life, there’s often more than one way to reach the truth, and this case is no exception. Today, we’ll use some “baby tools” to tackle this “baby theorem,” leaving the more advanced approaches to grown-up textbooks like A.W. Knapp's Advanced Algebra (see Lemma 2.45).

Lemma 2.45. Let B be a finite-dimensional simple algebra over a field F, and
write V for the algebra B considered as a vector space. For b in B and v in V ,
define members l(b) and r(b) of End_F (V ) by l(b)v = bv and r(b)v = vb. Then
the centralizer in End_F (V ) of l(B) is r(B).

So, let’s unwrap this mathematical gift and bring our story to a festive close!

I used the term "commutant" instead of "centralizer". From what I know those dealing with infinite-dimensional algebra (C* and von Neumann) use the term "commutant", those who deal mainly with finite-dimensional cases (pure algebra, no topology)  use the term "centralizer". The proof in the advanced algebra book is not that "instant" and uses previous lemmas. Here is a simple proof that I have produced for our baby case.

Proof  (of R(A) = L(A)')

We already know that R(A) ⊂ L(A)', therefore it is enough to show that L(A)' ⊂ R(A). So, let X be an operator in End(A), and assume that X commutes with L(u) for all u in A. We want to show that then X is necessarily in R(A). I will use Latin indices Wm,n,... instead of μ, ν as in the previous post. We know that X = x_mn L^mRⁿ.  Let us write L(u) = u_pL^p. Then [X,L(u)]=0 reads as

0 = u_px_mn [ L^p, L^m ] Rⁿ.

We used the fact that L's and R's commute.

Now, what do we know about the commutators  L^p, L^m ]? We know that L is a representation of A in End(A). We have defined L^p as L(e^p), where eⁱ (i=1,2,3) is an orthonormal basis in V, and e⁴=1. Since L is a representation, we have

[L^p,L^m]= L( [e^p,e^m]).

Exercise 1. Make sure that you really know why is it so. Since er form a basis in A, the commutator
[e^p,e^m] is a linear combination of e^r. We write it as

[e^p,e^m] = c^pm_r e^r.

The constants are called the structure constants of the Lie algebra. Now,

L([e^p,e^m]) = c^pm_r L(e^r) = c^pm_r L^r.

Therefore

0 =  c^pm_r u_px_mn L^r Rⁿ

for all u.

What do we know about the structure constants c^pm_r ? If p or m = 4, the structure constants are 0, because e⁴=1 commutes with every other basis vector. Thus the sums over p and m run, in fact, only through j,k = 1,2,3. On the other hand e¹e²= - e²e¹ = ie³ etc.  Thus [e¹,e²] = 2ie³ etc. While [e¹,e¹]=[e²,e²]=[e³,e³]=0.Therefore

[e^j,e^k] = 2i ε^jkl e^l.

So, we have

0 =  2i ε^pmr u_px_mn L^r Rⁿ

where p,m,r run only through 1,2,3. We know that L^rRⁿ are linearly independent, therefore
ε^pmr u_px_mn = 0. And this is true for any u, therefore

0 = ε^pmrx_{mn ,}

for all p,r = 1,2,3. To show that, for instance, x_1n =0, we choose p=2,r=3. We deduce this way  that x_mn=0 for m=1,2,3. The only possibly non-vanishing x_mn are x_4n. They stand in front of L⁴Rⁿ . But L⁴ is the identity. QED.

So, we are done. It was technical, but rather straightforward, and not scary at all - once you overcome the fear of flying!

I used the term "representation". Anna used it too in the comment under the previous post, when talking about the scary Shur's lemma. So, here comes the exercise that should help in overcoming the fear of flying:

Overcoming the fear of flying

Exercise 2: Is the representation L reducible or irreducible?

Exercise 3. Let ✶ denote the map from A to A defined by ✶(u) = u*. Then ✶ is real-linear, but complex anti-linear. Thus it is not an element of End(A), because by End(A) we have denoted the algebra of complex linear operators on A. Show that

L(u) = ✶∘R(u*)∘✶

Hint: don't be scared of flying. First try to understand what it is that you are supposed to prove. It only looks scary. 

P.S. 27-12.24 10:09 In a comment to Part 28 Anna asked for an explanation why the matrices  R^m are transposed to L^m?(Exercise 3),  One way to answer this question is by calculating them explicitly. But there is a way to see it without calculating explicitly. Suppose we accept the already discussed property that L and R matrices are Hermitian. Then we start with the defining relation for (L^m)_rⁿ:

L^meⁿ ₌ e^r (L^m)_rⁿ
_or

e^meⁿ ₌ e^r (L^m)_rⁿ.

We apply * to both sides. * is anti-linear, and (e^m)* = e^m. On the left we get

(e^meⁿ )* = eⁿe^m = R^meⁿ = e^r(R^m)_rⁿ.

On the right we get

e^r cc((L^m)_rⁿ),

where cc stands for complex conjugate. Comparing both sides we get

R^m = cc(L^m).

But L^m is Hermitian (conjugate transposed) , thus, for  L^m, cc is the same as transposed (why is it so?).

P.S. 29-12-24 8:07 This morning received the following email:

Dear Dr. Arkadiusz Jadczyk,

We are pleased to inform you about a recent citation to the following papers you have authored or co-authored. Your paper: The Explicit Form of the Unitary Representation of the Poincaré Group for Vector-Valued Wave Functions (Massive and Massless), with Applications to Photon Localization and Position Operators Mathematics 2024 was recently cited in: Note on rotational properties of position operators of massless particles Michał Dobrski Annals of Physics 2024, doi: 10.1016/j.aop.2024.169901 As of today, the paper received 1556 views and 256 downloads. For more article metrics, see: https://doi.org/10.3390/math12081140#metrics.

Although in the meantime I have almost forgotten about photon's localization problem, the phrase "light as foundation of being" is still in my mind. So, it is a good news.

P.S. 29-12-24 10:57 Anna, in her comment, mentioned the idea, supported by neuro-science research, that deep metaphysical questions exercise the most ancient parts of our brains. One such questions appeared in the comments to this blog: are we predetermined, or, perhaps, we are endowed with (necessarily limited) "free will"? How can we answer this question? I am applying my most ancient part, and I am reasoning, using it, as follows.

Whether we are predetermined or not, there are FACTS. One such fact is that we have senses, that these senses are limited, and we have brains, rather small compared to the size and complexity of the Universe. Thus our knowledge is limited, and our understanding is even more limited. There are many facts that we know about, but we do not understand them. Since our knowledge is limited, all conclusions are questionable. We can't really be sure of anything. What we know is the tip of an iceberg. So, how can we adhere to the conclusion that we are necessarily "predetermined". Such an idea is irrational. Of course someone may happen to be predetermined to hold to irrational ideas. But I choose to be rational, therefore open-minded. That is what my ancient part of my brain tells me. The newer part can find no fault in that kind of old-brain thinking.

Christmas Special - Hyperboloid of Engineer Garin

Christmas special 2024 - Ideals coming soon

 In 1929, Alexei Nikolayevich Tolstoy published a fantasy novel titled Hyperboloid of Engineer Garin.

 If you’ve never heard of it, don’t worry! You can find a synopsis and even some amusing mock-ups of Tolstoy on Wikipedia. But let’s cut to the chase—the story’s charm isn’t just in its sci-fi intrigue but in a delightful mix-up: Tolstoy mistook a hyperboloid for a paraboloid. Classic!

Engineer Garin, the story’s protagonist and a brilliant chemist, invents something eerily similar to a modern laser. His invention brings him all the usual drama: greedy tycoons, life-threatening escapades, and a healthy dose of existential dread. Tolstoy himself couldn’t decide how to wrap up this chaotic tale. His indecision is forgivable—after all, even his grandmother warned him that people like Garin either end up as tyrants or tragedies. So, to keep things flexible, Tolstoy gave his novel not one but two endings. Talk about a writer’s workaround! 

Now, let’s get nerdy for a moment. A hyperbola (or its three-dimensional sibling, the hyperboloid) does have foci, but—and this is important—they don’t behave like the focus of a parabola. A ray of light passing through one focus of a hyperbola reflects off its surface and reemerges as if it came from the second focus. Handy for confusion, right? This mix-up isn’t unique to Tolstoy. Mirror telescopes, for example, often feature a parabolic main mirror to concentrate light and a smaller hyperbolic mirror to redirect those rays toward your eye.

The result? Stunning starlit views… and maybe an unintended homage to Tolstoy’s hyperbolic adventures. Cheers to science, storytelling, and a little festive mix-up of geometry this holiday season!

P.S. 25-12-24 16:05 From the conversation at our Christmas dinner table while ago:

- Pi R squared.

- No, pies aren'tt squared, they are round.

Spin Chronicles Part 28: Left and Right Regular

As it is Sunday, and Christmas Eve is coming soon - it should be an easy talk today. In fact it is my intention that everything should be easy in my posts. By 'easy" I mean that even I myself can understand it. So, as Christmas is coming and light is a foundation of being, we decorated our home with light. The dedicated photographer in our extended family recorded it on a medium, for me to show you - there structure constants of the geometric algebra visible on the photo - for a trained eye:

Geometric algebra home

For this post I will denote our geometric Clifford algebra of space, Cl(V), by the bold letter A. It is an algebra over complex numbers, and we have a basis e⁰,e¹,e²,e³ in A. For calculation purposes, especially when dealing with matrices, it is more prudent to number the basis differently: e¹,e²,e³,e⁴, with e⁴ = e⁰ - the identity, the unit of A. And that is what I am going to use below. Thus every element of A can be written uniquely as

u = u₁e¹+... + u₄e⁴,

where u_μ are complex numbers.

A is endowed with involution "*", it is an involutive algebra. We notice that (e^μ)* = e^μ, μ - 1,...,4. For u,v in A we have (uv)*=v*u*.

A is also endowed with a positive definite scalar product

<u,v> = (u*v)₄.

We notice that the basis vectors form an orthonormal basis of A:

<e^μ, e^ν> = δ^μν.

Once we have a positive-definite scalar product, we have a norm, defined by ||u||² = <u,u>, and we notice that

||u*|| = ||u||.

We also know, from the previous post,  that A is a Hilbert algebra - we have

<ba,c> = <b,ca*>.

As any algebra, so A  acts on itself. It can act from the left, from the right, or from both sides at once. Let us denote these actions by L, R, and LR:

L(u)w = uw,

R(u)w = wu,

LR(u)w = uwu*.

From associativity of the algebra it follows then that left and right actions commute

L(u)R(v) = R(v)L(u),

and evidently

LR(u) = L(u)R(u*) = R(u*)L(u).

The map L from A to End(A) is a faithful representation of A. It is called the left regular representation. Similarly for R. Moreover, it is a *-representation, that is we have

L(u*) = L(u)*.

What the above equality means? On the left L(u*) - the meaning is clear. On the right we have L(u)*. What does that mean? It is the Hermitian adjoint of the operator L(u) with respect to the Hilbert space scalar product. How is the Hermitian adjoint operator defined? Here is the defining relation:

<L(u)*v,w> = <v,L(u)w>,

or, if you prefer:

<L(u)v,w> = <v,L(u)*w>

Exercise 1. Use the Hilbert algebra property to show that L is indeed a *-representation. Do the same for R.

Note. This is the right place for a side remark. We do not really need it, but, nevertheless, here it is: A is endowed with a norm ||u||. But we have a faithful representation L of A on the Hilbert space A. To each u in A we have the associated linear operator L(u) acting on a Hilbert space. This operator has a norm, like it is the case with every bounded linear operator.   We can therefore equip A with another norm, denoted ||u||', defined as

||u||' = ||L(u)||.

If we do this, we have a nice property:

||u*u||' = ||u||'²,

because operators in every Hilbert space have this property. *-algebras with such a norm are named C*-algebras. So A can be thought as a particularly simple case of a C*-algebra. There is a whole theory of abstract C*-algebras (in finite-dimensional case they are the same as von Neumann algebras). 

In the discussion under the last post Bjab calculated the matrix form of L(u) in a basis. Taking into account the change of indexing, index 0 replace d by index 4, L(u) is given by the matrix:

{{u₄, -iu₃, iu₂, u₁},
{iu₃, u₄, -iu₁, u₂},
{-iu₂, iu₁, u₄, u₃},
{u₁, u₂, u₃, u₄}}.

I have moved the first row to the end, the first column became the last, and replaced u0 by u4. Selecting u = e^μ, with (e^μ)_ν = δ^μν, we get the matrices L^μ calculated by Saša:

L¹ = {{0,0,0,1},{0,0,-i,0},{0,i,0,0},{1,0,0,0}},

L² = {{0,0,i,0},{0,0,0,1},{-i,0,0,0},{0,1,0,0}},

L³ = {{0,-i,0,0},{i,0,0,0},{0,0,0,1},{0,0,1,0}},

L⁴ = {{1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}}.

Matrices R^μ are transposed to the matrices L^μ.

Exercise 2. The matrices L^μ and R^μ are Hermitian. Why is it so?

Exercise 3. Why the matrices R^μ are simply transposed to the matrices L^μ?

The space End(A) - the space of all linear operators acting on A has complex dimension 16 (=4x4). We can build 16 matrices L^μR^ν. There are enough of these matrices to build a basis in End(A). But to be a basis the matrices should be linearly independent. Are they?

One way to address this question is that End(A) is also a Hilbert space with a natural scalar product - for A,B in End(A) the scalar product is given by the trace:

<X,Y> = Tr(X*Y).

So, if our basis happens to be orthonormal, then we automatically have linear independence. Using Mathematica I verified that

<L^μR^ν,L^σR^ρ> = 4 δ^μσ δ^νρ

Therefore indeed our 16 matrices L^μR^ν form a basis in End(A). Nice to know.

Let us concentrate now on L(A) - the image of A under the representation L. In other words: the set of all matrices L(u), u in A. L(A) is an algebra, a subalgebra of End(A). While End(A) is 16-dimensional, L(A) is only 4-dimensional. It is closed under Hermitian conjugation: if X is in L(A), then also  X* is in L(A). The same is true about R(A). We know that every elements in R(A) commutes with every element in L(A).

Exercise 4. Why is it so?

In algebra whenever we have a subalgebra S of an algebra T, we denote by S' the commutant of S in T:

S' = {X in T: XY = YX for all Y in S}.

The fact that every element in R(A) commutes with every element in L(A) can be expressed by the formulas:

R(A) ⊂ L(A)',

L(A) ⊂ R(A)'.

It would be cruel of me to ask the Reader, on Sunday, two days before Christmas Eve,  to prove that, in fact, we have

R(A) = L(A)',

L(A) = R(A)'.

So, I leave the proof for the next post. But, perhaps it is not so cruel to ask the following

Exercise 5. Show that L(A)∩R(A) = C, where C denotes here the algebra of cI, where c is a complex number and I is the identity matrix.

Spin Chronicles Part 27: Back to the roots

 We have to devote some space to Exercise 1 of the previous post. 

Back to the roots

The problems was: Prove that

<ba,c> = <b,ca*>,

where <u,v> is the scalar part of the product u*v, with   u* = τ(u). 

I have mentioned that it can be done in a hard way, using the definitions (and Anna did it), but there is also an easy way. It would be a pity not to examine the easy way, as it will be useful for us later on. The easy way involves using the matrix representation of Cl(V), that we have examined in Part 9. Specifically, we associate to u = (p₀,p) the 2x2 complex matrix m(u):

m(u) = p_μσ^μ = p₀σ⁰ + ... + p₃σ³,  

where σ⁰ is the identity matrix, and σⁱ (i=1,2,3) are the three Pauli Hermitian matrices. With this association we automatically have that

m(uv) = m(u)m(v),

and

m(u*) = m(u)*,

where u*= τ(u), and m(u)* is the Hermitian conjugate of m(u). We also have the fact that p₀, the scalar part of u, is given by

p₀ = (1/2) Tr(m(u)).

Thus

<u,v> = (1/2) Tr(m(u*v)) = (1/2) Tr(m(u)*m(v)).

We can now return to our problem - we have

<ba,c> = (1/2) Tr( m(ba)*m(c) ) = (1/2) Tr( m(a)*m(b)*m(c) ),

and

<b,ca*>, = (1/2) Tr( m(b)*m(c)m(a)* )

But

Tr( m(a)*m(b)*m(c) ) = Tr( m(b)*m(c)m(a)* ),

because in general Tr(XY) = Tr(YX) -  this is the fundamental property of the trace.

QED.

After this introduction we can start with the main topic of this post - the more natural representation of Cl(V), without using Pauli matrices.

In a comment to my previous post I quoted the famous mathematician Emil Artin:

 Emil Artin in "Geometric Algebra",  p. 13, Dover 2016, wrote:

"Mathematical education is still suffering from the enthusiasms which the discovery of the isomorphism (between the ring of endomorphisms and the ring of matrices A.J.) has aroused. The result has been that geometry has been eliminated and replaced by computations. Instead of the intuitive maps of a space preserving addition and multiplication by scalars (these maps have an immediate geometric meaning) matrices have been introduced. From the innumerable absurdities --from a pedagogical point of view--let me point out one example and contrast it with the direct description. (...)  It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out. Sometimes it can not be done; a determinant may have to be computed."

Here I will go against this warning, I will thus make an exception. Exceptions are sometimes even more important than rules, though it is bad when using an exception becomes a rule!

By the way, here is an example from Artin's biography in Wikipedia:

Emil Artin (photo by Natasha Artin

"On the orders of a Hamburg doctor whom he had consulted about a chronic cough, Artin had given up smoking years before. He had vowed not to smoke so long as Adolf Hitler remained in power. On May 8, 1945, at the news of Germany's surrender and the fall of the Third Reich, Natascha made the mistake of reminding him of this vow, and in lieu of a champagne toast, he indulged in what was intended to be the smoking of a single, celebratory cigarette. Unfortunately, the single cigarette led to a second, and another after that. Artin returned to heavy smoking for the rest of his life."

And so we have our geometric Clifford algebra of space Cl(V) (Cl(3) in the standard notation).  We have endowed V with the Euclidean metric and orientation, which enabled us to consider Cl(V) as a complex space of four dimensions. If e^j (j=1,2,3) is an oriented orthonormal basis in V, then we set 

i = e¹e²e³, 

 which is independent of the choice of such a basis. We select such a basis, and this choice enables us to realize Cl(V) as an algebra of complex matrices. In Part  9 we used the Pauli matrices, here we will do it in a more natural way.

Note: We will obtain the Pauli matrices also in a "natural way", but it will come later on, after we start discussing "observers", "reference frames", and ""measurements".

We are in the category of vector spaces and algebras. We have also decided to switch, for our case, from reals to complex numbers as the basic field. It came out naturally. Now Cl(V) is an algebra, we have associative multiplication defined for elements of Cl(V). But, first of all Cl(V) is a complex vector space. For vector spaces we have the concept of endomorphisms--in our case complex linear maps of Cl(V) into itself. Endomorphisms can be naturally composed. They form an associative algebra with unit, denoted End(Cl(V)), where the unit is the identity map. Notice that we are talking about endomorphisms of Cl(V), not about endomorphisms of V. V is a real vector space, and endomorphisms of V form a real algebra.

Now, every element u of Cl(V) defines an endomorphism of Cl(V), namely the left multiplication by u, we denote it by L(u):

L(u)v = uv  for all u,v in Cl(V).

Since Cl(V) is an associative algebra, we instantly get

L(u)L(v) = L(uv).

Also L(1) = Id.

Thus we have a homomorphism L between two algebras, Cl(V) and End(Cl(V)). L is "faithful", that is L(u)=0 if an only if u=0. This follows instantly by selecting v=1 in the formula  L(u)v = uv. In other words Cl(V), as an algebra, can be identified with its image in End(Cl(V)).

We do not have matrices yet, but we already have endomorphisms that Artin was talking about in the quote above. Now we will go for matrices. For this we select an oriented orthonormal basis eⁱ in V, i=1,2,3. Then automatically we have a complex basis E^μ (μ=0,1,2,3) in Cl(V):

E⁰ = 1, Eⁱ = eⁱ (i=,1,2,3).

Once we have a basis in Cl(V), its endomorphisms are naturally represented by matrices. We are going to find the complex 4x4 matrices L^μ representing the endomorphisms L(E^μ). Here I will assume that we already know how to find the matrix representing a given endomorphism in a given basis.

Back to the roots

To be continued in the next post (unless one of the Readers will have enough patience to do this calculation.