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.

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.