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.