The Spin Chronicles Part 31: Irreducibility

 Here is the beginning of the new post. I will be expanding it during the course of the day. At the same time I will be open to the feedback from my Readers to see if I need to change or adjust anything.

Happy 2025! 

This post will be about some basic stuff. I have ordered "Basic Algebra, Vol. II" by Nathan Jacobson, and yesterday it came by mail. It was on the kitchen table. Laura looked inside, opened it at the chapter "Primary Decomposition", looked at the symbols there, and noticed: "It so not so `basic'". Well the same will be with this post. Basic but not so basic.

We will be discussing "decompositions".  The general scenery is as follows:

- all spaces here are finite-dimensional and over the field of complex numbers

- we have an associative algebra A, with unit 1, and with an involution "*"

- we have a vector space E with a (positive defined) scalar product (u,v) (finite-dimensional Hilbert space, if you wish). We assume (u,v) is linear with respect to v, anti-linear with respect to u.

- we have a *-representation, let us call it ρ, of A on E. Thus for each u in A we have a linear operator ρ(u) acting on E (thus a member of End(E) such that ρ(1) is the identity operator and ρ(u*)=ρ(u)*, where * on the right hand side is the Hermitian conjugate of the operator ρ(u) with respect to the scalar product: (X* x,y) = (x,Xy) for all x,y in E, all X in End(E).

So far we have studied the case with A = Cl(V), E = A, and ρ = L or ρ = R - the left and right regular representations of A acting on A itself. But the reasoning below is the same for a general case. And it is more transparent at the same time, there are less possibilities for a confusion. In text below I will state certain things as evident: "it is so and so ...". In a good linear algebra course each of these things is being proven, or is left as an exercise - to be proved starting from definitions. I will also provide reasonings that sketch the proofs of less evident properties. 

We are interested in invariant subspaces of E. By a subspace I will always mean a linear (or vector-) subspace. "Invariant" means invariant for the representation ρ. Thus a subspace F⊂ E is invariant if ρ(A)F ⊂F or, explicitly

ρ(u)x∈F for all u∈A, x∈F.

The representation is said to be irreducible if there are no invariant subspaces other then the zero subspace (consisting of the zero vector alone) and the whole space E. These two subspaces are always trivially invariant.

If F is a subspace, its orthogonal complement F^⟂ is also a subspace, and they span the whole space:

F⊕F^⟂= E.

Thus every u in E E can be uniquely decomposed (orthogonal decomposition)

u = v+w,

where v is in F and w is in F^⟂.

We define P_F, the orthogonal projection on F, by

P_F u = v in the above orthogonal decomposition. Then P_F is a Hermitian idempotent P_F=P_F*=P_F², or "orthogonal projection", or "projection operator", or, simply,  "a projector".

To the decomposition F⊕F^⟂= E,  there corresponds the formula

P_F+P_F^⟂= I, or P_F^⟂= I-P_F,

where I stands for the identity operator.

It is an easy exercise to show that F is invariant if and only if P_F commutes with all ρ(u). If this happens, then, evidently, also P_F^⟂= I-P_F commutes with all ρ(u), thus F is invariant if and only if F^⟂is invariant. We then have a decomposition of E into the direct sum of two invariant subspaces:

E = F⊕F^⟂.

Note. In the case we have discussed in previous posts, we have E = A, and, for  ρ=L, ρ(u)x=ux. Therefore looking for a non-trivial invariant subspace is the same as looking for a non-trivial left ideal. For ρ=R, the right regular representation, looking for an invariant subspace is the same as looking for a right ideal. 

Reducibility in terms of a basis.

We can always choose an orthonormal basis e_i in E. If E is n-dimensional, then i=1,2,...,n. If F is a subspace of E, with dim(F)=m, 0<m<n, we can always choose a basis so that e₁,...,e_m are in F, while e_m+1,...,e_n are in F^⟂. Then e₁,...,e_m form a basis in F, while e_m+1,...,e_n form a basis in F^⟂. We call such a basis "adapted to the decomposition". So, let e_i be such a basis. Assume that F is invariant. The vectors e₁,...,e_m are in F, so, since  F is invariant, for any u in A, the vectors ρ(u)e_i (i=1,...,m) are also in F. But e₁,...,e_m form a basis in F. Therefore ρ(u)e_i are linear combinations of e₁,...,e_m. We write it as

ρ(u)e_i = ∑_j=1^m  e_j μ(u)^j_i (i=1,...,m).

The m² complex coefficients μ(u)^j_i form a matrix representation of A by m⨯m square matrices.We have

μ(uw) = μ(u)μ(w).

Now, since F^⟂is also invariant, we have

ρ(u)e_i = ∑_j=m+1ⁿ  e_j ν(u)^j_i (i=m+1,,,,,n).

Similarly we have

ν(uv) = ν(u)ν(v).

Thus we have another matrix representation of A with, this time, (n-m)⨯(n-m) square matrices. 

Now, all n basis vectors e₁,..., e_n form a basis for E. So, we get a matrix representation of ρ(u), call its matrices simply  ρ(u)ⁱ_j

ρ(u)e_i = ∑_j=1ⁿ  e_j ρ(u)^j_i , (i=1.,,,,n).

Now, in our adapted basis,  the n by n matrices ρ(u)^j_i are block diagonal: they are of the form

ρ(u) = {{μ(u),0},{0,ν(u}}

with blocks m⨯m, m⨯(n-m),(n-m)⨯m,(n-m)⨯(n-m). That is the matrix view of reducibility. A representation is reducible if there is an orthonormal basis in E in which the matrices of the representation are block diagonal.

Exercise 1. In the comments to the last post we have found a basis E1,E2 which spans a nontrivial left ideal of A. Is this basis orthonormal? Probably not, because we were  not taking care of the normalization. But are vectors E1,E2 orthogonal to each other in the scalar product of A? But then there should be another pair of vectors, say E1',E2' orthogonal to E1,E2 and to each other that span the orthogonal complement to our left ideal. It should also be an invariant subspace, thus a complementary left ideal. Can we find a simple form of E1',E2'?   Can we find ν? It is enough to find ν(ei), where ei is the basis of A.

It is now time to formulate a version of Shur's famous Lemma, adapted to our needs.

Shur's Lemma.

Given the *-representation ρ of A on E, ρ is irreducible if and only if the only linear operators acting on E and commuting with all ρ(u) are multiples of the identity operator.

In other words ρ is irreducible if and only if the commutant ρ(A)' is CI. That is if and only if the commutant is trivial. It is reducible if and only if the commutant is non-trivial.

In the proof below we use the standard results about eigenvalues and eigenspaces of Hermitian operators (matrices, if you wish).

Proof. One way is simple. Suppose ρ is reducible, then there is a nontrivial invariant subspace F. Then the projection P_F is non-trivial (different from 0 or I). But since F is invariant, P_F is in the commutant. Now the harder part: show that if the commutant is non-trivial, then the representation is reducible. So suppose ρ(A)' is non-trivial. Then there is an operator X commuting with all ρ(u), and X is not of the form cI, c being a complex number. Now, since ρ is a *-representation, it follows that also X* commutes with all ρ(u).

Exercise 2. Prove the last statement.

But then X=(X+X*)/2 +(X-X*)/2. The first term commutes with all ρ(u), the second term too. Also the second term multiplied by the imaginary i. Both X+X* and i(X-X*) are Hermitian. If X=(X+X*)/2 +(X-X*)/2 is not of the form cI, then at least one of X+X* and i(X-X*), call it Y, is not of the form cI. Now we have Y=Y*, Y commutes with all ρ(u), and Y is not of the form cI. Since Y is Hermitian, it has eigenvalues and eigenspaces. At least one of its eigenspaces must be nontrivial (different from 0 and E). Call it F. Since Y commutes with ρ(u), its eigenspaces are invariant. Thus F is a nontrivial invariant eigenspace. Thus ρ is reducible.

In practice

In practice we often choose an orthonormal basis in E, and we work with matrices. Then  ρ is irreducible if and only if the only matrix commuting with all ρ(u) matrices is a multiple of the identity matrix. But A is assumed to be finite-dimensional. Thus there is a basis, say ε_r in A, r =1,2,...,k, where k is dimension of the algebra A. Every element u of A can be written as u=∑_r=1^k  c_r ε_r, where c_r are complex numbers. For a matrix M to commute with all ρ(u) matrices, it is enough that M commutes with all ρ(ε_r), r=1,...,k. Thus ρ is irreducible if and only if from [M,ρ(ε_r)]=0, r=1,...,k, it follows that M is a multiple of the identity matrix.

Why do we care?

Why do we care about reducibility or irreducibility? Suppose ρ is reducible. So we have invariant subspaces F and F^⟂. Then ρ acts on F. We may ask if ρ acting on F is still reducible. The same with ρ acting on F^⟂. We proceed this way until we end up with irreducible representations. We get this way

E=F₁⊕...⊕F_p,

where each of F₁,...,F_p carries an irreducible representations. These are the building blocks of ρ, the "bricks", or "atoms". They can't be decomposed any further. Both physicists and mathematicians want to know these "atoms". And if a representation is reducible, there must be some "reason" for it. Perhaps it is a reducible representation for A, but irreducible for some bigger algebra B? Then what would be the "meaning" of this B? On the other hand atoms are build of nucleons and electrons. If ρ is an irreducible representation of A, perhaps it is reducible when restricted to a smaller algebra C? What would be the meaning of such a C?

There is another thing: suppose we have two irreducible representations of A, one on F₁, and one on F₂. Are they "essentially the same", or "essentially different"? Two representations are essentially the same (we say: they are "equivalent") if it is possible to choose some orthonormal basis in F1, and one orthonormal basis in F2, in such a way that the matrices of both representations are exactly the same. Of course it is enough that the matrices representing  ε_r are the same.

How to discover the atoms?

We have so far followed a geometrical way and looked for particular left ideals of our geometric algebra A. But once we know that it is a *-algebra, there is another way, more related to the concepts of quantum theory such as "states". Then there is a celebrated GNS (Gelfand-Naimark-Segal) construction of irreducible representations. This we discuss later on, and we will relate it to our adventure with ideals.

P.S. 02-01-25 12:56 In her comment 11:02 AM, Anna wrote

"(1) "It is an easy exercise to show that F is invariant if and only if PF commutes with all ρ(u)."
It is not so easy, for me anyhow. I console myself that it should not be easy since it is the key point to the Schur's lemma."

As it an essential point here is my reasoning. It depends on the fact that if F is invariant, also F^⟂ is invariant. This should be seen first. Suppose we accept it, if not, this is left as a separate exercise (it requires playing with the scalar product and *).

One way. Assume P_Fρ(u) =ρ(u)P_F for all u in A. Then, for x in F, we have

ρ(u)x = ρ(u)P_Fx  (because x in F) = P_Fρ(u)x (becasuse of assumed commutation) is in F, because P_F projects every vector on F.

So, one way is easy. Now the other way. Suppose F is invariant. Then F^⟂ is also invariant. We want to show that then

P_Fρ(u) =ρ(u)P_F. 

This is an equality of two linear operators. Since E = F⊕F^⟂, it is enough to show that it holds separately on vectors in F and on vectors in F^⟂. Suppose x is in F, then P_Fx=x, and the right-hand side is ρ(u)x.

But since F is invariant, ρ(u)x is also in F, therefore the left-hand-side becomes 
P_Fρ(u)x =  ρ(u)x. Left-hand-side and right hand side are thus equal. Now suppose x is in
F^⟂. Then the right hand side is 0. On the left we have P_Fρ(u)x. Since x is in F^⟂, also ρ(u)x is in F^⟂. But then P_Fρ(u)x = 0. Again left and right-hand sides are equal.

So it all depends crucially on the exercise mentioned above.

P.S. 02-01-25 12:56 In her comment 11:02 AM, Anna wrote:

(3) In attempt to find orthonormal projector PF⟂ i've got the following set of matrices complementary to the Pauli matrices:
{1/2(1, 1, 1, 1)}, {1/2(1, i, -i, 1)}, {1/2(0, 0, 0, 2)}
I checked that they are projectors, but how to extract the basis in explicit form is not clear yet.

Let us analyze the problem at hand. We have the algebra A, and we decide to work with its faithful representation as 2 by 2 complex matrices. So a general element of our algebra is now represented by a complex linear combination of the three Pauli matrices and the identity matrix. So, in our minds, we identify our algebra with the algebra of 2x2 complex matrices. We need to know how * operation (anti-automorphism tau) is implemented in this realization, and we know it is just the Hermitian conjugation of a matrix. We need to know how scalar product (u,v) is implemented in this realization, and we know it is 1/2 of the trace of the product u*v.

We have chosen p to be 1+e³, which in our realization becomes

E₁=I+σ₃.

Then Saša came with his E2 =e1-ie², that becomes

E₂=σ₁-iσ₂.

E₁ and E₂ span our F. We are looking now for F^⟂, another subspace of Mat(2,C), orthogonal to F. A general element of F^⟂will be a matrix X={{a,b},{c,d}} with four complex entries. We will impose now the condition that this matrix should be orthogonal to E₁ and to E₂. So we will have two conditions

Tr(E₁*X) = 0. and Tr(E₂*X)=0. These two conditions will enable us to eliminate two of the four unknowns a,b,c,d. If we do it cleverly, the basis in F should be easy to guess.

Or we can try to be brave and just guess: if we jut change signs and set

E₃=I-σ₃, E₄=σ₁+iσ₂

will then E₃ and E₄ span our F^⟂?